Jacaranda Maths Quest 10 + 10A for the Australian Curriculum [1 ed.] 9780730341789, 9780730341796, 9780730341819, 9780730341826

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maths Quest 10 + 10a for the australian Curriculum KylIe bOuCher debbIe KempFF lyN elms ruth baKOgIaNIs dOuglas sCOtt COral CONNOr tObIas COOper

CONtrIbutINg authOrs CarOl pattersON | rObert CahN | aNIta CaNN | jaCINta deylaN breNdaN OWeN | IreNe KIrOFF | rOsetta batsaKIs | eleNa IampOlsKy

First published 2012 by John Wiley & Sons Australia, Ltd 42 McDougall Street, Milton, Qld 4064 Typeset in 10/12pt Times LT © John Wiley & Sons Australia, Ltd 2012 The moral rights of the authors have been asserted. National Library of Australia Cataloguing-in-Publication data Title: Maths quest 10+10A for the Australian curriculum/ Kylie Boucher [et al.] ISBN: 978 0 7303 4178 9 (student ed. : pbk) 978 0 7303 4179 6 (student ed. : ebook) 978 0 7303 4181 9 (teacher ed. : pbk) 978 0 7303 4182 6 (teacher ed. : ebook) Series: Maths quest series. Notes: Includes index Target Audience: For secondary school age. Subjects: Mathematics—Textbooks. Mathematics—Study and teaching (Secondary) Other Authors/ Contributors: Boucher, Kylie. Dewey Number: 510 Reproduction and communication for educational purposes The Australian Copyright Act 1968 (the Act) allows a maximum of one chapter or 10% of the pages of this work, whichever is the greater, to be reproduced and/or communicated by any educational institution for its educational purposes provided that the educational institution (or the body that administers it) has given a remuneration notice to Copyright Agency Limited (CAL). Reproduction and communication for other purposes Except as permitted under the Act (for example, a fair dealing for the purposes of study, research, criticism or review), no part of this book may be reproduced, stored in a retrieval system, communicated or transmitted in any form or by any means without prior written permission. All inquiries should be made to the publisher. Internal design images: © Shutterstock/Nikolai Bird, 2010 Illustrated by Aptara and the Wiley Art Studio Typeset in India by Aptara Printed in Singapore by Craft Print International Ltd 10  9  8  7  6  5  4  3  2  1

Contents 3B Determining linear equations  64

Introduction  viii About eBookPLUS  x Acknowledgements  xi Chapter 1

Exercise 3B  67 3C The distance between two points on a Number and algebra Patterns and algebra

Indices  1 Are you ready?  2

1A Review of index laws  3

Exercise 1A  5 1B Negative indices  7 Exercise 1B  10 1C Fractional indices  12 Exercise 1C  14 1D Combining index laws  17 Exercise 1D  20 Summary  23 Chapter review  24 eBookPLUS activities  26

Chapter 2

straight line  68 Exercise 3C  70 3D The midpoint of a line segment  71 Exercise 3D  73 3E Parallel and perpendicular lines  74 Exercise 3E  79 Summary  83 Chapter review  85 eBookPLUS activities  88

Chapter 4

Simultaneous linear equations and inequations  89 Are you ready?  90

4A Graphical solution of simultaneous linear Number and algebra Patterns and algebra

Linear algebra  27 Are you ready?  28

4B

4C

2A Substitution  29

Exercise 2A  31 2B Adding and subtracting algebraic

fractions  33 Exercise 2B  36 2C Multiplying and dividing algebraic fractions  37 Exercise 2C  39 2D Solving linear equations  40 Exercise 2D  43 2E Solving equations with algebraic fractions and multiple brackets  45 Exercise 2E  48 Summary  50 Chapter review  52 eBookPLUS activities  54

Chapter 3

Number and algebra LINEAR AND NON-LINEAR RELATIONSHIPS

Number and algebra

4D

4E 4F 4G

equations  91 Exercise 4A  94 Solving simultaneous linear equations using substitution  96 Exercise 4B  98 Solving simultaneous linear equations using elimination  99 Exercise 4C  101 Problem solving using simultaneous linear equations  103 Exercise 4D  105 Solving linear inequations  106 Exercise 4E  108 Sketching linear inequations  110 Exercise 4F  113 Solving simultaneous linear inequations  116 Exercise 4G  118

Summary  123 Chapter review  125 eBookPLUS activities  130

Chapter 5

Measurement and geometry PYTHAGORAS AND TRIGONOMETRY

LINEAR AND NON-LINEAR RELATIONSHIPS

Coordinate geometry  55

Trigonometry I  131

Are you ready?  56

Are you ready?  132

3A Sketching linear graphs  57

5A Pythagoras’ theorem  133

Exercise 3A  62

Exercise 5A  137

5B Pythagoras’ theorem in three

7E Mixed factorisation  240

5C

Summary  242 Chapter review  244 eBookPLUS activities  246

5D

5E

5F 5G 5H

dimensions  140 Exercise 5B  143 Trigonometric ratios  145 Exercise 5C  149 Using trigonometry to calculate side lengths  151 Exercise 5D  154 Using trigonometry to calculate angle size  156 Exercise 5E  158 Angles of elevation and depression  161 Exercise 5F  163 Bearings and compass directions  165 Exercise 5G  169 Applications  172 Exercise 5H  173

Summary  177 Chapter review  179 eBookPLUS activities  182

Chapter 6

Measurement and geometry USING UNITS OF MEASUREMENT

Surface area and volume  183 Are you ready?  184

6A Area  185

Exercise 6A  189 6B Total surface area  193 Exercise 6B  199 6C Volume  203 Exercise 6C  208

Chapter 8

Quadratic equations  247 Are you ready?  248

8A Solving quadratic equations  249

Exercise 8A  253 8B The quadratic formula  255

Exercise 8B  257 8C Solving quadratic equations by inspecting

graphs  258 Exercise 8C  261 8D Finding solutions to quadratic equations by interpolation and using the discriminant  263 Exercise 8D  267 8E Solving a quadratic equation and a linear equation simultaneously  269 Exercise 8E  272 Summary  274 Chapter review  276 eBookPLUS activities  278

Functions  279 Are you ready?  280

9A Plotting parabolas  281

Exercise 9A  284 9B Sketching parabolas using the basic graph of Number and algebra

9C

Quadratic expressions  219 Are you ready?  220

9D

7A Expanding algebraic expressions  221

Exercise 7A  225 7B Factorising expressions with three terms  227 Exercise 7B  229 7C Factorising expressions with two or four terms  231 Exercise 7C  234 7D Factorising by completing the square  236 Exercise 7D  239 Contents

Number and algebra LINEAR AND NON-LINEAR RELATIONSHIPS

patterns and algebra

iv

Number and algebra LINEAR AND NON-LINEAR RELATIONSHIPS

Chapter 9

Summary  213 Chapter review  214 eBookPLUS activities  218

Chapter 7

Exercise 7E  240

9E 9F 9G

y = x2  287 Exercise 9B  291 Sketching parabolas in turning point form  292 Exercise 9C  296 Sketching parabolas of the form y = ax2 + bx + c  298 Exercise 9D  302 Exponential functions and their graphs  306 Exercise 9E  309 The hyperbola  312 Exercise 9F  314 The circle  315 Exercise 9G  317

Summary  319 Chapter review  321 eBookPLUS activities  324

Chapter 10

Measurement and geometry

Chapter 13

Statistics and probability data representation and interpretation

GEOMETRIC REASONING

Deductive geometry  325

Univariate data  429

Are you ready?  326

Are you ready?  430

13A Measures of central tendency  431

10A Congruence review  327 10B 10C 10D

10E

Exercise 13A  435

Exercise 10A  329 Similarity review  332 Exercise 10B  335 Congruence and proof  336 Exercise 10C  338 Quadrilaterals: definitions and properties  340 Exercise 10D  341 Quadrilaterals and proof  344 Exercise 10E  345

13B Measures of spread  439

Exercise 13B  442 13C Box-and-whisker plots  444

Exercise 13C  447 13D The standard deviation  449 Exercise 13D  451 13E Comparing data sets  454 Exercise 13E  455 13F Skewness  459 Exercise 13F  461

Summary  347 Chapter review  349 eBookPLUS activities  351

Summary  464 Chapter review  466 eBookPLUS activities  470

projects plus 

ICT activity

Chapter 14

pro-0099 Backyard flood  352

Statistics and probability DATA REPRESENTATION AND INTERPRETATION

Chapter 11

Problem solving

Are you ready?  472

Problem solving I  355 Chapter 12

14A Identifying related pairs of

Statistics and probability CHANCE

Probability  379 Are you ready?  380

12A Review of probability  381 12B

12C 12D

12E 12F

Exercise 12A  392 Complementary and mutually exclusive events  396 Exercise 12B  400 Two-way tables and tree diagrams  403 Exercise 12C  410 Independent and dependent events  413 Exercise 12D  415 Conditional probability  417 Exercise 12E  419 Subjective probability  420 Exercise 12F  421

Summary  423 Chapter review  425 eBookPLUS activities  428

Bivariate data  471 variables  474 Exercise 14A  476 14B Graphing bivariate data  477 Exercise 14B  481 14C Scatterplots  483 Exercise 14C  488 Summary  491 Chapter review  492 eBookPLUS activities  496

Chapter 15

Statistics and probability DATA REPRESENTATION AND INTERPRETATION

Statistics in the media  497 Are you ready?  498

15A Populations and samples  499

Exercise 15A  502 15B Primary and secondary

data  503 Exercise 15B  508 15C Evaluating inquiry methods and statistical reports  511 Exercise 15C  518 Contents

v

15D Statistical investigations

Exercise 15D

521

18F Logarithms

525

Summary 527 Chapter review 529 eBookPLUS activities 533

prOjeCts plus

ICt aCtIvIty

pro-0100 Climate change

534

Chapter 16

Financial maths

537

16C 16D 16E 16F

Chapter 19

polynomials

539

Are you ready?

Exercise 16A 540 Buying on terms 542 Exercise 16B 543 Successive discounts 546 Exercise 16C 547 Compound interest 549 Exercise 16D 551 Depreciation 553 Exercise 16E 554 Loan repayments 556 Exercise 16F 558

19B

19C 19D 19E

19F 19G

Chapter 17

prOblem sOlvINg

637 Exercise 19A 638 Adding, subtracting and multiplying polynomials 639 Exercise 19B 640 Long division of polynomials 641 Exercise 19C 646 Polynomial values 647 Exercise 19D 648 The remainder and factor theorems 649 Exercise 19E 650 Factorising polynomials 651 Exercise 19F 654 Solving polynomial equations 655 Exercise 19G 657

Number aNd algebra

Chapter 20

Are you ready?

vi

Contents

Are you ready?

590

Exercise 18A 594 595 Exercise 18B 597 18C Operations with surds 599 Exercise 18C 607 18D Fractional indices 609 Exercise 18D 612 18E Negative indices 614 Exercise 18E 616 18B Surds

Functions and relations

589

18A Number classification review

Number aNd algebra lINear aNd NON-lINear relatIONshIps

real Numbers

real numbers

636

Summary 659 Chapter review 661 eBookPLUS activities 662

565

10a Chapter 18

635

19A Polynomials

Summary 560 Chapter review 562 eBookPLUS activities 564

problem solving II

Number aNd algebra patterNs aNd algebra

538

16A Purchasing goods 16B

Summary 630 Chapter review 632 eBookPLUS activities 634

Number aNd algebra mONey aNd FINaNCIal mathematICs

Are you ready?

617 Exercise 18F 618 18G Logarithm laws 619 Exercise 18G 622 18H Solving equations 624 Exercise 18H 628

591

664

20A Functions and relations 20B 20C 20D 20E

663

665 Exercise 20A 669 Exponential functions 671 Exercise 20B 675 Cubic functions 679 Exercise 20C 682 Quartic functions 683 Exercise 20D 685 Transformations 686 Exercise 20E 692

22C  Area of triangles  745

Summary  694 Chapter review  696 eBookPLUS activities  698

Chapter 21

Measurement and geometry GEOMETRIC REASONING

Circle geometry  699 Are you ready?  700

21A Angles in a circle  701

Exercise 21A  706 21B Intersecting chords, secants and

tangents  708 Exercise 21B  713 21C Cyclic quadrilaterals  715 Exercise 21C  717 21D Tangents, secants and chords  718 Exercise 21D  720 Summary  724 Chapter review  726 eBookPLUS activities  730

Exercise 22C  747 22D The unit circle  749 Exercise 22D  752 22E Trigonometric functions  755 Exercise 22E  757 22F Solving trigonometric equations  759 Exercise 22F  761 Summary  762 Chapter review  764 eBookPLUS activities  766

Chapter 23

Statistics and probability DATA REPRESENTATION AND INTERPRETATION

Interpreting data  767 Are you ready?  768

23A Bivariate data  769

Exercise 23A  773 23B Lines of best fit  776

Exercise 23B  784 23C Time series  786

Chapter 22

Measurement and geometry PYTHAGORAS & TRIGONOMETRY

Trigonometry II  731 Are you ready?  732

22A The sine rule  733

Exercise 22A  739 22B The cosine rule  741

Exercise 22B  744

Exercise 23C  790 Summary  794 Chapter review  795 eBookPLUS activities  798

Answers  799 Glossary  889 Index  901

Contents

vii

Introduction Australian Mathematics education is entering a historic phase. A new curriculum offers new opportunities to engage future generations of students in the exciting and challenging world of Mathematics. The Australian Mathematics Curriculum provides students with essential mathematical skills and knowledge through the content strands of Number and algebra, Measurement and geometry and Statistics and probability. The Curriculum focuses on students becoming proficient in mathematical understanding, fluency, reasoning and problem solving. Maths Quest 10 + 10A for the Australian Curriculum is specifically written and designed to meet the requirements and aspirations of the Australian Mathematics Curriculum. This resource contains: a student textbook with accompanying eBookPLUS ■■ a teacher edition with accompanying eGuidePLUS ■■ a TI-Nspire CAS Calculator companion ■■ a Casio ClassPad CAS Calculator companion. ■■

Student textbook Full colour is used throughout to produce clearer graphs and headings, to provide bright, stimulating photos, and to make navigation through the text easier. Are you ready? sections at the start of each chapter provide introductory questions to establish students’ current levels of understanding. Each question is supported by a SkillSHEET that explains the concept involved and provides extra practice if needed. Clear, concise theory sections contain worked examples and highlighted important text and remember boxes. Icons appear for the eBookPLUS to indicate that interactivities and eLessons are available online to help with the teaching and learning of particular concepts. Worked examples in a Think/Write format provide clear explanation of key steps and suggest presentation of solutions. Exercises contain many carefully graded skills and application problems, including multiplechoice questions. Cross-references to relevant worked examples appear with the first ‘matching’ question throughout the exercises. Each chapter concludes with a summary and chapter review exercise containing examinationstyle questions (multiple-choice, short-answer and extended-response), which help consolidate students’ learning of new concepts. A glossary is provided to enhance students’ mathematical literacy. There are two problem-solving chapters designed to encourage students to apply their mathematical skills in non-routine situations.

Student website — eBookPLUS The accompanying eBookPLUS contains the entire student textbook in HTML plus additional exercises. Students may use the eBookPLUS on laptops, tablets, school or home computers, and cut and paste material for revision, assignments or the creation of notes for exams.

viii

Introduction

WorkSHEET icons link to editable Word documents, and may be completed on-screen, or printed and completed by hand. Individual pathway activity icons link to online activity sheets below, at and above the level presented in the text, for each exercise. These activities allow students to work at their own pace and to engage with the concepts being taught at an appropriate differentiated level. SkillSHEET icons link to printable pages designed to help students revise required concepts, and contain additional examples and problems. Interactivity icons link to dynamic animations, which help students to understand difficult concepts. eLesson icons link to videos or animations designed to elucidate concepts in ways that are more than what the teacher can achieve in the classroom. Hungry brain activities provide engaging, whole-class activities to introduce each chapter. Test yourself tests are also available. Answers are provided for students to receive instant feedback. Word searches and crosswords are available for each chapter. Two ProjectsPLUS activities provide students with the opportunity to work collaboratively and creatively, online, on a mathematics project.

Teacher website — eGuidePLUS The accompanying eGuidePLUS contains everything in the eBookPLUS and more. Two tests per chapter, fully worked solutions to WorkSHEETs, the work program and other curriculum advice in editable Word format are provided. Maths Quest is a rich collection of teaching and learning resources within one package. Maths Quest 10 + 10A for the Australian Curriculum provides ample material, such as exercises, problem-solving questions, projects, worksheets and technology files, from which teachers can assess their students.

Introduction

ix

About eBookPLUS

Next generation teaching and learning This book features eBookPLUS: an electronic version of the entire textbook and supporting multimedia resources. It is available for you online at the JacarandaPLUS website (www.jacplus.com.au).

Using the JacarandaPLUS website To access your eBookPLUS resources, simply log on to www.jacplus.com.au using your existing JacarandaPLUS login and enter the registration code. If you are new to JacarandaPLUS, follow the three easy steps below. Step 1. Create a user account The first time you use the JacarandaPLUS system, you will need to create a user account. Go to the JacarandaPLUS home page (www.jacplus.com.au), click on the button to create a new account and follow the instructions on screen. You can then use your nominated email address and password to log in to the JacarandaPLUS system. Step 2. Enter your registration code Once you have logged in, enter your unique registration code for this book, which is printed on the inside front cover of your textbook. The title of your textbook will appear in your bookshelf. Click on the link to open your eBookPLUS.

Using eBookPLUS references eBookPLUS logos are used throughout the printed books to inform you that a multimedia resource is available for the content you are studying. Searchlight IDs (e.g. INT-0001) give you instant access to multimedia resources. Once you are logged in, simply enter the searchlight ID for that resource and it will open immediately.

Minimum requirements • A modern internet browser such as Internet Explorer 7+, Mozilla Firefox 3+, Google Chrome 8+, Safari 3+ or Opera 9+ • Adobe Flash Player 10+ • Javascript must be enabled (most browsers are enabled by default).

Step 3. View or download eBookPLUS resources Your eBookPLUS and supporting resources are provided in a chapter-by-chapter format. Simply select the desired Troubleshooting • Go to the JacarandaPLUS help page at chapter from the drop-down list. Your eBookPLUS www.jacplus.com.au/jsp/help.jsp. contains the entire textbook’s content in easy-to-use HTML. The student resources panel contains supporting • Contact John Wiley & Sons Australia, Ltd. Email: [email protected] multimedia resources for each chapter. Phone: 1800 JAC PLUS (1800 522 7587) Once you have created your account, you can use the same email address and password in the future to register any JacarandaPLUS titles you own.

x

About eBookPLUS

Acknowledgements The authors and publisher would like to thank the following copyright holders, organisations and individuals for their assistance and for permission to reproduce copyright material in this book.

Images • Coral Connor: 353 (top left)/© Coral Connor • Copyright Agency Limited: 516/The Sunday Mail, 5 September, 2010, p. 15 • Corbis Australia: 55/Corbis/Ladislav Janicek • Corbis Royalty Free: 219, 268/© Corbis Corporation • Creatas Images: 494/© Creatas Images • Cultura: 131/ © Cultura • Digital Stock: 247 (bottom), 281 (left), 556/© Digital Stock/Corbis Corporation; 545 (centre right)/© Digital Stock • Digital Vision: 140/© Digital Vision; 207/© Digital Vision/Stephen Frink • Image Disk Photography: 164 (bottom)/© Image Disk Photography • iStockphoto: 32/© iStockphoto.com/Steve O’Connor; 171; 231; 484/© Vladimir Popovic/ iStockphoto.com; 671/© iStockphoto.com/Sebastian Kaulitzki; 767/© iStockphoto.com/technotr • John Wiley & Sons Australia: 311, 403, 458, 489, 530 (right)/© John Wiley & Sons Australia/ Photo by Kari-Ann Tapp; 366, 507/© John Wiley & Sons Australia/Photo by Renee Bryon • NASA: 535/© NASA • Photodisc: 105 (bottom), 138 (bottom), 138 (top), 139, 144 (bottom), 144 (centre), 144 (top), 164 (top), 172, 180, 198, 247 (top), 254 (bottom), 254 (top), 286 (bottom), 286 (top), 318, 322, 394, 411, 415, 416 (bottom), 416 (top), 492, 537, 545 (bottom left), 547, 555, 563, 637, 748, 749, 774, 785, 791 • Photolibrary: 490/Photolibrary/Science Photo Library/ROBERTO DE GUGLIEMO; 493/Photolibrary/Index Stock Imagery/Vohra Aneal • Photolibrary Royalty Free: 546/Photolibrary • Shutterstock (all images used under licence from Shutterstock): 1/© mattasbestos; 7/© ARTSILENSEcom, 2009; 16/© arbit , 2009; 89/ © Christina Richards/Shutterstock.com; 105 (top left)/© Angelo Gilardelli, 2009; 105 (top right)/© Hydromet, 2009; 160/© Aleix Ventayol Farrés; 174/© Jim Parkin/Shutterstock. com; 183/© Steven Frame/Shutterstock.com; 226/© Lord_Ghost, 2009; 235/© krechet, 2009; 236/© Benis Arapovic, 2009; 245; 279/© Alistair Michael Thomas, 2009; 281 (right)/© Samot/ Shutterstock.com; 305/© robcocquyt, 2009; 325/Tomasz Trojanowski; 352 (bottom)/© sevenke/ Shutterstock.com; 352 (top)/© sevenke/Shutterstock.com; 353 (bottom left)/© Dan Breckwoldt/ Shutterstock.com; 353 (right)/© Creative Illus/Shutterstock.com; 355/© Benis Arapovic; 374/ © Ronald Sumners; 375/© Steve Heap; 379/© Rafal Olkis, 2009; 413/© Neale Cousland/2009; 429/© Darren Baker; 435/© Omer N Raja; 436/© iofoto; 438/© Brasiliao; 439/© Lisa F. Young, 2010; 449/© Givaga, 2011; 452/© Mark Herreid; 456/© Brad Remy; 463/© Laurence Gough; 466/© Blend Images; 467/© Inger Anne Hulbækdal; 468/© Konstantin L; 471/ © Andresr, 2010; 477/© photobank.kiev.ua, 2010; 482/© AVAVA, 2010; 495/© Nagy-Bagoly Arpad, 2010; 497/© l i g h t p o e t/Shutterstock.com; 530 (left)/© Sonja Foos/Shutterstock.com; 531 (left)/© Nejron Photo/Shutterstock.com; 531 (right)/© Zina Seletskaya/Shutterstock.com; 534 (bottom)/© Mariia Sats/Shutterstock.com; 534 (top)/© Lysithee/Shutterstock.com; 539/ © Milevski Petar, 2009; 545 (bottom right)/© Sasha Radosavljevich, 2009 page 027/ © KOUNADEAS IOANNHS; 545 (top left)/© Marc Dietrich, 2009; 545 (top right)/© Losevsky Pavel, 2009; 565/© Kinetic Imagery; 566/© Harald Høiland Tjøstheim; 573/Kosarev Alexander; 575/© Julian W; 580/© Chris Hellyar; 583/© Simon Krzic; 588/© Serg64; 589/Markus Gebauer; 591 (left)/© Johan Larson; 591 (right)/© R. Gino Santa Maria; 614/Markus Gebauer; 629 (bottom)/© gmwnz; 629 (top)/EmiliaU; 635/© StudioSmart; 639/© BrunoRosa; 663/ © Neale Cousland; 699/© Egolii; 731/© Olivier Le Queinec • Viewfinder Australia Photo Library: 22/© Viewfinder Australia Photo Library • Jennifer Wright: 740/Creative Cohesions

Text • AFP, page 524/ ‘Single women earn more’, The Weekend Australian, 4–5 September 2010, p. 20 • Copyright Agency Limited: 516/The Sunday Mail, September 2010; 517/‘Sponges are Acknowledgements

xi

toxic’, The Sunday Mail, 5 September 2010, p. 36; 523/‘Word limit’, by Professor Emeritus Roland Sussex taken from The Courier Mail, 14–15 August 2010; 524/‘Egg Shortage’, by Paddy Hintz, The Courier Mail, 28–29 August 2010 • News Limited: 532/‘Taste Test’, The Sunday Mail, 4 April 2010, p. 26 Every effort has been made to trace the ownership of copyright material. Information that will enable the publisher to rectify any error or omission in subsequent editions will be welcome. In such cases, please contact the Permissions Section of John Wiley & Sons Australia, Ltd.

xii

Acknowledgements

number AnD AlgebrA • pAtterns AnD AlgebrA

1

       

1A  1B  1C  1D 

Review of index laws Negative indices Fractional indices Combining index laws

WhAt Do you knoW ?

Indices

1 List what you know about indices. Create a concept map to show your list. 2 Share what you know with a partner and then with a small group. 3 As a class, create a large concept map that shows your class’s knowledge of indices. eBook plus

Digital doc

Hungry brain activity Chapter 1 doc-5167

opening Question

If you open a new social networking account with a single friend and double the number of friends each day, how long would it take for you to have 1000 friends?

number AnD AlgebrA • pAtterns AnD AlgebrA

Are you ready?

Try■the■questions■below.■If■you■have■diffi■culty■with■any■of■them,■extra■help■can■be■obtained■by■ completing■the■matching■SkillSHEET.■Either■search■for■the■SkillSHEET■in■your■eBookPLUS■or■ ask■your■teacher■for■a■copy. eBook plus

Digital doc

SkillSHEET 1.1 doc-5168

eBook plus

Digital doc

SkillSHEET 1.2 doc-5169

eBook plus

Digital doc

SkillSHEET 1.3 doc-5170

eBook plus

Digital doc

SkillSHEET 1.4 doc-5171

eBook plus

Digital doc

SkillSHEET 1.5 doc-5172

eBook plus

Digital doc

SkillSHEET 1.6 doc-5173

eBook plus

Digital doc

SkillSHEET 1.7 doc-5174

Index form   1  State■the■base■and■power■for■each■of■the■following. a  34 b  25

c  157

Using a calculator to evaluate numbers given in index form   2  Calculate■each■of■the■following. a  24 b  53

c  46

Linking between squares and square roots   3  Complete■the■following■statements. a  If■32■=■9,■then■ 9 ■=■.■■.■■. c  If■172■=■289,■then■ 289 ■=■.■■.■■. Calculating square roots   4  Find■each■of■the■following.

Digital doc

SkillSHEET 1.8 doc-5175

2

64

a 

100

b 

25

c 

Linking between cubes and cube roots   5  Complete■the■following■statements. 3 a  If■23■=■8,■then■ 8 ■=■.■■.■■.

3 b  If■53■=■125,■then■ 125 ■=■.■■.■■.

3 c  If■93■=■729,■then■ 729 ■=■.■■.■■.

Calculating cube roots   6  Find■each■of■the■following. a 

3

64

b 

3

216

c 

3

1

Estimating square roots and cube roots   7  Estimate,■to■the■nearest■whole■number,■the■value■of■each■of■the■following.■(Do■not■use■a■

calculator.) 23

a  d 

eBook plus

b  If■112■=■121,■then■ 121 ■=■.■■.■■.

3

60

102

b  e 

3

11

40

c  f 

3

120

Using a calculator to evaluate square roots and cube roots   8  Use■a■calculator■to■fi■nd■the■value,■correct■to■4■decimal■places,■of■each■square■root■or■cube■root■ in■question■7.

maths Quest 10 for the Australian Curriculum

number AND algebra • patterns and algebra

Review of index laws

1A

■■

Index laws are the basis for exponential functions, which we will cover in a later chapter. A number in index form has two parts; the base and the index, and is written as: base

■■ ■■

ax

index (power or exponent)

Another name for an index is an exponent or a power. The first two index laws relate to multiplication and division of index expressions. First Index Law: When terms with the same base are multiplied, the indices are added. am ì an = am + n Second Index Law: When terms with the same base are divided, the indices are subtracted. am ó an = am - n

■■

Note: Constants or normal numbers should be treated normally when solving equations. Only apply the index laws to the indices themselves. This will become clearer in the following worked examples.

Worked Example 1

Simplify each of the following. a  m4n3p ì m2n5p3 b  2 a2b3 ì 3 a b4 2 x 5 y4 c 

10 x 2 y3

Think a

b

c

Write

1

Write the expression.

2

Multiply the terms with the same base by adding the indices. Note: p = p1.

1

Write the expression.

2

Simplify by multiplying the coefficients, then multiply the terms with the same base by adding the indices.

1

Write the expression.

2

Simplify by dividing both of the coefficients by the same factor, then divide terms with the same base by subtracting the indices.

■■

a m4n3p ì m2n5p3

= m4 + 2 n3 + 5 p1 + 3 = m6n8p4 b 2a2b3 ì 3ab4

= 2 ì 3 ì a2 + 1 ì b3 + 4 = 6a3b7

c

2x5 y4 10 x 2 y 3 1x 5 − 2 y 4 − 3 5 3 x y = 5

=

The Third Index Law is used in calculations when a zero index is involved. Third Index Law: Any term (excluding 0) with an index of 0, is equal to 1. a0 = 1 Chapter 1 Indices

3

number AND algebra • patterns and algebra

Worked Example 2

Simplify each of the following. a  (2  b3)0

b  -4(a2b5)0

Think a

b

Write a (2b3)0

1

Write the expression.

2

Apply the Third Index Law, which states that any term (excluding 0) with an index of 0, is equal to 1.

1

Write the expression.

2

The term inside the brackets has an index ■ of 0 so the bracket is equal to 1.

= -4 ì 1

3

Simplify.

= -4

■■

=1

b -4(a2b5)0

The Fourth, Fifth and Sixth Index Laws involve removing brackets from an index expression. Fourth Index Law: To remove brackets, multiply the indices inside the brackets by the index outside the brackets. Where no index is shown, assume that it is 1. (am)n = amn  ifth Index Law: To remove brackets containing a product, raise every part of the product to F the index outside the brackets. (ab)m = ambm  ixth Index Law: To remove brackets containing a fraction, multiply the indices of both the S numerator and denominator by the index outside the brackets. m

am  a  b  = m b ■■

Note: Do not forget to raise constants to the correct power as well.

Worked Example 3

Simplify each of the following. a  (2n4)3 b  (3 a2b7)3

 2 x3  c     y4 

Think a

b

4

4

d  (-4)3

Write a (2n4)3

1

Write the expression.

2

Apply the Fourth Index Law by multiplying the indices inside the brackets by the index outside the brackets. Simplify any constants raised to a power. Note: 2 = 21.

1

Write the expression.

2

Apply the Fifth Index Law by multiplying the indices inside the brackets by the index outside the brackets. Note: 3 = 31.

= 31 ì 3 ì a2 ì 3 ì b7 ì 3 = 33a6b21

3

Simplify.

= 27a6b21

Maths Quest 10 for the Australian Curriculum

= 21 ì 3n4 ì 3 = 23n12 = 8n12 b (3a2b7)3

number AnD AlgebrA • pAtterns AnD AlgebrA

c

Write■the■expression.■

2

Apply■the■Sixth■Index■Law■by■multiplying■ the■indices■of■both■the■numerator■and■ denominator■by■the■index■outside■the■ brackets.

3

d

 2x3    y4 

4

c 

1

=

Simplify.

=

21 × 4 × x 3 × 4 y4 × 4 16 x12 y16

d (-4)3

1

Write■the■expression.

2

Expand■the■brackets.

=■-4■ì■-4■ì■-4

3

Simplify,■taking■careful■note■of■the■sign.

= -64

Hint:■A■negative■number■raised■to■an■odd■power■will■ always■remain■negative;■a■negative■number■raised■to■ an■even■power■will■always■become■positive.■Why?

remember

To■simplify■expressions■with■constants■and/or■pronumerals■in■index■form,■the■following■ index■laws■are■used. 1.■ am■ì■an■=■am■+■n 2.■ am■ó■an■=■am■-■n 3.■ a0■=■1■(when■a■≠■0) 4.■ (am)n■=■amn 5.■ (ab)m■=■ambm m

am  a 6.■   =  b bm exerCise

1A inDiViDuAl pAthWAys eBook plus

Activity 1-A-1

Reviewing index operations doc-4948 Activity 1-A-2

Practising the index laws doc-4949

review of index laws FluenCy   1  We 1a, b ■Simplify■each■of■the■following. a  a3■ì■a4 b  a2■ì a3■ì■a 2 3 5 d  ab ■ì■a b e  m2n6■ì■m3n7 g  mnp■ì■m5n3p4 j 

1 2

k  4x2■ì■ xy3■ì■6x3y3

  2  We 1c ■Simplify■each■of■the■following. a  a4■ó■a3 b  a7■ó■a2 d 

Activity 1-A-3

Applying the index laws doc-4950

3m3■ì■2mn2■ì■6m4n5

h  2a■ì■3ab

g  j 

4a

7

3a3 m 7 n3 4 2

m n

7ab5c4■ó■ab2c4

e  h  k 

21b

6

7b 2 2x 4 y3 4

4x y 20 m 5 n3 p4 16m3 n3 p2

c  b■ì■b5■ì■b2 f  a2b5c■ì■a3b2c2 i  l 

4a2b3■ì■5a2b■ì■ 12b5 2x3y2■ì■4x■ì■ 12 x4y4

c  b6■ó■b3 f  i  l 

48m8

12m3 6x7y■ó■8x4 14 x 3 y 4 z 2 28 x 2 y 2 z 2 Chapter 1 Indices

5

number AND algebra • patterns and algebra 3   WE2  Simplify each of the following. a a0 d 3x0

 a 4

0

g 4a0 -    

b (2b)0 e 4b0

c (3m2)0 f -3 ì (2n)0

h 5y0 - 12

i

4   WE3  Simplify each of the following. a (a2)3

 2 n4    3 

g

 5m3  j    n2 

 m2  c    3 

e (a2b)3

f (3a3b2)2

 3m 2 n  h    4 

(2m3n5)4 4

 7x  k  5   2y 

m (-3)5

4

b (2a5)4 2

d 

5x0 - (5xy2)0

3

 a2  i    b3 

3

l

n (-7)2

2

 3a   3  5b

4

o (-2)5

5   MC  a   2m10n5 is the simplified form of: A m5n3 ì 2m4n2 D

2n(m5)2

ì

B

6m10 n4 3n

 2m 5  E    n3 

n4

b The value of 4 - (5a)0 is: A -1 D 3

B 9 E 5

6   MC  a  4 a3b ì b4 ì 5a2b3 simplifies to: A 9a5b8 B 20a5b7 5 7 D 9a b E 21a5b8 b

c

15 x 9 × 3 x 6 9 x10 × x 4 A 5x9 D 9x9 3 p7 × 8q 9 12 p3 × 4 q 4

d

B 9x E 5x

p4 q 4 2 4 q E 24

5a6 b 2 A

B

÷

7b 3 a 2 5b 5a 4

C 20a5b8

C 5x29

C

q4 2

simplifies to:

49a3 b 25

D ab3 6

C 1

simplifies to:

p4 q 4 24

7a 5 b 3

2

simplifies to:

A 2q4 D

C (2m5n2)2

Maths Quest 10 for the Australian Curriculum

25a3 b 49 25ab3 E 49 B

C a3b

number AnD AlgebrA • pAtterns AnD AlgebrA unDerstAnDing   7  Evaluate■each■of■the■following. a  23■ì■22■ì■2 b  2■ì■32■ì■22 d  g 

35 × 4 6 34 × 4 4 4 4 × 56 4 3 × 55

c  (52)2

 3

e  (23■ì■5)2

f     5

h  (33■ì■24)0

i 

  8  Simplify■each■of■the■following. a  (xy)3z b  ab■ì■(pq)0

 a2    b3 

x

d  

e 

n m

4(52■ì■35)0

c  ma■ì■nb■ì■(mn)0

n3 m 2 p

3

f  (am■+■n)■p

q

reAsoning 2

  9  Find■algebraically■the■exact■value■of■x■if■4x■+■1■=■2x .■Justify■your■answer.  10  Binary■numbers■(base■2■numbers)■are■used■in■computer■operations.■As■the■name■implies,■

binary■uses■only■two■types■of■numbers,■0■and■1,■to■express■all■numbers.■A■binary■ number■such■as■101■(read■one,■zero,■one)■means■ (1■ì■22)■+■(0■ì■21)■+■(1■ì■20)■=■4■+■0■+■1■=■5■ (in■base■10,■the■base■we■are■most■familiar■with). ■ The■number■1010■(read■one,■zero,■one,■zero)■ means■(1■ì■23)■+■(0■ì■22)■+■(1■ì■21)■+■(0■ì■20)■=■8■ +■0■+■2■+■0■=■10.■ ■ If■reading■the■binary■number■from■right■to■ left,■the■index■of■2■increases■by■one■each■time,■ beginning■with■a■power■of■zero. ■ Using■this■information,■write■out■the■numbers■ 1■to■10■in■binary■(base■2)■form. reFleCtion 

 

Why are these laws called index laws?

1b eBook plus

Interactivity Negative indices

int-2777

negative indices ■■

So■far■we■have■dealt■only■with■indices■that■are■positive■whole■numbers■or■zero.■To■extend■this,■ we■need■to■consider■the■meaning■of■an■index■that■is■a■negative■whole■number.■Consider■the a3 a3 expression■ 5 .■Using■the■Second■Index■Law,■ 5 ■=■a3■-■5 a a ■ =■a-2 a3 a×a×a Writing■terms■in■the■expanded■notation■we■have:■ 5 ■=■ a × a×a×a×a a 1 ■ =■ a×a 1 ■ =■ 2 a 1 By■equating■the■results■of■simplifi■cation,■using■the■two■methods,■we■have:■a-2■=■ 2 . a Chapter 1 Indices

7

number AND algebra • patterns and algebra

■■

In general terms,

1 a

n

=

a0 n

(1 = a0)

a = a0 - n (using the Second Index Law) = a-n 1 Seventh Index Law: a-n = n a The convention is that an expression should be written using positive indices and so we use the Seventh Index Law to do this.



■■

Worked Example 4

Express each of the following with positive indices. a  x-3 b  2m-4n2

c 

Think a

b

c

4 a−3

Write

1

Write the expression.

2

Apply the Seventh Index Law.

1

Write the expression.

2

Apply the Seventh Index Law to write the expression with positive indices.

1

Copy the expression and rewrite the fraction, using a division sign.

2

Apply the Seventh Index Law to write the expression with positive indices.

3

To divide the fraction, use the ‘multiply and flip’ method.

■■ ■■

a x-3

=

1 x3

b 2m-4n2

=

2 n2 m4

4

c

a −3

= 4 ó a-3 =4ó



=4ì



= 4a3

1 a3 a3 1

Worked Example 5

Simplify each of the following, expressing the answers with positive indices. −2  2 m3  2 x 4 y2 2 -3 -5 a  a b ì a b b  c   −2   n  3 xy5 Think a

8

1

= an. a −n All laws discussed in the previous section are applicable to the terms with negative indices. Part c from Worked example 4 demonstrates the converse of the Seventh Index Law

Write

1

Write the expression.

2

Apply the First Index Law. Multiply terms with the same base by adding the indices.

3

Express the answer with positive indices.

Maths Quest 10 for the Australian Curriculum

a a2 b-3 ì a-5b

= a2 + -5b-3 + 1 = a-3b-2 1 = 3 2 a b

number AND algebra • patterns and algebra

b

Write the expression.

2

Apply the Second Index Law. Divide terms with the same base by subtracting the indices.

=

2x 4 − 1y 2 − 5 3

=

2 x 3 y −3 3

Express the answer with positive indices.

=

3

c

4 2 b 2x y

1

3 xy 5

3y3

 2m3  c  −2   n 

1

Write the expression.

2

Apply the Sixth Index Law. Multiply the indices of both the numerator and denominator by the index outside the brackets. Remember that 2 = 21.

=

3

Express all terms with positive indices.

=

4

Simplify.

=

■■

2x3 −2

2 −2 m−6 n4

1 2

2 m6 n 4 1 4 m6 n 4

Numbers in index form can be easily evaluated if they are expressed with positive indices first. Consider the following example.

Worked Example 6

Evaluate 6 ì 3-3 without using a calculator. Think

Write

1

Write the multiplication.

6 ì 3-3

2

Express 3-3 with a positive index.

=6ì

3

Multiply the numerator of the fraction by the whole number.

=

4

Evaluate the denominator.

=

6 27

5

Cancel by dividing both the numerator and denominator by the same number.

=

2 9

1 33

6 33

Chapter 1 Indices

9

number AnD AlgebrA • pAtterns AnD AlgebrA

remember

1.■ A■term■with■a■negative■index■can■be■expressed■with■a■positive■index■using■the■Seventh■ Index■Law. 1 (a)■ a-n■=■ n ■ a 1 (b)■ −n ■=■an a 2.■ All■index■laws■apply■to■terms■with■negative■indices. 3.■ Always■express■answers■with■positive■indices■unless■otherwise■instructed. 4.■ Numbers■and■pronumerals■without■an■index■are■understood■to■have■an■index■of■1. exerCise

1b inDiViDuAl pAthWAys eBook plus

negative indices FluenCy   1  We4 ■Express■each■of■the■following■with■positive■indices. a  x-5 b  y-4 d 

Activity 1-B-1

g 

Negative indices doc-4951 Activity 1-B-2

Harder negative indices doc-4952 Activity 1-B-3

Tricky negative indices doc-4953

j 

4 -3 a 5 6a3b-1c-5

e  h 

6a

k 

3b −2

c  2a-9

3x2y-3

f  2-2m-3n-4

1

i 

−6

a 7a −4

l 

2b −3

2 3a −4 2m3 n −5 3a−2b 4

  2  We5 ■Simplify■each■of■the■following,■expressing■the■answers■with■positive■indices. a  a3b-2■ì■a-5b-1 b  2x-2y■ì■3x-4y-2 c  3m2n-5■ì■m-2n-3 d  4a3b2■ó■a5b7 g  j 

6m 4 n

h 

2 n3 m 6 (2a3m4)-5

 2 p2  m     3q3 

f  5x-2y3■ó■6xy2

e  2xy6■ó■3x2y5

4x2 y9

i 

x 7 y −3

k  4(p7q-4)-2

−3

 a− 4  n     2b −3 

l 

2

2m 2 n− 4 6m 5 n−1 3(a-2b-3)4

 6a 2  o     3b −2 

−3

  3  We6 ■Evaluate■each■of■the■following■without■using■a■calculator. a  2-3 d  g  j 

b  6-2

3-2■ì■23

e 

6

h 

−3

2

0

16 × 2

4

k 

82 × 2− 4

c  3-4

4-3■ì■22

f  5■ì■6-2

4 × 3−3

i 

−3

2 3

5 × 250

l 

252 × 5− 4

  4  Evaluate■each■of■the■following,■using■a■calculator. a  3-6 b  12-4 d   

 1  −8   2

 3   4

B  -5x

maths Quest 10 for the Australian Curriculum

× 5−2 × 34 34 × 4 2

123 × 150

c  7-5

−7

f  (0.045)-5

5  mC ■a■ ■x-5■is■the■same■as: A  -x5

10

e 

1 3

C  5x

D 

1 x

5

E 

1 x −5

number AND algebra • patterns and algebra

b

1

is the same as: a−4 a 4a 1

d c

1 8

a4

b -4a

c a4

e -a4

is the same as:

A 23

B 2-3

D 3-2

E

C 32

1 2−3

6   MC  a  Which of the following, when simplified, gives A D

3m − 4n−2 4 22 n−2

D

4 n2

?

−2 4 −2 B 3 × 2 × m × n

C

3n−2 2−2 m − 4

E 3m 4 × 22 n−2

3−1 m − 4

b When simplified, 3a-2b-7 ó A

3m 4

4

3 -4 6 a b is equal to: 4 9b B

a6 b13 4a2

E

b13

C

4 a6

9a 2 4b

4a2 b

c When (2x6y-4)-3 is simplified, it is equal to: A

D

2 x18

B

y12 8 y12

E

x18

x18

C

8 y12

y12 8 x18

x18 6 y12

3

 2a x  8b 9 , then x and y (in that order) are: is equal to  a6  by 

d If 

A -3 and -6 D -3 and -2

B -6 and -3 E -2 and -3

C -3 and 2

Understanding 7 Simplify, expressing your answer with positive indices. a c

m−3 n−2

b

m−5 n6 5(a3 b −3 )2 (ab− 4)−1

÷

(m3 n−2 )−7 (m −5 n3 )4

(5a −2 b)−1 (a − 4b)3

8 Simplify, expanding any expressions in brackets. a (r3 + s3) (r3 - s3) b (m5 + n5)2 c

( x a + 1 )b × x a + b x a ( b + 1) × x 2 b

 px + 1  d    px − 1 

−4

×

p8( x + 1) ( p2 x )4

×

p2 ( p12 x )0

Chapter 1 Indices

11

number AND algebra • patterns and algebra

 2r × 8r  ar + b.  in the form 2  22r × 16 

9 Write 

10 Write 2-m ì 3-m ì 62m ì 32m ì 22m as a power of 6. 11 Solve for x if 4x - 4x - 1 = 48. Reasoning 12 Look at the following pattern:

22 = 4 21 = 2 20 = 1 a What is changing on the left hand side of the equation each time? b What is the pattern shown on the right hand side of the equation? c How can this pattern be used to help display the

rule a-n =

1c

1

an

reflection 

 

Do any of the index laws from exercise 1A not apply to negative indices?

?

Fractional indices ■■

Terms with fractional indices can be written as surds, using the following laws: 1

1. a n = n a m

n 2. a n = a m

■■

=

( a) n

m

.

To understand how these laws are formed, consider the following numerical examples. Using 1

1

4 2 × 4 2 = 41 the First Index Law and we also know that 4 × 4 = 16 =4 1

If these two identities are true, then 4 2 = 4 . Similarly: 1

1

1

Using the First Index Law 8 3 × 8 3 × 8 3 = 81 3

and we also know that

8 × 3 8 × 3 8 = 3 512 =8 1

If these two identities are true, then 8 3 = 3 8 . 1

This can be generalised to a n = n a . m ■■

Now consider: a n = a



m×

1 n

m

=

n

×m

a m = ( n a )m m n

n Eighth Index Law:  a = a m = ( n a )m .

12

1

a n = an m 1  1 m n = (a ) =  a n    or

Maths Quest 10 for the Australian Curriculum

number AND algebra • patterns and algebra

■■

As can be seen from the above identities, the denominator of a fraction (n) indicates the power or type of root. That is, n = 3 implies cube root, n = 4 implies fourth root, and so on. Note that when n = 2 (square root), it is the convention not to write 2 at the square root sign.

Worked Example 7

Evaluate each of the following without using a calculator. 3

1

b  16 2

a  9 2 Think

Write 1

a

1

Rewrite the number using the Eighth Index Law.

2

Evaluate.

9

=3 m

b

a 92 = 3 16 2

3 = ( 16 )

1

Rewrite the number using a n = ( n a )m.

2

Evaluate the square root.



= 43

3

Evaluate the result.



= 64

b

Worked Example 8 1

Simplify each of the following. 1

2

a  m 5 × m 5

1 2 3 b  ( a b ) 6

 2 2  x3  c   3  y 4 

Think

Write 1

a

1

Write the expression.

2

a m5 × m5

=

3 m5

2

Apply the First Index Law to multiply terms with the same base by adding the indices.

1

Write the expression.

2

Use the Fourth Index Law to multiply each index inside the brackets by the index outside the brackets.

= a6 b6

3

Simplify the fractions.

= a3b2

1

b

b (a 2 b 3 ) 6 2 3

1 1

1

c

1

Write the expression.

 22 3 c x   3  y 4  1

2

Use the Sixth Index Law to multiply the index in both the numerator and denominator by the index outside the brackets.

=

x3 3

y8 Chapter 1 Indices

13

number AnD AlgebrA • pAtterns AnD AlgebrA

remember

1.■ Fractional■indices■are■those■which■are■expressed■as■fractions. 2.■ Terms■with■fractional■indices■can■be■written■as■surds,■using■the■following■identities: 1

an = n a m

a n = n a m = ( n a )m . 3.■ All■index■laws■are■applicable■to■fractional■indices. exerCise

1C inDiViDuAl pAthWAys

Fractional indices FluenCy   1  We7 ■Evaluate■each■of■the■following■without■using■a■calculator.

eBook plus

1

1

1

1

1

1

Activity 1-C-1

a  16 2

b  25 2

c  812

d  8 3

e  64 3

f  814

Fractional indices doc-4954

1 16 4

3 25 2

3

3

2

Activity 1-C-2

Harder fractional indices doc-4955

1

a  3 3

Activity 1-C-3

Tricky fractional indices doc-4956 eBook plus

Digital doc

SkillSHEET 1.9 doc-5176

7

g  h  i  36 2 j  100 2 k  16 4 l  27 3   2  Using■a■calculator,■evaluate■each■of■the■following.■Give■the■answer■correct■to■2■decimal■places.

d  g 

1 89

3  2 2   3

1

b 

1 52

e 

3 8 12

h 

c  7 5 4

f 

3  3 4   4

i 

(0.6) 5 2  4 3  5 

  3  We8a ■Simplify■each■of■the■following. 3

1

1

a  4 5 × 4 5 3

2

1

d  x 4 × x 5 g 

−

3

1

b  2 8 × 2 8

2

h 

3 2 8 a 5

f 

3 1 7 b 2

i 

5x3 × x 2

1

e  5m 3 × 2m 5

4y2 × y 9

1

c  a 2 × a 3

1

3

× 0.05a 4

2

× 4b 7

  4  Simplify■each■of■the■following. 2

3

1

3 2

3

a  a 3 b 4 × a 3 b 4 3

1

1 1

b  x 5 y 9 × x 5 y 3 1 1

2

eBook plus

Digital doc

SkillSHEET 1.10 doc-5177

1 1 1

1 4 5 3 e  x y 2 z 3 × x 6 y 3 z 2 m n 3   5  Simplify■each■of■the■following. d  6m 7 ×

1

d 

6 a7

÷

3 a7

2 3 1

f 

1

3 2 c  12 ÷ 12 2

2

e 

3 x2

÷

1 x4

4

f 

3

m5 5

g 

14

3 4x 5

maths Quest 10 for the Australian Curriculum

3

h 

7 n2 4 21n 3

i 

3 3

2a 5 b 8 c 4 × 4 b 4 c 4

m9 2x 4

3 4

b  5 3 ÷ 5 4

1

a  3 2 ÷ 3 3

1

c  2ab 3 × 3a 5 b 5

25b 5 1

20 b 4

number AnD AlgebrA • pAtterns AnD AlgebrA   6  Simplify■each■of■the■following. 4

3

5 2

3 2 a  x y ÷ x 3 y 5

d 

eBook plus

Digital doc

SkillSHEET 1.11 doc-5178

4 10 x 5 y

2 1 5x 3 y 4

÷

2

3 4

2

e 

3

c  m 8 n 7 ÷ 3n 8

b  a 9 b 3 ÷ a 5 b 5

7

3 3 4 5a b 5

f 

1 1 20 a 5 b 4

1

p8 q 4 2 1

7 p3 q6

  7  Simplify■each■of■the■following. 3

1

 35 a   2 4    d 

 24 b   5 3   

 1 c   7 5   

3

1

 48 e   m 9   

1 (a3 )10

f 

 13  2b 2   

i 

 a c  3m b   

n

14

b

 m p h   x n   

 3  15 g  4  p 7   

6

  8  We8b, c ■Simplify■each■of■the■following. 1

 1 12 a   a 2 b 3   

b  (a

1  1 3 33 3a 3 b 5 c 4

d  

e 

 



 4 5 g   m   7  n 8 

4

 3 7 c   x 5 y 8   

3 b) 4

2

1  1 2 22 5 x2y3z5

 

 

f 

 33  a4   b  

i 

 72  4x   3  2 y 4 

2

1

 33  b5  h   4   c 9 

2

2

2

  9    mC ■a■

y 5 ■is■equal■to:

 1 A   y 2   

5

2 B  y × 5

C 

1 (y5 ) 2

 1 5 E   y   

D  2 5 y

2

2

b  k 3 ■is■not■equal■to:

 1 A   k 3    1

c  5

g

2

2

B 

3

k

2

 1 C   k 2   

3

D 

( ) 3

k

−

5 2

2

1 2 E  (k ) 3

■is■equal■to: 2

5 A  g

B  g

−

2 5

5

C  g 2

D  g

1

E  2 g 5 Chapter 1 Indices

15

number AnD AlgebrA • pAtterns AnD AlgebrA m

1  3 n  10  mC  a■ ■If■ a 4  ■is■equal■to■a 4 ,■then m and n could■not■be:   A  1■and■3 B  2■and■6 C  3■and■8 D  4■and■9 E  both■C■and■D p

 mm an  ■is■equal■to: b  When■simplifi■ed,■   n  p  b m

A 

D 

eBook plus

Digital doc

SkillSHEET 1.12 doc-5179

ap

B 

n bm

a b

p

C 

n bm

a

E 

m

p n a

b

a

mp n n

bm

m2 np nm p2

 11  Simplify■each■of■the■following. a 

a8

b 

3

b9

c 

4

m16

d 

16 x 4

e 

3

8y 9

f 

4

16 x8 y12

27m 9 n15

h 

5

32 p5 q10

i 

3

216a6 b18

g 

3

unDerstAnDing  12  The■relationship■between■the■length■of■a■pendulum■(L)■in■a■grandfather■

clock■and■the■time■it■takes■to■complete■one■swing■(T)■in■seconds■is■ given■by■the■following■rule.■Note■that■g■is■the■acceleration■due■to■ gravity■and■will■be■taken■as■9.8. 1

 L 2 T = 2π    g a  Calculate■the■time■it■takes■a■1m■long■pendulum■to■complete■one■swing. b  Calculate■the■time■it■takes■the■pendulum■to■complete■10■swings. c  How■many■swings■will■be■completed■after■10■seconds? reAsoning eBook plus

Digital doc

WorkSHEET 1.1 doc-5180

16

( 7 )7

 13  Consider■the■term■ a . a  Use■the■Eighth■Index■Law■combined■with■the■

First■Index■Law■to■show■that■ ( 7 a ) ■=■a. b  Use■the■Eighth■Index■Law■combined■with■the■ 7

Fourth■Index■Law■to■show■that■ ( 7 a ) ■=■a.

maths Quest 10 for the Australian Curriculum

7

reFleCtion 

 

Why is it easier to perform operations with fractional indices than with expressions using surds?

number AND algebra • patterns and algebra

Combining index laws

1d

■■ ■■

In most practical situations, more than one index law is needed to simplify the expression. The following examples show simplification of expressions with indices, using several index laws.

Worked Example 9

Simplify. a  b 

( 2 a ) 4 b4 6 a 3 b2 3n − 2 × 9 n + 1 81n − 1

Think a

b

Write 4 4 a (2a) b

1

Write the expression.

2

Apply the Fourth Index Law to remove the bracket.

=

3

Apply the Second Index Law for each number and pronumeral to simplify.

=

4

Write the answer.

1

Write the expression.

2

Rewrite each term in the expression so that it has a base of 3.

=

3

Apply the Fourth Index Law to expand the brackets.

=

4

Apply the First and Second Index Laws to simplify.

=

6a 3 b 2

Write your answer.

6a 3 b 2 8ab 2 3

8ab 2 3 n −2 × 9 n +1 b 3

81n −1

= 5

16a 4 b 4

3n −2 × (32 ) n +1 (34 ) n −1 3n −2 × 32 n + 2 34 n − 4 33 n 34 n − 4 1 3n − 4 1

3

n−4

Chapter 1 Indices

17

number AND algebra • patterns and algebra

Worked Example 10

Simplify each of the following. a  (2a3b)4 ì 4a2b3

b 

7 xy3 ( 3 x 3 y2 )

2

c 

Think a

b

18

7 m3 n3 × mn2

Write a (2a3b)4 ì 4a2b3

1

Write the expression.

2

Apply the Fourth Index Law. Multiply each index inside the brackets by the index outside the brackets.

= 24a12b4 ì 4a2b3

3

Evaluate the number.

= 16a12b4 ì 4a2b3

4

Multiply coefficients and multiply pronumerals. Apply the First Index Law to multiply terms with the same base by adding the indices.

= 16 ì 4 ì a12 + 2b4 + 3 = 64a14b7

1

Write the expression.

2

Apply the Fourth Index Law in the denominator. Multiply each index inside the brackets by the index outside the brackets.

=

3

Apply the Second Index Law. Divide terms with the same base by subtracting the indices.

=

4

c

2 m 5 n × 3 m 7 n4

1 to express the answer am with positive indices. −m Use a =

b

7 xy 3 (3 x 3 y 2 )2

=

7 xy 3 9x6 y 4

7 x −5y −1 9 7 9x5 y

5 7 4 c 2m n × 3m n

1

Write the expression.

2

Simplify each numerator and denominator by multiplying coefficients and then terms with the same base.

=

3

Apply the Second Index Law. Divide terms with the same base by subtracting the indices.

=

6m8 n0 7

4

Simplify the numerator using a0 = 1.

=

6m8 × 1 7

=

6m8 7

Maths Quest 10 for the Australian Curriculum

7m3 n3 × mn2 6m12 n5 7m 4 n 5

number AND algebra • patterns and algebra

■■

■■

When more than one index law is used to simplify an expression, the following steps can be taken. Step 1:  If an expression contains brackets, expand them first. Step 2: If an expression is a fraction, simplify each numerator and denominator, then divide (simplify across then down). Step 3:  Express the final answer with positive indices. The following example illustrates the use of index laws for multiplication and division of fractions.

Worked Example 11

Simplify each of the following. a 

( 5 a 2 b3 ) 2 a10

×

a 2 b5 ( a 3 b)7



b 

8 m3 n−4 ( 6 mn2 ) 3

÷

4 m−2 n− 4 6 m−5 n

Think a

b

Write a

(5a 2 b3 )2

×

1

Write the expression.

2

Remove the brackets in the numerator of the first fraction and in the denominator of the second fraction.

=

3

Multiply the numerators and then multiply the denominators of the fractions. (Simplify across.)

=

4

Divide terms with the same base by subtracting the indices. (Simplify down.)

= 25a-25b4

5

Express the answer with positive indices.

=

1

Write the expression.

2

Remove the brackets.

=

3

Change the division sign to multiplication and flip the second fraction (multiply and flip).

=

4

Multiply the numerators and then multiply the denominators. (Simplify across.)

­=

5

Cancel common factors and divide pronumerals with the same base. (Simplify down.)

­=

6

Simplify and express the answer with positive indices.

­=

b

a10 25a 4 b6 a10

×

a2b5 (a 3 b) 7 a2b5 a 21b 7

25a6 b11 a31b 7

25b 4 a 25

8m3 n −4 (6mn2 )3

÷

8m3 n− 4 216m3 n6 8m3 n− 4 216m3 n6

4 m −2n −4 6m −5 n ÷ ×

4 m −2n− 4 6m −5n 6m −5n 4 m−2n − 4

48m −2n−3 864 mn2 m −3n−5 18 1 18m3 n5

Note that the whole numbers in part b of Worked example 11 could be cancelled in step 3. Chapter 1 Indices

19

number AnD AlgebrA • pAtterns AnD AlgebrA

remember

1.■ Simplifi■cation■of■expressions■with■indices■often■involves■application■of■more■than■one■ index■law. 2.■ If■an■expression■contains■brackets,■they■should■be■removed■fi■rst. 3.■ If■the■expression■contains■fractions,■simplify■across■then■down. 4.■ When■dividing■fractions,■change■ó■to■ì■and■fl■ip■the■second■fraction■(multiply■ and■fl■ip). 5.■ Express■the■fi■nal■answer■with■positive■indices. exerCise

1D inDiViDuAl pAthWAys

Combining index laws FluenCy   1  We10a ■Simplify■each■of■the■following. 2 2 3 4 3 a  (3a b ) × 2a b

5 2 3 6 b  (4 ab ) × 3a b

Activity 1-D-1

c  2m3 n−5 × (m 2 n−3) −6

3 2 2 4 3 d  (2 pq ) × (5 p q )

Review of indices doc-4957

7 2 2 3 3 2 e  (2a b ) × (3a b )

2 −2 3 5 −4 f  5(b c ) × 3(bc )

eBook plus

Activity 1-D-2

Indices practice doc-4958

g 

1 1 6x 2 y 3

Activity 1-D-3

Tricky indices doc-4959

1  3 42 4x 4 y 5

×  

 2 1 i  2  p 3 q 3   

−

3 4

 

 1 −3  × 3 p4 q 4   

−

1 3

 1 1 ×  m2 n4   

3 16m3 n4 4

)

h 

(

j 

 1 2 8p5q3   

−

1 3

3

2

1 33  ×  64 p 3 q 4   

  2  We10b ■Simplify■each■of■the■following. a 

5a 2 b3

b 

( 2 a 3 b )3

 4 x 3 y10  d     2x7 y4  g 

(

) ( )

1 5 p6 q 3

25

6

e 

4 x 5 y6 (2 xy 3 )4 (2a 7 b 4 )−3

 3b 2 c3  h     5b −3c − 4 

(3m 2 n3 )3 (2m 5 n 5 ) 7

 3g 2 h 5  f     2g4 h 

3a3 b −5

2

2 1 1 3 p2 q4

c 

−4

i 

(

(

3

1 1 1 x3 y4 z2

2 1 1 − x3 y 4z3

) )

2

−

3 2

  3  We10c ■Simplify■each■of■the■following. 2 3 4 a  2a b × 3a b

4 a3 b 5

d 

g 

6x3 y 2 × 4 x6 y 9 xy 5 × 2 x 3 y 6 a3 b 2 × 2(ab 5 )3 6(a 2 b3 )3 × a 4 b

6 3 5 b  4 m n × 12mn

6 m 7 n6

e 

h 

(6 x 3 y 2 )4 9 x 5 y 2 × 4 xy 7

c  f 

10 m6 n5 × 2m 2 n3 12m 4 n × 5m 2 n3 5 x 2 y 3 × 2 xy 5 10 x 3 y 4 × x 4 y 2

( p6 q 2 )−3 × 3 pq 2 p− 4q −2 × (5 pq 4 )−2

3 1

i 

2 20

maths Quest 10 for the Australian Curriculum

4

3

6x 2 y 2 × x 5 y 5

( )

1 1 5 x2y

1 1

× 3x 2 y 5

number AND algebra • patterns and algebra 4   WE 11a  Simplify each of the following. a

a3 b 2

×

5a 4 b 7

2a 6 b

b

a 9 b3 3

 2m3 n 2  6m 2 n 4 × d   4 m3 n10  3mn5   5 p6 q 4  ×   3 p5 

6 −5

5p q g 3q − 4

(2a 6 ) 2 7 3

10 a b

×

4 ab6 6a

c

3

4

 2 xy 2   x3 y 9  e  3 5  ×  10   3x y   2y 

−2

h

1 1 2a 2 b 3 1 1 6a 3 b 2

( m 4 n3 ) 2 6

(m n)

2

f

( ) ×

1 1 2 4a 4 b

i

1 b4a

4

4 x −5y−3 ( x 2 y 2 )−2

× ×

( m 3 n3 )3 (2mn)2 3x 5 y6 2−2 x −7y

2 1

1

3x 3 y 5

4x 2

1 1 9x 3 y 4

×

3 4 x y

5   WE 11b  Simplify each of the following. a

5a 2 b3 6a 7 b 5

÷

a9b4

b

3ab6

3

 4a9   3a 7  c  6  ÷  5   b   2b   x 5 y −3  e    2 xy 5 

g

1 3 4m 2 n 4

−4

÷

÷

 3ab  ÷  3a6 b 7  2a6 b 4 

4

4 5

3m3 n4

f

(3 x −2y 2 )−3

(2x y ) 2m−6 n−5

3

(4x y) ÷

3

6

5x 2 y6

d

4 x 6 y −10

2

10 xy 3

 2 m 4 n6  ÷  −1   m n 

−2

1

1 1 6m 3 n 4 3

7a 2 b 4

3 12  1 −2 3 3  − h  4 b c  ÷ 2b 3 c 5    1     6c 5 b 

1

8m 4 n 2 Understanding 6 Evaluate each of the following. 2 0 −3 0 5 6 −1 −3 a (5 × 2) × (5 × 2 ) ÷ (5 × 2 ) 3 3 −2 b (2 × 3 ) ÷

( 2 6 × 39 ) 0 26 × (3−2)−3

7 Evaluate the following for x = 8. (Hint: Simplify first.) 2

2x  x (2 x )−3 ×   ÷ 3 4  2 (2 ) 8 a  Simplify the following fraction.

a 2 y × 9b y × (5ab) y (a y )3 × 5(3b y )2

b Find the value of y if the fraction is equal to 125. 3

9   MC  Which of the following is not the same as (4 xy ) 2 ? 3 3

3 B ( 4 xy )

A 8 x 2 y 2 1

D

(2 x 3 y 3 ) 2 ( 32 )−1

1

C

64 x 3 y 3

1

2 E 4 xy 2 × (2 xy ) 2

Chapter 1 Indices

21

number AnD AlgebrA • pAtterns AnD AlgebrA

 10  The■expression■ A  D 

x2 y 2 3

(2 xy )

÷

xy 16 x 0

2

B 

x 2 y6 2 xy

■is■equal■to:

E 

6

2x2

C  2x2y6

b6 1 128 xy 5

 11  Simplify■the■following. a 

d 

3

2

m n ÷ mn

3 22

×4

−

1 4

3

× 16

−

b 

3 4

(

1

)

1

 1 2 g h ×  −3  n  −2

3

 a3 b −2  e   −3 −3  3 b 

−2

45 3

c 

3

3

9 4 × 15 2

 3−3 a −2 b  ÷  4 −2   a b 

2

f  (

5

3 2 2 d )

×(

3

1 5 5 d )

reAsoning  12  In■a■controlled■breeding■program■at■the■Melbourne■Zoo,■

the■population■(P)■of■koalas■at■t■years■is■modelled■by■ P■=■P0■ì■10kt.■The■initial■number■of■koalas■is■20■and■the■ population■of■koalas■after■1■year■is■40.■ a  Determine■the■value■of■P0■and■k.■ b  Calculate■the■number■of■koalas■after■2■years. c  When■will■the■population■be■equal■to■1000? eBook plus

Digital doc

WorkSHEET 1.2 doc-5181

 13  The■decay■of■uranium■is■modeled■by■D■=■D0■ì■2-kt.■If■it■takes■

6■years■for■the■mass■of■uranium■to■halve,■find■the■percentage■ remaining■after: a  2■years b  5■years c  10■years.

reFleCtion 

 

Do index laws need to be performed in a certain order?

22

maths Quest 10 for the Australian Curriculum

number AnD AlgebrA • pAtterns AnD AlgebrA

summary Review of index laws 

To■simplify■expressions■with■constants■and/or■pronumerals■in■index■form,■the■following■index■ laws■are■used. ■■ am■ì■an■=■am■+■n ■■ am■ó■an■=■am■-■n ■■ a0■=■1■(when■a■ò■0) ■■ (am)n■=■amn ■■ (ab)m■=■ambm m am  a ■■  b  = m b Negative indices ■■

■■ ■■ ■■

A■term■with■a■negative■index■can■be■expressed■with■a■positive■index■using■the■Seventh■Index■ Law. 1 1 (a)■ a-n■=■ n ■■ (b)■ −n ■=■an a a All■index■laws■apply■to■terms■with■negative■indices. Always■express■answers■with■positive■indices■unless■otherwise■instructed. Numbers■and■pronumerals■without■an■index■are■understood■to■have■an■index■of■1. Fractional indices

■■ ■■

Fractional■indices■are■those■which■are■expressed■as■fractions. Terms■with■fractional■indices■can■be■written■as■surds,■using■the■following■identities: ■ 1 an = n a m

■■

a n = n am = ( n a ) . All■index■laws■are■applicable■to■fractional■indices. m

Combining index laws ■■ ■■ ■■ ■■ ■■

Simplifi■cation■of■expressions■with■indices■often■involves■application■of■more■than■one■ index■law. If■an■expression■contains■brackets,■they■should■be■removed■fi■rst. If■the■expression■contains■fractions,■simplify■across■then■down. When■dividing■fractions,■change■ó■to■ì■and■fl■ip■the■second■fraction■(multiply■and■fl■ip). Express■the■fi■nal■answer■with■positive■indices.

MAPPING YOUR UNDERSTANDING

Homework  Book

Using■terms■from■the■summary,■and■other■terms■if■you■wish,■construct■a■concept■map■that■ illustrates■your■understanding■of■the■key■concepts■covered■in■this■chapter.■Compare■your■ concept■map■with■the■one■that■you■created■in■What do you know?■on■page■1. Have■you■completed■the■two■Homework sheets,■the■Rich task■and■two■Code puzzles■in■ your■Maths Quest 10 Homework Book?

Chapter 1 Indices

23

number AND algebra • patterns and algebra

Chapter review Fluency 1 3d10e4 is the simplified form of: 8

A d  6e2 ì 3d 4e3 B

6d 10 e 5

2e 2 C (3d 5e 2)2  d5  E 3  2  e 

A

2

4

4x-3

ó A 2 D 2x-1

D

9

E

12 x8 × 2 x 7

2

C 2x6

C 4x

B 2a6b13

A D

7 16

24

1 16

4p q 1 16

4p q −

3 4

÷

3 92

16m 7

E

5

C

a3 b6 2

22 16

p q

m7 7

10

5

32i 7 j 11 k 2 can be simplified to: 1

2

B

2 1 2 2i 7 j 11 k 5 2

50 25 2i 7 j 11 k10

1

2

10

2i 7 j 11 k 5 E D 5 10 Simplify each of the following.

b

26a 4 b6 c 5 12a3 b3 c3

 14 p7  d    21q3 

3

4

11 Evaluate each of the following. a

C

1 4 p8

E 22p16q

5a0

 2a  -   + 12  3

b -(3b)0 -

(4 b) 0 2

12 Simplify each of the following and express your

answer with positive indices. a 2a-5b2 ì 4a-6b-4 b 4x-5y-3 ó 20x12y-5 c (2m-3n2)-4 13 Evaluate each of the following without using a

calculator.

can be simplified to:

A 2

B

D 3 3 8

E

1 216 1 2

5

32i 7 j 11 k 2 C 5

0

( p5 q 2 )2 can be simplified to: 2 pq 5 B

8

l3

2l 3

 20 m 5 n2  6  

a3 b6 E 4 ÷

7

l3

4m7

c 

(2a 2 b) 2

a6 b13 4 a6 b13 D 2

5

C

3 5

simplifies to:

A

2m 7

a 5x3 ì 3x5y4 ì  x2y6

2 3 5 5 The expression (a b ) is equal to:

( 2 p 5 q 2 )3

11

2

x9

B 8x E 4x29

( p 2 q) 4

B

32i 7 j 11 k 5 A 5

is equal to:

A 4x5 D 8x5

6

8m 7

l3

B 2x0

6x9 × x5

can be simplified to:

2

l3

2 8m3n ì n4 ì 2m2n3 simplifies to: A 10m5n8 B 16m5n7 5 8 C 16m n D 10m5n7 5 8 E 17m n

8x3

−3

 1  8  lm −2  16 

(d  5)2 ì e3

D 3e

3

 2   2l 9 m −1   

C

8 27

Maths Quest 10 for the Australian Curriculum

a

 1   2 −3

c 4

−3

×

−3 b 2 × (3) × 5 8− 2

−5

 9   2

2

number AnD AlgebrA • pAtterns AnD AlgebrA  14  Simplify■each■of■the■following. a  b 

4 1 2a 5 b 2

×

1 3 3a 2 b 4

×

 18  Simplify■each■of■the■following■and■then■evaluate.

3 2 5a 4 b 5

a  (3 ×

3 1 43 x 4 y 9 4 1 16 x 5 y 3



1

1 × 814

2 22  b   125 3 − 27 3 

1 16 2

 16  Simplify. a 

3

a 9 + 4 16a8 b 2 − 3 ( 5 a )

b 

5

32 x 5 y10 +

15

3

b 

6

−62 × (3−3 )0

6a3m × 2b 2 m × (3ab)− m

(4 b) m × (9a 4 m ) 2   2  Answer■the■following■and■explain■your■reasoning 3 a  What■is■the■ten’s■digit■of■33 ? 309 b  What■is■the■one’s■digit■of■6 ? c  What■is■the■one’s■digit■of■81007? eBook plus

Interactivities

64 x 3 y 6

 17  Simplify■each■of■the■following. a 

1  1  32 × 63 

0

1

calculator.■Show■all■working.

6×

1  −  +  36 × 5 2 

  1  If m =■2,■determine■the■value■of:

 15  Evaluate■each■of■the■following■without■using■a■

a 

×5

−2

problem solVing

  b3 

3 16 4

×

3 32

b  (6 × 3−2 )−1 ÷

1 12 4a 3

c  

1 56 ) 2

Test yourself Chapter 1 int-2828

(5a −2b) −3 × 4 a6 b −2

Word search Chapter 1 int-2826

2a 2 b3 × 5−2 a −3 b − 6

Crossword Chapter 1 int-2827

2 x 4 y −5 3 y 6 x −2

 4 xy −2  ×  −6 3   3x y  1

−3

1   3 43  3 n −2 2 m n 4 m     c  ÷ 1  2    5− 3   5m 2 n 

−

1 2

Chapter 1 Indices

25

eBook plus

ACtiVities

Chapter opener Digital doc

•■ Hungry■brain■activity■Chapter■1■(doc-5167)■ (page 1) Are you ready? Digital docs (page 2) •■ SkillSHEET■1.1■(doc-5168):■Index■form •■ SkillSHEET■1.2■(doc-5169):■Using■a■calculator■to■ evaluate■numbers■given■in■index■form •■ SkillSHEET■1.3■(doc-5170):■Linking■between■ squares■and■square■roots •■ SkillSHEET■1.4■(doc-5171):■Calculating■square■ roots •■ SkillSHEET■1.5■(doc-5172):■Linking■between■cubes■ and■cube■roots •■ SkillSHEET■1.6■(doc-5173):■Calculating■cube■ roots •■ SkillSHEET■1.7■(doc-5174):■Estimating■square■roots■ and■cube■roots •■ SkillSHEET■1.8■(doc-5175):■Using■a■calculator■to■ evaluate■square■roots■and■cube■roots

1A   Review of index laws Digital docs (page 5) •■ Activity■1-A-1■(doc-4948):■Reviewing■index■ operations •■ Activity■1-A-2■(doc-4949):■Practising■the■index■ laws •■ Activity■1-A-3■(doc-4950):■Applying■the■index■ laws

1B   Negative indices

(page 10) •■ Activity■1-B-1■(doc-4951):■Negative■indices •■ Activity■1-B-2■(doc-4952):■Harder■negative■indices •■ Activity■1-B-3■(doc-4953):■Tricky■negative■indices Digital docs

Interactivity

•■ Negative■indices■(int-2777)■(page 7)

26

maths Quest 10 for the Australian Curriculum

1C   Fractional indices Digital docs

•■ Activity■1-C-1■(doc-4954):■Fractional■indices■ (page 14) •■ Activity■1-C-2■(doc-4955):■Harder■fractional■indices■ (page 14) •■ Activity■1-C-3■(doc-4956):■Tricky■fractional■indices■ (page 14) •■ SkillSHEET■1.9■(doc-5176):■Addition■of■fractions■ (page 14) •■ SkillSHEET■1.10■(doc-5177):■Subtraction■of■ fractions■(page 14) •■ SkillSHEET■1.11■(doc-5178):■Multiplication■of■ fractions■(page 15) •■ SkillSHEET■1.12■(doc-5179):■Writing■roots■as■ fractional■indices■(page 16) •■ WorkSHEET■1.1■(doc-5180):■Fractional■indices■ (page 16) 1D   Combining index laws Digital docs

•■ Activity■1-D-1■(doc-4957):■Review■of■indices■ (page 20) •■ Activity■1-D-2■(doc-4958):■Indices■practice■(page 20) •■ Activity■1-D-3■(doc-4959):■Tricky■indices■(page 20) •■ WorkSHEET■1.2■(doc-5181):■Combining■index■laws■ (page 22) Chapter review

(page 25) •■ Test■Yourself■Chapter■1■(int-2828):■Take■the■end-ofchapter■test■to■test■your■progress■ •■ Word■search■Chapter■1■(int-2826):■an■interactive■word■ search■involving■words■associated■with■this■chapter •■ Crossword■Chapter■1■(int-2827):■an■interactive■ crossword■using■the■defi■nitions■associated■with■the■ chapter Interactivities

To access eBookPLUS activities, log on to www.jacplus.com.au

number AnD AlgebrA • PAtterns AnD AlgebrA

2

2A Substitution 2B Adding and subtracting algebraic fractions 2C Multiplying and dividing algebraic fractions 2D Solving linear equations 2E Solving equations with algebraic fractions and multiple brackets WhAt Do you knoW ?

linear algebra

1 List what you know about linear equations. Create a concept map to show your list. 2 Share what you know with a partner and then with a small group. 3 As a class, create a large concept map that shows your class’s knowledge of linear equations. eBook plus

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oPening Question

Tod thinks of a number. If 5 is added to this number and then quadrupled, Tod’s answer is 224. What number did Tod start with?

number AnD AlgebrA • PAtterns AnD AlgebrA

Are you ready?

Try the questions below. If you have difficulty with any of them, extra help can be obtained by completing the matching SkillSHEET. Either search for the SkillSHEET in your eBookPLUS or ask your teacher for a copy. eBook plus

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SkillSHEET 2.1 doc-5183

Like terms 1 Select the like terms from each of the following lists. 1

a abc, 3acb, ab, 2 bc 1

b x2y, -3y, 4 yx2, xy c pq, -2q2p, 2pq2, p2q2

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SkillSHEET 2.2 doc-5184

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28

Collecting like terms 2 Simplify each of the following expressions. a 2x - 5 + 7 - 5x b -3a - 4 - 2a - 5 c 4p - 2q + 8 - 6p Finding the highest common factor 3 Find the highest common factor for each of the following pairs of terms. a 6x and 24y b 6ab and 9abc c -12pq and -20pqr Addition and subtraction of fractions 4 Calculate each of the following. a

2 3

+

3 4

b

7 8

−

5 12

c

4 15

+

3 20

Multiplication of fractions 5 Perform the following multiplications. a

3 4

×

1 3

b

5 12

×

4 15

c 2 × 1 2 15

1

1

b

14 15

÷

21 25

c 2 1 ÷ 3 3 4 8

Division of fractions 6 Calculate each of the following. a

5 8

÷

3 4

maths Quest 10 for the Australian Curriculum

number AND algebra • Patterns and algebra

2A

Substitution When the numerical values of pronumerals are known, they can be substituted into an algebraic expression and the expression can then be evaluated. It can be useful to place any substituted values in brackets when evaluating an expression.

Worked Example 1

If a = 4, b = 2 and c = -7, evaluate the following expressions. a  a - b b  a3 + 9b - c Think a

b

Write a a-b

1

Write the expression.

2

Substitute a = 4 and b = 2 into the expression.

=4-2

3

Simplify.

=2

1

Write the expression.

2

Substitute a = 4, b = 2 and c = -7 into the expression.

= (4)3 + 9(2) - (-7)

3

Simplify.

= 64 + 18 + 7 = 89

b a3 + 9b - c

Worked Example 2

If c = a2 + b2 , calculate c if a = 12 and b = -5. Think

Write

c=

a2 + b2

1

Write the expression.

2

Substitute a = 12 and b = -5 into the expression.

=

(12)2 + ( −5)2

3

Simplify.

=

144 + 25

= 169 = 13

Number laws ■■

■■

Recall from previous studies that when dealing with numbers and pronumerals, particular rules must be obeyed. Before progressing further, let us briefly review the Commutative, Associative, Identity and Inverse Laws. Consider any three pronumerals x, y and z, where x, y and z are elements of the set of Real numbers.

Commutative Law 1. x + y = y + x  (example: 3 + 2 = 5 and 2 + 3 = 5) 2. x - y ò y - x  (example: 3 - 2 = 1 but 2 - 3 = -1) Chapter 2 Linear algebra

29

number AND algebra • Patterns and algebra

3. x ì y = y ì x  (example: 3 ì 2 = 6 and 2 ì 3 = 6) 3

2

4. x ó y ò y ó x  (example: 3 ó 2 = 2 , but 2 ó 3 = 3 ) Therefore, the Commutative Law holds true for addition and multiplication, since the order in which two numbers or pronumerals are added or multiplied does not affect the result. However, the Commutative Law does not hold true for subtraction or division.

Associative Law 1. x + ( y + z) = (x + y) + z  [example: 2 + (3 + 4) = 2 + 7 = 9 and (2 + 3) + 4 = 5 + 4 = 9] 2. x - ( y - z) ò (x - y) - z  [example: 2 - (3 - 4) = 2 - -1 = 3 and (2 - 3) - 4 = -1 - 4 = -5] 3. x ì ( y ì z) = (x ì y) ì z  [example: 2 ì (3 ì 4) = 2 ì 12 = 24 and (2 ì 3) ì 4 = 6 ì 4 = 24] 4. x ó ( y ó z) ò (x ó y) ó z 3

4

8

2

2

1

2

1

[example: 2 ó (3 ó 4) = 2 ó 4 = 2 ì 3 = 3 but (2 ó 3) ó 4 = 3 ó 4 = 3 ì 4 = 12 = 6 ] The Associative Law holds true for addition and multiplication since grouping two or more numbers or pronumerals and calculating them in a different order does not affect the result. However, the Associative Law does not hold true for subtraction or division.

Identity Law The Identity Law states that in general: x+0=0+x=x xì1=1ìx=x In both of the examples above, x has not been changed (that is, it has kept its identity) when zero is added to it or it is multiplied by 1.

Inverse Law x + -x = -x + x = 0 1 1 xì = ìx=1 x x That is, when the additive inverse of a number or pronumeral is added to itself, it equals 0. When the multiplicative inverse of a number or pronumeral is multiplied by itself, it equals 1. The Inverse Law states that in general:

Closure Law A law that you may not yet have encountered is the Closure Law. The Closure Law states that, when an operation is performed on an element (or elements) of a set, the result produced must also be an element of that set. For example, addition is closed on natural numbers (that is, positive integers: 1, 2, 3,  .  .  .) since adding a pair of natural numbers produces a natural number. Subtraction is not closed on natural numbers. For example, 5 and 7 are natural numbers and the result of adding them is 12, a natural number. However, the result of subtracting 7 from 5 is -2, which is not a natural number.

Worked Example 3

Find the value of the following expressions, given the integer values x = 4 and y = -12. Comment on whether the Closure Law for integers holds for each of the expressions when these values are substituted. a  x + y      b  x - y      c  x ì y      d  x ó y Think a

30

Write a x + y = 4 + -12

1

Substitute each pronumeral into the expression.

2

Evaluate and write the answer.



3

Determine whether the Closure Law holds; that is, is the result an integer?

The Closure Law holds for these substituted values.

Maths Quest 10 for the Australian Curriculum

= -8

number AnD AlgebrA • PAtterns AnD AlgebrA b Repeat steps 1–3 of part a.

b x - y = 4 - -12

c Repeat steps 1–3 of part a.

c

d Repeat steps 1–3 of part a.

d x ó y = 4 ó -12

= 16 The Closure Law holds for these substituted values. x ì y = 4 ì -12 = -48 The Closure Law holds for these substituted values. 1

= -3 The Closure Law does not hold for these substituted values since the answer obtained is a fraction, not an integer. ■

It is important to note that, although a particular set of numbers may be closed under a given operation, for example multiplication, another set of numbers may not be closed under that same operation. For example, in part c of Worked example 3, integers were closed under multiplication. However, in some cases, the set of irrational numbers is not closed under multiplication, since 3 ì 3 = 9 = 3. In this example, two irrational numbers produced a rational number under multiplication. remember

1. When the numerical values of pronumerals are known, they can be substituted them into an algebraic expression and the expression can then be evaluated. 2. It is sometimes useful to place any substituted values in brackets when evaluating an expression. 3. When dealing with numbers and pronumerals, particular rules must be obeyed. (a) The Commutative Law holds true for addition and multiplication. (b) The Associative Law holds true for addition and multiplication. (c) The Identity Law states that, in general: x + 0 = x and x ì 1 = x. 1 (d) The Inverse Law states that, in general: x + -x = 0 and x ì = 1. x (e) The Closure Law states that, when an operation is performed on an element (or elements) of a set, the result produced must also be an element of that set. exerCise

2A inDiViDuAl PAthWAys eBook plus

Activity 2-A-1

Substitution doc-4960 Activity 2-A-2

Harder substitution doc-4961 Activity 2-A-3

Tricky substitution doc-4962

substitution fluenCy 1 We1 If a = 2, b = 3 and c = 5, evaluate the following expressions. a a + b b c - b c c - a - b d c - (a - b)

e 7a + 8b - 11c

g abc j c2 + a

h ab(c - b) k -a ì b ì -c

2 If d = -6 and k = -5, evaluate the following. a d + k b d - k d kd e -d(k + 1) g k3

h

k −1 d

a b c + + 2 3 5 i a2 + b2 - c2 l 2.3a - 3.2b f

c k - d f d 2 i

3k - 5d

Chapter 2 linear algebra

31

number AnD AlgebrA • PAtterns AnD AlgebrA

eBook plus

1

1

3 If x = 3 and y = 4 , evaluate the following.

Digital doc

a x + y

b y - x

SkillSHEET 2.7 doc-5189

c xy

d

e x2y3

f

x y 9x y2

4 We2 Calculate the unknown variable in the following real-life mathematical formulas.

a 2 + b 2 , calculate c if a = 8 and b = 15.

a If c = 1

b If A = 2 bh, determine the value of A if b = 12 and h = 5.

c The perimeter, P, of a rectangle is given by P = 2L + 2W. Calculate the perimeter, P, of a d e f g h i

rectangle, given L = 1.6 and W = 2.4. C If T = , determine the value of T if C = 20.4 and L = 5.1. L n +1 If K = , determine the value of K if n = 5. n −1 9C Given F = + 32, calculate F if C = 20. 5 If v = u + at, evaluate v if u = 16, a = 5, t = 6. The area, A, of a circle is given by the formula A = p r2. Calculate the area of a circle, correct to 1 decimal place, if r = 6. 1 If E = 2 mv2, calculate m if E = 40, v = 4.

A , evaluate A to 1 decimal place if r = 14.1. π 5 mC a If p = -5 and q = 4, then pq is equal to: A 20 B 1 j

Given r =

D -20

E

C -1

5 -4

b If c2 = a2 + b2, and a = 6 and b = 8, then c is equal to: A 28 B 100 D 14 E 44 c Given h = 6 and k = 7, then kh2 is equal to: A 294 B 252 D 5776 E 85

C 10

C 1764

unDerstAnDing 6 Knowing the length of two sides of a right-angled triangle, the third side can be calculated

using Pythagoras’ theorem. If the two shorter sides have lengths of 1.5 cm and 3.6 cm, calculate the length of the hypotenuse. 4 7 The volume of a sphere can be calculated using the formula 3 pr3. What is the volume of a sphere with a radius of 2.5 cm? Give your answer correct to 2 decimal places. 2.5 cm

8 A rectangular park is 200 m by 300 m. If Blake runs along the diagonal of the park, how far

will he run? Give your answer to the nearest metre. 32

maths Quest 10 for the Australian Curriculum

number AND algebra • Patterns and algebra Reasoning 9   WE 3  Determine the value of the following expressions, given the integer values x = 1, y = -2

and z = -1. Comment on whether the Closure Law for integers holds true for each of the expressions when these values are substituted. a x + y b y - z c y ì z d x ó z e z - x f x ó y 10 Find the value of the following expressions, given the natural number values x = 8, y = 2 and z = 6. Comment on whether the Closure Law for natural numbers holds true for each of the expressions. a x + y b y - z c y ì z d x ó z e z - x f x ó y 11 For each of the following, complete the relationship to illustrate the stated law. Justify your answer. a (a + 2b) + 4c = _______________ Associative Law b (x ì 3y) ì 5c = _______________ Associative Law c 2p ó q ò _______________ Commutative Law d 5d + q = _______________ Commutative Law reflection    e 3z + 0 = _______________ Identity Law f 2x ì _______ = _______________ Inverse Law Why is knowledge of the g (4x ó 3y) ó 5z ò _______________ Associative Law Commulative Law useful? h 3d - 4y ò _______________ Commutative Law

2B

Adding and subtracting algebraic fractions ■■

To add or subtract algebraic fractions, we perform the following steps. 1. Find the lowest common denominator (LCD) by finding the lowest common multiple (LCM) of the denominators. 2. Rewrite each fraction as an equivalent fraction with this common denominator. 3. Express as a single fraction. 4. Simplify the numerator.

Worked Example 4

Simplify the following expressions. 2x x a   - 2 3

b 

x+1 x+ 4 + 4 6

Think a

Write a

2x x − 3 2

1

Write the expression.

2

Rewrite each fraction as an equivalent fraction using the LCD. The LCM of 3 and 2 is 6.

=

3

Express as a single fraction.

=

4 x − 3x 6

4

Simplify the numerator.

=

x 6

2x 2 x 3 × − × 3 2 2 3 4 x 3x = − 6 6

Chapter 2 Linear algebra

33

number AND algebra • Patterns and algebra

b

x +1 x + 4 + 6 4

1

Write the expression.

2

Rewrite each fraction as an equivalent fraction using the LCD. The LCM of 6 and 4 is 12, not 24. Note that 24 is a common multiple but not the lowest common multiple. If 24 is used as the common denominator, then additional calculations will need to be performed to arrive at the final simplified answer.

=

3

Express as a single fraction.

=

4

Simplify the numerator by expanding brackets and collecting like terms.

=

■■

b

x +1 2 x + 4 3 × + × 6 2 4 3 2( x + 1) 3( x + 4) = + 12 12

2( x + 1) + 3( x + 4) 12

2 x + 2 + 3 x + 12 12 5 x + 14 = 12

If pronumerals appear in the denominator, we can treat these separately from their coefficients. This is demonstrated in the following worked example.

Worked Example 5

Simplify

2 1 . − 3x 4x

Think

Write

Write the expression.

2

Rewrite each fraction as an equivalent fraction using the LCD. The LCM of 3 and 4 is 12. The LCM of x and x is x. So the LCD is 12x, not 12x2. If you take care to ensure that you have found the LCM at this step, then the subsequent mathematics will be simpler.

=

3

Express as a single fraction.

=

8−3 12 x

4

Simplify the numerator.

=

5 12x

■■

34

2 1 − 3x 4 x

1

2 4 1 3 × − × 3x 4 4 x 3 8 3 = − 12 x 12 x

When there is an algebraic expression in the denominator of each fraction, we can obtain a common denominator by writing the product of the denominators. For example, if x + 3 and 2x - 5 are in the denominator of each fraction, then a common denominator of the two fractions will be (x + 3)(2x - 5).

Maths Quest 10 for the Australian Curriculum

number AND algebra • Patterns and algebra

Worked Example 6

Simplify

x + 1 2x − 1 by writing it first as a single fraction. + x+3 x+2

Think

Write

x + 1 2x − 1 + x+3 x+2

1

Write the expression.

2

Rewrite each fraction as an equivalent fraction using the LCD. The LCM of x + 3 and x + 2 is the product (x + 3)(x + 2).

=

( x + 1) ( x + 2) (2 x − 1) ( x + 3) × + × ( x + 3) ( x + 2) ( x + 2) ( x + 3))

=

( x + 1)( x + 2) (2 x − 1)( x + 3) + ( x + 3)( x + 2) ( x + 3)( x + 2)

3

Express as a single fraction.

=

( x + 1)( x + 2) + (2 x − 1)( x + 3) ( x + 3)( x + 2)

4

Simplify the numerator by expanding brackets and collecting like terms. Note: The denominator is generally kept in factorised form. That is, it is not expanded.

=

( x 2 + 2 x + x + 2) + (2 x 2 + 6 x − x − 3) ( x + 3)( x + 2)

=

x 2 + 3x + 2 + 2 x 2 + 5x − 3 ( x + 3)( x + 2)

=

3x 2 + 8x − 1 ( x + 3)( x + 2)

Worked Example 7

Simplify

x+2 x−1 by writing it first as a single fraction. + x − 3 ( x − 3)2

Think 1

Write the expression.

2

Rewrite each fraction as an equivalent fraction using the LCD. The LCM of x - 3 and (x - 3)2 is (x - 3)2 not (x - 3)3.

Write

x+2 x −1 + x − 3 ( x − 3)2 = = =

3

Express as a single fraction.

=

4

Simplify the numerator.

=

x+2 x−3 x −1 × + x − 3 x − 3 ( x − 3)2 ( x + 2)( x − 3) ( x − 3)

2

x2 − x − 6 ( x − 3)2

+

+

x −1 ( x − 3)2

x −1 ( x − 3)2

x2 − x − 6 + x − 1 ( x − 3)2 x2 − 7 ( x − 3)2

Chapter 2 Linear algebra

35

number AnD AlgebrA • PAtterns AnD AlgebrA

remember

1. Algebraic fractions contain pronumerals that may represent particular numbers or changing values. 2. To add or subtract algebraic fractions we perform the following steps. (a) Find the lowest common denominator (LCD) by finding the lowest common multiple (LCM) of the denominators. (b) Rewrite each fraction as an equivalent fraction with this common denominator. (c) Express as a single fraction. (d) Simplify the numerator.

exerCise

2b inDiViDuAl PAthWAys eBook plus

Activity 2-B-1

Introducing algebraic fractions doc-4963 Activity 2-B-2

Working with algebraic fractions doc-4964 Activity 2-B-3

Advanced algebraic fractions doc-4965

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Adding and subtracting algebraic fractions fluenCy 1 Simplify each of the following.

4 2 1 5 b + + 7 3 8 9 4 3 3 2 d e − − 9 11 7 5 5x 4 3 2x g − h − 9 27 8 5 2 We4 Simplify the following expressions. 2y y y y a − b − 3 4 8 5 8x 2 x 2w w d e + − 9 3 14 28 12 y y 10 x 2 x g h + + 5 7 5 15 x+2 x+6 2x − 1 2x + 1 j k + − 4 3 5 6 a

3 We5 Simplify the following. eBook plus

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SkillSHEET 2.8 doc-5190

2 + 4x 12 d + 5x 2 g 100 x a

1 8x 4 15 x +

7 20 x

3 1 − 4 x 3x 1 1 e + 6 x 8x 1 5 h + 10 x x b

4 We6, 7 Simplify the following by writing as single fractions.

36

a

2 3x + ( x + 4) ( x − 2)

b

2x 5 + ( x + 5) ( x − 1)

c

5 x + (2 x + 1) ( x − 2)

d

2x 3 − ( x + 1) (2 x − 7)

e

4x 3x + ( x + 7) ( x − 5)

f

x+2 x −1 + x +1 x + 4

maths Quest 10 for the Australian Curriculum

3 6 + 5 15 1 x f − 5 6 5 2 i − x 3 c

4x x − 3 4 y y f − 20 4 x +1 x + 3 i + 5 2 3x + 1 5x + 2 l + 2 3 c

5 1 + 3x 7 x 9 9 f − 4 x 5x 4 3 i − 3x 2 x c

number AND algebra • Patterns and algebra

x + 8 2x + 1 − x +1 x+2 x + 1 2x − 5 i − x + 2 3x − 1 4 3 + k 2 x +1 ( x + 1)

h

g

2c

j l

x+5 x −1 − x+3 x−2 2 3 − x −1 1− x 3 1 − x − 1 ( x − 1)2

reflection 

 

Why can't we just add the numerators and the denominators of fractions: a c a +c + = ? b d b +d

Multiplying and dividing algebraic fractions ■■

■■

When multiplying algebraic fractions, first cancel any common factors if possible, then multiply the numerators together and finally multiply the denominators together. Simplify the expression further if necessary. 2 x 4 xy 2 x × 4 xy For example, = × 3y × 7 3y 7 8x 2 (Cancel y from the numerator and denominator.) 21 When dividing algebraic fractions, change the division sign to a multiplication sign and write the following fraction as its reciprocal (swap the numerator and the denominator). 8x 4 x 8x 5 For example, = (The process then follows that for multiplication.) ÷ × 3 5 3 4x =



=

10 3

(Cancel 4x from the numerator and denominator.)

Worked Example 8

Simplify each of the following. 5 y 6z a  × 3x 7 y b 

2x x+1 × ( x + 1)(2 x − 3) x

Think a

Write a

5y 6z × 3x 7 y

1

Write the expression.

2

Examine the fractions and see if you can cancel any common factors in the numerator and denominator. The y can be cancelled in the denominator and the numerator. Also the 3 in the denominator can divide into the 6 in the numerator.

=

5 2z × x 7

3

Multiply the numerators, then multiply the denominators.

=

10 z 7x Chapter 2 Linear algebra

37

number AND algebra • Patterns and algebra

b

b

2x x +1 × ( x + 1)(2 x − 3) x

1

Write the expression.

2

Check for common factors in the numerator and the denominator. (x + 1) and the x are common in the numerator and the denominator and can therefore be cancelled.

=

2 1 × (2 x − 3) 1

3

Multiply the numerators, then multiply the denominators.

=

2 2x − 3

Worked Example 9

Simplify the following expressions. 3 xy 4 x a  ÷ 2 9y

b 

4 x−7 ÷ ( x + 1)(3 x − 5) x + 1

Think a

b

38

Write a

3 xy 4 x ÷ 2 9y

1

Write the expression.

2

Change the division sign to a multiplication sign and write the second fraction as its reciprocal.

=

3 xy 9 y × 2 4x

3

Check for common factors in the numerator and denominator and cancel. The pronumeral x is common to both the numerator and denominator and can therefore be cancelled.

=

3y 9 y × 2 4

4

Multiply the numerators, then multiply the denominators.

=

27 y 2 8

1

Write the expression.

2

Change the division sign to a multiplication sign and write the second fraction as its reciprocal.

=

4 x +1 × ( x + 1)(3 x − 5) x − 7

3

Check for common factors in the numerator and denominator and cancel. (x + 1) is common to both the numerator and denominator and can therefore be cancelled.

=

4 1 × 3x − 5 x − 7

4

Multiply the numerators, then multiply the denominators.

=

4 (3 x − 5)( x − 7)

Maths Quest 10 for the Australian Curriculum

b

4 x−7 ÷ ( x + 1)(3 x − 5) x + 1

number AnD AlgebrA • PAtterns AnD AlgebrA

remember

1. When multiplying algebraic fractions, first cancel any common factors if possible, then multiply the numerators together and finally multiply the denominators together. Simplify the expression further if necessary. 2. When dividing algebraic fractions, change the division sign to a multiplication sign and write the following fraction as its reciprocal (swap the numerator and the denominator). The process then follows that for multiplication.

exerCise

2C inDiViDuAl PAthWAys eBook plus

Activity 2-C-1

Learning operations with algebraic fractions doc-4966 Activity 2-C-2

Operations with algebraic fractions doc-4967

multiplying and dividing algebraic fractions fluenCy 1 We8a Simplify each of the following.

j

Activity 2-C-3

Advanced operations with algebraic fractions doc-4968

eBook plus

eBook plus

eBook plus

Digital doc

5y x × 3x 8y

k

c

−20 y −21z × 7x 5y

l

y x × −3w 2 y

a

2x x −1 × ( x − 1)(3 x − 2) x

b

5x 4x + 7 × ( x − 3)(4 x + 7) x

c

9x 5x + 1 × (5 x + 1)( x − 6) 2x

d

( x + 4) x +1 × ( x + 1)( x + 3) x + 4

e

2x x −1 × x + 1 ( x + 1)( x − 1)

f

2 x ( x + 1) × x (2 x − 3) 4

g

2x 3a × 4(a + 3) 15 x

h

15c 21d × 12(d − 3) 6c

j

7 x 2 ( x − 3) 3( x − 3)( x + 1) × 5 x ( x + 1) 14( x − 3)2 ( x − 1)

Digital doc

SkillSHEET 2.9 doc-5191

y 16 × 4 x 3w −7 f × −14 x x −9 z × i 3z 2y

b

2 We8b Simplify the following expressions.

Digital doc

SkillSHEET 2.5 doc-5187

x 12 × 4 y x −25 e × 10 2y −y 6z × h 3 x −7 y

x 20 × y 5 x 9 × d 2 2y 3 y 8z g × 4 x 7y a

i

6x2 20( x − 2)2

×

15( x − 2) 16 x 4

3 We9a Simplify the following expressions. a

3 5 ÷ x x

b

d

20 20 ÷ y 3y

e

g

3 xy 3 x ÷ 7 4y

j

8wx 3w ÷ 5 4y

SkillSHEET 2.10 doc-5192

2 9 ÷ x x

1 5 ÷ 5w w 2 xy 5 x h ÷ 5 y k

2 xy 3 xy ÷ 5 5

c

4 12 ÷ x x

7 3 ÷ 2 x 5x 6 y 3x ÷ i 9 4 xy f

l

10 xy 20 x ÷ 7 14 y

Chapter 2 linear algebra

39

number AnD AlgebrA • PAtterns AnD AlgebrA 4 We9b Simplify the following expressions. a

9 x+3 ÷ ( x − 1)(3 x − 7) x − 1

b

1 x−9 ÷ ( x + 2)(2 x − 5) 2 x − 5

refleCtion 

12( x − 3)2 4( x − 3) c ÷ ( x + 5)( x − 9) 7( x − 9) 13 3( x + 1) d ÷ 2 2 ( x − 4)( x − 1) 6( x − 4) ( x − 1)

eBook plus

Digital doc WorkSHEET 2.1 doc-5193

Is

3

x +2 Explain.

the same as

■ ■ ■

Solve the following equations. a a + 27 = 71

b

e = 0.87

c

40

1

Write the equation.

2

27 has been added to a to result in 71. The addition of 27 has to be undone. Therefore, subtract 27 from both sides of the equation to obtain the solution.

1

Write the equation.

2

Express 3 14 as an improper fraction.

3

The pronumeral d has been divided by 16 to result in 13 . Therefore 4 the division has to be undone by multiplying both sides of the equation by 16 to obtain d.

1

Write the equation.

2

The square root of e has been taken to result in 0.87. Therefore, the square root has to be undone by squaring both sides of the equation to obtain e.

maths Quest 10 for the Australian Curriculum

d 1 = 34 16

d f2 =

think

b

x +2

+

1

x +2

+

1

x +2

?

Equations are algebraic sentences that can be solved to give a numerical solution. An equation consists of two algebraic expressions joined by an equals sign. Remember, to solve any equation we need to isolate the pronumeral we wish to find; that is, we must undo all the operations that have been performed on the pronumeral.

WorkeD exAmPle 10

a

1

solving linear equations

2D

c

 

4 25

Write a

a + 27 = 71 a + 27 - 27 = 71 - 27 a = 44

b

d 1 = 34 16 d 13 = 16 4 d 13 ì 16 = ì 16 16 4 d = 52

c

e = 0.87

( e)

2

= 0.872 e = 0.7569

number AND algebra • Patterns and algebra

d

1

Write the equation.

2

The pronumeral f has been squared 4 to result in 25 . Therefore the squaring has to be undone by taking the square root of both sides of the equation to obtain f. Note that there are two possible solutions, one positive and one negative, since two negative numbers can also be multiplied together to produce a positive one.

■■ ■■

d

4 25

f2=

4 25

f=± 2

f = ê 5

Each of the equations in Worked example 10 was a one-step equation. Remember that in two-step equations, the reverse order of operations must be applied; that is, address addition and subtraction first, then multiplication and division, then exponents and roots and, lastly, any bracketed numbers.

Worked Example 11

Solve the following. a  5y - 6 = 79

b 

Think a

b

Write

1

Write the equation.

2

Add 6 to both sides of the equation.

3

Divide both sides of the equation by 5 to obtain y.

1

Write the equation.

2

Multiply both sides of the equation by 9.

3

Divide both sides of the equation by 4 to obtain x.

5y - 6 = 79

a

5y - 6 + 6 = 79 + 6 5y = 85

b



Express the improper fraction as a mixed number fraction.

5 y 85 = 5 5 y = 17 4x =5 9

4x ì9=5ì9 9 4x = 45

4

4x =5 9



4 x 45 = 4 4 45 x= 4 1

x = 11 4

Equations where the pronumeral appears on both sides ■■

In solving equations where the pronumeral appears on both sides add or subtract one of the pronumeral terms so that it is eliminated from one side of the equation. Chapter 2 Linear algebra

41

number AND algebra • Patterns and algebra

Worked Example 12

Solve the following equations. a  5h + 13 = 2h - 2 c  2(x - 3) = 5(2x + 4)

b  14 - 4 d = 27 - d

Think a

b

c

42

Write a 5h + 13 = 2h - 2

1

Write the equation.

2

Eliminate the pronumeral from the right-hand side by subtracting 2h from both sides of the equation. Note that it is also possible to instead subtract 5h from both sides, leaving -3h on the right-hand side. However, it simpler to work with positive pronumerals.

3

Subtract 13 from both sides of the equation.

3h = -15

4

Divide both sides of the equation by 3 and write your answer.

h = -5

1

Write the equation.

2

Create a single pronumeral term by adding 4d to both sides of the equation.

3

Subtract 27 from both sides of the equation.

-13 = 3d

4

Divide both sides of the equation by 3.

−

5

Express the improper fraction as a mixed number fraction.

-4 3 = d

6

Write your answer so that d is on the left-hand side.

1

Write the equation.

2

Expand the brackets on both sides of the equation.

2x - 6 = 10x + 20

3

Isolate the pronumeral on the righthand side by subtracting 2x from both sides of the equation.

2x - 2x - 6 = 10x - 2x + 20

4

Subtract 20 from both sides of the equation.

5

Divide both sides of the equation by 8.

6

Simplify and write your answer with the pronumeral on the left.

Maths Quest 10 for the Australian Curriculum

3h + 13 = -2

b 14 - 4d = 27 - d

14 = 27 + 3d

13 =d 3 1

1

d = -4 3 c

2(x - 3) = 5(2x + 4)

-6 - 20 = 8x + 20 - 20 -26 = 8x −

26 =x 8 x=−

13 4

number AnD AlgebrA • PAtterns AnD AlgebrA

remember

1. Equations are algebraic sentences that can be solved to give a numerical solution. 2. Equations are solved by undoing any operation that has been performed on the pronumeral. 3. When solving two-step equations, the reverse order of operations must be applied. exerCise

2D inDiViDuAl PAthWAys eBook plus

Activity 2-D-1

Simple puzzling equations doc-4969 Activity 2-D-2

Puzzling equations doc-4970 Activity 2-D-3

Advanced puzzling equations doc-4971

solving linear equations fluenCy 1 We10a Solve the following equations. a a + 61 = 85 b k - 75 = 46 d r - 2.3 = 0.7 e h + 0.84 = 1.1 g t - 12 = -7

h q +

1 3

=

1 2

c g + 9.3 = 12.2 f i + 5 = 3 i

x - 2 = -2

2 We10b Solve the following equations. a

f =3 4

d 9v = 63

b

i = -6 10

c 6z = -42

k 5 = 12 6 y 3 i =58 4

e 6w = -32

f

m 7 = 19 8 3 We10c, d Solve the following equations. g 4a = 1.7

a

t = 10

d f 2 = 1.44 g

g=

15 22

4 We11a Solve the following. a 5a + 6 = 26 d 7f - 18 = 45 g 6s + 46 = 75

h

b y2 = 289 e

h=

h j2 =

c

4 7

f p2 =

196 961

i

b 6b + 8 = 44 e 8q + 17 = 26 h 5t - 28 = 21

5 Solve the following.

f + 6 = 16 4 r c +6=5 10 n e + 5 = 8.5 8

q = 2.5 9 64 7

a2 = 2 9

c 8i - 9 = 15 f 10r - 21 = 33 i 8a + 88 = 28

g +4=9 6 m d - 12 = -10 9 p f - 1.8 = 3.4 12

a

b

6 Solve the following. a 6(x + 8) = 56 c 5(m - 3) = 7 e 5(3n - 1) = 80

b 7( y - 4) = 35 d 3(2k + 5) = 24 f 6(2c + 7) = 58

7 We11b Solve the following.

3k = 15 5 8u d = -3 11 a

9m = 18 8 11x e =2 4 b

7p = -8 10 4v f = 0.8 15 c

Chapter 2 linear algebra

43

number AND algebra • Patterns and algebra

p + 2 = 7 is: 5 A p = 5 B p = 25 D p = 10 E p = 1 b If 5h + 8 = 53, then h is equal to: 1 A 5 B 12.2

8   MC  a  The solution to the equation

C p = 45

C 225

D 10 E 9 c The exact solution to the equation 14x = 75 is: A x = 5.357 142 857

B x = 5.357 (to 3 decimal places)

D x = 5.4

E x = 5.5

9 Solve the following equations. a -x = 5 d -7 - x = 4

v =4 5 10 Solve the following equations. a 6 - 2x = 8 g −

d -3 - 2g = 1

b 2 - d = 3 e -5h = 10

c 5 - p = -2 f -6t = -30

h −

i

r 1 = 12 4

12

13

14

c 9 - 6l = -3

e -5 - 4t = -17

f −

3e = 14 5 4f i − +1=8 7

8j k =9 h − - 3 = 6 3 4   WE 12a  Solve the following equations. a 6x + 5 = 5x + 7 b 7b + 9 = 6b + 14 c 11w + 17 = 6w + 27 d 8f - 2 = 7f + 5 e 10t - 11 = 5t + 4 f 12r - 16 = 3r + 5 g 12g - 19 = 3g - 31 h 7h + 5 = 2h - 6 i 5a - 2 = 3a - 2   WE 12b  Solve the following equations. a 5 - 2x = 6 - x b 10 - 3c = 8 - 2c c 3r + 13 = 9r - 3 d k - 5 = 2k - 6 e 5y + 8 = 13y + 17 f 17 - 3g = 3 - g g 14 - 5w = w + 8 h 4m + 7 = 8 - m i 14 - 5p = 9 - 2p   WE 12c  Solve the following equations. a 3(x + 5) = 2x b 8( y + 3) = 3y c 6(t - 5) = 4(t + 3) d 10(u + 1) = 3(u - 3) e 12( f - 10) = 4( f - 5) f 2(4r + 3) = 3(2r + 7) g 5(2d + 9) = 3(3d + 13) h 5(h - 3) = 3(2h - 1) i 2(4x + 1) = 5(3 - x)   MC  a  The solution to 8 - 4k = -2 is: 1 1 1 A k = 2 2 B k = -2 2 C k = 1 2 1

D k = -1 2 b The solution to − 1

A n = 3 3

E k =

2 5

6n + 3 = -7 is: 5

1

D n = 8 3

1

B n = -3 3

A p =

2 5

2

E p =

4 5

Maths Quest 10 for the Australian Curriculum

1 3

1

4

B p = 2 5

C p = 4 3

C n =

E n = -8 3

c The solution to p - 6 = 8 - 4p is:

44

-4g = 3.2

b 10 - 3v = 7

g − 11

5

C x = 514

D p =

2 3

reflection 

 

Describe in one sentence what it means to solve linear equations.

number AnD AlgebrA • PAtterns AnD AlgebrA

solving equations with algebraic fractions and multiple brackets

2e eBook plus

Interactivity Solving euqations

int-2778

equations with multiple brackets Many equations need to be simplified by expanding brackets and collecting like terms before they are solved. Doing this reduces the equation to one of the basic types covered in the previous exercise.

WorkeD exAmPle 13

Solve each of the following linear equations. a 6(x + 1) - 4(x - 2) = 0 b 7(5 - x) = 3(x + 1) - 10 think a

b

Write a 6(x + 1) - 4(x - 2) = 0

1

Write the equation.

2

Expand all the brackets. (Be careful with the -4.)

3

Collect like terms.

4

Subtract 14 from both sides of the equation.

5

Divide both sides of the equation by 2 to find the value of x.

1

Write the equation.

2

Expand all the brackets.

35 - 7x = 3x + 3 - 10

3

Collect like terms.

35 - 7x = 3x - 7

4

Create a single pronumeral term by adding 7x to both sides of the equation.

35 = 10x - 7

5

Add 7 to both sides of the equation.

42 = 10x

6

Divide both sides of the equation by 10 to solve for x and simplify.

42 10

=x

21 5

=x

7

Express the improper fraction as a mixed number fraction.

45 = x

8

Rewrite the equation so that x is on the left-hand side.

6x + 6 - 4x + 8 = 0 2x + 14 = 0 2x = -14 x = -7 b 7(5 - x) = 3(x + 1) - 10

1

1

x = 45

equations involving algebraic fractions ■

■

To solve equations involving algebraic fractions, write every term in the equation as a fraction with the same lowest common denominator. Every term can then be multiplied by this common denominator. This has the effect of eliminating the fraction from the equation. Chapter 2 linear algebra

45

number AND algebra • Patterns and algebra

Worked Example 14

Solve each of the following linear equations. x 3x 1 a  − = 2 5 4 3 4 b  =1− 2x x Think a

b

46

Write

1

Write the equation.

2

The lowest common denominator of 2, 5, and 4 is 20. Write each term as an equivalent fraction with a denominator of 20.

3

Multiply both sides of the equation by 20. This is the same as multiplying each term by 20, which cancels out the 20 in the denominator and effectively removes it.

x 3x 1 = − 2 5 4

a

x 10 3 x 4 × − × = 2 10 5 4 10 x 12 x = − 20 20

1 5 × 4 5 5 20

 10 x 12 x  5  20 − 20  ì 20 = 20 ì 20 10 x 12 x 5 ì 20 ì 20 = ì 20 20 20 20 10x - 12x = 5 -2x = 5

4

Simplify the left-hand side of the equation by collecting like terms.

5

Divide both sides of the equation by -2 to solve for x.

x = - 5

6

Express the improper fraction as a mixed number fraction.

x = -2 12

1

Write the equation.

2

The lowest common denominator of 2x and x is 2x. Write each term as an equivalent fraction with a denominator of 2x.

3

Multiply each term by 2x. This effectively removes the denominator.

3 = 2x - 8

4

Add 8 to both sides of the equation.

11 = 2x

5

Divide both sides of the equation by 2 to solve for x.



6

Express the improper fraction as a mixed number.

7

Rewrite the equation so that x is on the lefthand side.

Maths Quest 10 for the Australian Curriculum

2

b



3 4 =1x 2x 3 1 2x 4 2 = × − × 2x 1 2 x x 2 3 2x 8 = − 2x 2 x 2 x

11 2

=x

1

52 = x 1

x = 52

number AND algebra • Patterns and algebra

Worked Example 15

Solve each of the following linear equations. 4 1 5( x + 3) 3( x − 1) = a  b  =4+ 3 ( x − 1 ) x +1 6 5 Think a

b

Write

5( x + 3) 3( x − 1) =4+ 6 5

a

1

Write the equation.

2

The lowest common denominator of 5 and 6 is 30. Write each term as an equivalent fraction with a common denominator of 30.

3

Multiply each term by 30. This effectively removes the denominator.

25(x + 3) = 120 + 18(x - 1)

4

Expand the brackets and collect like terms.

25x + 75 = 120 + 18x - 18 25x + 75 = 102 + 18x

5

Subtract 18x from both sides of the equation.

6

Subtract 75 from both sides of the equation.

7

Divide both sides of the equation by 7 to solve for x.

x=

8

Express the improper fraction as a mixed number.

x=

1

Write the equation.

2

The lowest common denominator of 3, x + 1 and x - 1 is 3(x - 1)(x + 1). Write each term as an equivalent fraction with a common denominator of 3(x - 1)(x + 1).

3

Multiply each term by the common denominator.



4(x + 1) = 3(x - 1)

4

Expand the brackets.



4x + 4 = 3x - 3

5

Subtract 3x from both sides of the equation.

6

Subtract 4 from both sides of the equation to solve for x.

25( x + 3) 120 18( x − 1) = + 30 30 30

7x + 75 = 102

7x = 27 27 7 6 37

4 1 = 3( x − 1) x + 1

b

4( x + 1) 3( x − 1) = 3( x − 1)( x + 1) 3( x − 1)( x + 1)

x + 4 = -3

x + 4 - 4 = -3 - 4 x = -7

remember

1. For equations involving brackets, expand the brackets and collect like terms. This will reduce the equation to a more basic type. 2. For complicated algebraic fraction equations, the following steps may be used. (a) Write each term in the equation as an equivalent fraction with the lowest common denominator. (b) Multiply each term by the common denominator. This has the effect of removing the fraction from the equation. (c) Continue to solve the equation using the same methods as for a basic algebraic equation. Chapter 2 Linear algebra

47

number AnD AlgebrA • PAtterns AnD AlgebrA

exerCise

2e inDiViDuAl PAthWAys eBook plus

Activity 2-E-1

Algebraic equations with fractions doc-4972 Activity 2-E-2

Harder algebraic equations with fractions doc-4973 Activity 2-E-3

Tridy algebraic equations with fractions doc-4974

solving equations with algebraic fractions and multiple brackets fluenCy 1 We13 Solve each of the following linear equations. a 6(4x - 3) + 7(x + 1) = 9 b 9(3 - 2x) + 2(5x + 1) = 0 c 8(5 - 3x) - 4(2 + 3x) = 3 d 9(1 + x) - 8(x + 2) = 2x e 6(4 + 3x) = 7(x - 1) + 1 f 10(4x + 2) = 3(8 - x) + 6 g 8(x + 4) + 2(x - 3) = 6(x + 1) h 7.2(3x - 1) + 2.3(5 - x) = -34.3 i 6(2x - 3) - 2(6 - 3x) = 7(2x - 1) j 9(2x - 5) + 5(6x + 1) = 100 k 5(2x - 1) - 3(6x + 1) = 8 l 7(2x + 7) - 5(2x + 1) = 2(4 - x) 2 Solve each of the following linear equations.

7x + 5 12 − 5 x = 11 b = -13 3 6 3x − 2 8x + 3 c = 5x d = 2x 5 4 2x − 1 x − 3 4x + 1 x + 2 e f = = 5 4 3 4 6 − x 2x − 1 8 − x 2x + 1 g h = = 3 5 9 3 5x − 3 1 i - =0 2 8 3 We14 Solve each of the following linear equations. x 4x 1 x x 3 a b − = + = 3 5 3 4 5 4 x 4x −3 x x 1 c d − =2 + = 4 7 5 8 4 2 x x −3 5x 2x e - = f -8= 3 6 4 8 3 2 x 3x 4 1 2 g h − = − = x 6 x 7 8 8 15 2 1 4 5 i j + = -4= x x 3 x x 2x − 4 x 4 x − 1 2x + 5 k l +6= =0 5 2 2 3 4 We15 Solve each of the following linear equations. 3( x + 1) 5( x + 1) 2( x + 1) 3(2 x − 5) a + =4 b + =0 2 3 7 8 2(4 x + 3) 6( x − 2) 1 8( x + 3) 3( x + 2) c d = = 5 2 2 5 4 a

48

maths Quest 10 for the Australian Curriculum

number AnD AlgebrA • PAtterns AnD AlgebrA

e g eBook plus

Digital doc WorkSHEET 2.2 doc-5194

i k

5(7 − x ) 2(2 x − 1) = +1 2 7 −5( x − 2) 6(2 x − 1) 1 = 3 5 3 1 3 8 + = x −1 x +1 x +1 1 3 −1 - = x − 1 x x −1

f h j l

2(6 − x ) 9( x + 5) 1 = + 3 6 3 9(2 x − 1) 4( x − 5) = 7 3 3 5 5 + = x +1 x − 4 x +1 4 5 −1 - = x 2x − 1 x

refleCtion 

 

Do the rules for the order of operations apply to algebraic functions? Explain.

Chapter 2 linear algebra

49

number AND algebra • Patterns and algebra

Summary Substitution ■■ ■■ ■■

When the numerical values of pronumerals are known, they can be substituted them into an algebraic expression and evaluated. It is sometimes useful to place any substituted values in brackets when evaluating an expression. When dealing with numbers and pronumerals, particular rules must be obeyed. (a) The Commutative Law holds true for addition and multiplication. (b) The Associative Law holds true for addition and multiplication. (c) The Identity Law states that, in general: x + 0 = x and x ì 1 = x. 1 (d) The Inverse Law states that, in general: x + -x = 0 and x ì = 1. x (e) The Closure Law states that, when an operation is performed on an element (or elements) of a set, the result produced must also be an element of that set. Adding and subtracting algebraic fractions

■■ ■■

Algebraic fractions contain pronumerals that may represent particular numbers or changing values. To add or subtract algebraic fractions we perform the following steps. (a) Find the lowest common denominator (LCD) by finding the lowest common multiple (LCM) of the denominators. (b) Rewrite each fraction as an equivalent fraction with this common denominator. (c) Express as a single fraction. (d) Simplify the numerator. Multiplying and dividing algebraic fractions

■■

■■

When multiplying algebraic fractions, first cancel any common factors if possible, then multiply the numerators together and finally multiply the denominators together. Simplify the expression further if necessary. When dividing algebraic fractions, change the division sign to a multiplication sign and write the following fraction as its reciprocal (swap the numerator and the denominator). The process then follows that for multiplication. Solving linear equations

■■ ■■ ■■

Equations are algebraic sentences that can be solved to give a numerical solution. Equations are solved by undoing any operation that has been performed on the pronumeral. When solving two-step equations, the reverse order of operations must be applied. Solving equations with algebraic fractions and multiple brackets

■■ ■■

50

For more complicated equations involving brackets, expand the brackets and collect like terms. This will reduce the equation to a more basic type. For complicated algebraic fraction equations, the following steps may be used. (a) Write each term in the equation as an equivalent fraction with the lowest common denominator. (b) Multiply each term by the common denominator. This has the effect of removing the fraction from the equation. (c) Continue to solve the equation using the same methods as for a basic algebraic equation.

Maths Quest 10 for the Australian Curriculum

number AnD AlgebrA • PAtterns AnD AlgebrA

MAPPING YOUR UNDERSTANDING

Homework Book

Using terms from the summary, and other terms if you wish, construct a concept map that illustrates your understanding of the key concepts covered in this chapter. Compare your concept map with the one that you created in What do you know? on page 27. Have you completed the two Homework sheets, the Rich task and two Code puzzles in your Maths Quest 10 Homework Book?

Chapter 2 linear algebra

51

number AND algebra • Patterns and algebra

Chapter review Fluency

10 Simplify the following.

1

1 Given E = 2 mv2 where m = 0.2 and v = 0.5, the

value of E is: A 0.000  625 C 0.005 E 0.0025

B 0.1 D 0.025

2 The expression -6d + 3r - 4d - r simplifies to: A 2d + 2r B -10d + 2r C -10d - 4r D 2d + 4r E -8dr 3 The expression 5(2f + 3) + 6(4f - 7) simplifies to: A 34f + 2 B 34f - 4 C 34f - 27 D 34f + 14 E 116f -14 4 The expression 7(b - 1) - (8 - b) simplifies to: A 8b - 9 B 8b - 15 C 6b - 9 D 6b - 15 E 8b + 1 5 If 14p - 23 = 6p - 7 then p equals: A -3 b -1 c 1 d 2 e 4 6 Simplify the following by collecting like terms. a 3c - 5 + 4c - 8 b -3k + 12m - 4k - 9m c -d + 3c - 8c - 4d d 6y2 + 2y + y2 - 7y 1

7 If A = 2 bh, determine the value of A if b = 10 and

h = 7.

8 For each of the following, complete the relationship

to illustrate the stated law. a (a + 3b) + 6c = _______ b 12a - 3b ò _______ c 7p ì _______ = _______ d (x ì 5y) ì 7z = _______ e 12p + 0 = _______ f (3p ó 5q) ó 7r ò _______ g 9d + 11e = _______ h 4a ó b ò _______

Associative Law Commutative Law Inverse Law Associative Law Identity Law Associative Law Commutative Law Commutative Law

9 Find the value of the following expressions given

the natural number values x = 12, y = 8 and z = 4. Comment on whether the Closure Law holds for each of the expressions when the values are substituted. a x ì y b z ó x c y - x

52

Maths Quest 10 for the Australian Curriculum

5y y x+4 x+2 − b + 3 2 5 2 x − 1 2x − 5 5 1 + c d − x+3 x+2 3x 5x 11 Simplify the following. y 32 a × 4 x 20 y 35z b × 7 x 16 y x+6 5( x + 1) × c ( x + 1)( x + 3) x+6 a

25 30 ÷ x x xy 10 x e ÷ y 5 2x 9x +1 ÷ f ( x + 8)( x − 1) x + 8 d

12 Solve the following equations. a p - 20 = 68 b s - 0.56 = 2.45

r = -5 7 f 2(x + 5) = -3

c 3b = 48

d

x = 12 y g - 3 = 12 4 i 5 - k = -7

h a2 = 36

e

13 Solve the following. a 42 - 7b = 14 b 12t - 11 = 4t + 5 c 2(4p - 3) = 2(3p - 5) 14 Solve each of the following linear equations. a 5(x - 2) + 3(x + 2) = 0 b 7(5 - 2x) - 3(1 - 3x) = 1 c 5(x + 1) - 6(2x - 1) = 7(x + 2) d 8(3x - 2) + (4x - 5) = 7x e 7(2x - 5) - 4(x + 20) = x - 5 f 3(x + 1) + 6(x + 5) = 3x + 40 15 Solve each of the following equations.

x x 3 + = 2 5 5 1 x x = − c − 21 7 6 2x − 3 3 x + 3 e − = 2 5 5 2( x + 2) 3 5( x + 1) = + f 3 7 3 a

x x − =3 3 5 3 2 5 + = d x 5 x b

number AnD AlgebrA • PAtterns AnD AlgebrA Problem solVing 1 A production is in town and many parents are

taking their children. An adult ticket costs $15 and a child’s ticket costs $8. Every child must be accompanied by an adult and each adult can have no more than 4 children with them. It costs the company $12 per adult and $3 per child to run the production. There is a seating limit of 300 people and all tickets are sold. a Determine how much profit the company makes on each adult ticket and on each child’s ticket. b To maximise profit, the company should sell as many children’s tickets as possible. Of the 300 available seats, determine how many should be allocated to children if there is a maximum of 4 children per adult. c Using your answer to part b, determine how many adults would make up the remaining seats. d Construct an equation to represent the profit that the company can make depending on the number of children and adults attending the production. e Substitute your values to calculate the maximum profit the company can make.

2 You are investigating prices for having business

cards printed for your new games store. A local printing company charges a flat rate of $250 for the materials used and $40 per hour for labour. a If h is the number of hours of labour required to print the cards, construct an equation for the cost of the cards, C. b You have budgeted $1000 for the printing job. How many hours of labour can you afford? Give your answer to the nearest minute. c The printer estimates that it can print 1000 cards per hour of labour. How many cards will be printed with your current budget? d An alternative to printing is photocopying. The company charges 15 cents per side for the first 10 000 cards and then 10 cents per side for the remaining cards. Which is the cheaper option for 18 750 single-sided cards and by how much? eBook plus

Interactivities

Test yourself Chapter 2 int-2831 Word search Chapter 2 int-2829 Crossword Chapter 2 int-2830

Chapter 2 linear algebra

53

eBook plus

ACtiVities

Chapter opener Digital doc

• Hungry brain activity Chapter 2 (doc-5182) (page 27) Are you ready?

(page 28) • SkillSHEET 2.1 (doc-5183): Like terms • SkillSHEET 2.2 (doc-5184): Collecting like terms • SkillSHEET 2.3 (doc-5185): Finding the highest common denominator • SkillSHEET 2.4 (doc-5186): Addition and subtraction of fractions • SkillSHEET 2.5 (doc-5187): Multiplication of fractions • SkillSHEET 2.6 (doc-5188): Division of fractions Digital docs

2A Substitution Digital docs

• Activity 2-A-1 (doc-4960): Substitution (page 31) • Activity 2-A-2 (doc-4961): Harder substitution (page 31) • Activity 2-A-3 (doc-4962): Tricky substitution (page 31) • SkillSHEET 2.7 (doc-5189): Order of operations (page 32) 2B Adding and subtracting algebraic fractions

(page 36) • Activity 2-B-1 (doc-4963): Introducing algebraic fractions • Activity 2-B-2 (doc-4964): Working with algebraic fractions • Activity 2-B-3 (doc-4965): Advanced algebraic fractions • SkillSHEET 2.4 (doc-5186): Addition and subtraction of fractions • SkillSHEET 2.8 (doc-5190): Writing equivalent algebraic fractions with the lowest common denominator Digital docs

2C Multiplying and dividing algebraic fractions Digital docs

• Activity 2-C-1 (doc-4966): Learning operations with algebraic fractions (page 39) • Activity 2-C-2 (doc-4967): Operations with algebraic fractions (page 39)

54

maths Quest 10 for the Australian Curriculum

• Activity 2-C-3 (doc-4968): Advanced operations with algebraic fractions (page 39) • SkillSHEET 2.5 (doc-5187): Multiplication of fractions • SkillSHEET 2.9 (doc-5191): Simplification of algebraic fractions (page 39) • SkillSHEET 2.10 (doc-5192): Division of fractions (page 39) • WorkSHEET 2.1 (doc-5193): Algebraic fractions (page 40) 2D Solving linear equations

(page 43) • Activity 2-D-1 (doc-4969): Simple puzzling equations • Activity 2-D-2 (doc-4970): Puzzling equations • Activity 2-D-3 (doc-4971): Advanced puzzling equations Digital docs

2E Solving equations with algebraic fractions and multiple brackets Digital docs

• Activity 2-E-1 (doc-4972): Algebraic equations with fractions (page 48) • Activity 2-E-2 (doc-4973): Harder algebraic equations with fractions (page 48) • Activity 2-E-3 (doc-4974): Tricky algebraic equations with fractions (page 48) • WorkSHEET 2.2 (doc-5194): Solving equations with fractions (page 49) Interactivity

• Solving equations (int-2778) (page 45) Chapter review

(page 53) • Test Yourself Chapter 2 (int-2831): Take the end-ofchapter test to test your progress. • Word search Chapter 2 (int-2829): an interactive word search involving words associated with this chapter • Crossword Chapter 2 (int-2830): an interactive crossword using the definitions associated with the chapter Interactivities

To access eBookPLUS activities, log on to www.jacplus.com.au

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships

3

  3a  Sketching linear graphs   3B  Determining linear equations   3C  The distance between two points on a straight line   3d  The midpoint of a line segment   3E  Parallel and perpendicular lines WhAt Do you knoW ?

Coordinate geometry

1 List what you know about linear graphs and their equations. Create a concept map to show your list. 2 Share what you know with a partner and then with a small group. 3 As a class, create a large concept map that shows your class’s knowledge of linear graphs and their equations. eBook plus

Digital doc

Hungry brain activity Chapter 3 doc-5195

opening Question

How can a knowledge of coordinate geometry help us design structures like this one?

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships

Are you ready?

Try■the■questions■below.■If■you■have■diffi■culty■with■any■of■them,■extra■help■can■be■obtained■by■ completing■the■matching■SkillSHEET.■Either■click■on■the■SkillSHEET■icon■next■to■the■question■ on■your■eBookPLUS■or■ask■your■teacher■for■a■copy. eBook plus

Digital doc

SkillSHEET 3.1 doc-5196

Measuring the rise and run   1  State■the■rise■and■the■run■for■each■of■the■following■straight■line■graphs. y y a  b  4 2 0 -2

eBook plus

Digital doc

2 2

x

0

5

Describing the gradient of a line   2  State■whether■each■line■in■question■1■has■a■positive,■zero,■negative■or■undefi■ned■gradient.

SkillSHEET 3.2 doc-5197

eBook plus

Digital doc

SkillSHEET 3.3 doc-5198

eBook plus

Digital doc

Plotting a line using a table of values   3  Draw■up■a■table■of■values■and■plot■the■graph■for■each■of■the■following■rules. a  y■=■x■+■3 b  y■=■x■-■2 c  y■=■2x

Stating the y-intercept from a graph   4  State■the■y-intercept■for■each■graph■shown■in■question■1.

SkillSHEET 3.4 doc-5199

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Digital doc

SkillSHEET 3.5 doc-5200

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Digital doc

SkillSHEET 3.6 doc-5201

Solving linear equations that arise when finding x- and y-intercepts   5  Consider■the■equation■3y■+■4x■=■12. a  Substitute■x■=■0■and■solve■to■fi■nd■the■value■of■y. b  Substitute■y■=■0■and■solve■to■fi■nd■the■value■of■x. Using Pythagoras’ theorem   6  Find■the■length■of■side■AB. a  B

b  B 13 cm

8m A

56

x

6m

C

maths Quest 10 for the Australian Curriculum

A

12 cm

C

number AND algebra • Linear and non-linear relationships

3A

Sketching linear graphs ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■

A linear graph has an equation that can be written in the standard form y = mx + c, where m is the gradient of the line, and c is the y-intercept. An alternative form of the linear equation is ax + by = k, where a, b and k are constants. The gradient (or slope) is a measure of the steepness of a graph. If the gradient, m, of the line is positive, the graph will have an upward slope to the right. If the gradient, m, of the line is negative, the graph will have a downward slope to the right. The greater the magnitude of the gradient, the steeper the linear graph will be. A linear graph is drawn on a Cartesian plane, with two axes (x and y) meeting at the origin (0, 0). The axes divide the plane into four regions, or quadrants. A point is specified by its x- and y-coordinates. A graph of the line y = 2x + 5 is shown in the figure at right. y y = 2x + 5 One method which can be used to draw a graph of an equation 10 is to simply plot the points on graph paper. Quadrant 2 Quadrant 1 5 The points can be plotted manually, or a graphing calculator can be used to plot the points. -10

0

-5

Quadrant 3

10 x

5

-5

Quadrant 4

-10

Worked Example 1

Plot the linear graph defined by the rule y = 2 x - 5 for the x-values -3, -2, -1, 0, 1, 2, 3. Think 1

2

3

Create a table of values using the given x-values. Find the corresponding y-values by substituting each x-value into the rule. Plot the points on a Cartesian plane and rule a straight line through them. Since the x-values have been specified, the line should only be drawn between the x-values of -3 and 3.

Write

x

-3

-2

-1

0

1

2

y

3  

x

-3

-2

-1

 0

 1

 2

  3 

y

-11

-9

-7

-5

-3

-1

1

y 2 1 0 -3 -2 -1 -1

(3, 1) 1 2 3x (2, -1)

-2 -3 (1, -3) -4 -5 (0, -5) -6 (-1, -7) -7 -8 (-2, -9) -9 -10 y = 2x - 5 (-3, -11) -11 -12 4

Label the graph.

Chapter 3 Coordinate geometry

57

number AND algebra • Linear and non-linear relationships

■■ ■■

A minimum of two points are necessary to plot a straight line. Two methods can be used to plot a straight line: •• Method 1: The x- and y-intercept method. •• Method 2: The gradient–intercept method.

Sketching a straight line using the x- and y-intercept method ■■

■■

As the name implies, this method involves finding the x- and y-intercepts, then joining them to sketch the straight line. If the equation is in the form y = mx + c, the value of c gives the y-intercept.

Worked Example 2

Sketch graphs of the following linear equations by finding the x- and y-intercepts. a  2x + y = 6 b  y = -3x - 12 Think a

Write/draw a 2x + y = 6

1

Write the equation.

2

Find the x-intercept by substituting y = 0.

x-intercept: when y = 0, 2x + 0 = 6 2x = 6 x=3 x-intercept is (3, 0).

3

Find the y-intercept by substituting x = 0.

y-intercept: when x = 0, 2(0) + y = 6 y=6 y-intercept is (0, 6).

4

Rule a straight line passing through both points that have been plotted.

y 2x + y = 6 (0, 6) 0 (3, 0)

b

58

x

5

Label the graph.

1

Write the equation.

2

Find the x-intercept by substituting y = 0.   i  Add 12 to both sides of the equation. ii Divide both sides of the equation by -3.

x-intercept: when y = 0, -3x - 12 = 0 -3x = 12 x = -4 x-intercept is (-4, 0).

3

Find the y-intercept. The equation is in the form y = mx + c, so compare this with our equation to find the y-intercept, c.

c = -12 y-intercept is (0, -12).

Maths Quest 10 for the Australian Curriculum

b y = -3x - 12

number AND algebra • Linear and non-linear relationships

4

y

Rule a straight line passing through both points that have been plotted. (-4, 0)

x

0

(0, -12) y = -3x - 12

5

Label the graph.

The gradient–intercept method ■■

■■

This method is often used if the equation is in the form y = mx + c, where m represents the gradient (slope) of the straight line, and c represents the y-intercept. The steps below outline how to use the gradient–intercept method to sketch a linear graph. Step 1:  Plot a point at the y-intercept. rise . (To write a whole number as a fraction, place Step 2: Write the gradient in the form m = run it over a denominator of 1.) Step 3: Starting from the y-intercept, move up the number of units suggested by the rise (move down if the gradient is negative). Step 4: Move to the right the number of units suggested by the run and plot the ■ second point. Step 5:  Rule a straight line through the two points.

Worked Example 3 2

Plot the graph of y = 5 x - 3 using the gradient-intercept method. Think

Write/DRAW 2

1

Write the equation of the line.

y = 5x - 3

2

Identify the value of c (that is, the y-intercept) and plot this point.

c = -3, so y-intercept: (0, -3).

3

Write the gradient, m, as a fraction. (In this case, it is a fraction already.)

m=

4

5

rise , interpret the numerator of the run fraction as the rise and the denominator as ■ the run. Since m =

Starting from the y-intercept at -3, move 2 units up and 5 units to the right to find the second point. We have still not found the x-intercept.

2 5

So, rise = 2; run = 5.

y x 0 -1 1 2 3 4 5 6 7 8 (5, -1) -2 -3 (0, -3) -4

Chapter 3 Coordinate geometry

59

number AND algebra • Linear and non-linear relationships

6

2

y = 5x - 3

To find the x-intercept, let y = 0.

2

0 = 5x - 3 2

3 = 5x 5

3ì2 =x x=

15 2

15

( 2 , 0) is the x-intercept. 7

Label the graph and draw a line through all the points found.

(152 , 0)

y

x 0 -1 1 2 3 4 5 6 7 8 (5, -1) -2 -3 (0, -3) y = 25 x - 3 -4

Sketching linear graphs of the form y = mx ■■ ■■ ■■

These graphs have the value of c = 0. They pass through the origin (0, 0), providing only one point to plot. This means that a second point must be determined by either: •• choosing any x-value, then calculating the corresponding y-value •• using the gradient–intercept method.

Worked Example 4

Sketch the graph of y = 3x. Think

Write/draw

1

Write the equation.

y = 3x

2

Find the x- and y-intercepts. Note: By recognising the form of this linear equation, y = mx you can simply state that the line passes through the origin, (0, 0).

x-intercept: when y = 0, 0 = 3x x=0 y-intercept: when x = 0, y = 0 Both the x- and y-intercepts are at (0, 0).

3

Find another point to plot by finding the y-value when x = 1.

When x = 1,  y = 3 ì 1         = 3 Another point on the line is (1, 3).

4

Plot the two points (0, 0) and (1, 3) and rule a straight line through them.

y

y = 3x

3 (0, 0)

5

60

Label the graph.

Maths Quest 10 for the Australian Curriculum

(1, 3) 1

x

number AND algebra • Linear and non-linear relationships

Sketching linear graphs of the form y = c and x = a ■■

■■ ■■

■■

It is possible to have an equation for a straight line that contains only an x-term or only a y-term. These equations can be written in the form y = c or x = a, where c and a are both constants. Lines of the form y = c are parallel to the x-axis, having a gradient of zero and a y-intercept of c. Lines of the form x = a are parallel to the y-axis, having an undefined (infinite) gradient and no y-intercept (since they do not cross the y-axis).

Worked Example 5

Sketch graphs of the following linear equations. a  y = -3 b  x = 4 Think a

1

Write the equation.

2

The y-intercept is -3. As x does not appear in the equation, the line is parallel to the x-axis, such that all points on the line have a y-coordinate equal to -3. That is, this line is the set of points (x, -3) where x is an element of the set of real numbers.

3

Sketch a horizontal line through (0, -3).

Write/draw a y = -3

y-intercept = -3

y

x

0

y = -3

(0, -3)

b

4

Label the graph.

1

Write the equation.

2

The x-intercept is 4. As y does not appear in the equation, the line is parallel to the y-axis, such that all points on the line have an x-coordinate equal to 4. That is, this line is the set of points (4, y) where y is an element of the set of real numbers.

3

Sketch a vertical line through (4, 0).

4

b x=4

x-intercept = 4

y

x=4

0

(4, 0)

x

Label the graph.

Chapter 3 Coordinate geometry

61

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships

remember

1.■ The■Cartesian■plane■is■a■grid,■consisting■of■two■axes■(x■and■y),■meeting■at■the■origin■ (0,■0). 2.■ A■location■(point)■is■specifi■ed■by■its■x-■and■y-coordinates. 3.■ A■linear■graph■consists■of■an■infi■nite■set■of■points■that■can■be■joined■to■form■a■straight■ line,■but■to■sketch■a■linear■graph,■the■coordinates■of■only■two■points■are■needed. 4.■ A■linear■rule■or■equation■can■be■used■to■obtain■the■coordinates■of■points■that■belong■to■ its■graph. 5.■ Linear■equations■may■be■written■in■several■different■forms.■The■two■most■common■ forms■are■y■=■mx■+■c■and■ax■+■by■=■k. 6.■ When■a■linear■equation■is■expressed■in■the■form■y■=■mx■+■c,■then■m■represents■the■ gradient■(slope)■of■the■straight■line■and■c■represents■the■y-intercept. 7.■ A■straight■line■with■a■positive■gradient■slopes■upward■to■the■right■and■a■straight■line■ with■a■negative■gradient■slopes■downward■to■the■right. 8.■ The■x-■and■y-intercept■method■allows■us■to■sketch■the■graph■of■any■linear■equation■by■ fi■nding■two■specifi■c■points:■the■x-intercept■and■y-intercept.■An■exception■is■the■case■of■ lines■passing■through■the■origin. 9.■ Graphs■of■equations■in■the■form■y■=■mx■pass■through■the■origin.■To■fi■nd■the■second■ point,■substitute■a■chosen■x-value■into■the■equation■to■fi■nd■the■corresponding■y-value. 10.■ Graphs■of■equations■in■the■form■y■=■c■have■a■gradient■of■zero■and■are■parallel■to■the■ x-axis. 11.■ Graphs■of■equations■in■the■form■x■=■a■have■an■undefi■ned■(infi■nite)■gradient■and■are■ parallel■to■the■y-axis. exerCise

3A inDiviDuAl pAthWAys eBook plus

Activity 3-A-1

Sketching linear graphs doc-4975 Activity 3-A-2

Graphs of linear equations doc-4976 Activity 3-A-3

More graphs of linear equations doc-4977

sketching linear graphs FluenCy   1  We 1 ■Using■a■graphing■calculator,■generate■a■table■of■values■and■then■plot■the■linear■graphs■

defi■ned■by■the■following■rules■for■the■given■range■of x-values. Rule x-values -5,■-4,■-3,■-2,■-1,■0,■1 a  y■=■10x■+■25 -1,■0,■1,■2,■3,■4 b  y■=■5x■-■12 -6,■-4,■-2,■0,■2,■4 c  y■=■-0.5x■+■10 ■ 0,■1,■2,■3,■4,■5 d  y■=■100x■-■240 -3,■-2,■-1,■0,■1,■2 e  y■=■-5x■+■3 -3,■-2,■-1,■0,■1,■2 f  y■=■7■-■4x   2  Plot■the■linear■graphs■defi■ned■by■the■following■rules■for■the■given■range■of x-values. Rule x-values a  y =■-3x■+■2 x -6 -4 -2 0 2 4

6

y b  y■=■-x■+■3

Digital doc

62

-3

-2

-1

0

1

2

3

-6

-4

-2

0

2

4

6

y

eBook plus

SkillSHEET 3.7 doc-5202

x

c  y■=■-2x■+■3

maths Quest 10 for the Australian Curriculum

x y

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships

eBook plus

Digital doc

SkillSHEET 3.5 doc-5200

  3  We2 ■Sketch■graphs■of■the■following■linear■equations■by■fi■nding■the■x-■and■y-intercepts. a  5x■-■3y■=■10 b  5x■+■3y■=■10 c  -5x■+■3y■=■10 d  -5x■-■3y■=■10 e  2x■-■8y■=■20 f  4x■+■4y■=■40 g  -x■+■6y■=■120 h  -2x■+■8y■=■-20 i  10x +■30y■=■-150 j  5x■+■30y■=■-150 k  -9x■+■4y■=■36 l  6x■-■4y■=■-24 m  y■=■2x■-■10 n  y■=■-5x■+■20 1 o  y■=■− 2 x■-■4

  4  We3 ■Sketch■graphs■of■the■following■linear■equations■using■the■gradient-intercept■method. a  y =■4x■+■1 b  y =■3x■-■7 c  y =■-2x■+■3 1 2

d  y =■-5x■-■4

e  y =■ x■-■2

g  y =■0.6x■+■0.5

h  y =■8x

f  y =■- 27 x■+■3 i 

y = x -■7

  5  We4 ■Sketch■the■graphs■of■the■following■linear■equations. a  y =■2x b  y =■5x c  y =■-3x 2

e  y =■ 3 x

1

d  y =■ 2 x 5

f  y =■− 2 x

  6  We5 ■Sketch■the■graphs■of■the■following■linear■equations. a  y =■10 b  y =■-10 c  x =■10 d  x =■-10 e  y =■100 f  y =■0 g  x =■0 h  x =■-100 i  y■=■-12 eBook plus

Digital doc

SkillSHEET 3.8 doc-5203

  7  Transpose■each■of■the■equations■to■standard■form■(that■is,■y■=■mx■+■c). State■the■x-■and■

y-intercept■for■each. a  5(y■+■2)■=■4(x■+■3) b  5(y■-■2)■=■4(x■-■3) c  2(y■+■3)■=■3(x■+■2) d  10(y■-■20)■=■40(x■-■2) e  4(y■+■2)■=■-4(x■+■2) f  2(y■-■2)■=■-(x■+■5) g  -5(y■+■1)■=■4(x■-■4) h  8(y■-■5)■=■-4(x■+■3) i  5(y■+■2.5)■=■2(x■-■3.5) j  2.5(y■-■2)■=■-6.5(x■-■1) unDerstAnDing   8  Find■the■x-■and■y-intercepts■of■the■following■lines. a  -y■=■8■-■4x b  6x■-■y■+■3■=■0 c  2y■-■10x■=■50

reFleCtion 

 

What types of straight lines have an x- and y-intercept of the same value?

Chapter 3 Coordinate geometry

63

number AND algebra • Linear and non-linear relationships

3B

Determining linear equations The equation of a straight line can be determined if given: •• two points through which the line passes •• the gradient of the line and one other point through which the line passes. ■■ The gradient of a straight line can be calculated from the coordinates of two points (x1, y1) and (x2, y2) which lie on the line. y rise y2 − y1 B = •• Gradient = m = run x2 − x1 (x2, y2) ■■

■■

■■

■■

The equation of the straight line can then be ■ found in the form y = mx + c, where c is the y-intercept. Once the gradient has been found, substitute ■ one pair of known x- and y-values into the standard linear equation y = mx + c to determine the value of c. If the y-intercept has already been given, then this is the value of c and no further calculation is required.

rise = y2 - y1 A (x1, y1) x-intercept y-intercept

Worked Example 6

Find the equation of the straight line shown in the graph. y 6

0

3

Think

x

Write

1

There are two points given on the straight line: the x-intercept (3, 0) and the y-intercept (0, 6). Therefore, c is 6.

c=6

2

We can now find the gradient of the line by using the rise y2 − y1 = , where (x1, y1) = (3, 0) formula m = run x2 − x1 and (x2, y2) = (0, 6).

m=

rise run y −y = 2 1 x2 − x1

6−0 0−3 6 = −3 = -2 The gradient m = -2. =

3

64

Substitute m = -2, and c = 6 into y = mx + c to find the equation.

Maths Quest 10 for the Australian Curriculum

y = mx + c y = -2x + 6

run = x2 - x1 x

number AND algebra • Linear and non-linear relationships

Worked Example 7 y

Find the equation of the straight line shown in the graph.  

1

(2, 1) 0

Think

2

x

Write

1

There are two points given on the straight line: the x- and y-intercept (0, 0) and another point (2, 1). The y-intercept, c, is 0.

c=0

2

We can now find the gradient of the line by using the rise y2 − y1 , where (x1, y1) = (0, 0) and = formula m = run x2 − x1 (x2, y2) = (2, 1).

m=

rise run y −y = 2 1 x2 − x1

1− 0 2−0 1 = 2 1 The gradient m = 2 . =

3

Substitute m = 12 and c = 0 into y = mx + c to determine the equation.



y = mx + c 1 y = 2x + 0



y = 12 x

Worked Example 8

Find the equation of the straight line passing through (-2, 5) and (1, -1). Think

Write

y = mx + c y −y m= 2 1 x2 − x1

1

Write the general equation of a straight line.

2

Write the formula for calculating the gradient of a line between two points.

3

Let (x1, y1) and (x2, y2) be the two points (-2, 5) and (1, -1) respectively. Substitute the values of the pronumerals into the formula to calculate the gradient.

m=

4

Substitute the value of the gradient into the general rule.

y = -2x + c

5

Select either of the two points, say (1, -1), and substitute its coordinates into y = -2x + c.

Point (1, -1): -1 = -2 ì 1 + c

6

Solve for c; that is, add 2 to both sides of the equation.

-1 = -2 + c 1=c

7

State the equation by substituting the value of c into y = -2x + c.

The equation of the line is y = -2x + 1.

−1 − 5 1 − −2 −6 = 3 = -2

Chapter 3 Coordinate geometry

65

number AND algebra • Linear and non-linear relationships

Finding the equation of a straight line using the gradient and another point (point–gradient method) Worked Example 9

Find the equation of the straight line with gradient of 2 and y-intercept of -5. Think

Write

1

Write the known information. In this instance, the other point is the y-intercept, which makes the calculation of c straightforward.

Gradient = 2, y-intercept = -5

2

State the values of m and c.

m = 2, c = -5

3

Substitute these values into y = mx + c to find the equation.

y = mx + c y = 2x - 5

■■ ■■

Sometimes the gradient and another point which is not the y-intercept is given. The value of c can then be found by substituting the coordinates of this point into y = mx + c.

Worked Example 10

Find the equation of the straight line passing through the point (5, -1) with a gradient of 3. Think

Write

1

Write the known information.

Gradient = 3, point (5, -1).

2

State the values of m, x and y.

m = 3, (x, y) = (5, -1)

3

Substitute these values into y = mx + c and solve to find c.

y = mx + c -1 = 3(5) + c -1 = 15 + c -16 = c

4

Substitute m = 3 and c = -16 into y = mx + c to determine the equation.

The equation of the line is y = 3x - 16.

remember

y2 − y1 rise or m = x − x . run 2 1 2. An equation of a straight line can be found if you are given either:■ (i)  two points that lie on the line or■ (ii)  the gradient of the line and another point (the point–gradient method).■ Note that alternative (i) can reduce to alternative (ii) since the gradient can be calculated using the two given points. 3. The equation of a straight line can be found by substituting the values of the gradient, m, into y = mx + c. The value of c can then be found by substituting the x- and y-values of a given point into y = mx + c. If one of the points given is the y-intercept then it is simply a matter of letting c = y-intercept. 1. The gradient of a straight line is equal to m =

66

Maths Quest 10 for the Australian Curriculum

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships

exerCise

3b inDiviDuAl pAthWAys eBook plus

Determining linear equations FluenCy   1  We 6 ■Determine■the■equation■for■each■of■the■straight■lines■shown. y a  b c y

Activity 3-B-1

x

-2 0

Activity 3-B-2

Linear equations doc-4979

5

12

4

Determining linear equations doc-4978

y

0

0

x

4

5

x

Activity 3-B-3

More complex linear equations doc-4980

d

y

e

y

f

y 3

eBook plus

0

x

4

x

-4

Digital doc

SkillSHEET 3.1 doc-5196

0

-16

x

0

-6

-8 g

h

y

0

y

-5

x

–5 7

-5

x

0

-15

  2  We 7 ■Determine■the■equation■of■each■of■the■straight■lines■shown. y a  b y (-4, 12) 12 6 (3, 6) x

0 3

c

eBook plus

Digital doc

SkillSHEET 3.9 doc-5204

0 -2

y

d

y

-4 (-4, -2)

x

-4 0

x

(-8, 6) -8

6 0

x

  3  We 8 ■Find■the■equation■of■the■straight■line■that■passes■through■each■pair■of■points. a  (1,■4)■and■(3,■6) b  (0,■-1)■and■(3,■5) c  (-1,■4)■and■(3,■2) d  (3,■2)■and■(-1,■0) e  (-4,■6)■and■(2,■-6) f  (-3,■-5)■and■(-1,■-7) Chapter 3 Coordinate geometry

67

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships   4  We 9 ■Find■the■linear■equation■given■the■information■in■each■case■below. a  Gradient■=■3,■y-intercept■=■3 b  Gradient■=■-3,■y-intercept■=■4 c  Gradient■=■-4,■y-intercept■=■2 d  Gradient■=■4,■y-intercept■=■2 e  Gradient■=■-1,■y-intercept■=■-4 f  Gradient■=■0.5,■y-intercept■=■-4 g  Gradient■=■5,■y-intercept■=■2.5 h  Gradient■=■-6,■y-intercept■=■3 i  Gradient■=■-2.5,■y-intercept■=■1.5 j  Gradient■=■3.5,■y-intercept■=■6.5   5  We 10 ■For■each■of■the■following,■fi■nd■the■equation■of■the■straight■line■with■the■given■gradient■

and■passing■through■the■given■point. a  Gradient■=■5,■point■=■(5,■6) b  Gradient■=■-5,■point■=■(5,■6) c  Gradient■=■-4,■point■=■(-2,■7) d  Gradient■=■4,■point■=■(8,■-2) e  Gradient■=■3,■point■=■(10,■-5) f  Gradient■=■-3,■point■=■(3,■-3) g  Gradient■=■-2,■point■=■(20,■-10) h  Gradient■=■2,■point■=■(2,■-0.5) i  Gradient■=■0.5,■point■=■(6,■-16) j  Gradient■=■- 0.5,■point■=■(5,■3)

eBook plus

Digital doc

WorkSHEET 3.1 doc-5205

3C

reFleCtion 

 

What problems might you encounter when calculating the equation of a line whose graph is actually parallel to one of the axes?

the distance between two points on a straight line Distance between two points ■■ ■■

The■distance■between■two■points■can■be■calculated■using■Pythagoras’■theorem. Consider■two■points■A(x1,■y1)■and■B(x2,■y2)■on■the■Cartesian■plane■as■shown. AC■=■x2■-■x1 y BC■=■y2■-■y1 y2

By■Pythagoras’■theorem: AB2■=■AC2■+■BC2 =■(x2■-■x1)2■+■(y2■-■y1)2

y1

A

B(x2, y2) C

(x1, y1)

Hence■ AB = ( x2 − x1 )2 + ( y 2 − y1 )2

x1

x2

x

The■distance■between■two■points■A(x1,■y1)■and■B(x2,■y2)■is: AB = ( x2 − x1 )2 + ( y 2 − y1 )2 ■■

This■distance■formula■can■be■used■to■calculate■the■distance■between■any■two■points■on■the■ Cartesian■plane.

WorkeD exAmple 11 y

Find the distance between the points A and B in the figure at right.

4 A think

68

Write

1

From■the■graph■fi■nd■points■A■and■B.

A(-3,■1)■and■B(3,■4)

2

Let■A■have■coordinates■(x1,■y1).

Let■(x1,■y1)■=■(-3,■1)

maths Quest 10 for the Australian Curriculum

-3

B

1 3

x

number AND algebra • Linear and non-linear relationships

3

Let B have coordinates (x2, y2).

Let (x2, y2) = (3, 4)

4

Find the length AB by applying the formula for the distance between two points.

AB = ( x2 − x1 )2 + ( y2 − y1 )2 = [3 − (−3)]2 + (4 − 1)2 = (6)2 + (3)2 = 36 + 9 = 45 =3 5 = 6.71 (correct to 2 decimal places)

Worked Example 12

Find the distance between the points P(-1, 5) and Q(3, -2). Think

Write

1

Let P have coordinates (x1, y1).

Let (x1, y1) = (-1, 5)

2

Let Q have coordinates (x2, y2).

Let (x2, y2) = (3, -2)

3

Find the length PQ by applying the formula for the distance between two points.

PQ = ( x2 − x1 )2 + ( y2 − y1 )2 = [3 − (−1)]2 + (−2 − 5)2 = (4)2 + (−7)2 = 16 + 49 = 65 = 8.06 (correct to 2 decimal places)

■■

The distance formula is useful in proving geometric properties of polygons.

Worked Example 13

Prove that the points A(1, 1), B(3, -1) and C(-1, -3) are the vertices of an isosceles triangle. Think 1

Plot the points. Note: For triangle ABC to be isosceles, two sides must have the same magnitude.

Write/draw y -1 C

2

Find the length AC. A(1, 1) = (x2, y2) C(-1, -3) = (x1, y1)

A

1 1

3 B

x

From the diagram, AC appears to have the same length as BC.

-3

AC = [1 − (−1)]2 + [1 − (−3)]2 = (2)2 + (4)2 = 20 Chapter 3 Coordinate geometry

69

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships

3

4

5

Find■the■length■BC. B(3,■-1)■=■(x2,■y2) C(-1,■-3)■=■(x1,■y1)

BC = [3 − (−1)]2 + [−1 − (−3)]2

Find■the■length■AB. A(1,■1)■=■(x1,■y1) B(3,■-1)■=■(x2,■y2)

AB = [3 − (1)]2 + [−1 − (1)]2

State■your■proof.

Since■AC■=■BC,■triangle■ABC■is■an■isosceles■triangle.

= (4)2 + (2)2 = 20

= (2)2 + (−2)2 = 4+4 = 8 =2 2

remember

The■distance■between■two■points■A(x1,■y1)■and■B(x2,■y2)■is: AB = ( x2 − x1 )2 + ( y2 − y1 )2 exerCise

3C inDiviDuAl pAthWAys eBook plus

Activity 3-C-1

Finding the distance between two points on a straight line doc-4981 Activity 3-C-2

Calculations of distance between two points doc-4982 Activity 3-C-3

Applications of distance between two points doc-4983

the distance between two points on a straight line FluenCy   1  We11 ■Find■the■distance■between■each■pair■of■points■     shown at■right.   2  We12 ■Find■the■distance■between■the■following■pairs■of■

points. (2,■5),■(6,■8) (-1,■2),■(4,■14) (-1,■3),■(-7,■-5) (5,■-1),■(10,■4) (4,■-5),■(1,■1) (-3,■1),■(5,■13) (5,■0),■(-8,■0) (1,■7),■(1,■-6) (a,■b),■(2a,■-b) (-a,■2b),■(2a,■-b)

a b c d e f g h i j

y

O 6 B 5 4 P 3 C 2A E H N L 1 -6 -5 -4-3-2-1 0 1 2 3 4 5 6 x -1 F -2 M -3 I J D -4 -5 -6 G

K

  3  We13 ■Prove■that■the■points■A(0,■-3),■B(-2,■-1)■and■C(4,■3)■are■the■vertices■of■an     isosceles■triangle. unDerstAnDing eBook plus

Digital doc

Spreadsheet 021 doc-5206

70

  4  The■points■P(2,■-1),■Q(-4,■-1)■and■R(-1,■3 3 − 1)■are■joined■to■form■a■triangle.■     Prove■that■triangle■PQR■is■equilateral.   5  Prove■that■the■triangle■with■vertices■D(5,■6),■E(9,■3)■and■F(5,■3)■is■a■right-angled■triangle.

maths Quest 10 for the Australian Curriculum

number AND algebra • Linear and non-linear relationships 6 The vertices of a quadrilateral are A(1, 4), B(-1, 8), C(1, 9) and D(3, 5). a Find the lengths of the sides. c What type of quadrilateral is it?

b Find the lengths of the diagonals.

Reasoning   7   MC  If the distance between the points (3, b) and (-5, 2) is 10 units, then the value of b is: a -8 B -4 C 4 d 0 E 2   8   MC  A rhombus has vertices A(1, 6), B(6, 6), C(-2, 2) and D(x, y). The coordinates of D are: a (2, -3) B (2, 3) C (-2, 3) d (3, 2) E (3, -2)   9 A rectangle has vertices A(1, 5), B(10.6, z), C(7.6, -6.2) and D(-2, 1). Find: a the length of CD reflection    b the length of AD c the length of the diagonal AC How could you use the distance d the value of z. formula to show that a series of 10 Show that the triangle ABC with coordinates

A(a, a), B(m, -a) and C(-a, m) is isosceles.

3D

points lay on the circumference of a circle with centre C ?

The midpoint of a line segment Midpoint of a line segment ■■ ■■

The midpoint of a line segment is the half-way point. The x- and y-coordinates of the midpoint are half-way between those of the coordinates of the end points.

Midpoint formula y Consider the line segment connecting the points B(x2, y2) A(x1, y1) and B(x2, y2). (y2 - y) Let P(x, y) be the midpoint of AB. AC is parallel to PD. P(x, y) (x2 - x) D PC is parallel to BD. (y - y1) AP is parallel to PB (collinear). A Hence triangle APC is similar to triangle PBD. (x1, y1) (x - x1) C But AP = PB (since P is the midpoint of AB). x Hence, triangle APC is congruent to triangle PBD. Therefore x - x1 = x2 - x 2x = x1 + x2 x + x2 x= 1 2 y +y Similarly it can be shown that y = 1 2 . 2 In general, the coordinates of the midpoint of a line segment joining y (x2, y2) the points (x1, y1) and (x2, y2) can be found by averaging the x- and y-coordinates of the end points, respectively. M x_____, y1 + y2 1 + x2 _____ The coordinates of the midpoint of the line segment joining

x +x y +y  (x1, y1) and (x2, y2) are:  1 2 , 1 2   2 2 

(x1, y1)

(

2

2

)

Chapter 3 Coordinate geometry

x 71

number AND algebra • Linear and non-linear relationships

Worked Example 14

Find the coordinates of the midpoint of the line segment joining (-2, 5) and (7, 1). Think 1

Label the given points (x1, y1) and (x2, y2).

2

Find the x-coordinate of the midpoint.

Write

Let (x1, y1) = (-2, 5) and (x2, y2) = (7, 1) x +x x= 1 2 2 −2 + 7 = 2 5 =2 1

3

Find the y-coordinate of the midpoint.

=22 y +y y= 1 2 2 5+1 = 2 6 =2 =3

4

Give the coordinates of the midpoint.

1

Hence, the coordinates of the midpoint are (2 2 , 3).

Worked Example 15

The coordinates of the midpoint, M, of the line segment AB are (7, 2). If the coordinates of A are (1, -4), find the coordinates of B. Think

2

Label the start of the line segment (x1, y1) and the midpoint (x, y). Find the x-coordinate of the end point.

3

Find the y-coordinate of the end point.

4

Give the coordinates of the end point.

5

Check that the coordinates are feasible.

1

Write/DRAW

Let (x1, y1) = (1, -4) and (x, y) = (7, 2) x1 + x2 2 1 + x2 7= 2 14 = 1 + x2 x2 = 13 y +y y= 1 2 2 − 4 + y2 2= 2 4 = -4 + y2 y2 = 8 Hence, the coordinates of the point B are (13, 8). x=

y 8 2

B (13, 8) M (7, 2)

1 7 -4 A (1, -4)

72

Maths Quest 10 for the Australian Curriculum

13

x

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships

remember

The■coordinates■of■the■midpoint■of■the■line■segment■ joining■(x1,■y1)■and■(x2,■y2)■are: ■

■

y (x2, y2)

 x1 + x2 y1 + y2   2 , 2 

M

1

(x1, y1)

exerCise

3D inDiviDuAl pAthWAys eBook plus

Activity 3-D-1

Finding the midpoint of a line segment doc-4984 Activity 3-D-2

Calculations — midpoint of a segment doc-4985 Activity 3-D-3

Applications — midpoint of a segment doc-4986

+ x _____ y +y (x_____, 2 2 ) 2

1

2

x

the midpoint of a line segment FluenCy   1  We14 ■Use■the■formula■method■to■fi■nd■the■coordinates■of■the■midpoint■of■the■line     segment joining■the■following■pairs■of■points. a  (-5,■1),■(-1,■-8) c  (0,■4),■(-2,■-2) e  (a,■2b),■(3a,■-b)

b  (4,■2),■(11,■-2) d  (3,■4),■(-3,■-1) f  (a■+■3b,■b),■(a■-■b,■a■-■b)

  2  We15 ■The■coordinates■of■the■midpoint,■M,■of■the■line■segment■AB■are■(2,■-3).■If■the■

coordinates■of■A■are■(7,■4),■fi■nd■the■coordinates■of■B. unDerstAnDing   3  Find: a  the■coordinates■of■the■centre■of■a■square■with■vertices■A(0,■0),■B(2,■4),■C(6,■2)■and■

D(4,■-2) b  the■side■length c  the■length■of■the■diagonals.

eBook plus

Digital doc

Spreadsheet 075 doc-5207

  4  mC ■The■midpoint■of■the■line■segment■joining■the■points■(-2,■1)■and■(8,■-3)■is: a  (6,■-2) B  (5,■2) C  (6,■2) d  (3,■-1) E  (5,■-2)   5  mC ■If■the■midpoint■of■AB■is■(-1,■5)■and■the■coordinates■of■B■are■(3,■8),■then■A■has■

coordinates: a  (1,■6.5) d  (4,■3)

B  (2,■13) E  (7,■11)

C  (-5,■2)

  6  a■ The■vertices■of■a■triangle■are■A(2,■5),■B(1,■-3)■and■C(-4,■3).■Find:   i  the■coordinates■of■P,■the■midpoint■of■AC   ii  the■coordinates■of■Q,■the■midpoint■of■AB   iii  the■length■of■PQ   iv  the■length■of■BC. b  Hence■show■that■BC■=■2PQ.   7  a■ ■A■quadrilateral■has■vertices■A(6,■2),■B(4,■-3),■C(-4,■-3)■and■D(-2,■2).■Find:   i  the■midpoint■of■the■diagonal■AC   ii  the■midpoint■of■the■diagonal■BD. b  Comment■on■your■fi■nding. Chapter 3 Coordinate geometry

73

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships   8  a■ ■■The■points■A(-5,■3.5),■B(1,■0.5)■and■C(-6,■-6)■are■the■vertices■of■a■triangle.■Find:   i  the■midpoint,■P,■of■AB   ii  the■length■of■PC   iii  the■length■of■AC   iv  the■length■of■BC. b  Describe■the■triangle.■What■could■PC■represent? reAsoning   9  Find■the■equation■of■the■straight■line■that■passes■through■the■midpoint■of■A(-2,■5)■and■ eBook plus

Digital doc

WorkSHEET 3.2 doc-5208

3e

B(-2,■3),■and■has■a■gradient■of■-3.  10  Find■the■equation■of■the■straight■line■that■ passes■through■the■midpoint■of■A(-1,■-3)■and■ 2 B(3,■-5),■and■has■a■gradient■of■ 3.

reFleCtion 

If the midpoint of a line segment is the origin, what are the possible values of the x- and y-coordinates of the end points?

parallel and perpendicular lines parallel lines ■■

 

eBook plus

Interactivity Parallel and perpendicular lines

Lines■which■have■the■same■gradient■are■parallel■lines.

int-2779

WorkeD exAmple 16

Show that AB is parallel to CD given that A has coordinates (-1, -5), B has coordinates (5, 7), C has coordinates (-3, 1) and D has coordinates (4, 15). think 1

Write

Find■the■gradient■of■AB.

Let■A(-1,■-5)■=■(x1,■y1)■and■B(5,■7)■=■(x2,■y2) y −y Since■ ■ m = 2 1 x2 − x1 7 − (−5) mAB = 5 − (−1) 12

= 6 =■2 2

Find■the■gradient■of■CD.

Let■C(-3,■1)■=■(x1,■y1)■and■D(4,■15)■=■(x2,■y2) 15 − 1 mCD = 4 − (−3) =

14 7

=■2 3

Compare■the■gradients■to■determine■ if■they■are■parallel.■(Note:■||■means■ ‘is■parallel■to’.)

■■

74

Since■parallel■lines■have■the■same■gradient■and■ mAB■=■mCD■=■2,■then■AB||CD.

Collinear■points■are■points■which■all■lie■on■the■same■straight■line.

maths Quest 10 for the Australian Curriculum

number AND algebra • Linear and non-linear relationships

Worked Example 17

Show that the points A(2, 0), B(4, 1) and C(10, 4) are collinear. Think 1

Write

Let A(2, 0) = (x1, y1) and B(4, 1) = (x2, y2) y −y Since m= 2 1 x2 − x1 1− 0 mAB = 4−2

Find the gradient of AB.

2

1 2

Let B(4, 1) = (x1, y1) and C(10, 4) = (x2, y2)

Find the gradient of BC.



3

=

Show that A, B and C are collinear.

mBC = 4 − 1 10 − 4



=



=

3 6 1 2 1

Since mAB = mBC = 2 then AB||BC Since B is common to both line segments, A, B and C must lie on the same straight line. That is, A, B and C are collinear.

Perpendicular lines ■■ ■■

There is a special relationship between the gradients of two perpendicular lines. Consider the diagram shown below where the line segment AB is perpendicular to the line segment BC, AC is parallel to the x-axis, and BD is the perpendicular height of the resulting triangle ABC. Let mAB = m1 y B a = b a q = tan (q  ) a Let mBC = m2 a C A q a c b =− D c x = -tan (a ) b =− a −1 = m1 Hence or

m2 = −

1 m1

m1m2 = -1 Chapter 3 Coordinate geometry

75

number AND algebra • Linear and non-linear relationships

Hence, if two lines are perpendicular to each other, then the product of their gradients is -1. Two lines are perpendicular if and only if:                   m1m2 = -1 −1 or               m2 = m1 Worked Example 18

Show that the lines y = -5x + 2 and 5y - x + 15 = 0 are perpendicular to one another. Think

Write

y = -5x + 2 m1 = -5

1

Find the gradient of equation 1.

Hence

2

Find the gradient of equation 2.

5y - x + 15 = 0 Rewrite in the form y = mx + c 5y = x - 15 x y= -3 5 m2 =

Hence 3

Test for perpendicularity. (The two lines are perpendicular if the product of their gradients is -1.)

1 5 1

m1m2 = -5 ì 5 = -1 Hence, the two lines are perpendicular to each other.

Determining the equation of a straight line parallel or perpendicular to another straight line ■■

The gradient properties of parallel and perpendicular straight lines can be used to determine the equations of other lines with particular attributes.

Worked Example 19

Find the equation of the straight line that passes through the point (3, -1) and is parallel to the straight line with equation y = 2x + 1. Think

Write

Write the general equation.



2

Find the gradient of the given line.

y = 2x + 1 has a gradient of 2 Hence m = 2

3

Substitute for m in the general equation.

so

4

Substitute the given point to find c.

5

76

y = mx + c

1

Substitute for c in the general equation.

Maths Quest 10 for the Australian Curriculum

y = 2x + c (x, y) = (3, -1) \ -1 = 2(3) + c =6+c c = -7

y = 2x - 7 or 2x - y - 7 = 0

number AND algebra • Linear and non-linear relationships

Worked Example 20

Find the equation of the line that passes through the point (0, 3) and is perpendicular to a straight line with a gradient of 5. Think

Write

m=5 1 m1 = − 5

1

Find the gradient of the perpendicular line.

Given

2

Substitute for m and (x1, y1) in the general equation.

Since y - y1 = m(x - x1) and (x1, y1) = (0, 3) 1

then y - 3 = − 5 (x - 0) x =− 5 5(y - 3) = -x 5y - 15 = -x x + 5y - 15 = 0

Horizontal and vertical lines ■■

Recall the following. •• Horizontal lines are parallel to the x-axis, have a gradient of zero, are expressed in the form y = c and have no x-intercept. •• Vertical lines are parallel to the y-axis, have an undefined (infinite) gradient, are expressed in the form x = a and have no y-intercept.

Worked Example 21

Find the equation of: a  the vertical line that passes through the point (2, -3) b  the horizontal line that passes through the point (-2, 6). Think

Write

a For a vertical line, there is no y-intercept so y does not

a x=2

b For a horizontal line, there is no x-intercept so x does not

b y=6

appear in the equation. The x-coordinate of the point is 2. appear in the equation. The y-coordinate of the point is 6.

Worked Example 22

Find the equation of the perpendicular bisector of the line joining the points (0, -4) and (6, 5). Think 1

Find the gradient of the line joining the given points using the general equation.

Write/draw

Let (0, -4) = (x1, y1) Let       (6, 5) = (x2, y2) y − y1 m= 2 x2 − x1 5 − (−4) m= 6−0 9 =6 =

3 2

Chapter 3 Coordinate geometry

77

number AND algebra • Linear and non-linear relationships

2

Find the gradient of the perpendicular line.

For lines to be perpendicular, m2 = − 2

m1 = − 3 3

1 m1

x1 + x2 2 0+6 = 2 =3 y + y2 y= 1 2 −4 + 5 = 2 1 = 2 x=

Find the midpoint of the line joining the given points.

1

Hence (3, 2 ) are the coordinates of the midpoint. 4

Substitute for m and (x1, y1) in the general equation.

Since y - y1 = m(x - x1) 2 1 and (x1, y1) = (3, 2 ) and m1 = − 3

5

Simplify by removing the fractions.

then y -

= − 3 (x - 3)

(a)  Multiply both sides by 3.



= -2(x - 3)

(b)  Multiply both sides by 2.

= -2x + 6 6y - 3 = -4x + 12 4x + 6y - 15 = 0

1 2 1 3(y - 2 ) 3 3y - 2

y 5

Note: The diagram at right shows the geometric situation.

2

(6, 5)

1 2 –2 1– 2

-4

3

6 x

-4

remember

1. The equation of a straight line may be expressed in the form: y = mx + c where m is the gradient of the line and c is the y-intercept, or y - y1 = m(x - x1) where m is the gradient and (x1, y1) is a point on the line. 2. The gradient can be calculated if two points, (x1, y1) and (x2, y2), are given by using y −y m= 2 1 x2 − x1 3. Parallel lines have the same gradient. 4. Collinear points lie on the same straight line. Two lines are perpendicular if and only if: m1m2 = -1 1 or m2 = − . m1 The equation of a straight line can be determined by two methods: 78

Maths Quest 10 for the Australian Curriculum

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships

5.■ The■y■=■mx■+■c■method. This■requires■the■gradient,■m,■and■a■given■point■to■be■known,■in■order■to■establish■the■ value■of■c. If■the■y-intercept■is■known,■then■this■can■be■directly■substituted■for■c. 6.■ Alternative■method:■y■-■y1■=■m(x■-■x1) This■requires■the■gradient,■m,■and■a■given■point■(x1,■y1)■to■be■known. 7.■ The■general■equation■for■a■vertical■line■is■given■by■x■=■a■and■a■horizontal■line■is■given■ by■y■=■c.

exerCise

3e inDiviDuAl pAthWAys eBook plus

Activity 3-E-1

Parallel and perpendicular lines doc-4987 Activity 3-E-2

More difficult parallel and perpendicular lines doc-4988 Activity 3-E-3

Complex parallel and perpendicular lines doc-4989

parallel and perpendicular lines FluenCy   1  We 16 ■Find■if■AB■is■parallel■to■CD■given■the■following■coordinates. a  b  c  d  e  f 

A(4,■13),■B(2,■9),■C(0,■-10),■D(15,■0). A(2,■4),■B(8,■1),■C(-6,■-2),■D(2,■-6). A(-3,■-10),■B(1,■2),■C(1,■10),■D(8,■16). A(1,■-1),■B(4,■11),■C(2,■10),■D(-1,■-5). A(1,■0),■B(2,■5),■C(3,■15),■D(7,■35). A(1,■-6),■B(-5,■0),■C(0,■0),■D(5,■-4).

  2  Which■pairs■of■the■following■straight■lines■are■parallel? a  2x■+■y■+■1■=■0■ c  2y■-■x■=■3■

b■ y■=■3x■-■1 d■ y■=■4x■+■3

x −1 ■ 2 g  3y■=■x■+■4■

e  y =

■

f■ 6x■-■2y■=■0 h■ 2y■=■5■-■x

  3  We 17 ■Show■that■the■points■A(0,■-2),■B(5,■1)■and■C(-5,■-5)■are■collinear.   4  Show■that■the■line■that■passes■through■the■points■(-4,■9)■and■(0,■3)■also■passes■through■

the■point■(6,■-6).   5  We 18 ■Show■that■the■lines■y■=■6x■-■3■and■x■+■6y■-■6■=■0■are■perpendicular■to■one■another. eBook plus

Digital doc

Spreadsheet 085 doc-5209

  6  Determine■if■AB■is■perpendicular■to■CD,■given■the■following■coordinates. a  A(1,■6),■B(3,■8),■C(4,■-6),■D(-3,■1)■ c  A(1,■3),■B(4,■18),■C(-5,■4),■D(5,■0)■ e  A(-4,■9),■B(2,■-6),■C(-5,■8),■D(10,■14)■

b■ A(2,■12),■B(-1,■-9),■C(0,■2),■D(7,■1) d■ A(1,■-5),■B(0,■0),■C(5,■11),■D(-10,■8) f■ A(4,■4),■B(-8,■5),■C(-6,■2),■D(3,■11)

  7  We 19 ■Find■the■equation■of■the■straight■line■that■passes■through■the■point■(4,■-1)■and■is■

parallel■to■the■straight■line■with■equation■y■=■2x■-■5. eBook plus

Digital doc

Spreadsheet 029 doc-5210

  8  We 20 ■Find■the■equation■of■the■line■that■passes■through■the■point■(-2,■7)■and■is■perpendicular 2

to■a■line■with■a■gradient■of■ 3.   9  Find■the■equations■of■the■following■straight■lines. a  Gradient■3■and■passing■through■the■point■(1,■5). b  Gradient■-4■and■passing■through■the■point■(2,■1). c  Passing■through■the■points■(2,■-1)■and■(4,■2). d  Passing■through■the■points■(1,■-3)■and■(6,■-5). e  Passing■through■the■point■(5,■-2)■and■parallel■to■x■+■5y■+■5■=■0. f  Passing■through■the■point■(1,■6)■and■parallel■to■x■-■3y■-■2■=■0. g  Passing■through■the■point■(-1,■-5)■and■perpendicular■to■3x■+■y■+■2■=■0. Chapter 3 Coordinate geometry

79

number AND algebra • Linear and non-linear relationships 10 Find the equation of the line which passes through the point (-2, 1) and is: a parallel to the straight line with equation 2x - y - 3 = 0 b perpendicular to the straight line with equation 2x - y - 3 = 0. 11 Find the equation of the line that contains the point (1, 1) and is: a parallel to the straight line with equation 3x - 5y = 0 b perpendicular to the straight line with equation 3x - 5y = 0. 12   WE 21  Find the equation of: a the vertical line that passes through the point (1, -8) b the horizontal line that passes through the point (-5, -7). 13   MC  a  The vertical line passing through the point (3, -4) is given by: A y = -4 B x = 3 C y = 3x - 4 D y = -4x + 3 E x = -4 b Which of the following points does the horizontal line given by the equation y = -5 pass

14

15 16 17

through? A (-5, 4) B (4, 5) C (3, -5) D (5, -4) E (5, 5) c Which of the following statements is true? A Vertical lines have a gradient of zero. B The y-coordinates of all points on a vertical line are the same. C Horizontal lines have an undefined gradient. D The x-coordinates of all points on a vertical line are the same. E A horizontal line has the general equation x = a. d Which of the following statements is false? A Horizontal lines have a gradient of zero. B The straight line joining the points (1, -1) and (-7, -1) is vertical. C Vertical lines have an undefined gradient. D The straight line joining the points (1, 1) and (-7, 1) is horizontal. E A horizontal line has the general equation y = c. The triangle ABC has vertices A(9, -2), B(3, 6) and C(1, 4). a Find the midpoint, M, of BC. b Find the gradient of BC. c Show that AM is the perpendicular bisector of BC. d Describe triangle ABC.   WE 22  Find the equation of the perpendicular bisector of the line joining the points (1, 2) and (-5, -4). Find the equation of the perpendicular bisector of the line joining the points (-2, 9) and (4, 0). ABCD is a parallelogram. The coordinates of A, B and C are (4, 1), (1, -2) and (-2, 1) respectively. Find: a the equation of AD b the equation of DC c the coordinates of D.

understanding 18 In each of the following, show that ABCD is a parallelogram. a A(2, 0), B(4, -3), C(2, -4), D(0, -1) b A(2, 2), B(0, -2), C(-2, -3), D(0, 1) c A(2.5, 3.5), B(10, -4), C(2.5, -2.5), D(-5, 5) 80

Maths Quest 10 for the Australian Curriculum

number AND algebra • Linear and non-linear relationships 19 In each of the following, show that ABCD is a trapezium. a A(0, 6), B(2, 2), C(0, -4), D(-5, -9) b A(26, 32), B(18, 16), C(1, -1), D(-3, 3) c A(2, 7), B(1, -1), C(-0.6, -2.6), D(-2, 3) 20   MC  The line that passes through the points (0, -6) and (7, 8) also passes through: A (4, 3) D (1, -8)

B (5, 4) E (1, 4)

C (-2, 10)

21   MC  The point (-1, 5) lies on a line parallel to 4x + y + 5 = 0. Another point on the same line

as (-1, 5) is: A (2, 9) B (4, 2) C (4, 0) E (3, -11) D (-2, 3) 22 Find the equation of the straight line given the following conditions: a passes through the point (-1, 3) and parallel to y = -2x + 5 b passes through the point (4, -3) and parallel to 3y + 2x = -3. 23 Determine which pairs of the following straight lines are perpendicular. a x + 3y - 5 = 0 d 2y = x + 1 g 2x + y = 6

b y = 4x - 7 e y = 3x + 2 h x + y = 0

c y = x f x + 4y - 9 = 0

24 Find the equation of the straight line that cuts the x-axis at 3 and is perpendicular to the line

with equation 3y - 6x = 12. 25 Calculate the value of m for which lines with the following pairs of equations are perpendicular to each other. a 2y - 5x = 7 and 4y + 12 = mx b 5x - 6y = -27 and 15 + mx = -3y 26   MC  The gradient of the line perpendicular to the line with equation 3x - 6y = 2 is: A 3 B -6 C 2 D

1 2

E -2

27   MC  Triangle ABC has a right angle at B. The vertices are A(-2, 9), B(2, 8) and C(1, z). The

value of z is: 1

A 8 4 3

D 7 4

B 4

C 12

E -4

Reasoning 28 The map shows the proposed course for a yacht race.

y

Scale: 1 unit «1 km

N

Buoys have been positioned at A(1, 5), B(8, 8) and 11 10 C(12, 6), but the last buoy’s placement, D(10, w), is yet 9 Buoy B to be finalised. 8 a How far is the first stage of the race, that is, from the 7 Buoy 6 A start, O, to buoy A? Buoy C M 5 b The race marshall boat, M, is situated halfway 4 between buoys A and C. What are the coordinates of E 3 Buoy D 2 H the boat? 1 O c Stage 4 of the race (from C to D) is perpendicular to (Start) 1 2 3 4 5 6 7 8 9 10 11 12 x stage 3 (from B to C). What is the gradient of CD? d Find the linear equation that describes stage 4. e Hence determine the exact position of buoy D. 2 f An emergency boat is to be placed at point E, 3 of the way from buoy A to buoy D. The coordinates of E are (7, 3). How far is the emergency boat from the hospital, located at H, 2 km North of the start? Chapter 3 Coordinate geometry

81

number AND algebra • Linear and non-linear relationships 29 Show that the following sets of points form the vertices of a right-angled triangle. a A(1, -4), B(2, -3), C(4, -7) b A(3, 13), B(1, 3), C(-4, 4) c A(0, 5), B(9, 12), C(3, 14) 30 Prove that the quadrilateral ABCD is a rectangle when A is (2, 5), B(6, 1), C(3, -2) and

D(-1, 2).

31 Prove that the quadrilateral ABCD is a rhombus, given A(2, 3), B(3, 5), C(5, 6) and D(4, 4).

Hint: The diagonals of a rhombus intersect at right angles. reflection 

 

How could you use coordinate geometry to design a logo for an organisation?

82

Maths Quest 10 for the Australian Curriculum

number AND algebra • Linear and non-linear relationships

Summary Sketching linear graphs ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■

■■ ■■ ■■

The Cartesian plane is a grid, consisting of two axes (x and y), meeting at the origin (0, 0). A location (point) is specified by its x- and y-coordinates. A linear graph consists of an infinite set of points that can be joined to form a straight line, but to sketch a linear graph, the coordinates of only two points are needed. A linear rule or equation can be used to obtain the coordinates of points that belong to ■ its graph. Linear equations may be written in several different forms. The two most common forms are y = mx + c and ax + by = k. When a linear equation is expressed in the form y = mx + c, then m represents the gradient (slope) of the straight line and c represents the y-intercept. A straight line with a positive gradient slopes upward to the right and a straight line with a negative gradient slopes downward to the right. The x- and y-intercept method allows us to sketch the graph of any linear equation by finding two specific points: the x-intercept and y-intercept. An exception is the case of lines passing through the origin. Graphs of equations in the form y = mx pass through the origin. To find the second point, substitute a chosen x-value into the equation to find the corresponding y-value. Graphs of equations in the form y = c have a gradient of zero and are parallel to the x-axis. Graphs of equations in the form x = a have an undefined (infinite) gradient and are parallel to the y-axis. Determining linear equations

■■ ■■

■■

y2 − y1 rise or m = x − x . run 2 1 An equation of a straight line can be found if you are given either:■ (i)  two points that lie on the line or■ (ii)  the gradient of the line and another point (the point–gradient method).■ Note that alternative (i) can reduce to alternative (ii) since the gradient can be calculated using the two given points. The equation of a straight line can be found by substituting the values of the gradient, m, into y = mx + c. The value of c can then be found by substituting the x- and y-values of a given point into y = mx + c. If one of the points given is the y-intercept then it is simply a matter of letting c = y-intercept. The gradient of a straight line is equal to m =

The distance between two points on a straight line

The distance between two points A(x1, y1) and B(x2, y2) is: AB = ( x2 − x1 )2 + ( y2 − y1 )2 The midpoint of a line segment

The coordinates of the midpoint of the line segment joining (x1, y1) and (x2, y2) are:

 x1 + x2 y1 + y2   2 , 2 

y (x2, y2) M

+ x _____ y +y (x_____, 2 2 ) 1

(x1, y1)

2

1

2

x

Chapter 3 Coordinate geometry

83

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships Parallel and perpendicular lines ■■

■■

■■ ■■

■■

■■

The■equation■of■a■straight■line■may■be■expressed■in■the■form: y■=■mx■+■c where■m■is■the■gradient■of■the■line■and■c■is■the■y-intercept,■or■ y■-■y1■=■m(x■-■x1)■ where■m■is■the■gradient■and■(x1,■y1)■is■a■point■on■the■line. The■gradient■can■be■calculated■if■two■points,■(x1,■y1)■and■(x2,■y2),■are■given■by■using y −y m= 2 1 x2 − x1 Parallel■lines■have■the■same■gradient. Collinear■points■lie■on■the■same■straight■line. Two■lines■are■perpendicular■if■and■only■if: m1m2■=■-1 1 or m2■=■− . m1 The■equation■of■a■straight■line■can■be■determined■by■two■methods: •■ The■y■=■mx■+■c■method. This■requires■the■gradient,■m,■and■a■given■point■to■be■known,■in■order■to■establish■the■value■ of■c. If■the■y-intercept■is■known,■then■this■can■be■directly■substituted■for■c. •■ Alternative■method:■y■-■y1■=■m(x■-■x1) This■requires■the■gradient,■m,■and■a■given■point■(x1,■y1)■to■be■known. The■general■equation■for■a■vertical■line■is■given■by■x■=■a■and■a■horizontal■line■is■given■ by■y■=■c.

MaPPING YOUR UNdERSTaNdING

Homework  Book

84

Using■terms■from■the■summary,■and■other■terms■if■you■wish,■construct■a■concept■map■that■ illustrates■your■understanding■of■the■key■concepts■covered■in■this■chapter.■Compare■your■ concept■map■with■the■one■that■you■created■in■What do you know?■on■page■55. Have■you■completed■the■two■Homework sheets,■the■Rich task■and■two■Code puzzles■in■ your■Maths Quest 10 Homework Book?

maths Quest 10 for the Australian Curriculum

number AND algebra • Linear and non-linear relationships

Chapter review Fluency

9 The equation of the line perpendicular to

1 The equation of the line drawn below is: y 3

2 x

0 A 3x + 2y = 6 C 2x + 3y = 6 E 2x - 3y = -6

B 3x - 2y = 6 D 2x - 3y = 6

2 The equation of a linear graph with gradient -3 and

x-intercept of 4 is: A y = -3x - 12 C y = -3x - 4 E y = 4x - 3

B y = -3x + 4 D y = -3x + 12

3 The equation of a linear graph which passes

through (2, -7) and (-2, -2) is: B 5x + 4y + 18 = 0 d 5x - 4y - 18 = 0

A 4x - 5y + 18 = 0 c 5x + 4y - 18 = 0 e 4x + 5y + 18 = 0

  4 The distance between the points (1, 5) and

(6, - 7) is: 53

A

B

C 13

29

193 E 12   5 The midpoint of the line segment joining the points (-4, 3) and (2, 7) is: A (-1, 5) B (-2, 10) C (-6, 4) D (-2, 4) E (-1, 2)   6 If the midpoint of the line segment joining the points A(3, 7) and B(x, y) has coordinates (6, 2), then the coordinates of B are: A (15, 3) B (0, -6) C (9, -3) D (4.5, 4.5) E (-9, 3) 7 If the points (-6, -11), (2, 1) and (x, 4) are collinear, then the value of x is: A 4 B 3.2 D

C

1 4

D

E 3

5 16

2x + y - 1 = 0 and passing through the point (1, 4) is: A 2x + y - 6 = 0 B 2x + y - 2 = 0 C x - 2y + 7 = 0 D x + 2y + 9 = 0 E x - 2y = 0

10 Produce a table of values, and sketch the graph of

the equation y = -5x + 15 for values of x between -10 and +10.

11 Sketch the graph of the following linear equations,

labelling the x- and y-intercepts. a y = 3x - 2 b y = -5x + 15 2

c y = − 3 x + 1 7 5

d y = x - 3 12 Find the x- and y-intercepts of the following

straight lines. a y = -7x + 6 3 8 4 x 7

b y = x - 5 c y =

- 43

d y = 0.5x + 2.8 13 Sketch graphs of the following linear equations by

finding the x- and y-intercepts. a 2x - 3y = 6 b 3x + y = 0 c 5x + y = -3 d x + y + 3 = 0 14 Sketch the graph of each of the following. 1 2

a y = x

b y = -4x

c x = -2

d y = 7

15 Sketch the graph of the equation

3(y - 5) = 6(x + 1).

16 Find the equations of the straight lines in the

following graphs. a

b

y

8 The gradient of the line perpendicular to

3x - 4y + 7 = 0 is: A

3 4

4 3

0

D 3

-2

B 4

C − 3 E -4

1

x

y -4

0

x

-4

Chapter 3 Coordinate geometry

85

number AND algebra • Linear and non-linear relationships c

y

d

2 0

e

6

x

y

y

x

0

f

28 Find the equation of the straight line joining the point

• (2, 8)

y

x

0 - –43

0

5

x

17 Find the linear equation given the information in

each case below. a gradient = 3, y-intercept = -4 b gradient = -2, y-intercept = -5 1 c gradient = , y-intercept = 5 2 d gradient = 0, y-intercept = 6 18 For each of the following, find the equation of the

straight line with the given gradient and passing through the given point. a gradient = 7, point (2, 1) b gradient = -3, point (1, 1) 1 c gradient = , point (-2, 5) 2 3 5

d gradient = , point (1, -3) 19 Find the distance between the points (1, 3) and

(7, -2).

20 Prove that triangle ABC is isosceles given A(3, 1),

B(-3, 7) and C(-1, 3). 21 Show that the points A(1, 1), B(2, 3) and C(8, 0)

are the vertices of a right-angled triangle. 22 The midpoint of the line segment AB is (6, -4).

If B has coordinates (12, 10), find the coordinates of A. 23 Show that the points A(3, 1), B(5, 2) and C(11, 5)

are collinear. 24 Show that the lines y = 2x - 4 and x + 2y - 10 = 0

are perpendicular to one another. 25 Find the equation of the straight line passing

through the point (6, -2) and parallel to the line x + 2y - 1 = 0.

26 Find the equation of the line perpendicular to

3x - 2y + 6 = 0 and having the same y-intercept.

27 Find the equation of the perpendicular bisector of

the line joining the points (-2, 7) and (4, 11). 86

Maths Quest 10 for the Australian Curriculum

(-2, 5) and the point of intersection of the straight lines with equations y = 3x - 1 and y = 2x + 5. 29 Using the information given in the diagram. a Find: y i the gradient of AD B(4, 9) 9 ii the gradient of AB iii the equation of BC C 4 A iv the equation of DC O D v the coordinates of C. x 45 9 b Describe quadrilateral ABCD. 30 In triangle ABC, A is (1, 5), B is (-2, -3) and C is (8, -2). a Find: i the gradient of BC ii the midpoint, P, of AB iii the midpoint, Q, of AC. b Hence show that: i PQ is parallel to BC ii PQ is half the length of BC. problem solving 1 John has a part-time job working as a gardener and

is paid $13.50 per hour. a Complete the following table of values relating the amount of money received to the number of hours worked. Number of hours

0

2

4

6

8

10

Pay ($) b Find a linear equation relating the amount of

money received to the number of hours worked. c Sketch the linear equation on a Cartesian plane

over a suitable domain. d Using algebra, calculate the pay that John will 3

receive if he works for 6 4 hours. 2 A fun park charges a $12.50 entry fee and an additional $2.50 per ride. a Complete the following table of values relating the total cost to the number of rides. Number of rides

0

2

4

6

8

10

Cost ($) b Find a linear equation relating total cost to the

number of rides. c Sketch the linear equation on a Cartesian plane

over a suitable domain. d Using algebra, calculate the cost for 7 rides.

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships

so■that■it■travels■in■a■straight■line■in■order■to■ displace■an■opponents■‘guard■balls’.■In■a■particular■ $22.50■per■hour. game,■player■X■has■2■guard■balls■close■to■the■ a  Sketch■a■graph■showing■the■total■cost■for■ jack.■The■coordinates■of■the■jack■are■(0,■0)■and■the■ between■0■and■12■hours. 4 b  State■the■equation■relating■cost■to■time■rented. coordinates■of■the■guard■balls■are■A(-1,■ 5 )■and■ c  Predict■the■cost■of■hiring■a■boat■for■12■hours■ 1 57 B(- 2 ,■ 40 ).■Player■Y■bowls■a■ball■so■that■it■travels and■15■minutes.     in■a■straight■line■toward■the■jack.■The■ball■is■   4  ABCD■is■a■quadrilateral■with■vertices■A(4,■9),■ bowled■from■the■position■S,■with■the■coordinates■ B(7,■4),■C(1,■2)■and■D(a,■10).■ (-30,■24). Given■that■the■diagonals■are■perpendicular■to■each■ y other,■fi■nd: a  the■equation■of■the■diagonal■AC 24 S(-30, 24) b  the■equation■of■the■diagonal■BD c  the■value■of■a.   5  An■architect■decides■to■design■a■building■with■a■ —) 57 B(- 1–2 , 57 – 40 14-metre-square■base■such■that■the■external■walls■ 40 are■initially■vertical■to■a■height■of■50■metres,■but■ taper■so■that■their■separation■is■8■metres■at■its■peak■ 4– A(-1, 4–5 ) height■of■90■metres.■A■profi■le■of■the■building■is■ 5 shown■with■the■point■(0,■0)■marked■as■a■reference■at■ 1 -30 x -1 -–2 the■centre■of■the■base.

  3  The■cost■of■hiring■a■boat■is■$160■plus■

y C

(Not to scale) a  Will■player■Y■displace■one■of■the■guard■balls?■

8m

If■so,■which■one? b  Due■to■bias,■the■displaced■guard■ball■is■

B 90 m 50 m

A

0 14 m

x

a  Write■the■equation■of■the■vertical■line■

connecting■A■and■B. b  Write■the■coordinates■of■B■and■C. c  Find■the■length■of■the■tapered■section■of■wall■

from■B■to■C.   6  In■a■game■of■lawn■bowls,■the■object■is■to■bowl■a■ biased■ball■so■that■it■gets■as■close■as■possible■to■ a■smaller■white■ball■called■a■jack.■During■a■game,■ a■player■will■sometimes■bowl■a■ball■quite■quickly■

knocked■so■that■it■begins■to■travel■in■a■straight■ line■(at■right■angles■to■the■path■found■in■part■a).■ Find■the■equation■of■the■line■of■the■guard■ball. c  Show■that■guard■ball■A■is■initially■heading■ directly■toward■guard■ball■B. d  Given■its■initial■velocity,■guard■ball■A■can■ travel■in■a■straight■line■for■1■metre■before■ its■bias■affects■it■path.■Calculate■and■explain■ whether■guard■ball■A■will■collide■with■guard■ ball■B. eBook plus

Interactivities

Test yourself Chapter 3 int-2834 Word search Chapter 3 int-2832 Crossword Chapter 3 int-2833

Chapter 3 Coordinate geometry

87

eBook plus

ACtivities

Chapter opener Digital doc

•■ Hungry■brain■activity■Chapter■3■(doc-5195)■(page 55) are you ready? Digital docs (page 56) •■ SkillSHEET■3.1■(doc-5196):■Measuring■the■rise■and■ the■run •■ SkillSHEET■3.2■(doc-5197):■Describing■the■gradient■ of■a■line •■ SkillSHEET■3.3■(doc-5198):■Plotting■a■line■using■a■ table■of■values •■ SkillSHEET■3.4■(doc-5199):■Stating■the■y-intercept■ from■a■graph •■ SkillSHEET■3.5■(doc-5200):■Solving■linear■ equations■that■arise■when■fi■nding■x-■and■y-intercepts •■ SkillSHEET■3.6■(doc-5201):■Using■Pythagoras’■ theorem

3a   Sketching linear graphs Digital docs

•■ Activity■3-A-1■(doc-4975):■Sketching■linear■graphs■ (page 62) •■ Activity■3-A-2■(doc-4976):■Graphs■of■linear■ equations■(page 62) •■ Activity■3-A-3■(doc-4977):■More■graphs■of■linear■ equations■(page 62) •■ SkillSHEET■3.7■(doc-5202):■Substitution■into■a■ linear■rule■(page 62) •■ SkillSHEET■3.5■(doc-5200):■Solving■linear■ equations■that■arise■when■fi■nding■x-■and■y-intercepts■ (page 63) •■ SkillSHEET■3.8■(doc-5203):■Transposing■linear■ equations■to■standard■form■(page 63) 3B   determining linear equations Digital docs

•■ Activity■3-B-1■(doc-4978):■Determining■linear■ equations■(page 67) •■ Activity■3-B-2■(doc-4979):■Linear■equations■(page 67) •■ Activity■3-B-3■(doc-4980):■More■complex■linear■ equations■(page 67) •■ SkillSHEET■3.1■(doc-5196):■Measuring■the■rise■and■ the■run■(page 67) •■ SkillSHEET■3.9■(doc-5204):■Finding■the■gradient■ given■two■points■(page 67) •■ WorkSHEET■3.1■(doc-5205):■Gradient■(page 68) 3C   The distance between two points on a straight line

(page 70) •■ Activity■3-C-1■(doc-4981):■Finding■the■distance■ between■two■points■on■a■straight■line Digital docs

88

maths Quest 10 for the Australian Curriculum

•■ Activity■3-C-2■(doc-4982):■Calculation■of■distance■ between■two■points •■ Activity■3-C-3■(doc-4983):■Applications■of■distance■ between■two■points •■ Spreadsheet■021■(doc-5206):■Distance■between■two■ points 3d   The midpoint of a line segment Digital docs

•■ Activity■3-D-1■(doc-4984):■Finding■the■midpoint■of■ a■line■segment■(page 73) •■ Activity■3-D-2■(doc-4985):■Calculations■—■ midpoint■of■a■segment■(page 73) •■ Activity■3-D-3■(doc-4986):■Applications■—■ midpoint■of■a■segment■(page 73) •■ Spreadsheet■075■(doc-5207):■Midpoint■of■a■segment■ (page 73) •■ WorkSHEET■3.2■(doc-5208):■Midpoint■of■a■line■ segment■(page 74) 3E   Parallel and perpendicular lines

(page 79) •■ Activity■3-E-1■(doc-4987):■Parallel■and■ perpendicular■lines •■ Activity■3-E-2■(doc-4988):■More■diffi■cult■parallel■ and■perpendicular■lines •■ Activity■3-E-3■(doc-4989):■Complex■parallel■and■ perpendicular■lines■ •■ Spreadsheet■085■(doc-5209):■Perpendicular■ checker •■ Spreadsheet■029■(doc-5210):■Equation■of■a■straight■ line Digital docs

Interactivity

•■ Parallel■and■perpendicular■lines■(int-2779)■ (page 74) Chapter review

(page 87) •■ Test■Yourself■Chapter■3■(int-2834):■Take■the■end-ofchapter■test■to■test■your■progress •■ Word■search■Chapter■3■(int-2832):■an■interactive■ word■search■involving■words■associated■with■this■ chapter •■ Crossword■Chapter■3■(int-2833):■an■interactive■ crossword■using■the■defi■nitions■associated■with■the■ chapter Interactivities

To access eBookPLUS activities, log on to www.jacplus.com.au

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships

4 simultaneous linear equations and inequations

4a Graphical solution of simultaneous linear equations 4b Solving simultaneous linear equations using substitution 4c Solving simultaneous linear equations using elimination 4d Problem solving using simultaneous linear equations 4e Solving linear inequations 4F Sketching linear inequations 4G Solving simultaneous linear inequations WhAt Do you knoW ? 1 List what you know about linear equations and linear inequations. Create a concept map to show your list. 2 Share what you know with a partner and then with a small group. 3 As a class, create a large concept map that shows your class’s knowledge of linear equations and linear inequations. eBook plus

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Hungry brain activity Chapter 4 doc-5211

opening Question

How could John decide which of the two concreting companies he should use — Angelico’s Concrete ($700 plus $20 per m2 of concrete) and Baux Cementing ($1200 plus $15 per m2 of concrete)?

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships

Are you ready?

Try■the■questions■below.■If■you■have■diffi■culty■with■any■of■them,■extra■help■can■be■obtained■by■ completing■the■matching■SkillSHEET.■Either■search■for■the■SkillSHEET■in■your■eBookPLUS■or■ ask■your■teacher■for■a■copy. eBook plus

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SkillSHEET 4.1 doc-5212

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SkillSHEET 4.2 doc-5213

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SkillSHEET 4.3 doc-5214

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SkillSHEET 4.4 doc-5215

Substitution into a linear rule 1 Substitute■-1■for■x■in■each■of■the■following■equations■to■calculate■the■value■of■y. a y■=■4x■-■2 b y■=■3■-■x c y■=■-2■-■5x

Solving linear equations that arise when finding x- and y-intercepts 2 For■each■of■the■following■equations,■substitute: i x■=■0■to■fi■nd■the■corresponding■value■of■y ii y■=■0■to■fi■nd■the■corresponding■value■of■x a 2x■+■3y■=■6 b x■-■3y■=■9 c 4y■=■3x■-■6 Transposing linear equations to standard form 3 Write■the■following■equations■in■the■form■y■=■mx■+■c. a 2y■+■4x■=■8 b 8x■-■2y■=■10

Measuring the rise and the run 4 Find■the■gradient■for■each■of■the■following■straight■lines. a b y y 10 20 5 10 -10 -5 0 -5

5 10 x

-10

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SkillSHEET 4.5 doc-5216

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SkillSHEET 4.6 doc-5217

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SkillSHEET 4.7 doc-5218

90

-10 -5 0 5 10 x -10 -20

c 2x■+■3y■+■5■=■0

c

y 10 5 -10 -5 0 -5

5 10 x

-10

Finding the gradient given two points 5 Find■the■gradient■of■the■line■passing■through■each■of■the■following■pairs■of■points. a (1,■2)■and■(3,■7) b (-1,■-4)■and■(2,■3) c (6,■-1)■and■(-2,■1)

Graphing linear equations using the x- and y-intercept method 6 Graph■each■line■with■the■following■equations■using■the■x-■and■y-intercept■method. a 5y■-■4x■=■20 b 4y■- 2x■=■5 c 3y■+■4x■=■-12

Checking whether a given point makes the inequation a true statement 7 For■each■of■the■following,■use■substitution■to■check■if■the■given■point■makes■the■inequality■a■

true■statement. a 3x■-■2y■ -8■(2,■-12)

number AND algebra • Linear and non-linear relationships

4A

Graphical solution of simultaneous linear equations Simultaneous linear equations ■■ ■■ ■■ ■■

Any two linear graphs will meet at a point, unless they are parallel. At this point, the two equations simultaneously share the same x- and y-coordinates. This point is referred to as the solution to the two simultaneous linear equations. Simultaneous equations can be solved graphically or algebraically.

Graphical solution ■■ ■■ ■■ ■■

This method involves drawing the graph of each equation on the same set of axes. The intersection point is the simultaneous solution to the two equations. An accurate solution depends on drawing an accurate graph. Graph paper or graphing software can be used.

Worked Example 1

Use the graph of the given simultaneous equations below to determine the point of intersection and, hence, the solution of the simultaneous equations. x + 2y = 4 y = 2x - 3

y 3

y = 2x - 3

2 1 -1 0 -1

x + 2y = 4 1

2

3

4

5 x

-2 -3

Think

Write

1

Write the equations and number them.

x + 2y = 4 y = 2x - 3

2

Locate the point of intersection of the two lines. This gives the solution.

Point of intersection (2, 1) Solution: x = 2 and y = 1

3

Check the solution by substituting x = 2 and y = 1 into the given equations. Comment on the results obtained.

Check equation [1]: LHS = x + 2y RHS = 4 = 2 + 2(1) =4 LHS = RHS Check equation [2]: LHS = y RHS = 2x - 3 = 1 = 2(2) - 3 =4-3 =1 LHS = RHS In both cases LHS = RHS, therefore the solution set (2, 1) is correct.

[1] [2]

Chapter 4 Simultaneous linear equations and inequations

91

number AND algebra • Linear and non-linear relationships

■■

It is always important to check the solution.

Worked Example 2

For the following simultaneous equations, use substitution to check if the given pair of coordinates, (5, -2), is a solution. 3x - 2y = 19 [1] 4y + x = -3 [2] Think

Write

1

Write the equations and number them.

3x - 2y = 19   4y + x = -3

2

Check by substituting x = 5 and y = -2 into equation [1].

Check equation [1]: LHS = 3x - 2y = 3(5) - 2(-2) = 15 + 4 = 19 LHS = RHS

Check by substituting x = 5 and y = -2 into equation [2].

Check equation [2]: LHS = 4y + x RHS = -3 = 4(-2) + 5 = -8 + 5 = -3 LHS = RHS In both cases, LHS = RHS. Therefore, the solution set (5, -2) is correct.

3

■■

[1] [2] RHS = 19

In order to obtain an accurate solution to a pair of simultaneous equations it is important to draw an accurate graph. This is demonstrated in the example below.

Worked Example 3

Solve the following pair of simultaneous equations using a graphical method. x + y = 6 2x + 4y = 20 Think 1

Write the equations, one under the other and ■ number them.

x + y = 6 2x + 4y = 20

2

Calculate the x- and y-intercepts for equation [1]. For the x-intercept, substitute y = 0 into equation [1].

Equation [1] x-intercept: when y = 0, x+0=6 x=6 The x-intercept is at (6, 0). y-intercept: when x = 0, 0+y=6 y=6 The y-intercept is at (0, 6).

For the y-intercept, substitute x = 0 into equation [1].

92

Write/draw

Maths Quest 10 for the Australian Curriculum

[1] [2]

number AND algebra • Linear and non-linear relationships

3

Calculate the x- and y-intercepts for equation [2]. For the x-intercept, substitute y = 0 into equation [2]. Divide both sides by 2.

For the y-intercept, substitute x = 0 into equation [2]. Divide both sides by 4. 4

Use graph paper to rule up a set of axes and label the x-axis from 0 to 10 and the y-axis from 0 to 6.

Equation [2] x-intercept: when y = 0, 2x + 0 = 20 2x = 20 x = 10 The x-intercept is at (10, 0). y-intercept: when x = 0, 0 + 4y = 20 4y = 20 y=5 The y-intercept is at (0, 5). y 6 5 4 3 2 1

(2, 4)

5

Plot the x- and y-intercepts for each equation.

6

Produce a graph of each equation by ruling a straight line through its intercepts.

7

Label each graph.

8

Locate the point of intersection of the lines.

The point of intersection is (2, 4).

9

Check the solution by substituting x = 2 and y = 4 into each equation.

Check [1]: LHS = x + y =2+4 =6 LHS = RHS Check [2]: LHS = 2x + 4y = 2(2) + 4(4) = 4 + 16 = 20 LHS = RHS

10

State the solution.

■■

x 0 1 2 3 4 5 6 7 8 910 -3-2-1 -1 -2 x+y=6 -3

RHS = 6

RHS = 20

In both cases, LHS = RHS. Therefore, the solution set (2, 4) is correct. The solution is x = 2, y = 4.

A CAS calculator can be used to obtain a graphical (as well as an algebraic) solution to simultaneous linear equations.

Parallel lines It is possible for two simultaneous linear equations to have no solution. ■■ This occurs when the graphs of the two equations do not cross because they have the same gradient. ■■ In other words, the two graphs are parallel. ■■ Consider the following pair of simultaneous equations. 2x - y = 5 [1] 4x - 2y = 6 [2] They can be graphed to show two parallel lines. ■■

2x + 4y = 20

6 5 4 3 2 1 -1 -10 -2 -3 -4 -5 -6

y 4x - 2y = 6 x 1

2

3

4

5

2x - y = 5

Chapter 4 Simultaneous linear equations and inequations

93

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships

■■

■■

We■can■confi■rm■that■the■two■lines■are■in■fact■parallel■by■rearranging■each■equation■into■the■ form■y■=■mx■+■c■and■checking■the■gradient■of■each. ■ 2x■-y■=■5■ [1] -y■=■5■-■2x■ -y=■-2x■+■5■  y=■2x■-■5■ gradient■m■=■2■ ■ 4x■-■2y=■6■■ [2] ■ -2y=■6■-■4x ■ -2y=■-4x■+■6 ■ y=■2x■-■3 ■gradient■m■=■2 It■is■also■possible■for■two■simultaneous■linear■equations■to■have■many■solutions.■This■occurs■ when■the■two■linear■equations■are,■in■fact■the■same■equation,■simply■expressed■in■a■different■ form.■For■example, y■=■2x■-■5■ ■ [1] 6x■-■3y■=■15■ ■ [2] Simplifying■equation■[2]■by■dividing■by■3■gives■2x■-■y■=■5. Rearranging■it■in■the■same■form■as■equation■[1]■gives■y■=■2x■-■5. A■word■of■caution■here:■Make■sure■that■the■signs■are■exactly■the■same■in■both■equations.■They■ will■not■represent■the■same■equation■if■this■is■not■the■case. remember

1.■ When■solving■simultaneous■equations■graphically,■obtaining■an■accurate■solution■ depends■on■drawing■accurate■graphs. 2.■ The■solution■to■linear■simultaneous■equations■is■the■point■where■their■graphs■ intersect. 3.■ Lines■that■have■the■same■gradient■are■parallel. 4.■ If■the■graphs■of■the■two■simultaneous■equations■are■parallel■lines,■then■the■simultaneous■ equations■have■no■solution,■as■they■have■no■point■of■intersection. exerCise

4A inDiviDuAl pAthWAys eBook plus

Activity 4-A-1

Investigating graphs of simultaneous equations doc-4990 Activity 4-A-2

Graphing simultaneous equations doc-4991

94

graphical solution of simultaneous linear equations fluenCy 1 We1 ■Use■the■graphs■below■of■the■given■simultaneous■equations■to■write■the■point■of■

intersection■and,■hence,■the■solution■of■the■simultaneous■equations. a x■+y=■3 b x■+y=■2 x■-y=■1 3x■-y=■2

y 5 4 3 2 1 -3 -2-1 0 -1 -2 -3

y

x-y=1

1 2 3 4 5

x

x+y=3

maths Quest 10 for the Australian Curriculum



6 5 4 3 2 1 -0.5 -10 -2 -3 -4

3x - y = 2

x 0.5

1.0

1.5

2.0

2.5

y+x=2

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships c y■-x■=■4

inDiviDuAl pAthWAys

d y■+■2x■=■3

3x■+■2y■=■8

eBook plus

3x + 2y = 8

Activity 4-A-3

Further graphing of simultaneous equations doc-4992

2y■+x■=■0 6

y

4

-3

y + 2x = 3

2

x

x

-1 0 -2

-2

y

1

2 -4

3

y-x=4

2

1

-1 0 -1

3

-4

-2

-6

-3

■

1

2

3

4

5

2y + x = 0

e y■-■3x■=■2

f 2y■-■4x■=■5

x■-y=■2

4y■+■2x■=■5 6

y

y

y - 3x = 2

6

4

-3

-2

-1

0

1

2

2

3

0

-1.0 -0.5

-2

4y + 2x = 5 0.5

1.0

1.5 2.0

x

-2

-4

-4

-6

■

2y - 4x = 5

4

x-y=2 x

2

-6



2 We2 ■For■the■following■simultaneous■equations,■use■substitution■to■check■if■the■given■pair■of■

coordinates■is■a■solution. a (7,■5)■ 3x■+■2y■=■31 ■ ■ 2x■+■3y■=■28 c (9,■1)■ x■+■3y■=■12 ■ ■ 5x■-■2y■=■43 e (4,■-3)■ y■=■3x■-■15 ■ ■ 4x■+■7y■=■-5 g (4,■-2)■ 2x■+y=■6 ■ ■ x■-■3y■=■8 i (-2,■-5)■ 3x■-■2y■=■-4 ■ ■ 2x■-■3y■=■11 eBook plus

Digital doc

SkillSHEET 4.6 doc-5217

b (3,■7)■ d f h j

■ ■ (2,■5)■ ■ ■ (6,■-2)■ ■ ■ (5,■1)■ ■ ■ (-3,■-1)■ ■ ■

y■-x■=■4 2y■+x■=■17 x■+y=■7 2x■+■3y■=■18 x■-■2y■=■2 3x■+y=■16 y■-■5x■=■-24 3y■+■4x■=■23 y■-x■=■2 2y■-■3x■=■7

3 We3 ■Solve■each■of■the■following■pairs■of■simultaneous■equations■using■a■graphical■method. a x■+y=■5 b x■+■2y■=■10

2x■+y=■8 c 2x■+■3y■=■6

2x■-y=■-10 e 6x■+■5y■=■12

5x■+■3y■=■10

3x■+y=■15 d x■-■3y■=■-8

2x■+y=■-2 f y+■2x■=■6

2y■+■3x■=■9

g y=■3x■+■10

h y=■8

y=■2x■+■8 i 4x■-■2y■=■-5 x■+■3y■=■4 k 3x■+■4y■=■27 x■+■2y■=■11

3x■+y=■17 j 3x■+y=■11 4x■-y=■3 l 3y■+■3x■=■8 3y■+■2x■=■6 Chapter 4 simultaneous linear equations and inequations

95

number AND algebra • Linear and non-linear relationships 4 Solve each of the following pairs of simultaneous equations. a y = 8 - x b y = 3x + 10

y=x+2 d y = 3 + 4x y = 1 + 3x

y = 2x + 8 e y = 16 - 3x y = 11 - 2x

g y = 7

h y =

y = 2x + 15

c y = 2x - 3

x=5 f 3y + x = 0 2y = 3x - 22

2x +2 3 y = 2x - 2

Understanding 5 Using technology, determine which of the following pairs of simultaneous equations have no

solutions. Confirm by finding the gradient of each line. a y = 2x - 4 b 5x - 3y = 13 3y - 6x = 10 4x - 2y = 10 d y = 4x + 5 e 3y + 2x = 9 2y - 10x = 8 6x + 4y = 22 g 4y + 3x = 7 h 2y - x = 0 12y + 9x = 22 14y - 6x = 2

c x + 2y = 8

5x + 10y = 45

f y = 5 - 3x

3y = -9x + 18

Reasoning 6 Two straight lines intersect at the point (3, -4). One of the lines has a y-intercept of 8. The

second line is a mirror image of the first in the line x = 3. Determine the equation of the second line. (Hint: Draw a graph of both lines) 7 At a well-known beach resort it is possible to hire a jet-ski by the hour in two different locations. On the Northern beach the cost is $20 plus $12 per hour, while on the Southern beach the cost is $8 plus $18 per hour. The jet-skis can be rented for up to 5 hours. a Write the rules relating cost to the length of rental. b On the same set of axes sketch a graph of cost (y-axis) against length of rental (x-axis) for 0–5 hours. reflection    c For what rental times, if any, is the Northern beach What do you think is the major rental cheaper than the Southern beach rental? Use error made when solving your graph to justify your answer. simultaneous equations d For what length of rental time are the two rental graphically? schemes identical? Use the graph and your rules ■ to justify your answer.

4b

Solving simultaneous linear equations using substitution ■■ ■■

There are two algebraic methods which can be used to solve simultaneous equations. They are the substitution method and the elimination method.

Substitution method ■■ ■■ ■■

96

This method is particularly useful when one (or both) of the equations is in a form where ■ one of the two variables is the subject. This variable is then substituted into the other equation, producing a third equation with only one variable. This third equation can then be used to determine the value of the variable.

Maths Quest 10 for the Australian Curriculum

number AND algebra • Linear and non-linear relationships

Worked Example 4

Solve the following simultaneous equations using the substitution method. y = 2x - 1 and 3x + 4y = 29 Think

Write

1

Write the equations, one under the other and number them.

y = 2x - 1 3x + 4y = 29

2

Substitute the expression (2x - 1) for y from equation [1] into equation [2].

Substituting (2x - 1) into [2]: 3x + 4(2x - 1) = 29

3

Solve for x. (i) Expand the brackets on the LHS of the equation. (ii) Collect like terms. (iii) Add 4 to both sides of the equation. (iv) Divide both sides by 11.

3x + 8x - 4 = 29

4

Substitute the value of x into any of the equations, say [1], to find the value of y.

Substituting x = 3 into [1]: y = 2(3) - 1 =6-1 =5

5

Write your answer.

Solution: x = 3, y = 5 or (3, 5)

6

Check the answer by substituting the point of intersection into equation [2].

Check: Substitute into 3x + 4y = 29. LHS = 3(3) + 4(5) RHS = 29 = 9 + 20 = 29 As LHS = RHS, the solution is correct.

■■ ■■ ■■

[1] [2]

[3]

11x - 4 = 29 11x = 33 x=3

In some cases, both equations may be written with the same variable as the subject. They can then be made equal to each other. This produces a third equation with only one variable.

Worked Example 5

Solve the following pair of simultaneous equations using the substitution method. y = 5x - 8 and y = -3x + 16 Think

Write

y = 5x - 8 y = -3x + 16

1

Write the equations, one under the other and number them.



2

Both equations are written with y as the subject, so equate them.

5x - 8 = -3x + 16

3

Solve for x. (i) Add 3x to both sides of the equation. (ii) Add 8 to both sides of the equation. (iii) Divide both sides of the equation by 8.

4

Substitute the value of x into either of the original equations, say [1], and solve for y.

[1] [2]

8x - 8 = 16 8x = 24 x=3 Substituting x = 3 into [1]: y = 5(3) - 8 = 15 - 8 =7

Chapter 4 Simultaneous linear equations and inequations

97

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships 5

Write■your■answer.

Solution:■x■=■3,■y■=■7■or■(3,■7)

6

Check■the■answer■by■substituting■the■point■of■ intersection■into■equation■[2].

Check:■Substitute■into■y■=■-3x■+■16. LHS■=■y ■ =■7■ RHS■=■-3x■+■16 ■ =■-3(3)■+■16 ■ =■-9■+■16 ■ =■7 As■LHS■=■RHS,■the■solution■is■correct.

remember

When■using■the■substitution■method■to■solve■simultaneous■equations: 1.■ choose■the■equation■in■which■one■of■the■variables■is■the■subject 2.■ substitute■this■expression■for■the■variable■into■the■other■equation■and■solve 3.■ substitute■the■value■you■have■found■into■the■rearranged■equation■to■solve■for■the■ other■variable 4.■ check■your■solution. exerCise

4b inDiviDuAl pAthWAys eBook plus

Activity 4-B-1

Learning substitution doc-4993 Activity 4-B-2

Practising substitution doc-4994 Activity 4-B-3

Tricky substitution doc-4995

98

solving simultaneous linear equations using substitution fluenCy 1 We4 ■Solve■the■following■simultaneous■equations■using■the■substitution■method.■Check■your■

solutions■using■technology. a x■=■-10■+■4y

b 3x■+■4y■=■2 3x■+■5y■=■21 x■=■7■+■5y c 3x■+y=■7 d 3x■+■2y■=■33 x■=■-3■-■3y y=■41■-■5x e y=■3x■-■3 f 4x■+y=■9 -5x■+■3y■=■3 y=■11■-■5x g x■=■-5■-■2y h x=■-4■-■3y 5y■+■x=■-11 -3x-■4y=■12 i x■=7■+■4y j x■=■14■+■4y 2x■+■y■=-4 -2x■+■3y■=■-18 k 3x+■2y■=■12 l y■=2x■+■1 x■=■9■-■4y -5x■-■4y■=■35 2 We5 ■Solve■the■following■pairs■of■simultaneous■equations■using■the■substitution■method.■ Check■your■solutions■using■technology. a y■=■2x■-■11■and■y■=■4x■+■1 b y■=■3x■+■8■and■y■=■7x■- 12 c y■=■2x■-■10■and■y■=■-3x d y■=■x■-■9■and■y■=■-5x e y■=■-4x■-■3■and■y■=■x■-■8 f y■=■-2x■-■5■and■y■=■10x■+■1 g y■=■-x■-■2■and■y■=■x■+■1 h y■=■6x■+■2■and■y■=■-4x i y■=■0.5x■and■y■=■0.8x■+■0.9

maths Quest 10 for the Australian Curriculum

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships j y■=■0.3x■and■y■=■0.2x■+■0.1 2 4 7 7 3 1 y■=■-x■and■y■=■- 4 x■- 4

k y■=■-x■and■y■=■- x■+ l

unDerstAnDing 3 A■small■farm■has■sheep■and■chickens.■There■are■twice■

as■many■chicken■as■sheep,■and■there■are■104■legs■ between■the■sheep■and■the■chickens.■How■many■ chickens■are■there?

4C eBook plus

Interactivity Simultaneous equations

int-2780

refleCtion 

 

When would you choose the substitution method in solving simultaneous equations?

solving simultaneous linear equations using elimination ■■ ■■ ■■

Elimination■is■best■used■when■the■two■equations■are■given■in■the■form■ax■+■by■=■k. The■method■involves■combining■the■two■equations■so■that■one■of■the■variables■is■eliminated. Addition■or■subtraction■can■be■used■to■reduce■the■two■equations■with■two■variables■into■ one■equation■with■only■one■variable.

WorkeD exAmple 6

Solve the following pair of simultaneous equations using the elimination method. -2x - 3y = -9 2x + y = 7 think

Write

1

Write■the■equations,■one■under■the■other■and■number■ them.

-2x■-■3y■=■-9■ 2x■+y=■7■

2

Look■for■an■addition■or■subtraction■that■will■eliminate■ eitherx■or■y. Note:■Adding■equations■[1]■and■[2]■in■order■will■ eliminatex.

[1]■+■[2]: -2x■-■3y■+■(2x■+■y)■=■-9■+■7 -2x■-■3y■+■2x■+■y■=■-2 -2y■=■-2

3

Solve■foryby■dividing■both■sides■of■the■equation■by■-2.

4

Substitute■the■value■ofyinto■equation■[2].■ Note:y=■1■may■be■substituted■into■either■equation.

5

Solve■forx. ■(i)■ Subtract■1■from■both■sides■of■the■equation. (ii)■ Divide■both■sides■of■the■equation■by■2.

6

Answer■the■question.

Solution:x■=■3,y=■1■or■(3,■1)

7

Check■the■answer■by■substituting■the■point■of■ intersection■into■equation■[1]■since■equation■[2]■was■ used■to■fi■nd■the■value■ofx.

Check:■Substitute■into■-2x■-■3y■=■-9. LHS■=■-2(3)■-■3(1) =■-6■-■3 =■-9 RHS■=■-9 LHS■=■RHS,■so■the■solution■is■correct.

■■

[1] [2]

y=■1 Substitutingy=■1■into■[2]: 2x■+■1■=■7 2x■=■6 x■=■3

When■the■like■terms■do■not■have■the■same■coeffi■cient,■multiply■one■or■both■equations■by■a■ constant■so■as■to■create■the■same■coeffi■cient. Chapter 4 simultaneous linear equations and inequations

99

number AND algebra • Linear and non-linear relationships

Worked Example 7

Solve the following pair of simultaneous equations using the elimination method. x - 5y = -17    2 x + 3y = 5 Think

Write

1

Write the equations, one under the other and number them.

  x - 5y = -17 2x + 3y = 5

[1] [2]

2

Look for a single multiplication that will create the same coefficient of either x or y. Multiply equation [1] by 2 and call the new equation [3].

[1] ì 2:   2x - 10y = -34

[3]

3

Subtract equation [2] from [3] in order to eliminate x.

[3] - [2]: 2x - 10y - (2x + 3y) = ‑34 - 5 2x - 10y - 2x - 3y = ‑39 -13y = -39

4

Solve for y by dividing both sides of the equation by -13.

5

Substitute the value of y into equation [2].

6

Solve for x. (i) Subtract 9 from both sides of the equation. (ii)  Divide both sides of the equation by 2.

y=3 Substituting y = 3 into [2]: 2x + 3(3) = 5 2x + 9 = 5 2x = -4 x = -2

7

Write your answer.

Solution: x = -2, y = 3 or (-2, 3)

8

Check the answer by substituting into equation [1].

Check: Substitute into x - 5y = -17. LHS = (-2) - 5(3) = -2 - 15 = -17 RHS = -17 LHS = RHS, so the solution is correct.

Note: In this example, equation [1] could have been multiplied by -2 (instead of by 2), then the two equations added (instead of subtracted) to eliminate x. ■■ Sometimes it is necessary to multiply both equations by a constant in order to achieve the same coefficient for one of the variables. Worked Example 8

Solve the following pair of simultaneous equations using the elimination method. 6x + 5y = 3    5x + 4y = 2 Think

100

Write

1

Write the equations, one under the other and number them.

6x + 5y = 3 5x + 4y = 2

[1] [2]

2

Decide which variable to eliminate, say y. Multiply equation [1] by 4 and call the new equation [3]. Multiply equation [2] by 5 and call the new equation [4].

Eliminate y. [1] ì 4:   24x + 20y = 12 [2] ì 5:   25x + 20y = 10

[3] [4]

Maths Quest 10 for the Australian Curriculum

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships

3

Subtract■equation■[3]■from■[4]■in■order■to■eliminate■y.■

[4]■-■[3]: 25x■+■20y■-■(24x■+■20y)■=■10■-■12 25x■+■20y■-■24x■-■20y■=■-2 x■=■-2

4

Substitute■the■value■ofx■into■equation■[1].

Substitutingx■=■-2■into■[1]: 6(-2)■+■5y■=■3 -12■+■5y■=■3

5

Solve■for■y. ■(i)■ Add■12■to■both■sides■of■the■equation (ii)■ Divide■both■sides■of■the■equation■by■5.

5y■=■15 y■=■3

6

Write■your■answer.

Solution:x■=■-2,y=■3■or■(-2,■3)

7

Check■the■answer■by■substituting■the■solution■into■ equation■[2].

Check:■Substitute■into■5x■+■4y■=■2. LHS■=■5(-2)■+■4(3) =■-10■+■12 =■2 RHS■=■2 LHS■=■RHS,■so■the■solution■is■correct.

Note:■Equation■[1]■could■have■been■multiplied■by■-4■(instead■of■by■4),■then■the■two■equations■ added■(instead■of■subtracted)■to■eliminate■y. remember

1.■ Simultaneous■equations■of■the■form■ax■+■by■=■k■can■be■solved■by■the■elimination■ method■by■looking■for■an■addition■or■subtraction■of■the■equations■that■will■eliminate■ one■of■the■variables. 2.■ For■like■terms■with■the■same■coeffi■cient■but■opposite■signs,■add■the■equations.■For■like■ terms■with■the■same■coeffi■cient■and■the■same■sign,■subtract■the■equations. 3.■ If■the■terms■do■not■have■the■same■coeffi■cient,■multiply■one■or■both■equations■by■ a■constant■to■create■the■same■coeffi■cient.■Remember■to■multiply■both■sides■of■the■ equation■to■keep■it■balanced. 4.■ Once■one■variable■has■been■eliminated,■solve■the■single■variable■equation■formed.■ Substitute■the■solution■back■into■one■of■the■original■equations■to■fi■nd■the■value■of■the■ variable■that■was■originally■eliminated. 5.■ Check■your■solution■by■substitution. exerCise

4C inDiviDuAl pAthWAys eBook plus

Activity 4-C-1

Elimination practice doc-4996

solving simultaneous linear equations using elimination fluenCy 1 We 6 ■Solve■the■following■pairs■of■simultaneous■equations■by■adding■equations■to■eliminate■

eitherx■or■y.

b 5x■+■4y■=■2 c -2x■+y=■10 -x■+■4y■=■1 5x■-■4y■=■-22 2x■+■3y■=■14 2 Solve■the■following■pairs■of■equations■by■subtracting■equations■to■eliminate■eitherx■or■y. a 3x■+■2y■=■13 b 2x■-■5y■=■-11 c -3x■-y=■8 5x■+■2y■=■23 2x■+y=■7 -3x■+■4y■=■13 a x■+■2y■=■5

Chapter 4 simultaneous linear equations and inequations

101

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships

inDiviDuAl pAthWAys eBook plus

3 Solve■each■of■the■following■equations■using■the■elimination■method. a x■+■2y■=■12 b 3x■+■2y■=■-23 c

Activity 4-C-2

Let’s eliminate doc-4997 Activity 4-C-3

More elimination doc-4998

e g i

3x■-■2y■=■12 6x■+■5y■=■-13 -2x■+■5y■=■-29 x■-■4y■=■27 3x■-■4y■=■17 -5x■+■3y■=■3 -5x■+y=■-4 4x■-■3y■-■1■=■0 4x■+■7y■-■11■=■0

5x■+■2y■=■-29 d 6x■-■5y■=■-43 6x■-y=■-23 f -4x■+y=■-10 4x■-■3y■=■14 h 5x■-■5y■=■1 2x■-■5y■=■-5

4 We7 ■Solve■the■following■pairs■of■simultaneous■equations. a x■+■2y■=■4 b 3x■+■2y■=■19

3x■-■4y■=■2■

6x■-■5y■=■-7

c -2x■+■3y■=■3

d 6x■+y=■9

e

f

g i k

5x■-■6y■=■-3 x■+■3y■=■14 3x■+y=■10■ -6x■+■5y■=■-14 -2x■+y=■-6 -3x■+■2y■=■6 x■+■4y■=■-9 2x■+■3y■=■9 4x■+y=■-7

h j l

-3x■+■2y■=■3 5x■+y=■27 4x■+■3y■=■26 2x■+■5y■=■14 3x■+y=■-5 3x■-■5y■=■7 x■+y=■-11 -x■+■5y■=■7 5x■+■5y■=■19

5 We8 ■Solve■the■following■pairs■of■simultaneous■equations. a 2x■+■3y■=■16 b 5x■-■3y■=■6 c e g i k

3x■+■2y■=■19 3x■+■2y■=■6 4x■+■3y■=■10 2x■-■3y■=■14 3x■-■5y■=■21 -4x■+■5y■=■-9 2x■+■3y■=■21■ 2x■-■2y■=■-4 5x■+■4y■=■17 x y + =2 2 3 x y + =4 4 3

d f h j l

3x■-■2y■=■3■ 2x■+■7y■=■3 3x■+■2y■=■13 -3x■+■7y■=■-2 4x■+■2y■=■14 2x■+■5y■=■-6 3x■+■2y■=■2■ 2x■-■3y■=■6 4x■-■5y■=■9 x y 3 + = 3 2 2 x y 1 + =− 2 5 2

6 Solve■the■following■simultaneous■equations■using■an■appropriate■method.■Check■your■answer■

using■technology. a 7x■+■3y■=■16 y■=■4x■-■1 c -3x■+■2y■=■19 4x■+■5y■=■13 e -4x■+■5y■=■-7 x■=■23■-■3y 102

maths Quest 10 for the Australian Curriculum

b 2x■+■y■=■8

4x■+■3y■=■16 d -3x■+■7y■=■9 4x■-■3y■=■7 f y■=■-x 1 2 y■= - 5 x - 5

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships reAsoning 7 Ann,■Beth■and■Celine■wanted■to■weigh■themselves■on■a■coin■weighing■machine.■The■problem■

was■they■only■had■enough■money■for■one■weighing.■They■decided■to■weigh■themselves■in■ pairs,■one■stepping■off■as■another■stepped■on. • Ann■and■Beth■weighed■119■kg refleCtion    • Beth■and■Celine■weighed■112■kg How does eliminating • Celine■and■Ann■weighed■115■kg one variable help to solve How■much■did■each■of■the■girls■weigh? simultaneous equations?

eBook plus

Digital doc

WorkSHEET 4.1 doc-5220

4D

problem solving using simultaneous linear equations ■■ ■■

Many■word■problems■can■be■solved■using■simultaneous■linear■equations. Follow■these■steps. •■ Defi■ne■the■unknown■quantities■using■appropriate■pronumerals. •■ Use■the■information■given■in■the■problem■to■form■two■equations■in■terms■of■these■ pronumerals. •■ Solve■these■equations■using■an■appropriate■method. •■ Write■the■solution■in■words. •■ Check■the■solution.

WorkeD exAmple 9

Ashley received better results for his Maths test than for his English test. If the sum of the two marks is 164 and the difference is 22, calculate the mark he received for each subject. think

Write

1

Defi■ne■the■two■variables.

Letx■=■the■maths■mark. Lety=■the■English■mark.

2

Formulate■two■equations■from■the■information■given■and■ number■them. Note:■Sum■means■to■add■and■difference■means■to■subtract.

x■+y=■164 x■-y=■22

3

Use■the■elimination■method■by■adding■equations■[1]■and■ [2]■to■eliminate■y.■

[1]■+■[2]:■ ■ 2x■=■186

4

Solve■forx■by■dividing■both■sides■of■the■equation■by■2.

5

Substitute■the■value■ofx■into■equation■[1].

6

Solve■foryby■subtracting■93■from■both■sides■of■the■ equation.

7

Answer■the■question.

Solution: Maths■mark■(x)■=■93 English■mark■(y)■=■71

8

Check■the■answer■by■substituting■x■=■93■andy=■71■into■ equation■[1].

Check:■Substitute■intox■+y=■164. LHS■=■93■+■71■ RHS■=■164 =■164 As■LHS■=■RHS,■the■solution■is■correct.

[1] [2]

x■=■93 Substitutingx■=■93■into■[1]: x■+y=■164 93■+y=■164 y=■71

Chapter 4 simultaneous linear equations and inequations

103

number AND algebra • Linear and non-linear relationships

Worked Example 10

To finish a project, Genevieve buys a total of 25 nuts and bolts from a hardware store. If each nut costs 12 cents, each bolt costs 25 cents and the total purchase price is $4.30, how many nuts and how many bolts does Genevieve buy? Think

Write

1

Define the two variables.

Let x = the number of nuts. Let y = the number of bolts.

2

Formulate two equations from the information given and number them. Note: The total number of nuts and bolts is 25. Each nut cost 12 cents, each bolt cost 25 cents and the total cost is 430 cents ($4.30).

x + y = 25 12x + 25y = 430

3

Solve simultaneously using the substitution method since equation [1] is easy to rearrange.

4

Rearrange equation [1] to make x the subject by subtracting y from both sides of equation [1].

Rearrange equation [1]: x + y = 25 x = 25 - y

5

Substitute the expression (25 - y) for x into equation [2].

Substituting (25 - y) into [2]: 12(25 - y) + 25y = 430

6

Solve for y.

300 - 12y + 25y = 430 300 + 13y = 430 13y + 300 = 430 13y = 130 y = 10

7

Substitute the value of y into the rearranged equation x = 25 - y from step 4.

Substituting y = 10 into x = 25 - y: x = 25 - 10 x = 15

8

Answer the question.

Solution: The number of nuts (x) = 15. The number of bolts (y) = 10.

9

Check the answer by substituting x = 15 and y = 10 into equation [1].

Check: Substitute into x + y = 25. LHS = 15 + 10    RHS = 25 = 25 As LHS = RHS, the solution is correct.

[1] [2]

remember

1. To solve worded problems, read the question carefully and define the two variables using appropriate pronumerals. 2. Formulate two equations from the information given and number them. 3. Use either the elimination method or the substitution method to solve the two equations simultaneously. 4. Check your answer by substituting the values obtained for each variable into the original equations. 104

Maths Quest 10 for the Australian Curriculum

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships

exerCise

4D inDiviDuAl pAthWAys eBook plus

Activity 4-D-1

Problem solving doc-4999

problem solving using simultaneous linear equations fluenCy 1 We 9 ■Rick■received■better■results■for■his■Maths■test■than■for■his■English■test.■If■the■sum■of■his■

two■marks■is■163■and■the■difference■is■31,■fi■nd■the■mark■for■each■subject. 2 We 10 ■Rachael■buys■some■nuts■and■bolts■to■fi■nish■a■project.■She■does■not■buy■the■same■number■

of■nuts■and■bolts,■but■buys■30■items■in■total.■If■each■nut■costs■10■cents,■each■bolt■costs■20■cents■ and■the■total■purchase■price■is■$4.20,■how■many■nuts■and■how■many■bolts■does■she■buy?

Activity 4-D-2

Harder problem solving doc-5000 Activity 4-D-3

Tricky problem solving doc-5001

unDerstAnDing 3 Find■two■numbers■whose■difference■is■5■and■whose■sum■is■11. 4 The■difference■between■two■numbers■is■2.■If■three■times■the■larger■number■minus■double■the■

smaller■number■is■13,■fi■nd■the■two■numbers. 5 One■number■is■9■less■than■three■times■a■second■number.■If■the■fi■rst■number■plus■twice■the■ second■number■is■16,■fi■nd■the■two■numbers. 6 A■rectangular■house■has■a■perimeter■of■40■metres■and■the■length■is■4■metres■more■than■the■ width.■What■are■the■dimensions■of■the■house? 7 Mike■has■5■lemons■and■3■oranges■in■his■shopping■basket.■The■cost■of■the■fruit■is■$3.50.■Voula,■ with■2■lemons■and■4■oranges,■pays■$2.10■for■her■fruit.■How■much■does■each■type■of■fruit■cost?

■ 8 A■surveyor■measuring■the■dimensions■of■a■block■of■land■fi■nds■that■the■length■of■the■block■is■

three■times■the■width.■If■the■perimeter■is■160■metres,■what■are■the■dimensions■of■the■block? 9 Julie■has■$3.10■in■change■in■her■pocket.■If■she■has■only■50■cent■and■20centpieces■and■the■total■

number■of■coins■is■11,■how■many■coins■of■each■type■does■she■have? 10 Mr■Yang’s■son■has■a■total■of■twenty-one■$1■and■$2■coins■in■his■moneybox.■When■he■counts■his■ money,■he■fi■nds■that■its■total■value■is■$30.■How■many■coins■of■each■type■does■he■have? 11 If■three■Magnums■and■two■Paddlepops■cost■$8.70■and■the■difference■in■price■between■a■ Magnum■and■a■Paddlepop■is■90■cents,■how■much■does■each■type■of■ice-cream■cost? 12 If■one■Redskin■and■4■Golden■roughs■cost■$1.65,■whereas■2■Redskins■and■3■Golden■roughs■cost■ $1.55,■how■much■does■each■type■of■sweet■cost? reAsoning 13 A■catering■fi■rm■works■out■its■pricing■based■on■a■fi■xed■cost■

for■overheads■and■a■charge■per■person.■It■is■known■that■a■party■ for■20■people■costs■$557,■whereas■a■party■for■35■people■costs■ $909.50.■Use■this■information■to■work■out■the■fi■xed■cost■and■the■ cost■per■person■charged■by■the■company. 14 The■difference■between■Sally’s■PE■mark■and■Science■mark■is■ 12,■and■the■sum■of■the■marks■is■154.■If■the■PE■mark■is■the■higher■ mark,■what■did■Sally■get■for■each■subject?

Chapter 4 simultaneous linear equations and inequations

105

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships 15 Mozza’s■cheese■supplies■sells■six■Mozzarella■cheeses■and■eight■Swiss■cheeses■to■Munga’s■deli■

for■$83.60,■and■four■Mozzarella■cheeses■and■four■Swiss■cheeses■to■Mina’s■deli■for■$48.■How■ much■does■each■type■of■cheese■cost? 16 If■the■perimeter■of■the■triangle■in■the■diagram■is■12■cm■and■the■length■of■the■rectangle■is■1■cm■ more■than■the■width,■fi■nd■the■value■ofx■and■y.

2x cm

y cm

x cm m

5c

(y + 3) cm

17 Mr■and■Mrs■Waugh■want■to■use■a■caterer■for■a■birthday■party■for■their■twin■sons.■The■manager■

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says■the■cost■for■a■family■of■four■would■be■$160.■However,■the■sons■want■to■invite■8■friends,■ making■12■people■in■all.■The■cost■for■this■would■be■$360.■If■the■total■cost■in■each■case■is■made■up■ of■the■same■cost■per■person■and■the■same■fi■xed■cost,■fi■nd■the■cost■per■person■and■the■fi■xed■cost. 18 Joel■needs■to■buy■some■blank■DVDs■and■zip■disks■to■ back■up■a■large■amount■of■data■that■has■been■generated■ refleCtion    by■an■accounting■fi■rm.■He■buys■6■DVDs■and■3■zip■disks■ How do you decide which for■$96.■He■later■realises■these■are■not■suffi■cient■and■so■ method to use when solving buys■another■5■DVDs■and■4■zip■disks■for■$140.■How■ word problems using much■did■each■DVD■and■each■zip■disk■cost?■(Assume■ simultaneous linear equations? the■same■rate■per■item■was■charged■for■each■visit.)

solving linear inequations ■■ ■■

■■ ■■

An■equation■is■a■statement■of■equality■such■as■x■=■2;■an■inequation■is■a■statement■of■ inequality■such■as■x■■2

x■is■greater■than■2

x■í■2

x■is■greater■than■or■equal■to■2

x■ > to < Ç to í í to Ç Chapter 4 Simultaneous linear equations and inequations

107

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships

WorkeD exAmple 12

Solve each of the following linear inequations. a -3m + 5 < -7 b 5(x - 2) í 7(x + 3) think a

b

Write

1

Write■the■inequation.

2

Subtract■5■from■both■sides■of■the■inequation.■ (No■change■to■the■inequality■sign.)

3

Obtain■m■by■dividing■both■sides■of■the■ inequation■by■-3.■Reverse■the■inequality■sign,■ since■you■are■dividing■by■a■negative■number.

1

Write■the■inequation.

2

Expand■both■brackets.

3

Combine■the■pronumeral■terms■by■subtracting■ 7x■from■both■sides■of■the■inequation.

4

Add■10■to■both■sides■of■the■inequation.

5

Obtain■x■by■dividing■both■sides■of■the■ inequation■by■-2.■Since■we■need■to■divide■by■ a■negative■number,■reverse■the■direction■of■the■ inequality■sign.

-3m■+■5■■to■■3■ b■ a■+■2■>■1■ d m■-■1■í■3■ e■ p■+■4■■-3 2 Solve■each■of■the■following■inequations.■Check■your■solutions■by■substitution. a 3m■>■9■ b■ 5p■Ç■10■ c■ 2a■■-25■ f■ 3x■Ç■-21

maths Quest 10 for the Australian Curriculum

c■ f■ i■ l■

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships

g 2m■í■-1■

inDiviDuAl pAthWAys

Activity 4-E-3

Puzzling inequations 3 doc-5004

3

4

eBook plus

x a k■ Ç■-2■ ■■40■-■2x■ g 7b■+■5■■a■+■13■ j 5(2m■-■3)■Ç■3m■+■6■ k■ 3(5b■+■2)■Ç■-10■+■4b■ Solve■each■of■the■following■inequations. x−2 x +1 ■í■-4■ a b■ ■Ç■4■ 2 5 2x + 3 3x − 1 ■>■6■ ■í■2■ d e■ 4 7 We12 ■Solve■each■of■the■following■inequations. a -2m■>■4■ b■ -5p■Ç■15■ d -p■-■3■Ç■2■ e■ 10■-■y■í■13■ g 1■-■6p■>■1■ h■ 2■-■10a■Ç■0■ j -4(a■+■9)■í■8■■ k■ -15■Ç■-3(2■+■b)■■ m k■+■5■■6 3 m l■ ■í■5 5 i■

c■ f■ i■ l■

5p■-■9■>■11 8y■-■2■>■14 3b■+■2■■7■+■2x i (3,■0)■ y■Ç■-3x j (0,■4)■ y■+■2x■>■4 2 Use■the■graphs■of■the■equations■given■below■to■sketch■the■graph■of■the■given■inequations.■ (Remember■to■shade■the■region■required.) a x■+y>■3 b x■+■2y■Ç■6■ 3

Activity 4-G-3

Further simultaneous linear inequations doc-5010

y

3

2

2

x+y=3

1 -1 0 -1

2

3

4

x + 2y = 6

1

x 1

y

x

-2 0 -1

5

2

-2

-2

-3 c 3x■-■2y■>■12 y 6

-3 d 4x■+yí■-8■ 12 10 8 6 4 2

4 2 -1 0 -2

x 1

2

3

4

-4

5

-6 e y■íx■+■4■ 6

-4

y

-2

118

8

10

y

x 1

2

3

4x + y = -8

y

2

2 0

6

0 -1 -2 -4 -6 -8 -10 -12

3 1

x 2

4

6

-2



-2

f y■ x + 3

x+yÇ4 a

6 5 4 3 2 1

y

6 5 4 3 2 1

3

4

d

2

3

4

6 5 4 3 2 1

1

3

4

5

y

x 2

1

-2

3

4

5

Region required

–4

y Region required

x

0

-1

2

Region required

0

-1

5

Region required

–4

E

6 5 4 3 2 1

x 1

x

-2

y

-2

y

0

-1

5

Region required

0

-1

2

1

-2

c

6 5 4 3 2 1

x

0

-1

b

2

1

3

4

5

-2 –4

b x Ç 5

2y + x > 2 a

3

y

b

Region required

2

y

Region required

2

1 -1 0 -1

3

x 1

2

3

4 5

1 -1 0 -1

-2

-2

-3

-3

x 1

2

3

4 5

Chapter 4 Simultaneous linear equations and inequations

119

number AND algebra • Linear and non-linear relationships c

3

d

y

1

1

x

-1 0 -1

1

2

x

-1 0 -1

3 4 5

1

2

3 4 5

-2

-2 -3 3

y

2

2

E

3

-3

Region required

y

Region required

Region required

2 1

x

-1 0 -1

1

2

3

4 5

-2 -3 c y í 3 - x

2x + 3y Ç 6 a

3

b

y

1 1

2

3

4

y

d

Region required

3

4

5

3

y

Region required

2

1

x

-1 0 -1

1

2

3

4

1 -1 0 -1

5

-2

-2

-3

-3 y

Region required

2 1

x 1

2

3

4

-2 -3 120

2

-3

Region required

2

-1 0 -1

1

-2

-3

3

x

-1 0 -1

5

-2

E

Region required

1

x

-1 0 -1

3

y

2

2

c

3

Maths Quest 10 for the Australian Curriculum

5

x 1

2

3

4

5

number AND algebra • Linear and non-linear relationships d x - y < 3

x - 2y í 4 a

3

b

y

1 1

2

3

4

5

1

2

3

4

5

-2

-3

-3

Region required

y

d

Region required

3

2

2

1

1

-1 0 -1

1

2

3

4

5

x

–1 0 -1

-2

-2

-3

-3

3

x

-1 0 -1

-2

E

Region required

1

x

-1 0 -1

3

y

2

2

c

3

y

y

Region required

1

2

3

4

5

x

Region required

2 1 -1 0 -1

1

2

3

4

5

x

-2 -3

understanding 4   WE 14  Use a graphical technique to solve the following simultaneous inequations. a x + y < 3 b 3x + 2y > 12 c 2y > x - 2

2x - y í 4

d y > 2x + 4

y < 4 - 2x g x + 2y í 10 3x + y > 15 j y - x > 4 2x + 3y Ç 6 m x + y > 7 2x - 3y í 18 5 a b

x + 5y Ç 10

e y - 2x Ç 5

x+y>4 h y > 2x - 3 x 3 y < 2x n y > 4 y í 2x

y 17 i 3y - 2x < 6 y í 2x - 2 l y - 2x í 9 x+yÇ4

Sketch the half plane represented by the region:  i y Ç x + 2 i i y í 4 - 2x.

Show the region where both the inequations y Ç x + 2 and y í 4 - 2x hold true. 6 Show the region where the inequations 2x + y < 0 and x - 2y > 0 simultaneously hold true. Chapter 4 Simultaneous linear equations and inequations

121

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships 7 Natsuko■is■starting■to■plan■a■monthly■budget■by■classifying■expenditures■such■as■rent■and■other■

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122

expenses■(r)■and■savings■(x).■Her■total■net■income■is■$2000■per■month.■She■can■spend■no■more■ than■30■per■cent■of■her■income■on■rent. a Write■an■inequation■to■express■the■constraint■that■Natsuko■can■spend■no■more■than■ $2000■per■month. b Write■an■inequation■to■express■the■30■per■cent■rent■and■other■expenses■limitation. c Do■any■other■inequations■apply■to■this■situation?■Explain. d Sketch■a■graph■of■the■region■that■applies■for■all■your■inequations. e State■three■possible■solutions■of■allocating■rent■and■other■expenses/savings. 8 Monica■wants■to■take■a■minimum■of■450■units■of■vitamin■C■and■300■units■of■vitamin■E■per■ day.■Each■brand■A■tablet■provides■100■units■of■vitamin■C■and■50■units■of■vitamin■E,■while■each■ brand■B■tablet■provides■75■units■of■vitamin■C■and■75■units■of■vitamin■E. a Write■an■inequation■which■indicates■the■combination■needed■of■each■brand■of■vitamin■ tablet■to■meet■the■daily■requirement■of■vitamin■C. (Hint:■Let■a■=■the■number■of■brand■A■tablets■and■b=■the■number■of■brandBtablets.) b Write■an■inequation■which■indicates■the■combination■needed■of■each■brand■of■vitamin■ tablet■to■meet■the■daily■requirement■of■vitamin■E. c Graph■the■two■inequations■and■indicate■the■region■which■provides■a■solution■to■both■the■ vitamin■C■and■vitamin■E■requirements. d Recommend■to■Monica■two■different■vitamin■plans■that■fi■t■the■restrictions. 9 Maria■is■making■some■high-energy■sweets■using■peanuts■and■chocolate■chips.■She■wanted■ to■make■a■maximum■of■400■g■of■the■sweets,■but■wanted■them■to■contain■at■least■180■g■of■ carbohydrates. a Let■the■mass■of■peanuts■be■p■and■the■mass■of■chocolate■chips■be■c.■Write■an■inequation■to■ represent■the■constraint■that■the■total■mass■must■be■at■most■400■■g. b On■a■Cartesian■plane,■sketch■the■region■defi■ned■by■the■inequation■obtained■in■part■a. (Hint:■Consider■only■the■positive■axes■as■the■values■of■both■p■and■c■must■be■positive.) c The■peanuts■provide■30%■of■their■mass■in■carbohydrates■and■the■chocolate■chips■provide■ 60%■of■their■mass■in■carbohydrates.■Write■an■inequation■that■represents■the■constraint■that■ the■mass■of■carbohydrates■must■be■at■least■180■■g. d On■a■Cartesian■plane,■sketch■the■region■defi■ned■by■the■inequation■obtained■in■part■c . e On■a■Cartesian■plane,■show■the■region■where■the■inequations■sketched■in■parts■b■and■d■ both■hold■true. f The■region■obtained■in■part■e■shows■all■possible■ refleCtion    masses■of■peanuts■and■chocolate■chips■that■meet■ Maria’s■requirements.■List■fi■ve■sets■of■possible■ How do the solutions from a masses■of■peanuts■and■chocolate■chips■that■would■ system of equations differ from a system of inequations? meet■her■requirements.

maths Quest 10 for the Australian Curriculum

number AND algebra • Linear and non-linear relationships

Summary Graphical solution of simultaneous linear equations ■■

■■ ■■ ■■

When solving simultaneous equations graphically, obtaining an accurate solution depends on drawing accurate graphs. The solution to linear simultaneous equations is the point where their graphs intersect. Lines that have the same gradient are parallel. If the graphs of the two simultaneous equations are parallel lines, then the simultaneous equations have no solution, as they have no point of intersection. Solving simultaneous linear equations using substitution

When using the substitution method to solve simultaneous equations: choose the equation in which one of the variables is the subject ■■ substitute this expression for the variable into the other equation and solve ■■ substitute the value you have found into the rearranged equation to solve for the other ■ variable ■■ check your solution. ■■

Solving simultaneous linear equations using elimination ■■

■■

■■

■■

■■

Simultaneous equations of the form ax + by = k can be solved by the elimination method by looking for an addition or subtraction of the equations that will eliminate one of the variables. For like terms with the same coefficient but opposite signs, add the equations. For like terms with the same coefficient and the same sign, subtract the equations. If the terms do not have the same coefficient, multiply one or both equations by a constant to create the same coefficient. Remember to multiply both sides of the equation to keep it balanced. Once one variable has been eliminated, solve the single variable equation formed. Substitute the solution back into one of the original equations to find the value of the variable that was originally eliminated. Check your solution by substitution. Problem solving using simultaneous linear equations

■■

■■ ■■

■■

To solve worded problems, read the question carefully and define the two variables using appropriate pronumerals. Formulate two equations from the information given and number them. Use either the elimination method or the substitution method to solve the two equations simultaneously. Check your answer by substituting the values obtained for each variable into the original equations. Solving linear inequations

■■

■■

■■

The solution to an inequation is a portion of the number line. (That is, there are an infinite number of solutions to any given inequation.) When solving an inequation, imagine an equals sign in place of the inequality sign and solve as if it was a linear equation. Remember to keep writing the original inequality sign between the two sides of each step unless the special case applies. Special case: if in the process of the solution you need to multiply or divide both sides of the inequation by a negative number, reverse the inequality sign. That is, change < to >, > to ■(greater■than),■í■(greater■than■and■equal■to),■ ■or■■mx■+■c,■then■the■region■above■the■line■is■the■ required■region. If■the■inequation■is■of■the■form■y■Ç■mx■+■c■or■y■ 3x - 12 d y < 5x e x í 7 1 f y Ç 2 x + 1 g 2x + y í 9 h 4x - 3y í 48 i y > -12

2

22 Use substitution to check if the given pair of



d gradient = 5, point (1, -3)

20 Sketch the half plane given by each of the

y 6 5 4 3 2 1

the substitution method. a y = 3x + 1 b x + 2y = 16 c 2x + 5y = 6 d 3 y = 2x + 5 e y = 3x - 11

y = 5x + 17

y = 2x + 7 3y - 4x = 11 y = -x y = 8x + 21

f y = 4x - 17

y = 6x - 22

25 Solve the following simultaneous equations using

the elimination method. a 3x + y = 17 b 7x - y = 33 c 3x - 7y = -2 d -2x - 7y = 13 e 5x + 2y = 6 f 4x + 3y = 2

4x + 3y = 1 -4x + y = 11 4y - 3x = 9 y + 3x = 6 x - 4y = -4 4x - 2y = 12

Chapter 4 Simultaneous linear equations and inequations

127

number AND algebra • Linear and non-linear relationships 26 Solve the following simultaneous equations using

an appropriate method. a 3x + 2y = 6 b 6x - 4y = -6 3y + 5x = 9 7x + 3y = -30 c 6x + 2y = 14 x = -3 + 5y

equations and solve. a Find two numbers whose difference is 5 and whose sum is 23. b A rectangular house has a total perimeter of 34 metres and the width is 5 metres less than the length. What are the dimensions of the house? c If two Chupa Chups and three Wizz Fizzes cost $2.55, but five Chupa Chups and seven Wizz Fizzes cost $6.10, find the price of each type of lolly.

27 Solve the following simultaneous inequations. a y Ç x + 4 b 2y - 3x í 12

yí3

y + 3x > 0

c 5x + y < 10

x + 2y < 11

5 Laurie buys milk and bread for his family on the

problem solving 1 John has a part-time job working as a gardener and

is paid $13.50 per hour. a Complete the following table of values relating the amount of money received to the number of hours worked. Number of hours

0

2

4

6

8

10

Pay ($) b Find a linear equation relating the amount

of money received to the number of hours worked. c Sketch the linear equation on a Cartesian plane over a suitable domain. d Using algebra, calculate the pay that John will 3 receive if he works for 6 4 hours. 2 A fun park charges a $12.50 entry fee and an

additional $2.50 per ride. a Complete the following table of values relating the total cost to the number of rides. Number of rides

0

2

4

6

8

10

Cost ($) b Find a linear equation relating total cost to the

number of rides. c Sketch the linear equation on a Cartesian plane

over a suitable domain. d Using algebra, calculate the cost of entry and

7 rides. 3 The cost of hiring a boat is $160 plus

$22.50 per hour. a Sketch a graph showing the total cost for 0 to

12 hours. b State the equation relating cost to time rented. c Predict the cost of hiring a boat for 12 hours

and 15 minutes. 128

4 Write the following as a pair of simultaneous

Maths Quest 10 for the Australian Curriculum

way home from school each day, paying with a $10 note. If he buys three cartons of milk and two loaves of bread, he receives 5 cents in change. If he buys two cartons of milk and one loaf of bread, he receives $4.15 in change. How much does each item cost? 6 A paddock contains some cockatoos (2-legged)

and kangaroos (4-legged). The total number of animals is 21 and they have 68 legs in total. Using simultaneous equations, determine how many cockatoos and kangaroos there are in the paddock. 7 At a fun park, the cost of a rollercoaster ride and

a Ferris wheel ride is $10. The cost of a Gravitron ride and a Ferris wheel ride is $12. The cost of a Rollercoaster ride and a Gravitron ride is $14. What is the cost of each ride? 8 There are two sections to a concert hall. Seats in the

‘Dress circle’ are arranged in rows of 40 and cost $140 each. Seats in the ‘Bleachers’ are arranged in rows of 70 and cost $60 each. There are 10 more rows in the ‘Dress circle’ than in the ‘Bleachers’ and the capacity of the hall is 7000. a If d represent the number of rows in the ‘Dress circle’ and b represents the number of rows in the ‘Bleachers’ then write an equation in terms of these two variables based on the fact that there are 10 more rows in the ‘Dress circle’ than in the ‘Bleachers’. b Write an equation in terms of these two variables based on the fact that the capacity of the hall is 7000 seats. c Solve the two equations from a and b simultaneously using the method of your choice to find the number of rows in each section. d Now that you have the number of rows in each section, calculate the number of seats in each section. e Hence, calculate the total receipts for a concert where all tickets are sold.

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships 9 John■is■comparing■two■car■rental■companies,■

10 Frederika■has■$24■■000■saved■for■a■holiday■and■a■ Golden■Ace■Rental■Company■and■Silver■Diamond■ new■stereo.■Her■travel■expenses■are■$5400■and■her■ Rental■Company.■Golden■Ace■Rental■Company■ daily■expenses■are■$260. charges■a■fl■at■rate■of■$38■per■day■and■$0.20■per■ a Write■down■an■equation■for■the■cost■of■her■ kilometre.■The■Silver■Diamond■Rental■Company■ holiday■if■she■stays■for■d■days. charges■a■fl■at■rate■of■$30■per■day■plus■$0.32■per■ ■■ Upon■her■return■from■holidays■Frederika■wants■ kilometre. to■purchase■a■new■stereo■system■that■will■cost■ a Write■an■algebraic■equation■for■the■cost■of■ her■$2500. renting■a■car■for■three■days■from■the■Golden■ b How■many■days■can■she■spend■on■her■holiday■ Ace■Rental■Company■in■terms■of■the■number■of■ if■she■wishes■to■purchase■a■new■stereo■upon■her■ kilometres■travelled,■k. return? b Write■an■algebraic■equation■for■the■cost■of■ 11 Mick■the■painter■has■fi■xed■costs■(e.g.■insurance,■ renting■a■car■for■three■days■from■the■Silver■ equipment,■etc)■of■$3400■per■year.■His■running■cost■ Diamond■Rental■Company■in■terms■of■the■ to■travel■to■jobs■is■based■on■$0.75■per■kilometre.■ number■of■kilometres■travelled,■k. Last■year■Mick■had■costs■that■were■less■than■ c How■many■kilometres■would■John■have■to■ $16■■000. travel■so■that■the■cost■of■hiring■from■each■ a Write■an■inequality■to■show■this■information■ company■for■three■days■is■the■same? and■solve■it■to■fi■nd■how■many■kilometres■Mick■ d Write■an■inequation■that,■when■solved,■will■tell■ travelled■for■the■year.■ you■the■number■of■kilometres■for■which■it■is■ b Explain■the■information■you■have■found. cheaper■to■use■Golden■Ace■Rental■Company■ when■renting■for■three■days. eBook plus e For■what■number■of■kilometres■will■it■be■ Interactivities cheaper■to■use■Silver■Diamond■Rental■ Test yourself Chapter 4 Company■for■three■days’■hire? int-2837 Word search Chapter 4 int-2835 Crossword Chapter 4 int-2836

Chapter 4 simultaneous linear equations and inequations

129

eBook plus

ACtivities

chapter opener Digital doc

•■ Hungry■brain■activity■Chapter■4■(doc-5211)■(page89) are you ready? Digital docs

(page90)

•■ SkillSHEET■4.1■(doc-5212):■Substitution■into■a■linear■rule •■ SkillSHEET■4.2■(doc-5213):■Solving■linear■equations■ that■arise■when■fi■nding■x-■and■ y-intercepts •■ SkillSHEET■4.3■(doc-5214):■Transposing■linear■ equations■to■standard■form •■ SkillSHEET■4.4■(doc-5215):■Measuring■the■rise■and■ the■run •■ SkillSHEET■4.5■(doc-5216):■Finding■the■gradient■given■ two■points •■ SkillSHEET■4.6■(doc-5217):■Graphing■linear■equations■ using■the■x-■and■y-intercept■method •■ SkillSHEET■4.7■(doc-5218):■Checking■whether■ a■given■point■makes■the■inequation■a■true■statement 4a Graphical solution of simultaneous linear equations

•■ Activity■4-A-1■(doc-4990):■Investigating■graphs■of■ simultaneous■equations■(page94) •■ Activity■4-A-2■(doc-4991):■Graphing■simultaneous■ equations■(page94) •■ Activity■4-A-3■(doc-4992):■Further■graphing■of■ simultaneous■equations■(page95) •■ SkillSHEET■4.6■(doc-5217):■Graphing■linear■equations■ using■the■x-■and■y-intercept■method■(page95) 4b Solving simultaneous linear equations using substitution

(page98)

•■ Activity■4-B-1■(doc-4993):■Learning■substitution■ •■ Activity■4-B-2■(doc-4994):■Practising■substitution■ •■ Activity■4-B-3■(doc-4995):■Tricky■substitution■ 4c Solving simultaneous linear equations using elimination Digital docs

•■ Activity■4-C-1■(doc-4996):■Elimination■practice■ (page101) •■ Activity■4-C-2■(doc-4997):■Let’s■eliminate■(page102) •■ Activity■4-C-3■(doc-4998):■More■elimination■(page102) •■ WorkSHEET■4.1■(doc-5220):■Simultaneous■equations■ I■(page103) Interactivity

•■ Simultaneous■linear■equations■(int-2780)■(page99) 4d Problem solving using simultaneous linear equations Digital docs

•■ Activity■4-D-1■(doc-4999):■Problem■solving■ (page105) 130

4e Solving linear inequations Digital docs

•■ Activity■4-E-1■(doc-5002):■Puzzling■inequations■1■ (page108) •■ Activity■4-E-2■(doc-5003):■Puzzling■inequations■2■ (page108) •■ Activity■4-E-3■(doc-5004):■Puzzling■inequations■3■ (page109) •■ SkillSHEET■4.7■(doc-5218):■Checking■whether■a■ given■point■makes■the■inequation■a■true■statement■ (page109) •■ SkillSHEET■4.8■(doc-5219):■Writing■equations■from■ worded■statements■(page109) 4F Sketching linear inequations Digital docs

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•■ Activity■4-D-2■(doc-5000):■Harder■problem■solving■ (page105) •■ Activity■4-D-3■(doc-5001):■Tricky■problem■solving■ (page105) •■ WorkSHEET■4.2■(doc-5221):■Simultaneous■ equations■II■(page106)

maths Quest 10 for the Australian Curriculum

•■ Activity■4-F-1■(doc-5005):■Understanding■linear■ inequations■(page113) •■ Activity■4-F-2■(doc-5006):■Graphing■linear■ inequations■(page114) •■ Activity■4-F-3■(doc-5007):■Interpreting■linear■ inequation■graphs■(page114) 4G Solving simultaneous linear inequations Digital docs

•■ Activity■4-G-1■(doc-5008):■Introducing■ simultaneous■linear■inequations■(page118) •■ Activity■4-G-2■(doc-5009):■Practising■simultaneous■ linear■inequations■(page118) •■ Activity■4-G-3■(doc-5010):■Further■simultaneous■ linear■inequations■(page118) •■ WorkSHEET■4.3■(doc-5222):■Simultaneous■ equations■III■(page122) chapter review

(page129) •■ Test■Yourself■Chapter■4■(int-2837):■Take■the■end-ofchapter■test■to■test■your■progress •■ Word■search■Chapter■4■(int-2835):■an■interactive■ word■search■involving■words■associated■with■this■ chapter •■ Crossword■Chapter■4■(int-2836):■an■interactive■ crossword■using■the■defi■nitions■associated■with■the■ chapter Interactivities

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measuremenT anD geomeTry • PyThagoras anD TrigonomeTry

5 Trigonometry I

5A Pythagoras’ theorem 5B Pythagoras’ theorem in three dimensions 5C Trigonometric ratios 5D Using trigonometry to calculate side lengths 5E Using trigonometry to calculate angle size 5F Angles of elevation and depression 5G Bearings and compass directions 5H Applications WhaT Do you knoW ? 1 List what you know about trigonometry. Create a concept map to show your list. 2 Share what you know with a partner and then with a small group. 3 As a class, create a large concept map that shows your class’s knowledge of trigonometry. eBook plus

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Hungry brain activity Chapter 5 doc-5223

oPening QuesTion

How can Raylene find her way to the finish line of the orienteering course?

measuremenT anD geomeTry • PyThagoras anD TrigonomeTry

are you ready?

Try the questions below. If you have difficulty with any of them, extra help can be obtained by completing the matching SkillSHEET. Either search for the SkillSHEET in your eBookPLUS or ask your teacher for a copy. eBook plus

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SkillSHEET 5.1 doc-5224

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SkillSHEET 5.2 doc-5225

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SkillSHEET 5.3 doc-5226

Rounding to a given number of decimal places 1 Round the following numbers to 3 decimal places. a 0.6845 b 1.3996

c 0.7487

Rounding the size of an angle to the nearest minute and second 2 Round the following angles: i to the nearest minute ii to the nearest second. a 15è32Å40.5ë b 63è15Å32.4ë c 27è10Å15.8ë Labelling the sides of a right-angled triangle 3 Label the sides of the following right-angled triangles using the letters H (for hypotenuse),

O (for opposite) and A (for adjacent) with respect to angle q. a

b

c

q

q q

eBook plus

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SkillSHEET 5.5 doc-5227

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Digital doc

SkillSHEET 5.6 doc-5228

Rearranging formulas 4 Rearrange each of the following formulas to make x the subject. a tan 15è =

x 30

4.2 x

c

x = 5.3 tan 64°

Drawing a diagram from given directions 5 Draw a diagram for each of the following situations. a Kate’s bushwalking route took her from A to B, a distance of 5 km at a bearing of 25èT

then to C, a further distance of 7.5 km at a bearing of 120èT. b A ship steamed S20èE for a distance of 180 km, then the ship travelled N60èW for a

further 70 km.

132

b tan 28è =

maths Quest 10 for the australian Curriculum

measurement AND geometry • Pythagoras and trigonometry

5A

Pythagoras’ theorem Similar right-angled triangles In the two similar right-angled triangles shown below, the angles are the same and the corresponding sides are in the same ratio. D

A

6 cm

3 cm

B

10 cm

5 cm

4 cm

C

E

8 cm

F

The corresponding sides are in the same ratio. AB AC BC = = . DE DF EF To write this using the side lengths of the triangles gives: AB 3 1 = = DE 6 2 AC 5 1 = = DF 10 2 BC 4 1 = = EF 8 2 This means that for right-angled triangles, when the angles are fixed, the ratios of the sides in the triangle are constant. We can examine this idea further by completing the following activity. Using a protractor and ruler, draw an angle of 70è, measuring horizontal distances of 3 cm, 7 cm and 10 cm as demonstrated in the diagram below.

c b a 70è 3 cm 7 cm 10 cm Note: Diagram not drawn to scale.

Measure the perpendicular heights a, b and c. a ö 8.24 cm    b ö 19.23 cm    c ö 27.47 cm Chapter 5 Trigonometry I

133

measurement AND geometry • Pythagoras and trigonometry

To test if the theory that for right-angled triangles, when the angles are fixed, the ratios of the sides in the triangle are constant is correct, calculate the ratios of the side lengths. a 8.24 ≈ ≈ 2.75 3 3 b 19.23 ≈ ≈ 2.75 7 7 c 27.47 ≈ ≈ 2.75 10 10 The ratios are the same because the triangles are similar. This important concept forms the basis of trigonometry.

Review of Pythagoras’ theorem ■■

■■ ■■

Pythagoras’ theorem states that in any right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. The rule is written as c2 = a2 + b2 where a and b are the two shorter sides and c is the hypotenuse. The hypotenuse is the longest side of a right-angled triangle and is always the side that is opposite the right angle. Pythagoras’ theorem gives us a way of finding the length of the third side in a triangle, if we know the lengths of the two other sides.

c

a

b x

4 7

Finding the hypotenuse ■■

To calculate the length of the hypotenuse when given the length of the two shorter sides, substitute the known values into the formula c2 = a2 + b2.

Worked Example 1

For the triangle at right, calculate the length of the hypotenuse, x, correct to 1 decimal place.

x

4 7 Think 1

Write/draw

Copy the diagram and label the sides a, b and c. Remember to label the hypotenuse as c.

c=x

a=4

b=7 3

Substitute the values of a, b and c into this rule and simplify.

4

Calculate x by taking the square root of 65. Round the answer correct to 1 decimal place.

x = 65 x = 8.1

Maths Quest 10 for the Australian Curriculum

a2

+

x2

2

134

=

Write Pythagoras’ theorem.

c2

b2

= + 72 = 16 + 49 = 65 42

measurement AND geometry • Pythagoras and trigonometry

Finding a shorter side Sometimes a question will give you the length of the hypotenuse and ask you to find one of the shorter sides. In such examples, we need to rearrange Pythagoras’ formula. Given that c2 = a2 + b2, we can rewrite this as: a2 = c2 - b2 or b2 = c2 - a2. ■■

Worked Example 2

Calculate the length, correct to 1 decimal place, of the unmarked side of the triangle at right. 14 cm 8 cm Think 1

Write/draw

Copy the diagram and label the sides a, b and c. Remember to label the hypotenuse as c; it does not matter which side is a and which side is b.

a c = 14 b=8

2

Write Pythagoras’ theorem.

3

Substitute the values of a, b and c into this rule and simplify.

4

Find a by taking the square root of 132. Round to 1 decimal place.

■■

c2 = a2 + b2 142 = a2 + 82 196 = a2 + 64 a2 = 196 - 64 = 132 a = 132 = 11.5  cm

In many cases we are able to use Pythagoras’ theorem to solve practical problems.   First model the problem by drawing a diagram, then use Pythagoras’ theorem to solve the right-angled triangle. Use the result to give a worded answer.

Worked Example 3

A ladder that is 4.5  m long leans up against a vertical wall. The foot of the ladder is 1.2  m from the wall. How far up the wall does the ladder reach? Give your answer correct to 1 decimal place. Think 1

Write/draw

Draw a diagram and label the sides a, b and c. Remember to label the hypotenuse as c.

c = 4.5 m

a

b = 1.2 m Chapter 5 Trigonometry I

135

measurement AND geometry • Pythagoras and trigonometry

2

Write Pythagoras’ theorem.

c2 = a2 + b2

3

Substitute the values of a, b and c into this rule and simplify.

4.52 = a2 + 1.22 20.25 = a2 + 1.44 a2 = 20.25 - 1.44 = 18.81

4

Find a by taking the square root of 18.81. Round to 1 decimal place and include the unit of measurement (m).

5

Answer the question in a sentence.

■■

a = 18.81 = 4.3  m The ladder will reach a height of 4.3  m up the wall.

Sometimes the unknown length involves more than one side.   In these cases, substitute into Pythagoras’ theorem, then solve the following equation for the unknown.

Worked Example 4

Calculate the value of the pronumeral, correct to 2 decimal places, in the triangle at right.

3x 78 2x

Think 1

Copy the diagram and label the sides a, b and c.

Write/draw b = 3x c = 78 a = 2x

136

c2 = a2 + b2

2

Write Pythagoras’ theorem.

3

Substitute the values of a, b and c into this rule and simplify.

782 = (3x)2 + (2x)2 6084 = 9x2 + 4x2 6084 = 13x2

4

Rearrange the equation so that the pronumeral is on the left-hand side of the equation.

13x2 = 6084

5

Divide both sides of the equation by 13.

6

Find x by taking the square root. Round the answer correct to 2 decimal places.

Maths Quest 10 for the Australian Curriculum

13 x 2 6084 = 13 13 x2 = 468 x = 468 = 21.63

measuremenT anD geomeTry • PyThagoras anD TrigonomeTry

rememBer

1. The hypotenuse is the longest side of the triangle and is opposite the right angle. 2. On your diagram, check whether you are finding the length of the hypotenuse or one of the shorter sides. 3. The length of a side can be found if we are given the length of the other sides by using the formula c2 = a2 + b2. 4. When using Pythagoras’ theorem, always check the units given for each measurement. 5. If necessary, convert all measurements to the same units before using the rule. 6. Worded problems can be solved by drawing a diagram and using Pythagoras’ theorem to solve the problem. 7. Worded problems should be answered in a sentence. exerCise

5a inDiViDual PaThWays eBook plus

Activity 5-A-1

Pythagoras’ theorem FluenCy 1 We1 For each of the following triangles, calculate the length of the hypotenuse, giving

answers correct to 2 decimal places. 4.7

a

b

Review of Pythagoras’ theorem doc-5011

804 6.3

Activity 5-A-2

27.1

Practising Pythagoras’ theorem doc-5012 Activity 5-A-3

More of Pythagoras’ theorem doc-5013

c

19.3

562 0.9

e

d

152

f

7.4

87 10.3

eBook plus

2.7

Digital doc

SkillSHEET 5.1 doc-5224

2 We2 Find the value of the pronumeral, correct to 2 decimal places. a b s c 1.98 30.1 47.2

u

8.4

2.56

17.52 t

d

e

0.28 v 0.67

f

2870

w

1920

468 x

114 Chapter 5 Trigonometry I

137

measurement AND geometry • Pythagoras and trigonometry 3   WE3  The diagonal of the rectangular sign at right is 4

5

6

7

34  cm. If the height of this sign is 25 cm, find the width. A right-angled triangle has a base of 4  cm and a height of 12  cm. Calculate the length of the hypotenuse to 2 decimal places. Calculate the lengths of the diagonals (to 2 decimal places) of squares that have side lengths of: a 10  cm b 17  cm c 3.2  cm. The diagonal of a rectangle is 120  cm. One side has a length of 70  cm. Determine: a the length of the other side b the perimeter of the rectangle c the area of the rectangle.   WE4  Find the value of the pronumeral, correct to 2 decimal places for each of the following. a

c

b 25

3x

2x

3x

4x 18 x

30

6x

understanding 8 An isosceles triangle has a base of 30  cm and a height of 10  cm. Calculate the length of the two

equal sides. 9 An equilateral triangle has sides of length 20  cm. Find the height of the triangle. 10 A right-angled triangle has a height of 17.2  cm, and a base that is half the height. Calculate the

length of the hypotenuse, correct to 2 decimal places. 11 The road sign shown below is in the form of an equilateral triangle. Find the height of the sign

and, hence, find its area.

76 cm

12 A flagpole, 12  m high, is supported by three wires, attached from the top of the pole to the

ground. Each wire is pegged into the ground 5  m from the pole. How much wire is needed to support the pole? 13 Ben’s dog ‘Macca’ has wandered onto a frozen pond, and is too frightened to walk back. Ben estimates that the dog is 3.5  m from the edge of the pond. He finds a plank, 4  m long, and thinks he can use it to rescue Macca. The pond is surrounded by a bank that is 1  m high. Ben uses the plank to make a ramp for Macca to walk up. Will he be able to rescue his dog? 14 Sarah goes canoeing in a large lake. She paddles 2.1  km to the 3.8 km north, then 3.8  km to the west. Use the triangle at right to find out how far she must then paddle to get back to her starting point in 2.1 km the shortest possible way. Starting point 138

Maths Quest 10 for the Australian Curriculum

measurement AND geometry • Pythagoras and trigonometry 15 A baseball diamond is a square of side length 27  m. When a runner on first base tries to steal

second base, the catcher has to throw the ball from home base to second base. How far is that throw?

Second base

27 m First base

Home base Catcher 16 Penny, a carpenter, is building a roof for a new house. The roof has

17 18

19 20

21 22

a gable end in the form of an isosceles triangle, with a base of 6  m and sloping sides of 7.5  m. She decides to put 5 evenly spaced vertical 7.5 m 7.5 m strips of wood as decoration on the gable as shown at right. How many metres of this decorative wood does she need? Calculate the length, in mm, of the hypotenuse of a right-angled 6m triangle, if the two shorter sides are 5  cm and 12  cm. The hypotenuse and one other side of a right-angled triangle are given for each case below. Find the length of the third side in the units specified. Give your answers correct to 2 decimal places. a Sides 46  cm and 25  cm, third side in mm b Sides 843  mm and 1047  mm, third side in cm c Sides 4500  m and 3850  m, third side in  km d Sides 20.3  cm and 123  mm, third side in cm e Sides 6420  mm and 8.4  m, third side in cm f Sides 0.358  km and 2640  m, third side in m g Sides 491  mm and 10.8  cm, third side in mm h Sides 379  000  m and 82  700  m, third side in  km A rectangle measures 35  mm by 4.2  cm. Calculate the length of its diagonal in millimetres to 2 decimal places. A rectangular envelope has a length of 21  cm and a diagonal measuring 35  cm. Calculate: a the width of the envelope b the area of the envelope. A sheet of A4 paper measures 210  mm by 297  mm. Calculate the length of the diagonal in centimetres to 2 decimal places. A right-angled triangle has a hypotenuse of 47.3  cm and one other side of 30.8  cm. Calculate the area of the triangle. Chapter 5 Trigonometry I

139

measurement AND geometry • Pythagoras and trigonometry 23 A swimming pool is 50  m by 25  m. Peter is bored by his usual training routine, and decides

24

25 26 27

28

to swim the diagonal of the pool. How many diagonals must he swim to complete his normal distance of 1200  m? Give your answer to 2 decimal places. Sarah is making a gate that has to be 1200  mm wide. It must be braced with a diagonal strut made of a different type of timber. She has only 2  m of this kind of timber available. What is the maximum height of the gate that she can make? A hiker walks 4.5  km west, then 3.8  km south. How far in metres is she from her starting point? Give your answer to 2 decimal places. A square has a diagonal of 10  cm. What is the length of each side? Wally is installing a watering system in his garden. The pipe is to go all around the edge of the rectangular garden, and have a branch diagonally across the garden. The garden measures 5  m by 7.2  m. If the pipe costs $2.40 per metre (or part thereof), what will be the total cost of the pipe? The size of a rectangular television screen is given by the length of its diagonal. What is the size of the screen at right to the nearest centimetre if its dimensions are 158  cm wide and 96  cm deep?

Reasoning 29 During a recent earthquake, Helen’s large

bookshelf fell over. The bookshelf is x metres wide and 2.5 metres high. The ceiling is 3 metres high. Show that if the bookshelf is lying on its side next to the wall and is able to be stood up directly, then x is less than 1.658 metres

5B

 

The square root of a number usually gives us both a positive and negative answer. Why do we only take the positive answer when using Pythagoras’ theorem?

Pythagoras’ theorem in three dimensions ■■

Many real-life situations involve 3-dimensional (3-D) shapes: shapes with length, width and height. Some common 3-D shapes used in this section include boxes, pyramids and rightangled wedges.

Box ■■

140

reflection 

Pyramid

Right-angled wedge

The important thing about 3-D shapes is that in a diagram, right angles may not look like right angles, so it is important to redraw sections of the diagram in two dimensions, where the right angles can be seen accurately.

Maths Quest 10 for the Australian Curriculum

measurement AND geometry • Pythagoras and trigonometry

Worked Example 5

Determine the length AG in this box.

A

B 6 cm

C

D

F

E

5 cm 10 cm

H Think 1

G

Write/draw A

Draw the diagram in three dimensions.

B 6 cm

C

D

F

E

5 cm H 2

3

Draw in two dimensions, a right-angled triangle that contains AG and label the sides. Only 1 side is known, so we need to find another right-angled triangle to use. Draw EFGH in two dimensions and show the diagonal EG. Label the side EG as x. We have two of the three side lengths so we can calculate the unknown.

10 cm

A

6 E

G

E

F

5

x

5

H

10

G

4

Use Pythagoras’ theorem to calculate EG.

c2 = a2 + b2 x2 = 52 + 102 = 25 + 100 = 125 x = 125 = 11.18  cm

5

Place this information on triangle AEG. Label the side AG as y. Now we have two of the three side lengths.

A 6 E

6

Use Pythagoras’ theorem to find AG.

G

y

11.18

G

c2 = a2 + b2 y2 = 62 + ( 125 )2 = 36 + 125 = 161 y = 161 = 12.69

7

Answer the question in a sentence.

The length of AG is 12.69  cm.

Chapter 5 Trigonometry I

141

measurement AND geometry • Pythagoras and trigonometry

Worked Example 6

A piece of cheese in the shape of a right-angled wedge sits on a table. It has a rectangular base measuring 14  cm by 8  cm, and is 4  cm high at the thickest point. An ant crawls diagonally across the sloping face. How far, to the nearest millimetre, does the ant walk? Think 1

Draw a diagram in three dimensions and label the vertices. Mark BD, the path taken by the ant, with a dotted line.

Write/draw B E A

14 cm

D

C 4 cm F 8 cm

x 2

Draw in two dimensions a right-angled triangle that contains BD, and label the sides. Only one side is known, so we need to find another rightangled triangle to use.

B 4 D

E 3

Draw EFDA in two dimensions, and show the diagonal ED. Label the side ED as x.

4

Use Pythagoras’ theorem to calculate ED.

5

Place this information on triangle BED. Label the side BD as y.

E

F

8

x

8

A

14

D

c2 = a2 + b2 x2 = 82 + 142 = 64 + 196 = 260 x = 260 = 16.12  cm B y

4 E

142

6

Solve this triangle for BD.

7

Check the answer’s units. We need to convert cm to mm, so multiply by 10.

8

Answer the question in a sentence.

Maths Quest 10 for the Australian Curriculum

16.12

D

c2 = a2 + b2 2 y2 = 42 + ( 260 ) = 16 + 260 = 276 y = 276 = 16.61  cm = 166.1  mm The ant walks 166  mm, correct to the nearest millimetre.

measuremenT anD geomeTry • PyThagoras anD TrigonomeTry

rememBer

1. Pythagoras’ theorem can be used to solve problems in three dimensions. 2. Some common 3-D shapes include boxes, pyramids and right-angled wedges. 3. To solve problems in three dimensions it is helpful to draw sections of the original shape in two dimensions. exerCise

5B inDiViDual PaThWays eBook plus

Activity 5-B-1

Pythagoras in 3-dimensions doc-5014

Pythagoras’ theorem in three dimensions Where appropriate in this exercise, give answers correct to 2 decimal places. FluenCy 1 We5 Calculate the length, AG. a b A B C

D

10

A

B

c

C

D

A

B C

D

10.4

Activity 5-B-2

Pythagoras in 3-D figures doc-5015

E

eBook plus

Digital doc

E

F 7.3

10

Activity 5-B-3

Investigating triangles in 3-D figures doc-5016

10

F

H

10

G

H

F

E

G

8.2

5 H

5

G

2 Calculate the length of CE in the wedge at right and, hence,

A

obtain AC.

E

SkillSHEET 5.4 doc-5229

D 3 If DC = 3.2 m, AC = 5.8 m, and CF = 4.5 m in the figure at right,

7

C

10

B 4 F

B

A

calculate the length of AD and BF.

F D

C

4 Calculate the length of BD and, hence, the height of the pyramid

V

at right.

8 A 8

D 5 The pyramid ABCDE has a square base. The pyramid is 20 cm high.

B

8

C

E

Each sloping edge measures 30 cm. Calculate the length of the sides of the base.

EM = 20 cm

A D

B M C

Chapter 5 Trigonometry I

143

measuremenT anD geomeTry • PyThagoras anD TrigonomeTry 6 The sloping side of a cone is 10 cm and the height is 8 cm.

What is the length of the radius of the base? 10 cm

8 cm 7 An ice-cream cone has a diameter across the top of 6 cm, and

sloping side of 13 cm. How deep is the cone?

r

8 We6 A piece of cheese in the shape of a right-angled wedge

B

sits on a table. It has a base measuring 20 mm by 10 mm, and is 4 mm high at the thickest point, as shown in the figure. A fly crawls diagonally across the sloping face. How far, to the nearest millimetre, does the fly walk?

E A

20 mm

C 4 mm F D 10 mm

unDersTanDing 9 Jodie travels to Bolivia, taking with her a suitcase as shown in the photo. She buys a carved

walking stick 1.2 m long. Will she be able to fit it in her suitcase for the flight home? 30

cm

65 cm

90 cm

10 A desk tidy is shaped like a cylinder, height 18 cm and diameter

10 cm. Pencils that are 24 cm long rest inside. What lengths of the pencils are above the top of the cylinder?

11 A 10-m high flagpole is in the corner of a rectangular park 10 m that measures 240 m by 150 m. A 240 m a Calculate: i the length of the diagonal of the park 150 m ii the distance from A to the top of the pole B iii the distance from B to the top of the pole. b A bird flies from the top of the pole to the centre of the park. How far does it fly? 144

maths Quest 10 for the australian Curriculum

measuremenT anD geomeTry • PyThagoras anD TrigonomeTry 12 A candlestick is in the shape of two cones, joined at the vertices

as shown. The smaller cone has a diameter and sloping side of 7 cm, and the larger one has a diameter and sloping side of 10 cm. How tall is the candlestick? 13 The total height of the shape below is 15 cm. Calculate the length of the sloping

side of the pyramid. 15 cm 6 cm 14 cm 14 cm 14 A sandcastle is in the shape of a truncated cone as shown. Calculate the length of the diameter

of the base. 20 cm

30 cm

eBook plus

32 cm

15 A tent is in the shape of a triangular prism, with a height of

120 cm

120 cm as shown at right. The width across the base of the door is 1 m and the tent is 2.3 m long. Calculate the length of each sloping side, in metres. Then calculate the area of fabric used in the construction of the sloping rectangles which form the sides.

Digital doc

WorkSHEET 5.1 doc-5230

2.3 m 1m

reasoning 16 Stephano is renovating his apartment, which is at the end of two corridors. The corridors of the

apartment building are 2 m wide with 2 m high ceilings, and the first corridor is at right angles to the second. Show that he can carry lengths of timber up to 6 m long to his apartment. reFleCTion 

 

The diagonal distance across a rectangle of 2

2

dimensions x by y is x + y . What would be the rule to find the length of a diagonal across a cuboid of dimensions x by y by z ? Use your rule to check your answers to question 1.

5C

Trigonometric ratios angles and the calculator ■ ■

Last year you were shown that each angle has specific values for its sine, cosine and tangent. These values are needed for practically every trigonometry problem and can be obtained with the aid of a calculator. Chapter 5 Trigonometry I

145

measurement AND geometry • Pythagoras and trigonometry

Worked Example 7

Calculate the value of each of the following, correct to 4 decimal places, using a calculator. a  cos 65è57Å b  tan 56è45Å30ë Think

Write

a Write your answer to the correct number

of decimal places. b Write your answer to the correct number

of decimal places.

a cos 65è57Å = 0.4075 b tan 56è45Å30ë = 1.5257

Worked Example 8

Calculate the size of angle q, correct to the nearest degree, given sin q  = 0.6583. Think

Write

1

Write the given information.

sin q = 0.6583

2

To find the size of the angle, we need to undo sine with its inverse, sin-1. Ensure your calculator is in degrees mode.

q = sin-1 (0.6583)

3

Write your answer to the nearest degree.

q = 41è

■■ ■■

Sometimes, we need to be able to find an angle correct to either the nearest minute or nearest second. When we use an inverse trigonometric function, the angle is expressed in degrees as a decimal. It should be converted to degrees, minutes and seconds (DMS).

Worked Example 9

Calculate the value of q : a  correct to the nearest minute, given that cos q  = 0.2547 b  correct to the nearest second, given that tan q  = 2.364. Think a

b

146

1

Write the equation.

2

Write your answer, rounding to the nearest minute. Remember there are 60 minutes in 1 degree and 60 seconds in 1 minute. Hence, for the nearest minute, we round up at 30ë or higher.

1

Write the equation.

2

Write your answer, rounding to the nearest second.

Maths Quest 10 for the Australian Curriculum

Write a cos q = 0.2547

cos-1 0.2547 = 75è15Å

b tan q = 2.364

tan-1 2.364 = 67è4Å16ë

measurement AND geometry • Pythagoras and trigonometry

Review of SOH CAH TOA We are able to find a side length in a right-angled triangle if we are given one other side length and the size of one of the acute angles. These sides and angle are related using one of the three trigonometric ratios. ■■ The sine ratio The sine ratio is defined as the ratio of the length of the side opposite angle q (O) to the length of the hypotenuse (H). This is O written as sin q  = . H The sine of an angle is not dependent on the size of the right-angled triangle as all these triangles are similar in shape. q ot en

yp ot en us e

H yp

The cosine ratio The cosine ratio is defined as the ratio of the length of the adjacent side (A) to the length of the hypotenuse (H) and is written as A cos q  = . H The cosine of an angle also does not depend on the size of the right-angled triangle.

H

■■

Opposite

us e

■■

q Adjacent

■■

Opposite

The tangent ratio O The tangent ratio is defined as tan q  = , where O is the length of A the side opposite angle q and A is the length of the side adjacent to it. Again, the tangent ratio does not depend on the size of the right-angled triangle. q Adjacent

■■

■■

Having defined the three trigonometric ratios, we need to decide in each case which of the three to use. We do this by labelling the sides relative to the angle we have been given. We then select the ratio that contains both the side we are finding and the side we have been given. The three ratios can be remembered easily by using the mnemonic or abbreviation SOH CAH TOA: •  SOH stands for ‘Sine, Opposite, Hypotenuse’. •  CAH stands for ‘Cosine, Adjacent, Hypotenuse’. •  TOA stands for ‘Tangent, Opposite, Adjacent’.

Worked Example 10

For the triangle shown, write the expressions for the sine, cosine and tangent ratios of the given angle.

c a q b Chapter 5 Trigonometry I

147

measurement AND geometry • Pythagoras and trigonometry

Think

Write/draw

Label the diagram using the symbols O, A, H with respect to the given angle (angle q ).

1

c=H a=O q

b=A 2

From the diagram, identify the values of O (opposite side), A (adjacent side) and H (the hypotenuse).

O = a, A = b, H = c

3

Write the formula for each of the sine, cosine and tangent ratios.

sin q  =

4

Substitute the values of A, O and H into each formula.

O A O , cos q  = , tan q  = H H A

a b a sin q  = , cos q  = , tan q  = c c b

Worked Example 11

Write the trigonometric ratio which must be used in order to find the value of the pronumeral in each of the following triangles. Set up a suitable equation. a               b  18 15

x

6

50è b

Think a

1

Label the sides of the triangle whose lengths are given, using the appropriate symbols.

Write/draw a 15 = H

6=O

b

b

148

O H

2

We are given the lengths of the opposite side (O) and the hypotenuse (H). Write the ratio that contains both of these sides.

sin q  =

3

Identify the values of the pronumerals.

O = 6, H = 15

4

Substitute the values of the pronumerals into the ratio. (Since the given angle is denoted with the letter b, replace q with b.)

sin b =

1

Label the sides of the triangle whose lengths are either given, or need to be found, using the appropriate symbols.

Maths Quest 10 for the Australian Curriculum

b

6 2 = 15 5

18 = A 50è

x=O

measuremenT anD geomeTry • PyThagoras anD TrigonomeTry

O A

2

The length of the adjacent side (A) is given and the length of the opposite side (O) needs to be found. Write the ratio that contains these sides.

tan q  =

3

Identify the values of the pronumerals.

4

Substitute the values of the pronumerals into the ratio.

O = x, A = 18, q = 50è x tan 50è = 18

rememBer

1. When using the calculator to find values of sine, cosine and tangent, make sure the calculator is in Degree mode. 2. To find the size of an angle whose sine, cosine or tangent is given, perform an inverse operation; that is, sin-1, cos-1 or tan-1. 3. Use the calculator’s conversion function to convert between decimal degrees and degrees, minutes and seconds. 4. There are 60 minutes in 1 degree and 60 seconds in 1 minute. 5. The three trigonometric ratios, sine, cosine and tangent, are defined as: O A O sin q  = , cos q  = and tan q  = , H H A where H is the hypotenuse, O is the opposite side and A is the adjacent side. 6. The three ratios are abbreviated to the useful mnemonic SOH CAH TOA. 7. To determine which trigonometric ratio to use, follow these steps. (a) Label the sides of the right-angled triangle that are either given, or need to be found, using the symbols O, A, H with respect to the angle in question. (b) Consider the sides that are involved and write the trigonometric ratio containing both of these sides. (Use SOH CAH TOA to assist you.) (c) Identify the values of the pronumerals in the ratio. (d) Substitute the given values into the ratio.

exerCise

5C inDiViDual PaThWays eBook plus

Activity 5-C-1

Review of trigonometry doc-5017 Activity 5-C-2

Using trigonometry doc-5018 Activity 5-C-3

Applying trigonometry doc-5019

Trigonometric ratios FluenCy 1 Calculate each of the following, correct to 4 decimal places. a sin 30è b cos 45è d sin 57è e tan 83è

c tan 25è f cos 44è

2 We7 Calculate each of the following, correct to 4 decimal places. a sin 40è30Å b cos 53è57Å c d tan 123è40Å e sin 92è32Å f g cos 35è42Å35ë h tan 27è42Å50ë i j sin 23è58Å21ë k cos 8è54Å2ë l m tan 420è n cos 845è o

tan 27è34Å sin 42è8Å cos 143è25Å23ë sin 286è sin 367è35Å 3 We8 Find the size of angle q, correct to the nearest degree, for each of the following. a sin q  = 0.763 b cos q  = 0.912 c tan q  = 1.351 d cos q  = 0.321 e tan q  = 12.86 f cos q  = 0.756 Chapter 5 Trigonometry I

149

measuremenT anD geomeTry • PyThagoras anD TrigonomeTry 4 We9a Find the size of the angle q in each of the following, correct to the nearest minute. a sin q  = 0.814 b sin q  = 0.110 c tan q  = 0.015 d cos q  = 0.296 e tan q  = 0.993 f sin q  = 0.450 5 We9b Find the size of the angle q in each of the following, correct to the nearest second. a tan q  = 0.5 b cos q  = 0.438 c sin q  = 0.9047 d tan q  = 1.1141 e cos q  = 0.8 f tan q  = 43.76 6 Find the value for each of the following, correct to 3 decimal places. a 3.8 cos 42è b 118 sin 37è c 2.5 tan 83è

eBook plus

Digital doc

SkillSHEET 5.3 doc-5226

2 cos 23° 5 sin 18° 55.7 i cos 89è21Å 3.2 cos 34è52Å l 0.8 sin 12è48Å

220 cos 14° 18.7 h sin 35è25Å42ÅÅ 2.5 sin 27è8Å k 10.4 cos 83è2Å

2 sin 45° 12.8 g tan 60è32Å 3.8 tan 1è51Å44ÅÅ j 4.5 sin 25è45Å d

e

f

7 We10 For each of the following triangles, write the expressions for ratios of each of the given

angles: i sine iii tangent a d q

ii cosine b

c k

j

a

e

f

b

h

i g

l d

e

o g

f a

n

b

b

m

u

v

c

g t eBook plus

Digital doc

SkillSHEET 5.7 doc-5231

8 We11 Write the trigonometric ratio which must be used in order to find the value of the

pronumeral in each of the following triangles. a

b

12

c

25

5

q

15

4

30 q

d

2.7

e

p q

t

17

14.3

f

35è

17.5 a

150

maths Quest 10 for the australian Curriculum

q

measuremenT anD geomeTry • PyThagoras anD TrigonomeTry

g

h

7

i

q 20

9.8

31

a 3.1

x 15è

reasoning 9 Consider the right-angled triangle shown at right. a Label each of the sides using the letters O, A, H with b c

d e

f g h

eBook plus

Interactivity Using trigonometry

int-1146

using trigonometry to calculate side lengths ■

Consider this right-angled triangle. Labelling the sides with respect to the 42è angle, we can see that the unknown side is opposite and we are given the hypotenuse. x From the diagram at right, sin 42è = . Using a 24 calculator, we know that the sine ratio of a 42è angle is approximately 0.6691. We can now solve this equation to find the value of x. We are therefore able to calculate a side length if we are given the size of an angle and one other side.

x

24 m 42è

Opposite

5D

a

respect to the 41è angle. Measure the side lengths (to the nearest millimetre). Determine the value of each trigonometric ratio. (Where applicable, answers should be given correct to 2 decimal places.) i sin 41è 41è ii cos 41è iii tan 41è What is the value of the unknown angle, a ? Determine the value of each of these trigonometric ratios, correct to 2 decimal places. i sin a ii cos a iii tan a (Hint: First relabel the sides of the triangle with respect to angle a.) reFleCTion    What do you notice about the relationship between sin 41è and cos a ? How do we determine which of sin, What do you notice about the relationship cos or tan to use in a trigonometry question? between sin a and cos 41è? Make a general statement about the two angles.

x

Hy po 24 tenus m e 42è Adjacent

Chapter 5 Trigonometry I

151

measurement AND geometry • Pythagoras and trigonometry

■■

■■

The solution to the above problem is: O the sine ratio formula sin q  = H x the result of substituting into the formula sin 42è = 24 x = 24 ì sin 42è rearranging the formula to make x the subject x ö 16.06  m the result of the calculation. We need to apply this method using any of the three trigonometric ratios to find a side length. The steps used in solving the problem are as follows. Step 1. Label the sides of the triangle, which are either given, or need to be found, with respect to the given angle. Step 2. Consider the sides involved and determine which of the trigonometric ratios is required. (Use the mnemonic SOH  CAH  TOA to help you.) (a) Use the sine ratio if the hypotenuse (H) and the opposite side (O) are involved. (b) Use the cosine ratio if the hypotenuse (H) and the adjacent side (A) are involved. (c) Use the tangent ratio if the opposite (O) and the adjacent (A) sides are involved. Step 3. Substitute the values of the pronumerals into the ratio. Step 4. Solve the resultant equation for the unknown side length.

Worked Example 12

Find the value of the pronumeral for each of the following. Give answers correct to 3 decimal places. a         b  6 cm

a

32è 0.346 cm

f

35è

Think a

1

Label the sides of the triangle, which are either given, or need to be found.

Write/draw a 6 cm

H

a

35è

152

2

Identify the appropriate trigonometric ratio to use.

3

Substitute O = a, H = 6, q = 35è.

4

Make a the subject of the equation.

5

Calculate and round the answer, correct to 3 decimal places.

Maths Quest 10 for the Australian Curriculum

O

sin q  =

O H

sin 35è =

a 6

6 sin 35è = a a = 6 sin 35è a ö 3.441  cm

measurement AND geometry • Pythagoras and trigonometry b

1

Label the sides of the triangle, which are either given, or need to be found.

b H

32è

0.346 cm

2

Identify the appropriate trigonometric ratio to use.

3

Substitute A = f, H = 0.346 and q = 32è.

4

Make f the subject of the equation.

5

Calculate and round the answer, correct to 3 decimal places.

A f

A H

cos q  =

f 0.346 0.346 cos 32è = f cos 32è =

f = 0.346 cos 32è ö 0.293  cm



Worked Example 13

Find the value of the pronumeral in the triangle shown. Give the answer correct to 2 decimal places. 120 m 5è P Think 1

Write/draw

Label the sides of the triangle, which are either given, or need to be found.

5è

A

2

Identify the appropriate trigonometric ratio to use.

tan q  =

O A

3

Substitute O = 120, A = P and q = 5è.

tan 5è =

120 P

4

Make P the subject of the equation. (i)  Multiply both sides of the equation by P. (ii)  Divide both sides of the equation by tan 5è.

5

Calculate and round the answer, correct to 2 decimal places.

O 120 m

H P

P ì tan 5è = 120 120 P= tan 5° P ö 1371.61  m

remember

The trigonometric ratios can be used to find a side length in a right-angled triangle when we are given one other side length and one of the acute angles. Chapter 5 Trigonometry I

153

measuremenT anD geomeTry • PyThagoras anD TrigonomeTry

exerCise

5D inDiViDual PaThWays eBook plus

Activity 5-D-1

Calculating side lengths using trigonometry doc-5020

using trigonometry to calculate side lengths FluenCy 1 We12 Find the length of the unknown side in each of the following, correct to 3 decimal

places. a

c

b 8

x 10 cm

Activity 5-D-2

Applying trigonometry to simple figures doc-5021

25è

a

31è

a

14

60è

Activity 5-D-3

Practical applications of trigonometry doc-5022

2 We13 Find the length of the unknown side in each of the following triangles, correct to

2 decimal places. a

b

71è

4.6 m 13è

m

n 2.3 m

c 94 mm 68è t 3 Find the length of the unknown side in each of the following, correct to 2 decimal places. b

a

8è5

c

P

2'4

14 m

5''

11.7 m

43.95 m

2'

t

1 8è

1

x

40è26'

d

f

e

80.9 cm

x 75.23 km ' 42 11.2 mm

154

maths Quest 10 for the australian Curriculum

21è25'34"

x

è 34

6è25' x

measurement AND geometry • Pythagoras and trigonometry 4 Find the value of the pronumeral in each of the following, correct to 2 decimal places. a

b

x

23.7 m 36è42' y

43.9 cm

46è

c

d

34

z

2 è1

p

15.3 m

'

12.3 m

13è12' e

f

0.732 km

q p

a

73è5'

63è11' 47.385 km

b

understanding 5 Given that the angle q is 42è and the length of the hypotenuse is 8.95  m in a right-angled

triangle, find the length of: a the opposite side b the adjacent side. Give each answer correct to 1 decimal point. 6 A ladder rests against a wall. If the angle between the ladder and the ground is 35è and the foot of the ladder is 1.5  m from the wall, how high up the wall does the ladder reach? Reasoning 7 Tran is going to construct an enclosed rectangular desktop that is at an incline of 15è. The

diagonal length of the desktop is 50 cm. At one end, the desktop will be raised 8 cm. The desktop will be made of wood. The diagram below represents this information. Side view of the desktop x 15è

Top view of the desktop 8 cm

y

a Determine the values (in centimetres) of x, y

and z of the desktop. Write your answers correct to 2 decimal places. b Using your answer from part a determine the minimum area of wood, in cm2, Tran needs to construct his desktop. Write your answer correct to 2 decimal places.

z 50 cm

reflection 

 

How does solving a trigonometric equation differ when we are finding the length of the hypotenuse side compared to when finding the length of a shorter side?

Chapter 5 Trigonometry I

155

measurement AND geometry • Pythagoras and trigonometry

Using trigonometry to calculate angle size

5E

■■

To find the size of an angle using the trigonometric ratios, we need to be given the length of any two sides.

Worked Example 14

For each of the following, find the size of the angle, q, correct to the nearest degree. a  b  5m 5 cm

3.5 cm

q 11 m

q Think a

1

Label the sides of the triangle, which are either given, or need to be found.

Write/draw a H

O

5 cm

3.5 cm

q

b

2

Identify the appropriate trigonometric ratio to use. We are given O and H, so choose the sine ratio.

sin q  =

3

Substitute O = 3.5 and H = 5 and evaluate the expression.

sin q  =

4

Make q the subject of the equation using inverse sine.

5

Evaluate q and round the answer, correct to the nearest degree.

1

Label the sides of the triangle, which are either given, or need to be found.

O H

3.5 5 = 0.7

q = sin-1 0.7 = 44.427  004è

q ö 44è

b

O 5m q

2

156

Identify the appropriate trigonometric ratio to use. We are given O and A, so choose the tangent ratio.

Maths Quest 10 for the Australian Curriculum

tan q  =

11 m

O A

A

measurement AND geometry • Pythagoras and trigonometry

3

Substitute O = 5 and A = 11. As the value of tan (q ) is a simple fraction, we do not need to evaluate the expression.

4

Make q the subject of the equation using inverse tangent.

5

5

tan q  = 11 5

q = tan-1 11 = 24.443  954  78è

Evaluate q and round the answer, correct to the nearest degree.

■■

q ö 24è



When asked for a more accurate measurement of an angle, we are able to use the calculator to find an angle correct to the nearest minute or nearest second.

Worked Example 15

Find the size of angle q  in each of the triangles shown below. a  b  3.1 m q 55 cm

7.2 m

q 42 cm

(Answer correct to the nearest minute.)

(Answer correct to the nearest second.)

Think a

1

Label the sides of the triangle, which are either given, or need to be found.

Write/draw a

3.1 m A q O 7.2 m

O A

2

Identify the appropriate trigonometric ratio to use.

tan q  =

3

Substitute O = 7.2 and A = 3.1 and evaluate the expression.

tan q  =

4

Make q the subject of the equation using inverse tangent.

q = tan-1 2.322  580  645

5

Evaluate q and write the calculator display.

q = 66.705  436  75è

6

Use the calculator to convert the answer to degrees, minutes and seconds and round the answer to the nearest minute.

7.2 3.1 = 2.322  580  645



= 66è42Å19.572ë q ö 66è42Å Chapter 5 Trigonometry I

157

measuremenT anD geomeTry • PyThagoras anD TrigonomeTry b

1

b

Label the sides of the triangle, which are either given, or need to be found.

H 55 cm

q 42 cm A 2

Identify the appropriate trigonometric ratio to use.

cos q  =

A H

3

Substitute A = 42 and H = 55.

cos q  =

42 55

4

Make q the subject of the equation using inverse cosine.

q = cos-1

5

Evaluate q and write the calculator display.

q = 40.214 171 02è

6

Use the calculator to convert the answer to degrees, minutes and seconds and round the answer to the nearest second.

= 40è12Å51.016ë q ö 40è12Å51ë

42 55

rememBer

1. The trigonometric ratios can be used to find the size of the acute angles in a right-angled triangle when we are given the length of two sides. 2. To find an angle size we need to use the inverse trigonometric functions. 3. Answers may be given correct to the nearest degree, minute or second, or as decimal degrees.

exerCise

5e inDiViDual PaThWays eBook plus

Activity 5-E-1

using trigonometry to calculate angle size FluenCy 1 We14 Find the size of the angle, q, in each of the following. Give your answer correct to the

nearest degree.

5.2

Activity 5-E-2

Calculation angles using trigonometry doc-5024

158

c

b

a

Review of angle calculations doc-5023

4.8

q

maths Quest 10 for the australian Curriculum

4.7 q 3.2

8

q 3

measuremenT anD geomeTry • PyThagoras anD TrigonomeTry

inDiViDual PaThWays eBook plus

2 We15a Find the size of the angle marked with the pronumeral in each of the following. Give

your answer correct to the nearest minute. b

a

c

7.2 m

b

12

Activity 5-E-3

17

Applying trigonometry to angle calculations doc-5025

4m

q

q

10

12 3 We15b Find the size of the angle marked with the pronumeral in each of the following. Give

your answer correct to the nearest second. a

b

a

5m

8 3m q eBook plus

Digital doc

SkillSHEET 5.8 doc-5232

2 c 2.7 a 3.5 4 Find the size of the angle marked with the pronumeral in each of the following, giving your

answer correct to the nearest degree. a

b a

13.5

89.4

15.3 c

c

77.3

d

106.4

d 43.7

18.7

92.7 b

e

f

12.36 13.85

e

7.3 cm

12.2 cm

18.56 9.8 cm

a

Chapter 5 Trigonometry I

159

measuremenT anD geomeTry • PyThagoras anD TrigonomeTry

eBook plus

Digital doc

WorkSHEET 5.2 doc-5233

5 Find the size of each of the angles in the following, giving your answers correct to the nearest

minute. b

a

c

d

a

x

0.798

5.7

2.3

56.3

y 0.342

b

e

27.2

unDersTanDing 6 a Calculate the length of the sides r, l and h. Write

your answers correct to 2 decimal places. b Calculate the area of ABC, correct to the nearest square centimetre. c Calculate ±BCA.

A h D

l 20 cm

r 125è 30 cm

B

C

reasoning 7 In the sport of air racing, small planes have to travel between two large towers (or pylons). The

gap between a pair of pylons is smaller than the wing-span of the plane, so the plane has to go through on an angle with one wing ‘above’ the other. The wing-span of a competition airplane is 8 metres.

a Determine the angle, correct to 1 decimal place, that the plane has to tilt if the gap

between pylons is: i 7 metres ii 6 metres b Because the plane has rolled away from the

horizontal as it travels between the pylons it loses speed. If the plane’s speed is below 96 km/h it will stall and possibly crash. For each degree of ‘tilt’ the speed of the plane is reduced by 0.98 km/h. What is the minimum speed the plane must go through each of the pylons in part a? Write your answer correct to 2 decimal places. 160

maths Quest 10 for the australian Curriculum



iii 5 metres.

reFleCTion 

 

How is finding the angle of a right-angled triangle different to finding a side length?

measuremenT anD geomeTry • PyThagoras anD TrigonomeTry

5F eBook plus

eLesson Height of a satellite

angles of elevation and depression ■

■

eles-0173

Many people use trigonometry at work. It is particularly important in careers such as the building trades, surveying, architecture, navigation and engineering. Trigonometric ratios have a variety of applications, some of which will be discussed in this section. Trigonometric ratios can be used to solve problems. When solving a problem, the following steps can be of assistance. 1. Sketch a diagram to represent the situation described in the problem. 2. Label the sides of the right-angled triangle with respect to the angle involved. 3. Identify what is given and what needs to be found. 4. Select an appropriate trigonometric ratio and use it to find the unknown measurement. 5. Interpret your result by writing a worded answer.

angles of elevation and depression ■

■

When we need to look up or down in order to see a certain object, our line of vision (that is, the straight line from the observer’s eye to the object) is inclined. The angle of inclination of the line of vision to the horizontal when looking up is referred to as the angle of elevation, and when looking down it is referred to as the angle of depression. The angle of elevation is measured up from the horizontal line to the line of vision.

q

Angle of elevation Horizontal

■

The angle of depression is measured down from the horizontal line to the line of vision. Horizontal q

■

Angle of depression

For any two objects, A and B, the angle of elevation of B, as seen from A, is equal to the angle of depression of A as seen from B. Angle of depression q of A from B

q

B

Angle of elevation of B from A

A Chapter 5 Trigonometry I

161

measurement AND geometry • Pythagoras and trigonometry

Worked Example 16

From an observer, the angle of elevation of the top of a tree is 50è. If the observer is 8 metres from the tree, find the height of the tree. Think 1

Write/draw

Sketch a diagram and label the sides of the triangle with respect to the given angle. Let the height of the tree be h. h O 50è 8m

A

2

Identify the appropriate trigonometric ratio. We are given A and need to find O, so choose the tangent ratio.

tan q  =

O A

3

Substitute O = h, A = 8 and q = 50è.

tan 50è =

h 8

4

Rearrange to make h the subject.



5

Calculate and round the answer to 2 decimal places.



6

Give a worded answer.

The height of the tree is 9.53  m.

h = 8 tan 50è ö 9.53

remember

1. To solve a problem involving trigonometric ratios, follow these steps: (a)  Draw a diagram to represent the situation. (b) Label the diagram with respect to the angle involved (either given or that needs to be found). (c)  Identify what is given and what needs to be found. (d) Select an appropriate trigonometric ratio and use it to find the unknown side or angle. (e)  Interpret the result by writing a worded answer. 2. The angle of elevation is measured up and the angle of depression is measured down from the horizontal line to the line of vision. Horizontal q

q

Angle of elevation Horizontal

3. For any two objects, A and B, the angle of elevation of B, as seen from A, is equal to the angle of depression of A as seen from B.

Angle of depression q of A from B

A

162

Angle of depression

Maths Quest 10 for the Australian Curriculum

q

Angle of elevation of B from A

B

measuremenT anD geomeTry • PyThagoras anD TrigonomeTry

exerCise

5F inDiViDual PaThWays eBook plus

Activity 5-F-1

Identifying elevation and depression doc-5026 Activity 5-F-2

Calculating elevation and depression doc-5027 Activity 5-F-3

Applications of elevation and depression doc-5028

eBook plus

Digital doc

SkillSHEET 5.6 doc-5228

angles of elevation and depression FluenCy 1 We16 The angle of elevation from an observer to the top of a tree is 54è22Å. If the tree is

known to be 12.19 m high, how far is the observer from it? 2 From the top of a cliff 112 m high, the angle of depression to a boat is 9è15Å. How far is the

boat from the foot of the cliff? 3 A person on a ship observes a lighthouse on the cliff, which is 830 metres away from the ship. The angle of elevation of the top of the lighthouse is 12è. a How far above sea level is the top of the lighthouse? b If the height of the lighthouse is 24 m, how high is the cliff? 4 At a certain time of the day a post, 4 m tall, casts a shadow of 1.8 m. What is the angle of elevation of the sun at that time? 5 An observer, who is standing 47 m from a building, measures the angle of elevation of the top of the building as 17è. If the observer’s eye is 167 cm from the ground, what is the height of the building? unDersTanDing 6 A surveyor needs to determine the height of a building. She measures the angle of elevation of

the top of the building from two points, 38 m apart. The surveyor’s eye level is 180 cm above the ground.

h 47è12

'

x

35è5

0' 38 m

180 cm

a Find two expressions for the height of the building, h, in terms of x using the two

angles. b Solve for x by equating the two expressions obtained in a. c Find the height of the building. 7 The height of another building needs to be determined but cannot be found directly. The

surveyor decides to measure the angle of elevation of the top of the building from different sites, which are 75 m apart. The surveyor’s eye level is 189 cm above the ground.

h 43è35

'

x

32è1

8'

75 m

189 cm

a Find two expressions for the height of the building, h, in terms of x using the two

angles. b Solve for x. c Find the height of the building. Chapter 5 Trigonometry I

163

measurement AND geometry • Pythagoras and trigonometry 8 A lookout tower has been erected on top of a cliff. At a distance of 5.8  km from the foot of the

cliff, the angle of elevation to the base of the tower is 15.7è and to the observation deck at the top of the tower is 16è respectively as shown in the figure below. How high from the top of the cliff is the observation deck?

  

16è 15.7è 5.8 km

9 Elena and Sonja were on a camping trip to the

Angle of depression

1.3 km Grampians, where they spent their first day 20è hiking. They first walked 1.5  km along a path 1.5 km 150 m inclined at an angle of 10è to the horizontal. 10è Then they had to follow another path, which 1.4 km was at an angle of 20è to the horizontal. They walked along this path for 1.3  km, which brought them to the edge of the cliff. Here Elena spotted a large gum tree 1.4  km away. If the gum tree is 150  m high, what is the angle of depression from the top of the cliff to the top of the gum tree?

10 a Find the height of a telegraph

pole in the photograph at right if the angle of elevation to the top of the pole is 8è from a point at the ground level 60  m from the base of the pole. b Find the height of the light pole in the figure below.

43.3è

8è

60 m 60 m

11 From a point on top of a cliff, two boats are observed.

If the angles of depression are 58è and 32è and the cliff is 46  m above sea level, how far apart are the boats?

32è 58è 46 m

164

Maths Quest 10 for the Australian Curriculum

measuremenT anD geomeTry • PyThagoras anD TrigonomeTry 12 Joseph is asked to obtain an estimate of the height

eBook plus

Digital doc

WorkSHEET 5.3 doc-5234

of his house using any mathematical technique. He decides to use an inclinometer and basic trigonometry. Using the inclinometer, Joseph determines the angle of elevation, q, from his eye level to the top of his house to be 42è. The point h from which Joseph measures the angle of elevation is 15 m away from his house and the distance from Joseph’s eyes to the ground is 1.76 m. a Fill in the given information on the diagram provided (substitute values for the pronumerals). b Determine the height of Joseph’s house. 13 The competitors of a cross-country run are nearing the finish line. From a lookout 100 m above the track, the angles of depression to the two leaders, Nathan and Rachel, are 40è and 62è respectively. How far apart are the two competitors? 14 A 2.05 m tall man, standing in front of a street light 3.08 m high, casts a 1.5 m shadow. a What is the angle of elevation from the ground to the source of light? b How far is the man from the bottom of the light pole?

x q d

40è 62è 100 m

3.08 m

2.05 m 1.5 m

reasoning 15 The angle of elevation of a hot air balloon changes

reFleCTion    from 27è at 7.00 am to 61è at 7.03 am, according to an observer who is 300 m away from the take-off point. What is the difference between an angle of elevation and an a Assuming a constant speed, calculate that speed angle of depression? (in m/s and km/h) at which the balloon is rising, correct to 2 decimal places. b The balloon then falls 120 metres. What is the angle of elevation now? Write your answer correct to 1 decimal place.

5g

Bearings and compass directions ■

A bearing can be expressed as either a true bearing or a compass direction.

Compass directions ■

■

Compass (conventional) bearings are directions measured from the north–south line in either a clockwise or anticlockwise direction. To identify the compass direction of an object we need to state: 1. whether the angle is measured from north (N) or south (S) 2. the size of the angle and 3. whether the angle is measured in the direction of west (W) or east (E). Chapter 5 Trigonometry I

165

measurement AND geometry • Pythagoras and trigonometry

■■

For example, the compass direction of S20èE means the direction is 20è from south towards east, while the compass direction N40èW means the direction that is 40è from north towards west. N

N

N40èW

40è

W

E

W

E

20è

S

S20èE

S

      

True bearings ■■ ■■

True bearings are measured from north in a clockwise direction. They are always expressed in 3 digits. The diagrams below show the bearings of 025è true and 250è true respectively. (These true bearings are more commonly written as 025èT and 250èT.) N

N

025è true

25è

W

E

W

E

250è

250èT S

S

      

Worked Example 17

A boat travels a distance of 5  km from P to Q in a direction of 035èT. a  How far east of P is Q? b  How far north of P is Q? c  What is the true bearing of P from Q?

a

1

Draw a diagram to represent the situation. Label the hypotenuse and the opposite and adjacent sides.

Write/draw a

N

O x

5k m

Think

Ay

35è P 2

166

To determine how far Q is east of P, we need to find the value of x. We are given the length of the hypotenuse (H) and need to find the length of the opposite side (O). Choose the sine ratio.

Maths Quest 10 for the Australian Curriculum

sin q  =

O H

Q H

measurement AND geometry • Pythagoras and trigonometry

b

c

3

Substitute O = x, H = 5 and q = 35è.

4

Make x the subject of the equation.

5

Evaluate and round the answer, correct to 2 decimal places.

6

Write the answer in words.

1

To determine how far Q is north of P, we need to find the value of y. This can be done in several ways, namely: using the cosine ratio, the tangent ratio, or Pythagoras’ theorem. Let’s use the cosine ratio.

2

Substitute P = y, H = 5 and q = 35è.

3

Make y the subject of the equation.

4

Evaluate and round the answer, correct to 2 decimal places.

5

Write the answer in words.

1

To find the bearing of P from Q, we need to draw the compass directions through Q and then measure the angle in the clockwise direction from the north line through Q to the line PQ. Show the required angle on the diagram.

sin 35è =

x 5

x = 5 sin 35è = 2.87 Point Q is 2.87  km east of P. b

cos q  =

A H

cos 35è =

y 5

y = 5 cos 35è = 4.10 Point B is 4.10  km north of A. c

N

N Q

a 35è P

2

Study the diagram. The angle that represents the true bearing is the sum of 180è (from north to south) and the angle, labelled a. Now the north lines through P and Q are parallel and so the line PQ is a transversal. Therefore angle 35è and angle a are equal (being alternate angles). Calculate the true bearing.

True bearing = 180è + a a = 35è True bearing = 180è + 35è = 215è

3

Write the answer in words.

The bearing of P from Q is 215èT.

■■

Sometimes a person or an object (for example, a ship) changes direction during their journey. (This can even happen more than once.) In situations like this we are usually interested in the total distance the object has moved and its final bearing from the starting point.   The following worked example shows how to deal with such situations.

Worked Example 18

A boy walks 2  km on a true bearing of 090è and then 3  km on a true bearing of 130è. a How far east of the starting point is the boy at the completion of his walk? (Answer correct to 1 decimal place.) b How far south of the starting point is the boy at the completion of his walk? (Answer correct to 1 decimal place.) c What is the bearing of the boy (from the starting point), in degrees and minutes, at the completion of his walk? Chapter 5 Trigonometry I

167

measurement AND geometry • Pythagoras and trigonometry

Think

Write/draw

Draw a diagram of the boy’s journey.

N

N 130è

2 km

3 km

a

1

The first leg of the journey is due east so we find the eastern component of the second leg. Construct a triangle about the second leg of the journey. We can calculate one of the missing angles by using the rule of supplementary angles: 180è - 130è = 50è.

a

N

N 130è

2 km Ay

50è H

E

3 km

x O

b

2

We need to find the eastern component of the journey, x, which is the opposite side and have been given the hypotenuse. Choose the sine ratio.

3

Substitute O = x, H = 3 and q = 50è.

4

Make x the subject of the equation.

5

Evaluate and round correct to 1 decimal place.

6

Add to this the 2  km east that was walked in the first leg of the journey and give a worded answer.

1

In the first part of the journey the boy has not moved south at all. Thus the distance that he moved south of the starting point is the southern component of the second leg, labelled y. (See the diagram in part a.)

2

To find y we can use Pythagoras’ theorem, as we know the lengths of two out of three sides in the right-angled triangle. Note that the hypotenuse, c, is 3 and one of the sides is 2.3, as found in part a. Round the answer correct to 1 decimal place. Note: Alternatively, the cosine ratio could have been used.

3

168

Write the answer in words.

Maths Quest 10 for the Australian Curriculum

sin q  =

O H

x 3 x = 3 sin 50è

sin 50è =

= 2.3  km Total distance east = 2 + 2.3 = 4.3  km The boy walked a total of 4.3  km east of the starting point. b Distance south = y  km

a2 = c2 - b2 y2 = 32 - 2.32 = 9 - 5.29 = 3.71 y = 3.71 = 1.9  km

y 3 y = 3 cos 50è = 1.9  km

cos 50è =

The boy walked a total of 1.9 km south of the starting point.

measurement AND geometry • Pythagoras and trigonometry c

1

Draw a diagram of the journey and write in the distances found in parts a and b. The bearing of the boy from the starting point is represented by the angle a (that is, the angle measured in a clockwise direction from north to the line joining the starting and the finishing points of the journey).

c

N

a

N 130è

2 km

q 3 km

A 1.9 km 4.3 km O

2

The size of angle a cannot be found directly. Find the size of the supplementary angle labelled q.

3

We have the lengths of the opposite side and the adjacent side, so choose the tangent ratio.

tan q  =

4

Substitute O = 4.3 and A = 1.9 and evaluate.

tan q  =

5

Make q the subject of the equation using the inverse tangent function.

q = tan-1 2.263  157  895

6

Evaluate and round to the nearest minute.

7

Find the angle a.

a = 180è - 66è10Å = 113è50Å

8

Write the answer in words.

The bearing of the boy from his starting point is 113è50Å T.

O A

4.3 1.9 = 2.263  157  895

= 66.161  259  82è = 66è9Å40.535ë = 66è10Å

remember

1. To identify the compass direction of an object we need to state (in this order): (a) whether the angle is measured from north (N) or south (S) (b) the size of the angle and (c) whether the angle is measured in the direction of west (W) or east (E). 2. True bearings are measured from north in a clockwise direction and expressed as 3 digits and with a T. 3. When solving problems involving bearings or compass directions, always draw a clear diagram prior to attempting the problem. Exercise

5G

Bearings and compass directions fluency 1

Change each of the following compass directions to true bearings. a N20èE d S28èE

b N20èW e N34èE

c S35èW f S42èW Chapter 5 Trigonometry I

169

measuremenT anD geomeTry • PyThagoras anD TrigonomeTry

inDiViDual PaThWays eBook plus

Activity 5-G-1

Bearings doc-5029

2 Change each of the following true bearings to compass directions. a 049èT d 330èT

b 132èT e 086èT

c 267èT f 234èT

3 Describe the following paths using true bearings. a

b

N

N

Activity 5-G-2

3k

Calculations involving bearings doc-5030

m

35è W

Activity 5-G-3

22è 2.5 km

Applications involving bearings doc-5031

E

S c

d

N

N 35è 2.5 km

W

4 km

E

35è m 8k S

f

N 12 km 65è

N

N 50è

m 7k

50 0m

e

40è

30 0m

N 50è

4 Show each of the following by drawing the paths. a A ship travels 040èT for 40 km and then 100èT for 30 km. b A plane flies for 230 km in a direction 135èT and a further 140 km in a direction 240èT. c A bushwalker travels in a direction 260èT for 0.8 km, then changes direction to 120èT for

1.3 km, and finally travels in a direction of 32è for 2.1 km. d A boat travels N40èW for 8 km, then changes direction to S30èW for 5 km and then

S50èE for 7 km. e A plane travels N20èE for 320 km, N70èE for 180 km and S30èE for 220 km. 5 We17 a You are planning a trip on your yacht. If you travel 20 km from A to B on a bearing

of 042èT: i how far east of A is B? ii how far north of A is B? iii what is the bearing of A from B? b In the next part of the journey you decide to travel 80 km from B to C on a bearing of 130èT. i Show the journey to be travelled using a diagram. ii How far south of B is C? iii How far east of B is C? iv What is the bearing of B from C? 170

maths Quest 10 for the australian Curriculum

measurement AND geometry • Pythagoras and trigonometry

In the next part of the journey you decide to travel 45  km from C to D on a bearing of 210èT. Show the journey to be travelled using a diagram. How far south of C is D? How far west of C is D? What is the bearing of C from D? 6 If a farmhouse is situated 220  m N35èE from a shed, what is the true bearing of the shed from the house? c

  i  ii iii iv

Understanding 7 A pair of hikers travel 0.7  km on a true bearing of 240è and then 1.3  km on a true bearing of

300è. How far west have they travelled from their starting point? 8   WE 18  A boat travels 6  km on a true bearing of 120è and then 4  km on a true bearing of 080è. a How far east is the boat from the starting point on the completion of its journey? b How far south is the boat from the starting point on the completion of its journey? c What is the bearing of the boat from the starting point on the completion of its journey? 9 A plane flies on a true bearing of 320è for 450  km. It then flies on a true bearing of 350è for 130  km and finally on a true bearing of 050è for 330  km. How far north of its starting point is the plane? 10 Find the final bearing for each of the following. Express your answer in true bearings, correct to the nearest minute. a A boat travels due east for 4  km and then travels N20èE for 3  km. What is the final bearing of the boat from the starting point? b A bushwalker travels due north for 3  km, then due east for 8  km. What is the final bearing of the bushwalker from the starting point? c A car travels due south for 80  km, then travels due west for 50  km, and finally due south for a further 30  km. What is the final bearing of the car from the starting point? Reasoning 11 A yacht is sailing around islands in the Pacific

y

Ocean. The sailor sees a mountain range on an island that is on a bearing of 330èT. The 29è yacht sails at a rate of 5 km/h for 30 minutes N N x due West, such that the mountain range is 30è now on a bearing of 050èT. From this new 50è a position, the sailor determines that the angle q of elevation to the highest point on the d km 330è mountain range is 29è, as shown in the Note: Diagram is not drawn to scale. diagram at right. a Determine the exact values of the angles labelled q and a. b Determine the distance d, in kilometres, the yacht sailed on a bearing of due West before another bearing of the mountain range was taken. c Using any suitable method, determine the value of x, in kilometres. Write your answer correct to 3 decimal places. reflection    d Using the value of x from part b determine the What is the difference between value of y, the height above sea level of highest true bearings and compass point on the mountain range, in kilometres. Write directions? your answer correct to 2 decimal places. Chapter 5 Trigonometry I

171

measuremenT anD geomeTry • PyThagoras anD TrigonomeTry

applications

5h eBook plus

Interactivity Applying trigonometry to drafting problems

■

■

int-2781

■

Many people use trigonometry at work. It is particularly important in careers such as the building trades, surveying, architecture and engineering. Trigonometric ratios can be used to find angles of elevation and depression, as well as to calculate distances which we could not otherwise easily measure. When solving a problem, remember the following steps. 1. Sketch a diagram to represent the situation described in the problem. 2. Label the sides of the right-angled triangle with respect to the angle involved. 3. Identify what is given and what needs to be found. 4. Select an appropriate trigonometric ratio and use it to find the unknown measurement. 5. Interpret your result by writing a worded answer.

WorkeD examPle 19

A ladder of length 3 m makes an angle of 32è with the wall. a How far is the foot of the ladder from the wall? b How far up the wall does the ladder reach? c What angle does the ladder make with the ground? Think

WriTe/DraW

Sketch a diagram and label the sides of the right-angled triangle with respect to the given angle.

(wall)

32è y

3m H a

a

172

1

We need to find the distance of the foot of the ladder from the wall (O) and are given the length of the ladder (H). Choose the sine ratio.

2

Substitute O = x, H = 3 and q = 32è.

3

Make x the subject of the equation.

4

Evaluate and round the answer to 2 decimal places.

5

Write the answer in words.

maths Quest 10 for the australian Curriculum

A

x O

a sin q =

O H

sin 32è =

x 3

x = 3 sin 32è ö 1.59 m The foot of the ladder is 1.59 m from the wall.

measuremenT anD geomeTry • PyThagoras anD TrigonomeTry

b

c

b cos q =

A H

1

We need to find the height the ladder reaches up the wall (A) and are given the hypotenuse (H). Choose the cosine ratio.

2

Substitute A = y, H = 3 and q = 32è.

3

Make y the subject of the equation.

y = 3 cos 32è

4

Evaluate and round the answer to 2 decimal places.

y ö 2.54 m

5

Write the answer in words.

1

To find the angle that the ladder makes with the ground, we could use any of the trigonometric ratios, as the lengths of all three sides are known. However, it is quicker to use the angle sum of a triangle.

2

cos 32è =

y 3

The ladder reaches 2.54 m up the wall. c a  + 90è + 32è = 180è

a + 122è = 180è a  = 180è - 122è a  = 58è

Write the answer in words.

The ladder makes a 58è angle with the ground.

rememBer

To solve a problem involving trigonometric ratios, follow these steps: (a) Draw a diagram to represent the situation. (b) Label the diagram with respect to the angle involved (either given or that needs to be found). (c) Identify what is given and what needs to be found. (d) Select an appropriate trigonometric ratio and use it to find the unknown side or angle. (e) Interpret the result by writing a worded answer. exerCise

5h inDiViDual PaThWays eBook plus

Activity 5-H-1

Trigonometry applications 1 doc-5032 Activity 5-H-2

Trigonometry applications 2 doc-5033

applications FluenCy 1 We19 A 3 m-long ladder is placed against a wall so that it reaches 1.8 m up the wall. a What angle does the ladder make with the ground? b What angle does the ladder make with the wall? c How far from the wall is the foot of the ladder? 2 Jamie decides to build a wooden pencil box. He wants his ruler to be able to lie across

the bottom of the box, so he allows 32 cm along the diagonal. The width of the box is to be 8 cm. 32 cm q

Activity 5-H-3

Trigonometry applications 3 doc-5034

8 cm

Calculate: a the size of angle q b the length of the box. Chapter 5 Trigonometry I

173

measurement AND geometry • Pythagoras and trigonometry 3 A carpenter wants to make a roof pitched at 29è30Å, as shown in the diagram. How long should

he cut the beam, PR?

R

P

29è30'

Q

10.6 m

4 The sloping sides of a gable roof are each 7.2  m long. They rise to a height of 2.4  m in the

centre. What angle do the sloping sides make with the horizontal?

5 The mast of a boat is 7.7  m high. A guy wire from the top of the mast is fixed to the deck 4  m

from the base of the mast. Determine the angle the wire makes with the horizontal. understanding 6 A desk top of length 1.2  m and width 0.5  m rises to 10  cm.

0.5 m

E

F 10 cm

C

D

A

B

1.2 m

Calculate: a ±DBF b ±CBE. 7 A cuboid has a square end. H

G X

D

C O

E

F

45 cm

A 25 cm B a If the length of the cuboid is 45  cm and its height and width are 25  cm each, calculate: i the length of BD   ii  the length of BG iii the length of BE iv  the length of BH v ±FBG vi  ±EBH. 174

Maths Quest 10 for the Australian Curriculum

measurement AND geometry • Pythagoras and trigonometry

If the midpoint of FG is X and the centre of the rectangle ABFE is O calculate: i the length OF    ii the length FX iii ±FOX iv the length OX. 8 In a right square-based pyramid, the length of the side of the base is 12  cm and the height is 26  cm.

26 cm

b

12 cm

Determine: a the angle the triangular face makes with the base b the angle the sloping edge makes with the base c the length of the sloping edge. 9 In a right square-based pyramid, the length of the side of the square base is 5.7  cm.

68è 5.7 cm

If the angle between the triangular face and the base is 68è, determine: a the height of the pyramid b the angle the sloping edge makes with the base c the length of the sloping edge. 10 In a right square-based pyramid, the height is 47  cm. If the angle between a triangular face and

the base is 73è, calculate: a the length of the side of the square base b the length of the diagonal of the base c the angle the sloping edge makes with the base. 11 The height of a vertical cone is 24.5  cm. 48è37'10"

24.5 cm

If the angle at the apex is 48è37Å10ë, determine: a the length of the slant edge of the cone b the radius of the cone. Chapter 5 Trigonometry I

175

measurement AND geometry • Pythagoras and trigonometry Reasoning 12 Aldo the carpenter is lost in a rainforest. He comes across a large river and he knows that he

can not swim across it. Aldo intends to build a bridge across the river. He draws some plans to calculate the distance across the river as shown in the diagram below.

72è Tree

River

4.5 cm 88è

a Aldo used a scale of 1 cm to represent 20 m. Find the real-life distance represented by

4.5 cm in Aldo’s plans. b Use the diagram below to write an equation for h in terms of d and the two angles.

h J1

J2

d-x

x d

c Use your equation from b to find the distance across the river, correct to the nearest

metre. reflection 

 

What are some real-life applications of trigonometry?

176

Maths Quest 10 for the Australian Curriculum

measurement AND geometry • Pythagoras and trigonometry

Summary Pythagoras’ theorem ■■ ■■ ■■ ■■ ■■ ■■ ■■

The hypotenuse is the longest side of the triangle and is opposite the right angle. On your diagram, check whether you are finding the length of the hypotenuse or one of the shorter sides. The length of a side can be found if we are given the length of the other sides by using the formula c2 = a2 + b2. When using Pythagoras’ theorem, always check the units given for each measurement. If necessary, convert all measurements to the same units before using the rule. Worded problems can be solved by drawing a diagram and using Pythagoras’ theorem to solve the problem. Worded problems should be answered in a sentence. Pythagoras’ theorem in three dimensions

■■ ■■ ■■

Pythagoras’ theorem can be used to solve problems in three dimensions. Some common 3-D shapes include boxes, pyramids and right-angled wedges. To solve problems in three dimensions it is helpful to draw sections of the original shape in two dimensions. Trigonometric ratios

When using the calculator to find values of sine, cosine and tangent, make sure the calculator is in Degree mode. ■■ To find the size of an angle whose sine, cosine or tangent is given, perform an inverse operation; that is, sin-1, cos-1 or tan-1. ■■ Use the calculator’s conversion function to convert between decimal degrees and degrees, minutes and seconds. ■■ There are 60 minutes in 1 degree and 60 seconds in 1 minute. ■■ The three trigonometric ratios, sine, cosine and tangent, are defined as: O A O sin q  = , cos q  = and tan q  = , H H A where H is the hypotenuse, O is the opposite side and A is the adjacent side. ■■ The three ratios are abbreviated to the useful mnemonic SOH  CAH  TOA. ■■ To determine which trigonometric ratio to use, follow these steps. (a) Label the sides of the right-angled triangle that are either given, or need to be found, using the symbols O, A, H with respect to the angle in question. (b) Consider the sides that are involved and write the trigonometric ratio containing both of these sides. (Use SOH  CAH  TOA to assist you.) (c) Identify the values of the pronumerals in the ratio. (d) Substitute the given values into the ratio. ■■

Using trigonometry to calculate side lengths ■■

The trigonometric ratios can be used to find a side length in a right-angled triangle when we are given other side length and one of the acute angles. Using trigonometry to calculate angle size

■■ ■■ ■■

The trigonometric ratios can be used to find the size of the acute angles in a right-angled triangle when we are given the length of two sides. To find an angle size we need to use the inverse trigonometric functions. Answers may be given correct to the nearest degree, minute or second, or as decimal degrees. Chapter 5 Trigonometry I

177

measuremenT anD geomeTry • PyThagoras anD TrigonomeTry Angles of elevation and depression ■

The angle of elevation is measured up and the angle of depression is measured down from the horizontal line to the line of vision. Horizontal q

q

■

Angle of depression

Angle of elevation Horizontal

For any two objects, A and B, the angle of elevation of B, as seen from A, is equal to the angle of depression of A as seen from B. Angle of depression q of A from B

q

B

Angle of elevation of B from A

A Bearings and compass directions ■

■

■

To identify the compass direction of an object we need to state (in this order): (a) whether the angle is measured from north (N) or south (S) (b) the size of the angle (c) whether the angle is measured in the direction of west (W) or east (E). True bearings are measured from north in a clockwise direction and expressed as 3 digits and with a T. When solving problems involving bearings or compass directions, always draw a clear diagram prior to attempting the problem. Applications

■

To solve a problem involving trigonometric ratios, follow these steps: (a) Draw a diagram to represent the situation. (b) Label the diagram with respect to the angle involved (either given or that needs to be found). (c) Identify what is given and what needs to be found. (d) Select an appropriate trigonometric ratio and use it to find the unknown side or angle. (e) Interpret the result by writing a worded answer.

MAPPING YOUR UNDERSTANDING

Homework Book

178

Using terms from the summary, and other terms if you wish, construct a concept map that illustrates your understanding of the key concepts covered in this chapter. Compare your concept map with the one that you created in What do you know? on page 131. Have you completed the two Homework sheets, the Rich task and two Code puzzles in your Maths Quest 10 Homework Book?

maths Quest 10 for the australian Curriculum

measurement AND geometry • Pythagoras and trigonometry

Chapter review Fluency

7 Which of the following statements is correct? A sin 55è = cos 55è B sin 45è = cos 35è C cos 15è = sin 85è D sin 30è = sin 60è E sin 42è = cos 48è

1 The most accurate measure

for the length of the third side in the triangle at right is: A 4.83  m B 23.3  cm C 3.94  m D 2330  mm E 4826  mm

5.6 m 2840 mm

8 Which of the following can be used to find the

value of x in the diagram below?

2 What is the value of x in

28.7

this figure? a 5.4 5 b 7.5 c 10.1 d 10.3 e 4 3 What is the closest length of AG of the cube at right? D A 10 B 30 C 20 D 14 E 17

x

2

35è 7

A

B C

10

E

F 10

H

G

10

x

A 28.7 sin 35è B 28.7 cos 35è C 28.7 tan 35è D

28.7 sin 35°

E

28.7 cos 35°

9 Which of the following expressions can be used to

find the value of a in the triangle shown?

4 If sin 38è = 0.6157, which of the following will

also give this result? sin 218è sin 322è sin 578è sin 682è sin 142è 5 The angle 118è52Å34ë is also equal to: 52 A 118.5234è B 118 ° A B C D E

75

a

34

C 118.861è E 118.786è

A 35 sin 75è

D 118.876è

6 Which trigonometric ratio for the triangle shown at

right is incorrect? b A sin a = c a B sin a = c a C cos a = c b D tan a = a a E tan q  = b

35

C sin-1

75 35

35

B sin-1 75 35

D cos-1 75

75

E cos-1 35

10 If a school is 320  m S42èW from the police station, a a

b c

q

what is the true bearing of the police station from the school? A 042èT B 048èT C 222èT D 228èT E 312èT Chapter 5 Trigonometry I

179

measuremenT anD geomeTry • PyThagoras anD TrigonomeTry 11 Calculate x, correct to 2 decimal places. a

ProBlem solVing 1 A surveyor needs to determine the height of a

building. She measures the angle of elevation of the top of the building from two points, 64 m apart. The surveyor’s eye level is 195 cm above the ground.

x 123.1 cm 48.7 cm b

117 mm 82 mm

x

12 Calculate the value of the pronumeral, correct to

2 decimal places. 13.4 cm x

x

13 Calculate the height of this pyramid.

10 mm

8 mm

8 mm

14 A person standing 23 m away from a tree observes

15

16

17

18

180

the top of the tree at an angle of elevation of 35è. If the person is 1.5 m tall, what is the height of the tree? A man of height 1.8 m stands at the window of a tall building. He observes his young daughter in the playground below. If the angle of depression from the man to the girl is 47è and the floor on which the man stands is 27 m above the ground, how far from the bottom of the building is the child? A plane flies 780 km in a direction of 185èT. How far west has it travelled from the starting point? A hiker travels 3.2 km on a bearing of 250èT and then 1.8 km on a bearing of 320èT. How far west has she travelled from the starting point? If a 4 m ladder is placed against a wall and the foot of the ladder is 2.6 m from the wall, what angle does the ladder make with the wall? maths Quest 10 for the australian Curriculum

h 47è48

x

'

36è2

4' 64 m

195 cm

a Find the expressions for the height of the

building, h, in terms of x using the two angles. b Solve for x by equating the two expressions obtained in part a. c Find the height of the building. 2 The height of a right square-based pyramid is 13 cm. If the angle the face makes with the base is 67è, find: a the length of the edge of the square base b the length of the diagonal of the base c the angle the slanted edge makes with the base.

measuremenT anD geomeTry • PyThagoras anD TrigonomeTry 3 A boat sails on a compass direction of E12èS for

10 km then changes direction to S27èE for another 20 km. The boat then decides to return to its starting point. A 12è

10 km

B

27è

20 km

C a How far, correct to 2 decimal places, is the boat

4 A car is travelling northwards on an elevated

expressway 6 m above ground at a speed of 72 km/h. At noon another car passes under the expressway, at ground level, travelling west, at a speed of 90 km/h. a How far apart, in metres, are the two cars 40 seconds after noon? b At this time the first car stops, while the second car keeps going. At what time will they be 3.5 km apart? Write your answer correct to the nearest tenth of a second. 5 Two towers face each other separated by a distance, d, of 20 metres. As seen from the top of the first tower, the angle of depression of the second tower’s base is 59è and that of the top is 31è. What is the height, in metres correct to 2 decimal places, of each of the towers?

from its starting point? b On what bearing should the boat travel to

return to its starting point? Write the angle correct to the nearest degree.

eBook plus

Interactivities

Test yourself Chapter 5 int-2840 Word search Chapter 5 int-2838 Crossword Chapter 5 int-2839

Chapter 5 Trigonometry I

181

eBook plus

aCTiViTies

Chapter opener Digital doc

• Hungry brain activity Chapter 5 (doc-5223) (page 131) Are you ready? Digital docs

(page 132)

• SkillSHEET 5.1 (doc-5224): Rounding to a given number of decimal places • SkillSHEET 5.2 (doc-5225): Rounding the size of an angle to the nearest minute and second • SkillSHEET 5.3 (doc-5226): Labelling the sides of a right-angled triangle • SkillSHEET 5.5 (doc-5227): Rearranging formulas • SkillSHEET 5.6 (doc-5228): Drawing a diagram from given directions 5A Pythagoras’ Theorem Digital docs

(page 137)

• Activity 5-A-1 (doc-5011): Review of Pythagoras’ theorem • Activity 5-A-2 (doc-5012): Practising Pythagoras’ theorem • Activity 5-A-3 (doc-5013): More of Pythagoras’ theorem • SkillSHEET 5.1 (doc-5224): Rounding to a given number of decimal places 5B Pythagoras’ theorem in three dimensions Digital docs

• Activity 5-B-1 (doc-5014): Pythagoras in 3-dimensions (page 143) • Activity 5-B-2 (doc-5015): Pythagoras in 3-D figures (page 143) • Activity 5-B-3 (doc-5016): Investigating triangles in 3-D figures (page 143) • SkillSHEET 5.4 (doc-5229): Drawing 3-D shapes (page 143) • WorkSHEET 5.1 (doc-5230): Pythagoras’ theorem (page 145) 5C Trigonometric ratios Digital docs

• Activity 5-C-1 (doc-5017): Review of trigonometry (page 149) • Activity 5-C-2 (doc-5018): Using trigonometry (page 149) • Activity 5-C-3 (doc-5019): Applying trigonometry (page 149) • SkillSHEET 5.3 (doc-5226): Labelling the sides of a right-angled triangle (page 150) • SkillSHEET 5.7 (doc-5231): Selecting an appropriate trigonometric ratio based on the given information (page 150) 5D Using trigonometry to calculate side lengths Interactivity

• Using trigonometry (int-1146) (page 151) Digital docs

(page 154)

• Activity 5-D-1 (doc-5020): Calculating side lengths using trigonometry • Activity 5-D-2 (doc-5021): Applying trigonometry to simple figures 182

maths Quest 10 for the australian Curriculum

• Activity 5-D-3 (doc-5022): Practical applications of trigonometry 5E Using trigonometry to calculate angle size Digital docs

• Activity 5-E-1 (doc-5023): Review of angle calculations (page 158) • Activity 5-E-2 (doc-5024): Calculating angles using trigonometry (page 158) • Activity 5-E-3 (doc-5025): Applying trigonometry to angle calculations (page 159) • SkillSHEET 5.8 (doc-5232): Rounding angles to the nearest degree (page 159) • WorkSHEET 5.2 (doc-5233): Using trigonometry (page 160) 5F Angles of elevation and depression eLesson

• Height of a satellite (eles-0173) (page 161) Digital docs

• Activity 5-F-1 (doc-5026): Identifying elevation and depression (page 163) • Activity 5-F-2 (doc-5027): Calculating elevation and depression (page 163) • Activity 5-F-3 (doc-5028): Applications of elevation and depression (page 163) • SkillSHEET 5.6 (doc-5228): Drawing a diagram from given directions (page 163) • WorkSHEET 5.3 (doc-5234): Elevation and depression (page 165) 5G Bearings and compass directions Digital docs

(page 170)

• Activity 5-G-1 (doc-5029): Bearings • Activity 5-G-2 (doc-5030): Calculations involving bearings • Activity 5-G-3 (doc-5031): Applications involving bearings 5H Applications Interactivity

(page 172)

Digital docs

(page 173)

• Applying trigonometry to drafting problems (int-2781) • Activity 5-H-1 (doc-5032): Trigonometry calculations 1 • Activity 5-H-2 (doc-5033): Trigonometry calculations 2 • Activity 5-H-3 (doc-5034): Trigonometry calculations 3 Chapter review Interactivities

(page 181)

• Test Yourself Chapter 5 (int-2840): Take the end-ofchapter test to test your progress • Word search Chapter 5 (int-2838): an interactive word search involving words associated with this chapter • Crossword Chapter 5 (int-2839): an interactive crossword using the definitions associated with the chapter To access eBookPLUS activities, log on to www.jacplus.com.au

measurement anD geometry • using units of measurement

6 surface area and volume

6a Area 6b Total surface area 6c Volume What Do you knoW ? 1 List what you know about measurement. Create a sunshine wheel to show your list. 2 Share what you know with a partner and then with a small group. 3 As a class, create a large sunshine wheel that shows your class’s knowledge of measurement. eBook plus

Digital doc

Hungry brain activity Chapter 6 doc-5235

opening Question

How can you determine the most efficient size of a cylindrical water tank of a particular volume?

measurement anD geometry • using units of measurement

are you ready?

Try the questions below. If you have difficulty with any of them, extra help can be obtained by completing the matching SkillSHEET. Either search for the SkillSHEET in your eBookPLUS or ask your teacher for a copy. eBook plus

Digital doc

SkillSHEET 6.1 doc-5236 eBook plus

Digital doc

SkillSHEET 6.2 doc-5237

Conversion of area units 1 Convert each of the following to the units given in brackets. Write the answer using scientific

notation (standard form). a 3.6 m2 (mm2)

b 20 000 cm2 (km2)

Using a formula to find the area of a common shape 2 Find the area of each of the following plane figures. a

b

Digital doc

SkillSHEET 6.3 doc-5238

c

8 cm

3m

3 cm

8m

eBook plus

c 5.2 ha (m2)

2.5 cm

12 cm

Total surface area of cubes and rectangular prisms 3 Find the total surface area of each of the following prisms. a b 10 cm 5 cm

c

10 cm 3m

10 cm

2 cm 8 cm

eBook plus

Digital doc

SkillSHEET 6.4 doc-5239

eBook plus

Digital doc

Conversion of volume units 4 Convert each of the following to the units given in brackets. Give the answers in scientific

notation (standard form). a 3.4 m3 (cm3)

b 250 000 mm3 (m3)

Volume of cubes and rectangular prisms 5 Find the volume of each of the prisms in question 3.

SkillSHEET 6.5 doc-5240

184

maths Quest 10 for the australian Curriculum

c 6.5 cm3 (mm3)

measurement anD geometry • using units of measurement

6a

area ■ ■

eBook plus

■

Digital doc

SkillSHEET 6.1 doc-5236

■

The area of a figure is the amount of surface or flat space within the boundaries of the figure. The units used for area are mm2, cm2, m2, km2 or ha (hectares), depending upon the size of the figure. l ha = 10 000 (or 104) m2 There are many real-life situations that require an understanding of the area concept. Some are, ‘the area to be painted’, ‘the floor area of a room or house’, and ‘how much land one has’, ‘how many tiles are needed for a wall’. It is important that you are familiar with converting units of area. This can be revised by completing SkillSHEET 6.1.

using area formulas ■

The area of many plane figures can be found by using a formula. The table below shows the formula for the area of some common shapes. Shape

Formula A = l 2, where l is a side length.

1. Square l

2. Rectangle

A = lw, where l is the length and w is the width.

l w

1

A = 2 bh, where b is the base length and h the height.

3. Triangle h b a

4. Trapezium h

1

A = 2 (a + b)h, where a and b are lengths of parallel sides and h the height.

b

A = p r2, where r is the radius.

5. Circle r

A = bh, where b is the base length and h the height.

6. Parallelogram h b

(continued) Chapter 6 surface area and volume

185

measurement anD geometry • using units of measurement

Shape

Formula

θ ì p r2, where q is the sector angle 360° in degrees and r is the radius.

A=

7. Sector

q

r

A=

8. Rhombus

1 2

xy, where x and y are diagonals.

x

y

9. Ellipse

A = p ab, where a and b are the lengths of the semi-major and semi-minor axes respectively.

b a

■

Measurements must be in the same unit of length before substituting into an area formula.

alternative way to find the area of a triangle eBook plus

eLesson Heron’s formula

eles-0177

■

If the lengths of all three sides of a triangle are known, its area, A, can be found by using Heron’s formula: A = s( s − a )( s − b )( s − c ) where a, b and c are the lengths of a+b+c the three sides and s is the semi-perimeter or s = . 2

c

WorkeD example 1

Find the areas of the following plane figures, correct to 2 decimal places. a

b 3 cm

5 cm

c

2 cm

15 cm

5 cm 40è

6 cm think a

186

Write

1

The shape is a triangle with three side lengths given, but not the height. In this case Heron’s formula is used.

2

Identify the values of a, b and c.

3

Calculate the value of s, the semi-perimeter of the triangle.

maths Quest 10 for the australian Curriculum

a

a

b

A = s(s − a)(s − b)(s − c)

a = 3, b = 5, c = 6 a+b+c s= 2 3+ 5+6 = 2 14 = 2 =7

measurement AND geometry • Using units of measurement

b

c

A = 7(7 − 3)(7 − 5)(7 − 6)

Substitute the values of a, b, c and s into Heron’s formula and evaluate, correct to 2 decimal places.

4

= 7 × 4 × 2 ×1 = 56 = 7.48  cm2 b A = p ab

1

The shape shown is an ellipse. Write the appropriate area formula.

2

Identify the values of a and b (the semimajor and semi-minor axes).

a = 5, b = 2

3

Substitute the values of a and b into the formula and evaluate, correct to 2 decimal places.

A=pì5ì2 = 31.42  cm2

1

The shape shown is a sector. Write the formula for finding the area of a sector.

2

Write the value of q and r.

q = 40è, r = 15

3

Substitute and evaluate the expression, correct to 2 decimal places.

A=

c

A=

θ ì p r2 360°

40° ì p ì 152 360° = 78.54  cm2

Areas of composite figures ■■ ■■

A composite figure is a figure made up of a combination of simple figures. The area of a composite figure can be calculated by: •• calculating the sum of the areas of the simple figures that make up the composite figure •• calculating the area of a larger shape and then subtracting the extra area involved.

Worked Example 2

Find the area of each of the following composite shapes. C

a 

AB = 8 cm EC = 6 cm FD = 2 cm

A

E

F

Think

B 9 cm D

C

2 cm

B

D

a

b  A

E 5 cm H

10 cm

F G

Write a Area ACBD = Area ABC + Area ABD

1

ACBD is a quadrilateral that can be split into two triangles: ABC and ABD.

2

Write the formula for the area of a triangle containing base and height.

Atriangle = 2 bh

3

Identify the values of b and h for ABC.

ABC: b = AB = 8, h = EC = 6

1

Chapter 6 Surface area and volume

187

measurement AND geometry • Using units of measurement

1 2 1 2

Substitute the values of the pronumerals into the formula and, hence, calculate the area of ABC.

Area of ABC =

5

Identify the values of b and h for ABD.

ABD: b = AB = 8, h = FD = 2

6

Calculate the area of ABD.

Area of ABD =

4

=

ì AB ì EC ì8ì6

= 24  cm2

=

1 2 1 2

AB ì FD ì8ì2

= 8  cm2

b

7

Add the areas of the two triangles together to find the area of the quadrilateral ACBD.

Area of ACBD = 24  cm2 + 8  cm2 = 32  cm2

1

One way to find the area of the shape shown is to find the total area of the rectangle ABGH and then subtract the area of the smaller rectangle DEFC.

b Area = Area ABGH - Area DEFC

2

Write the formula for the area of a rectangle.

Arectangle = l ì w

3

Identify the values of the pronumerals for the rectangle ABGH.

Rectangle ABGH: l = 9 + 2 + 9 = 20 w = 10

4

Substitute the values of the pronumerals into the formula to find the area of the rectangle ABGH.

Area of ABGH = 20 ì 10 = 200  cm2

5

Identify the values of the pronumerals for the rectangle DEFC.

Rectangle DEFC: l = 5, w = 2

6

Substitute the values of the pronumerals into the formula to find the area of the rectangle DEFC.

Area of DEFC = 5 ì 2 = 10  cm2

7

Subtract the area of the rectangle DEFC from the area of the rectangle ABGH to find the area of the given shape.

Area = 200 - 10 = 190  cm2

remember

1. Area is a measure of the amount of surface within the boundaries of a figure. 2. The units for measuring area are mm2, cm2, m2 and km2. 3. Land area is usually measured in hectares (ha) where 1 ha = 10  000 (or 104) m2. 4. Areas can be calculated by using formulas that are specific to the given plane figure. 5. Areas of composite figures can be calculated by adding the areas of the simple figures making the composite figure or by calculating the area of an extended figure and subtracting the extra area covered.

188

Maths Quest 10 for the Australian Curriculum

measurement anD geometry • using units of measurement

exerCise

6a inDiViDual pathWays eBook plus

Activity 6-A-1

Review of area doc-5035

area Where appropriate, give answers correct to 2 decimal places. fluenCy 1 Find the areas of the following shapes. a

c

b 4 cm

Activity 6-A-2

4 cm

Area problems doc-5036

15 cm

12 cm

Activity 6-A-3

Tricky area problems doc-5037

10 cm

eBook plus

d

e

12 cm

f

Digital doc

SkillSHEET 6.2 doc-5237

8 cm

15 cm

13 mm

8 mm

18 cm 7 mm

g

h

i 15 cm

6m

10 cm

7m

18 cm

2 Express the area in questions 1e and 1g in terms of p. 3 We 1a Use Heron’s formula to find the area of the following triangles. a

b 3 cm

8 cm

5 cm

16 cm 6 cm 12 cm

Chapter 6 surface area and volume

189

measurement AND geometry • Using units of measurement 4   WE 1b  Find the area of the following ellipses. Answer correct to 1 decimal place.



a

b

9 mm

12 mm

4 mm 5 mm

5   WE 1c  Find the area of the following shapes, i  stating the answer exactly; that is in terms of p and ii correct to 2 decimal places. a

b

c

30è

70è

6 mm

18 cm

12 cm 345è

6   MC  A figure has an area of about 64  cm2. Which of the following cannot possibly represent

the figure? a A triangle with base length 16  cm and height 8  cm b A circle with radius 4.51  cm c A rectangle with dimensions 16  cm and 4  cm d A square with side length 8  cm e A rhombus with diagonals 16  cm and 4  cm 7   MC  The area of the quadrilateral shown below right is to be calculated. Which of the following lists all the lengths required to calculate the area? a AB, BC, CD and AD b AB, BE, AC and CD B c BC, BE, AD and CD d AC, BE and FD e AC, CD and AB

C F

E D

A 8   WE 2  Find the area of the following composite shapes. a

b

20 cm

15 cm

190

Maths Quest 10 for the Australian Curriculum

40 m 28 m

measurement AND geometry • Using units of measurement c

d

8 cm 2 cm

3 cm 4 cm

3.8 m

e

2.1 m

f 28 cm 18 cm

5 cm 12 cm 9 Find the shaded area in each of the following. a

b

16 m 8m

2m

2m

r = 7 cm

3 cm

c

d

8m

e

3m 40è

f

8m

5m

15 m 5m

2m

7.5 m

13 m 7 m

5m Chapter 6 Surface area and volume

191

measurement AND geometry • Using units of measurement understanding 10 A sheet of cardboard is 1.6  m by 0.8  m. The following shapes are cut from the cardboard:

•  a circular piece with radius 12  cm •  a rectangular piece 20  cm by 15  cm •  2 triangular pieces with base 30  cm and height 10  cm •  a triangular piece with side length 12  cm, 10  cm and 8  cm. What is the area of the remaining piece of cardboard? 11 A rectangular block of land, 12  m by 8  m, is surrounded by a concrete path 0.5  m wide. Find the area of the path. 12 Concrete slabs 1  m by 0.5  m are used to cover a footpath 20  m by 1.5  m. How many slabs are needed? 13 A city council builds a 0.5  m wide concrete path around the garden as shown below. 12 m 5m 8m

3m

Find the cost of the job if the workman charges $40.00 per m2. 14 A game of tennis can be played with 4 people using the whole court or it can be played with 2 people using the singles court, which excludes the edge on either side, as shown in the diagram.

1.8 m

x

8.23 m

6.40 m

10.97 m

11.89 m

a b c d e f

What is the total area of the whole tennis court? What is the area of the singles court? What area can one person use when playing doubles? What area can one person use when playing singles? What percentage of the total tennis court area is used by one person for singles? How far is the ball served before it bounces if it follows the path indicated in the diagram? That is, calculate x.

15 Ron the excavator operator has 100 metres of barricade mesh and needs to enclose an area to

work in safely. He chooses to make a rectangular region with dimensions x and y. a Write an equation that connects x, y and the perimeter. b Write y in terms of x. c Write an equation for the area of the region in terms of x. d Fill in the table for different values of x. x

0

5

10

Area 192

Maths Quest 10 for the Australian Curriculum

15

20

25

30

35

40

45

50

measurement anD geometry • using units of measurement

Can x have a value more than 50? Why? Sketch a graph of area against x. Determine the value of x that makes the area a maximum. What is the value of y for maximum area? What shape encloses the maximum area? Calculate the maximum area. Ron decides to choose to make a circular area with the barricade mesh. k What is the radius of this circular region? l What area is enclosed in this circular region? m How much extra area does Ron now have compared to his rectangular region? e f g h i j

reasoning 16 Dan has purchased a country property with layout and dimensions

eBook plus

Digital doc

WorkSHEET 6.1 doc-5241

6b eBook plus

Digital doc

SkillSHEET 6.3 doc-5238

as shown in the diagram. a Show that the property has a total area of 987.5 ha. b Dan wants to split the property in half (in terms of area) 1500 m 5000 m by building a straight-lined fence either vertically or 2000 m horizontally through the property. Assuming the cost of the fencing is a fixed amount per linear metre, justify where the 1000 m fence should be built (that is, how many metres from the top left-hand corner and in which direction), to minimise the cost. 17 In question 15, Ron the excavator operator could choose to enclose a rectangular or circular area with 100 m of barricade mesh. In this case, the circular region resulted in a larger safe work area. a Show that for 150 m of barricade mesh, a circular refleCtion region again results in a larger safe work area as opposed to a rectangular region. How are perimeter and area b Show that for n metres of barricade mesh, a circular different but fundamentally related? region will result in a larger safe work area as opposed to a rectangular region.

total surface area ■

The total surface area (TSA) of a solid is the sum of the areas of all the faces (outside surfaces) of that solid. It can be found by calculating the area of the net of the solid.

tsa of rectangular prisms and cubes ■

The following formulas have been introduced in previous years.

h

Rectangular prism (cuboid): w

TSA = 2(lh + lw + wh) l ■

A special case of the rectangular prism is the cube where all sides are equal (l = w = h). Cube: TSA = 6l2 l

To see a worked example and revise finding the total surface area of cubes and rectangular prisms, complete the SkillSHEET shown. Chapter 6 surface area and volume

193

measurement anD geometry • using units of measurement

tsa of spheres and cylinders

eBook plus

Sphere:

Interactivity TSA-sphere

TSA = 4p r2

int-2782

r

Note: The mathematics required to obtain the rule for the total surface area of a sphere is beyond the scope of Year 10. Cylinder: h

TSA = 2p r(r + h) or 2p r2 + 2p rh r

■

The formula for the TSA of the cylinder is found from the area of the net as shown. TSA = p r 2 + p r 2 + 2p rh r = 2p r 2 + 2p rh A = pr 2 = 2p r(r + h) 2p r

A = 2prh

h

r A = pr 2

WorkeD example 3

Find the total surface area of the solids below, correct to 1 decimal place. a

r = 7 cm

b

r

think a

b

194

50 cm 1.5 m

Write a TSA = 4p r2

1

Write the formula for the TSA of a sphere.

2

Identify the value for r.

3

Substitute and evaluate.

1

Write the formula for the TSA of a cylinder.

2

Identify the values for r and h. Note that the units will need to be the same.

r = 50 cm, h = 1.5 m = 150 cm

3

Substitute and evaluate.

TSA = 2 ì p ì 50 ì (50 + 150) = 62 831.9 cm2

maths Quest 10 for the australian Curriculum

r=7 TSA = 4 ì p ì 72 = 615.8 cm2 b TSA = 2p r(r + h)

measurement AND geometry • Using units of measurement

TSA of cones ■■

The total surface area of a cone can be found by considering its net. SA = Acircular base + Acurved surface, = p r2 + Asector of radius, s

s s

r

r

The sector is a fraction of the full circle of radius, s, with circumference, 2p s. The sector has arc length, equivalent to the circumference of the base of the cone, 2p r. ■■ The fraction of the full circle represented by the sector can be found by writing the arc length as a fraction of the circumference of the full circle, 2π r = r . 2π s s Area of a sector = fraction of the circle ì p r2 r = × π s2 s = p rs ■■ ■■

Therefore, SA = Acircular base + Acurved surface = p r2 + p rs



= p r(r + s) Cone: TSA = p r(r + s) or p r2 + p rs

Worked Example 4

Find the total surface area of the cone shown.

15 cm 12 cm

Think

Write

1

Write the formula for the TSA of a cone.

TSA = p r (r + s)

2

State the values of r and s.

r = 12, s = 15

3

Substitute and evaluate.

TSA = p  ì 12 ì (12 + 15) = 1017.9  cm2

TSA of other solids ■■ ■■ ■■

TSA can be found by summing the areas of each face. The areas of each face may need to be calculated separately. Always consider top, bottom, front, back, left and right faces. Chapter 6 Surface area and volume

195

measurement AND geometry • Using units of measurement

Worked Example 5

Find the total surface area of the square-based pyramid shown. 5 cm 6 cm

Think

Write/draw

1

There is no formula, so write the components of the TSA. These are the square base and four identical triangles.

TSA = Area of square base + Area of four triangular faces

2

Find the area of the square base.

Area of base = l2, where l = 6 Area of base = 62 = 36  cm2

3

The four side faces are isosceles triangles. Draw one face and write the formula for finding its area.

h

5 cm

3 cm 1

Area of a triangular face = 2 bh; b = 6 4

Find the height of the triangle using Pythagoras’ theorem.

a2 = c2 - b2, where a = h, b = 3, c = 5 h2 = 52 - 32 h2 = 25 - 9 h2 = 16 h = 4  cm

5

Calculate the area of the triangular face by substituting b = 6 and h = 4.

Area of triangular face =

Calculate the TSA by adding the area of the square base and the area of four triangular faces together.

TSA = 36 + 4 ì 12 = 36 + 48 = 84  cm2

6

■■

1 2

ì6ì4

= 12  cm2

Note that the area of a triangular face of the square-based pyramid in the previous example could also be calculated using Heron’s formula, as the lengths of all three sides were given.

TSA of composite solids ■■ ■■

196

The TSA of a composite solid is calculated by summing the areas of the solid’s faces. When two smaller solids are joined, the surfaces involved in the join will not be part of the surface of the composite solid. For example, if a square-based pyramid is stacked on top of a cube of the same base dimensions, the two faces in the join (one of the cube’s faces and the pyramid’s base) are inside and therefore not part of the surface of the solid.

Maths Quest 10 for the Australian Curriculum

measurement AND geometry • Using units of measurement

Worked Example 6

Find the total surface area of the object shown. 6 cm

10 cm 10 cm

Think

Write/draw

1

The solid shown is made up of a cube and a square pyramid. Only five faces of the cube are on the surface. Likewise, only triangular faces of the pyramid are on the surface. Thus, the TSA of the solid consists of five identical squares and four identical triangles.

TSA = 5 ì area of a square + 4 ì area of a triangle

2

Find the area of the square face with the side length 10  cm.

Asquare = l2, where l = 10 A = 102 A = 100  cm2

3

Draw a triangular face and work out its height using Pythagoras’ theorem.

6 cm

h

5 cm

a2 = c2 - b2, where a = h, b = 5, c = 6 h2 = 62 - 52 h2 = 36 - 25 h2 = 11 h = 3.32  cm 4

5

1 2 1 2

Find the area of the triangular face with the base of 10  cm and the height of 3.32  cm.

Atriangle =

Find the TSA of the solid by adding the area of 5 squares and 4 triangles together.

TSA = 5 ì 100 + 4 ì 16.6 = 500 + 66.4 = 566.4  cm2

■■

bh, where b = 10, h = 3.32

= ì 10 ì 3.32 = 16.6  cm2

Applications of surface area are commonly seen when calculating the amount of material needed for building structures such as silos, tanks, swimming pools, or painting or tiling surfaces. Chapter 6 Surface area and volume

197

measurement AND geometry • Using units of measurement

Worked Example 7

The silo shown at right is to be built from metal. The top portion of the silo is a cylinder of diameter 4  m and height 8  m. The bottom part of the silo is a cone of slant height 3  m. The silo has a circular opening of radius 30  cm on the top. a What area of metal (to the nearest m2) is required to build the silo? b If it costs $12.50 per m2 to cover the surface with an anti-rust material, how much will it cost to cover the silo completely?

Think a

b

198

1

8m

4m

3m

Write

The surface area of the silo consists of the circle (the top face), the curved part of the cylinder and the curved part of the cone. The circular opening is cut out from the top face and thus its area must be subtracted.

a TSA = area of a large circle



- area of a small circle + area of curved section of a cylinder + area of curved section of a cone

2

To find the area of the top face, subtract the area of the small circle from the area of the larger circle. Let R = radius of small circle.

Area of top face = Alarge circle - Asmall circle = p r2 - p R2 4 where r = 2 = 2  m and R = 30  cm = 0.3  m. Area of top face = p ì 22 - p ì 0.32 = 12.28  m2

3

The middle part of the silo is the curved part of a cylinder. Find its area. (Note that in the formula TSAcylinder = 2p r 2 + 2p rh, the curved part is represented by 2p rh.)

Area of curved section of cylinder = 2p rh where r = 2, h = 8. Area of curved section of cylinder =2ìpì2ì8 = 100.53  m2

4

The bottom part of the silo is the curved section of a cone. Find its area. (Note that in the formula TSAcone = p r 2 + p rs, the curved part is given by prs.)

Area of curved section of cone = p rs where r = 2, s = 3. Area of curved section of cone = p ì 2 ì 3 = 18.85  m2

5

Find the total surface area of the silo by finding the sum of the surface areas calculated above.

TSA = 12.28 + 100.53 + 18.85 = 131.66  m2

6

Write the worded answer. (Remember that we were asked to give the answer to the nearest square metre.)

The area of metal required is 132  m2.

To find the total cost, multiply the total surface area of the silo by the cost of an anti-rust material per m2 ($12.50).

Maths Quest 10 for the Australian Curriculum

b Cost = 132 ì $12.50

= $1650.00

measurement anD geometry • using units of measurement

remember

1. The total surface area (TSA) of a figure is the sum of the areas of all its outside faces. 2. TSA of a cube with the length of the edge, l, is given by the formula TSA = 6l 2 3. TSA of a rectangular prism with dimensions l, w and h is TSA = 2(lw + lh + wh) 4. TSA of a closed cylinder of radius, r, and height, h, is TSA = 2p rh + 2p r 2 5. TSA of a sphere of radius, r, is TSA = 4p r 2 6. TSA of a closed cone with radius, r, and slant height, s, is TSA = p rs + p r 2 7. TSA of a pyramid = area of base + area of triangular faces 8. TSA of a composite shape can be found by calculating the areas of individual faces that are on the surface and then adding them together

exerCise

6b inDiViDual pathWays eBook plus

Activity 6-B-1

total surface area fluenCy

Note: Where appropriate, give the answers correct to 1 decimal place. 1 Find the total surface areas of the solids shown. b

a

Introducing surface area doc-5038 Activity 6-B-2

Surface area problems doc-5039 Activity 6-B-3

Tricky surface area problems doc-5040

10 cm c

8 cm d

12 cm

2m 1.5 m

15 cm

3m

20 cm eBook plus

Digital doc

SkillSHEET 6.3 doc-5238

2 We3 Find the total surface area of the solids shown below. a b 21 cm r=3m

r

c

30 cm

d

0.5 m

12 cm 2.1 m

Chapter 6 surface area and volume

199

measurement AND geometry • Using units of measurement 3   WE 4  Find the total surface area of the cones below. a b

8 cm

20 cm 12 cm

14 cm

4   WE5  Find the total surface area of the solids below. a

b 12 cm

2.5 m

15 cm c

1.5 m d

9.1 cm 8 cm

14 cm

6 cm

7.2 cm

5.1 cm

10 cm 7 cm

5 Find the surface areas of the following. a A cube of side length 1.5  m b A rectangular prism 6  m ì 4  m ì 2.1  m c A cylinder of radius 30  cm and height 45  cm, open at one end d A sphere of radius 28  mm e An open cone of radius 4  cm and slant height 10  cm f A square pyramid of base length 20  cm and slant edge 30  cm 6   WE 6  Find the total surface area of the objects shown. a

10 cm

b

8 cm 5 cm

12 cm 5 cm

20 cm

20 cm 35 cm 12 cm

c

d 2 cm

m

2.5 c

5 cm

3 cm

3 cm

200

Maths Quest 10 for the Australian Curriculum

measurement AND geometry • Using units of measurement e

f

5 cm

3.5 cm

20 cm

10 cm 12 cm

15 cm 7   MC  A cube has a total surface area of 384  cm2. The length of the edge of the cube is: a 9  cm b 8  cm c 7  cm d 6  cm e 5  cm understanding 8 Open cones are made from nets cut from a large sheet of paper 1.2  m ì 1.0  m. If a cone has

a radius of 6  cm and a slant height of 10  cm, how many cones can be made from the sheet? (Assume there is 5% wastage of paper.) 9 A steel girder is to be painted. Calculate the area of the surface to be painted. 2 cm

2 cm



5 cm 20 cm 2 cm

120 cm

12 cm

10   WE 7  The greenhouse shown below is to be built using shade cloth. It has a wooden door of

dimensions 1.2  m ì 0.5  m. a Find the total area of shade cloth needed to complete the greenhouse. b Find the cost of the shade cloth at $6.50 per m2.

5m

2.5 m 3m

11 A cylinder is joined to a hemisphere to make a cake holder, as shown below. The surface of the

cake holder is to be chromed at 5.5 cents per cm2. a Find the total surface area to be chromed. b Find the cost of chroming the cake holder.

15 cm

10 cm

Chapter 6 Surface area and volume

201

measurement anD geometry • using units of measurement 12 A soccer ball is made up of a number of hexagons sewn together

2 cm

on its surface. Each hexagon can be considered to have dimensions as shown in the diagram. y a Calculate q o. x b Calculate the values of x and y exactly. c Calculate the area of the trapezium in the diagram. d Hence, determine the area of the hexagon. e If the total surface area of the soccer ball is 192 3 cm2, how q many hexagons are on the surface of the soccer ball? 13 a Determine the exact total surface area of a sphere with radius 2 metres. An inverted cone with side length 4 metres is placed on top of the sphere so that the centre of its base is 0.5 metres above the centre of the sphere. b Find the radius of the cone exactly. c Find the area of the curved surface of the cone exactly. d What are the exact dimensions of a box that could precisely fit the cone connected to the sphere? Complete the following question without the aid of a calculator. 14 The table shown below is to be varnished (including the base of each leg). The table top has a

thickness of 180 mm and the cross-sectional dimension of the legs is 50 mm by 50 mm. 80 cm 60 cm

70 cm

A friend completes the calculation as shown. Assume there are no simple calculating errors. Analyse the working presented and justify if the TSA calculated is correct. Table top

0.96

2 ì (0.8 ì 0.6)

Legs

0.416

16 ì (0.52 ì 0.05)

Table top edging

0.504

TSA

0.18 ì (2(0.8 + 0.6)) 2

1.88 m

15 A shower recess with dimensions 1500 mm (back wall) by 900 mm (side wall) needs to have

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Digital doc

WorkSHEET 6.2 doc-5242

202

the back and two side walls tiled to a height of 2 m. a Calculate the area to be tiled in m2. b Justify that 180 tiles (including those that need to be cut) of dimension 20 cm by 20 cm will be required. Disregard the grout. c Evaluate the cheapest option of tiling; $1.50/tile or $39.50/box, where a box covers 1 m2 or tiles of dimension 30 cm by 30 cm costing $3.50/tile. refleCtion 16 If the surface area of a sphere to a cylinder is in the Why is calculating the total surface ratio 4 : 3 and the sphere has a radius of 3a, show area of a composite solid more that if the radius of the cylinder is equal to its height, then the radius of the cylinder is 3 3a . 2

maths Quest 10 for the australian Curriculum

difficult than for a simple solid such as a rectangular prism or cylinder?

measurement anD geometry • using units of measurement

6C eBook plus

Digital doc

SkillSHEET 6.4 doc-5239

Volume ■ ■ ■

Volume of prisms and other shapes ■

■

■

■

eBook plus

The volume of a 3-dimensional figure is the amount of space it takes up. The units for volume are mm3, cm3 and m3. To revise the technique of converting from one unit to another, complete SkillSHEET 6.4.

The volume of any solid with a uniform cross-sectional area is given by the formula: V = AH, where A is the cross-sectional (or base) area and H is the height of the solid. The height of a prism simply means the dimension perpendicular to a solid’s cross-sectional base. This is often the physical height, depth or length. Prisms are the most recognisable solids with uniform cross-sectional areas. A prism is a solid shape with identical opposite ends joined by straight edges, forming a congruent cross-section. In some cases a special formula can be developed from the formula V = AH. Volume = AH = area of a square ì height = l2 ì l = l3

Cube

Interactivity Maximising the volume of a cuboid

l

int-1150

Rectangular prism h

Volume = AH = area of a rectangle ì height = lwh

w l

Cylinder

r h

Volume = AH = area of a triangle ì height 1 = 2 bh ì H

Triangular prism

H

h b

eBook plus

Digital doc

SkillSHEET 6.5 doc-5240

Volume = AH = area of a circle ì height = p r 2h

■

To see a worked example and revise the volume of a cube and rectangular prism, complete SkillSHEET 6.5. Chapter 6 surface area and volume

203

measurement AND geometry • Using units of measurement

Worked Example 8

Find the volumes of the following shapes. a 14 cm

b

4 cm

20 cm

10 cm

Think a

b

5 cm

Write a V = pr 2h

1

Write the formula for the volume of the cylinder.

2

Identify the value of the pronumerals.

r = 14, h = 20

3

Substitute and evaluate.

V = p ì 142 ì 20 ö 12 315.04  cm3

1

Write the formula for the volume of a triangular prism.

2

Identify the value of the pronumerals. (Note h is the height of the triangle and H is the depth of the prism.)

b = 4, h = 5, H = 10

3

Substitute and evaluate.

V = 2 ì 4 ì 5 ì 10 = 100  cm3

b

1

V = 2 bh ì H

1

Worked Example 9 a What effect will doubling each of the side lengths of a cube have on its volume (in comparison

with the original shape)?

b What effect will halving the radius and doubling the height of a cylinder have on its volume

(in comparison with the original shape)?

Think a

204

Write a V = l3

1

Write the formula for the volume of the cube.

2

Identify the value of the pronumeral. Note: Doubling is the same as multiplying by 2.

lnew = 2l

3

Substitute and evaluate.

Vnew = (2l)3

4

Compare the answer obtained in step 3 with the volume of the original shape.

5

Write your answer.

Maths Quest 10 for the Australian Curriculum

= 8l 3 Doubling each side length of a cube will increase the volume by a factor of 8; that is, the new volume will be 8 times as large as the original volume.

measurement AND geometry • Using units of measurement b

1

Write the formula for the volume of the cylinder.

2

Identify the value of the pronumerals. Note: Halving is the same as dividing by 2.

3

Substitute and evaluate.

b V = p r2h

r rnew = , hnew = 2h 2 2

Vnew = p  r  2h  2  =π× =

4

Compare the answer obtained in step 3 with the volume of the original shape.

5

Write your answer.

r2 × 2h 24

π r 2h 2 1

= 2 pr2h Halving the radius and doubling the height of a cylinder will decrease the volume by a factor of 2; that is, the new volume will be half as large as the original volume.

Volume of spheres ■■

4

Volume of a sphere of radius, r, can be calculated using the formula: V = 3 p r 3.

Worked Example 10

Find the volume of a sphere of radius 9  cm. Answer correct to 1 decimal place. Think

Write 4

1

Write the formula for the volume of a sphere.

V = 3 pr3

2

Identify the value of r.

r=9

3

Substitute and evaluate.

V=

4 3

ì p ì 93 = 3053.6  cm3

Volume of pyramids ■■

Pyramids (including cones) are not prisms as the cross-section changes from the base upwards.

■■

It has been found that the volume of a pyramid is one-third the volume of an equivalent prism with the same base area and height.



1

Volume of a pyramid = 3  AH

H Area of base = A base Chapter 6 Surface area and volume

205

measurement AND geometry • Using units of measurement

■■

Since a cone is a pyramid with a circular cross-section, the volume of a cone is one-third the volume of a cylinder with the same base area and height.

1

Volume of a cone = 3  AH



1

= 3 p r 2h



h r

Worked Example 11

Find the volume of each of the following solids. a b  12 cm

10 cm 8 cm

8 cm Think a

b

Write 1

a V = 3 p r2h

1

Write the formula for the volume of a cone.

2

Identify the values of r and h.

r = 8, h = 10

3

Substitute and evaluate.

V = 3 ì p ì 82 ì 10 = 670.21  cm3

1

Write the formula for the volume of a pyramid.

2

Find the area of the square base.

A = l2 where l = 8 A = 82 = 64  cm2

3

Identify the value of H.

H = 12

4

Substitute and evaluate.

V = 3 ì 64 ì 12 = 256  cm3

1

b V=

1 3

AH

1

Volume of composite figures ■■ ■■ ■■

206

A composite solid is made from smaller solids. The volume of each smaller solid component can be calculated separately. The volume of a composite solid is calculated by summing the volumes of each of the smaller solid components.

Maths Quest 10 for the Australian Curriculum

measurement AND geometry • Using units of measurement

Worked Example 12

Calculate the volume of the composite solid shown.

3m 1.5 m

Think

Write

1

The given solid is a composite figure, made up of a cube and a square-based pyramid.

V = Volume of cube + Volume of pyramid

2

Find the volume of the cube.

Vcube = l  3 where l = 3 Vcube = 33 = 27  m3

3

Write the formula for finding the volume of a square-based pyramid.

Vsquare-based pyramid = 3 AH

4

Find the area of the square base.

A = l2 = 32 = 9  m2

5

Identify the value of H.

H = 1.5

6

Substitute and evaluate the volume of the pyramid.

Vsquare-based pyramid = 3 ì 9 ì 1.5 = 4.5  m3

7

Find the total volume by adding the volume of the cube and pyramid.

V = 27 + 4.5 = 31.5  m3

1

1

Capacity ■■ ■■ ■■

Capacity measures the amount of liquid that will fit in a 3-dimensional figure. The units for capacity are: mL, L and kL. Volume and capacity are fundamentally related: 1  cm3 = 1  mL 1000  cm3 = 1  L 1  m3 = 1000  L = 1  kL

Worked Example 13

Find the capacity (in litres) of a rectangular aquarium, which is 50  cm long, 30  cm wide and 40  cm high.

Chapter 6 Surface area and volume

207

measurement anD geometry • using units of measurement

think

Write

1

Write the formula for the volume of a rectangular prism.

V = lwh

2

Identify the values of the pronumerals.

l = 50, w = 30, h = 40

3

Substitute and evaluate.

V = 50 ì 30 ì 40 = 60 000 cm3

4

State the capacity of the container in millilitres, using 1 cm3 = 1 mL.

60 000 cm3 = 60 000 mL

5

Since 1 L = 1000 mL, to convert millilitres to litres divide by 1000.

6

Give a worded answer.

= (60 000 ó1000) L = 60 L The capacity of the fish tank is 60 L.

remember

1. Volume of a 3-dimensional figure is the amount of space it takes up 2. Volume is measured in cubic units 3. Volume of a prism = AH, where A is the cross-sectional area (or base) and H is the height of the prism 4. The height of a prism is the dimension perpendicular to the prism’s cross-section. 4 5. Volume of a sphere = 3 p r3 1

6. Volume of a cone = 3 p r2h 1

7. Volume of a pyramid = 3 AH 8. Capacity of a 3-dimensional figure is the amount of liquid that will fit in that figure. 9. The relationship between the volume of a solid and the capacity (amount of liquid it can hold) is: 1 cm3 = 1 mL, 1000 cm3 = 1 L. 10. 1 m3 = 1000 L = 1 kL exerCise

6C inDiViDual pathWays eBook plus

Volume fluenCy 1 Find the volumes of the following prisms. a

b

Activity 6-C-1

Review of volume and capacity doc-5041 Activity 6-C-2

Volume and capacity problem doc-5042

3 cm c

4.2 m d

12 cm

Activity 6-C-3

Tricky volume and capacity problems doc-5043

15 cm

4.2 cm

20 cm

7.5 cm 3 cm

208

maths Quest 10 for the australian Curriculum

measurement anD geometry • using units of measurement

eBook plus

2 Calculate the volume of each of these solids. a

b

Digital doc

SkillSHEET 6.5 doc-5240

18 mm

15 cm

[Base area: 25 mm2]

[Base area: 24 cm2]

3 We8 Find the volume of each of the following. Give each answer correct to 1 decimal place

where appropriate. a

b

14 cm

2.7 m

12 cm

1.5 m

c

d

10 cm 7 cm

12 mm

8 mm 6 mm

8 cm

4 We10 Find the volume of a sphere (correct to 1 decimal place) with a radius of: a 1.2 m b 15 cm c 7 mm d 50 cm. 5 We11a Find the volume of each of the following cones, correct to 1 decimal place. a b 20 mm

10 cm

22 mm

6 cm 6 We11b Find the volume of each of the following pyramids. a

b

12 cm

42 cm 24 cm 10 cm

30 cm

c 12 cm

m

16 cm

18 c

Chapter 6 surface area and volume

209

measurement AND geometry • Using units of measurement 7   WE 12  Calculate the volume of each of the following composite solids correct to 2 decimal

places where appropriate. a

b

8 cm

10 cm

5 cm

12 cm 5 cm





20 cm

20 cm 35 cm 12 cm

d

c

2 cm

m

2.5 c

5 cm

3 cm 3 cm e

f

5 cm

3.5 cm

20 cm

10 cm



12 cm



15 cm

Understanding 8   WE 9  a What effect will tripling each of the side lengths of a cube have on its volume

(in comparison with the original shape)? b What effect will halving each of the side lengths of a cube have on its volume

(in comparison with the original shape)? c What effect will doubling the radius and halving the height of a cylinder have

on its volume (in comparison with the original shape)? d What effect will doubling the radius and dividing the height of a cylinder by 4 have on its

volume (in comparison with the original shape)? e What effect will doubling the length, halving the width and tripling the height of a

rectangular prism have on its volume (in comparison with the original shape)? 9   MC  A hemispherical bowl has a thickness of 2  cm and an outer 2 cm diameter of 25  cm. If the bowl is filled with water to its full capacity, the volume of the water will be: a 1526.04  cm3 b 1308.33  cm3 c 3052.08  cm3 25 cm d 2616.66  cm3 3 e 2424.52  cm 210

Maths Quest 10 for the Australian Curriculum

measurement AND geometry • Using units of measurement 10 Tennis balls of diameter 8  cm are packed in a box 40  cm ì 32  cm ì 10  cm,

as shown. What is unused space in the box?

11   WE 13  A cylindrical water tank has a diameter of 1.5  m and a height of 2.5  m. What is the

capacity (in litres) of the tank? 12 A monument in the shape of a rectangular pyramid (base length of 10  cm, base width of 6  cm,

height of 8  cm), a spherical glass ball (diameter of 17  cm) and conical glassware (radius of 14  cm, height of 10  cm) are packed in a rectangular prism of dimensions 30  cm by 25  cm by 20  cm. The extra space in the box is filled up by a packing material. What volume of packing material is required? 13 A swimming pool is being constructed so that it is the 8m upper part of an inverted square-based pyramid. a Calculate H. b Calculate the volume of the pool. 3m c How many 6  m3 bins will be required to take the dirt away? 4m d How many litres of water are required to fill this pool? H e How deep is the pool when it is half-filled? 14 A soft drink manufacturer is looking to repackage cans of soft drink to minimise the cost of packaging while keeping the volume constant.   Consider a can of soft drink with a capacity of 400  mL. a If the soft drink was packaged in a spherical can: i find the radius of the sphere ii find the total surface area of this can. b If the soft drink was packaged in a cylindrical can with a radius of 3  cm: i find the height of the cylinder ii find the total surface area of this can. c If the soft drink was packaged in a square-based pyramid with a base side length of 6  cm: i find the height of the pyramid ii find the total surface area of this can. d Which can would you recommend the soft drink manufacturer use for its repackaging? Why? Reasoning 15 Marion has mixed together ingredients for a cake. The recipe requires a baking tin that is

cylindrical in shape with a diameter of 20  cm and a height of 5  cm. Marion only has a tin as shown and a muffin tray consisting of 24 muffin cups. Each of the muffin cups in the tray is a portion of a cone as shown in the diagram. Should Marion use the tin or muffin tray? Explain. 8 cm

12 cm

4 cm

4 cm

10 cm

15 cm

8 cm

Chapter 6 Surface area and volume

211

measurement anD geometry • using units of measurement 16 Nathaniel and Andrew are going to the snow for

survival camp. They plan to construct an igloo, consisting of an entrance and hemispherical structure, as shown. Nathaniel and Andrew are asked to redraw their plans and increase the size of the liveable region (hemispherical structure) so that the total volume (including the entrance) is doubled. How can this be achieved?

1.5 m

1m

1.5 m

17 Sam is having his 16th birthday party and wants to make an ice trough to keep drinks cold. He

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WorkSHEET 6.3 doc-6753

has found a square piece of sheet metal with a side length of 2 metres. He cuts squares, of side length x metres, from each corner then bends the sides of the remaining sheet. When four squares of the appropriate side length are cut from the corners the capacity of the trough can be maximised at 588 litres. Explain how Sam should proceed to maximise the capacity of the trough. 18 The Hastings family house has a rectangular roof with dimensions 17 m ì 10 m providing water to three water tanks, each with a radius of 1.25 m and a height of 2.1 m. When rain falls it is measured in millimetres. This means that this is the depth to which the water would fill if it were captured. Show that approximately 182 millimetres of rain must fall on the roof to fill the tanks. refleCtion Volume is measured in cubic units. How is this reflected in the volume formula?

212

maths Quest 10 for the australian Curriculum

measurement anD geometry • using units of measurement

summary area ■ ■ ■ ■ ■

Area is a measure of the amount of surface within the boundaries of a figure. The units for measuring area are mm2, cm2, m2 and km2. Land area is usually measured in hectares (ha) where 1 ha = 10 000 (or 104) m2. Areas can be calculated by using formulas that are specific to the given plane figure. Areas of composite figures can be calculated by adding the areas of the simple figures making the composite figure or by calculating the area of an extended figure and subtracting the extra area covered. Total surface area

■ ■ ■ ■ ■ ■ ■ ■

The total surface area (TSA) of a figure is the sum of the areas of all its outside faces. TSA of a cube with the length of the edge, l, is given by the formula TSA = 6l 2 TSA of a rectangular prism with dimensions l, w and h is TSA = 2(lw + lh + wh) TSA of a closed cylinder of radius, r, and height, h, is TSA = 2p rh + 2p r 2 TSA of a sphere of radius, r, is TSA = 4p r 2 TSA of a closed cone with radius, r, and slant height, s, is TSA = p rs + p r 2 TSA of a pyramid = area of base + area of triangular faces TSA of a composite shape can be found by calculating the areas of individual faces that are on the surface and then adding them together Volume

■ ■ ■

■ ■ ■ ■ ■ ■

■

Volume of a 3-dimensional figure is the amount of space it takes up Volume is measured in cubic units Volume of a prism = AH, where A is the cross-sectional area (or base) and H is the height of the prism The height of a prism is the dimension perpendicular to the prism’s cross-section. Volume of a sphere = 43 πr 3 Volume of a cone = 13 πr 2h

Volume of a pyramid = 13 AH Capacity of a 3-dimensional figure is the amount that will fit in that figure The relationship between the volume of a solid and the capacity (amount of liquid it can hold) is: 1 cm3 = 1 mL, 1000 cm3 = 1 L. 1 m3 = 1000 L = 1 kL

MaPPING YOUR UNdeRSTaNdING

Homework book

Using terms from the summary, and other terms if you wish, construct a concept map that illustrates your understanding of the key concepts covered in this chapter. Compare your concept map with the one that you created in What do you know? on page 183. Have you completed the two Homework sheets, the Rich task and two Code puzzles in your Maths Quest 10 Homework Book?

Chapter 6 surface area and volume

213

measurement AND geometry • Using units of measurement

Chapter review Fluency

5 Find the areas of the following plane figures. All

1 If all measurements are in cm, the area of the figure

below is: 7

measurements are in cm. a

3 14

A 16.49  cm2 C 9.81  cm2 E 30  cm2

12

B 39.25  cm2 D 23.56  cm2

2 If all measurements are in

b

10

6

centimetres, the area of the figure at right is: A 50.73  cm2 B 99.82  cm2 C 80.18  cm2 D 90  cm2 E 119.45  cm2

8 5 15 5 c 5

3 7

3 If all measurements are in centimetres, the shaded

area of the figure below is:

5

30è d 3

2 7

A 3.93  cm2 C 388.77  cm2 E 129.59  cm2

B 11.52  cm2 D 141.11  cm2

6

e 10

4 The total surface area of the solid below is:

12

28 mm

f

40 mm

80è 10 A 8444.6  mm2 C 14146.5  mm2 E 16609.5  mm2 214

B 9221  mm2 D 50271.1  mm2

Maths Quest 10 for the Australian Curriculum

measurement AND geometry • Using units of measurement 6 Find the areas of the following figures. All

8 Find the total surface area of each of the following

measurements are in cm.

solids.

a

a

35 cm

15 50 cm 20 b 12

b

14 mm

10 20 mm 8 c c

10 6

10

8 cm 5

20 7 Find the shaded area in each of the following. All

measurements are in cm. a

Q

14 cm 12 cm

d

QO = 15 cm SO = 8 cm PR = 18 cm

18 cm

O

7 cm P

10 cm

R

S

e b

10 mm 10 mm

14 mm

12.5

4 mm

[closed at both ends] f 12 cm

c

5 10 cm 10 cm 10 cm Chapter 6 Surface area and volume

215

measurement AND geometry • Using units of measurement 9 Find the volume of each of the following.

g

11 cm

a

9 cm h

7 cm

30 cm b 7 cm

20 cm 8 cm 12 cm

c

42 cm i

35 cm

12 mm 40 cm

problem solving 1 A rectangular block of land 4  m ì 25  m is

d

surrounded by a concrete path 1  m wide. a Calculate the area of the path. b Calculate the cost of concreting at $45 per 3.7 m

square metre. 2 What effect will tripling the radius and dividing

the height of a cylinder by 6 have on its volume (in comparison with the original shape)?

1m

3 What effect will halving the length, tripling the

e

width and doubling the height of a rectangular prism have on its volume (in comparison with the original shape)?

10 cm

30 cm 12 cm f 12 cm

10 cm

216

Maths Quest 10 for the Australian Curriculum

4 A cylinder of radius 14  cm and height 20  cm is

joined to a hemisphere of radius 14  cm to form a bread holder. a Find the total surface area. b Find the cost of chroming the bread holder on the outside at $0.05 per cm2. c What is the storage volume of the bread holder? d How much more space is in this new bread holder than the one it is replacing, which had a quarter circle end with a radius of 18  cm and a length of 35  cm?

measurement anD geometry • using units of measurement 5 Bella Silos has two rows of silos for storing wheat.

Each row has 16 silos and all the silos are identical, with a cylindrical base (height of 5 m, diameter of 1.5 m) and conical top (diameter of 1.5 m, height of 1.1 m). a What is the slant height of the conical tops? b What is the total surface area of all the silos? c What will it cost to paint the silos if one litre of paint covers 40 m2 at a bulk order price of $28.95 per litre?

d How much wheat can be stored altogether in

these silos? e Wheat is pumped from these silos into cartage

trucks with rectangular containers 2.4 m wide, 5 m long and 2.5 m high. How many truckloads are necessary to empty all the silos? f If wheat is pumped out of the silos at 2.5 m3/min, how long will it take to fill one truck? eBook plus

Interactivities

Test yourself Chapter 6 int-2843 Word search Chapter 6 int-2841 Crossword Chapter 6 int-2842

Chapter 6 surface area and volume

217

eBook plus

aCtiVities

chapter opener Digital doc

• Hungry brain activity Chapter 6 (doc-5235) (page 183) are you ready? Digital docs (page 184) • SkillSHEET 6.1 (doc-5236): Conversion of area units • SkillSHEET 6.2 (doc-5237): Using a formula to find the area of a common shape • SkillSHEET 6.3 (doc-5238): Total surface area of cubes and rectangular prisms • SkillSHEET 6.4 (doc-5239): Conversion of volume units • SkillSHEET 6.5 (doc-5240): Volume of cubes and rectangular prisms

6a area Digital docs

• Activity 6-A-1 (doc-5035): Review of area (page 189) • Activity 6-A-2 (doc-5036): Area problems (page 189) • Activity 6-A-3 (doc-5037): Tricky area problems (page 189) • SkillSHEET 6.1 (doc-5236): Conversion of area units (page 185) • SkillSHEET 6.2 (doc-5237): Using a formula to find the area of a common shape (page 189) • WorkSHEET 6.1 (doc-5241): Area (page 193) eLesson

• Heron’s formula (eles-0177) (page 186) 6b Total surface area Interactivity

• TSA-sphere (int-2782) (page 194) Digital docs

• Activity 6-B-1 (doc-5038): Introducing surface area (page 199)

218

maths Quest 10 for the australian Curriculum

• Activity 6-B-2 (doc-5039): Surface area problems (page 199) • Activity 6-B-3 (doc-5040): Tricky surface area problems (page 199) • SkillSHEET 6.3 (doc-5238): Total surface area of cubes and rectangular prisms (page 193, 199) • WorkSHEET 6.2 (doc-5242): Surface area (page 202) 6c Volume Digital docs

• Activity 6-C-1 (doc-5041): Review of volume and capacity (page 208) • Activity 6-C-2 (doc-5042): Volume and capacity problem (page 208) • Activity 6-C-3 (doc-5043): Tricky volume and capacity problems (page 208) • SkillSHEET 6.4 (doc-5239): Conversion of volume units (page 203) • SkillSHEET 6.5 (doc-5240): Volume of cubes and rectangular prisms (page 203, 209) • WorkSHEET 6.3 (doc-6753): Volume (page 212) Interactivity

• Maximising the volume of a cuboid (int-1150) (page 203) chapter review Interactivities (page 217) • Test Yourself Chapter 6 (int-2843): Take the end-ofchapter test to test your progress • Word search Chapter 6 (int-2841): an interactive word search involving words associated with this chapter • Crossword Chapter 6 (int-2842): an interactive crossword using the definitions associated with the chapter

To access eBookPLUS activities, log on to www.jacplus.com.au

number AnD AlgebrA • PAtterns AnD AlgebrA

7

7A Expanding algebraic expressions 7B Factorising expressions with three terms 7C Factorising expressions with two or four terms 7D Factorising by completing the square 7E Mixed factorisation WhAt Do you knoW ?

Quadratic expressions

1 List what you know about quadratic expressions. Create a concept map to show your list. 2 Share what you know with a partner and then with a small group. 3 As a class, create a large concept map that shows your class’s knowledge of quadratic expressions. eBook plus

Digital doc

Hungry brain activity Chapter 7 doc-5243

oPening Question

What distance does the dolphin cover in one leap?

number AnD AlgebrA • PAtterns AnD AlgebrA

Are you ready?

Try■the■questions■below.■If■you■have■diffi■culty■with■any■of■them,■extra■help■can■be■obtained■by■ completing■the■matching■SkillSHEET.■Either■search■for■the■SkillSHEET■in■your eBookPLUS■or■ ask■your■teacher■for■a■copy. eBook plus

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SkillSHEET 7.1 doc-5244

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SkillSHEET 7.2 doc-5245

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SkillSHEET 7.3 doc-5246

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SkillSHEET 7.4 doc-5247

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SkillSHEET 7.7 doc-5248

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SkillSHEET 7.8 doc-5249

220

Expanding brackets 1 Expand■each■of■the■following. a 4(3x■+■5)

c -4x(3■-■2x)

b 5x(2x■-■3)

Expanding a pair of brackets 2 Expand■each■of■the■following■expressions. a (x■+■2)(x■-■2) b (2x■-■3)2

c (3x■+■2)(2x■-■5)

Factorising by taking out the highest common factor 3 Factorise■each■of■the■following■expressions. a 4x2■+■8x b -15x2■-■9x

c 6x2■-■x

Factorising by taking out a common binomial factor 4 Factorise■each■of■the■following. a 3x(x■+■2)■+■4(x■+■2) b 4x(x■-■1)■-■(x■-■1)

c -2x(x■+■3)■-■(x■+■3)

Simplifying algebraic fractions 5 Write■each■of■the■following■fractions■in■simplest■form. a

( x + 3)( x − 2) ( x − 2)

b

x+7

( x − 3)( x + 7)

2

c

3 x ( x + 2)2 ( x + 3) 6 x( x + 3)2 ( x + 2)

Simplifying surds 6 Simplify■each■of■the■following. a

24

maths Quest 10 for the Australian Curriculum

b 3 12

c 4 243

number AND algebra • Patterns and algebra

7A

Expanding algebraic expressions Binomial expansion ■■ ■■

When an expression contains two sets of brackets, expansion is known as binomial expansion. Consider the expression (a + b)(c + d). Its expansion can be represented visually by this area model. a

+ b

c +

ac

bc

d

ad

bd

This diagram visually shows that (a + b)(c + d) = ac + ad + bc + bd

■■ ■■

factorised expanded ■ form form Generally, binomial expressions contain only one variable, together with constants. Expansion of the binomial expression (x + 3)(x + 2) can be shown visually by this area model. x x

+

3

x ì x = x2

3ìx = 3x

2 ì x = 2x

3ì2 =6

+ 2

Expressed mathematically this is: (x + 3)(x + 2) = x2 + 2x + 3x + 6 = x2 + 5x + 6

■■

factorised expanded ■ form form It is not necessary to draw this area model when expanding binomial expressions. The shortcut is to simply multiply each term in the first bracket by each term in the second bracket.

Worked Example 1

Expand each of the following. a (x + 3)(x + 2)

b (x - 7)(6 - x)

Think a

Write a (x + 3)(x + 2)

1

Write the expression.

2

Multiply the terms in the second bracket by the first term in the first bracket and then the second term in the first bracket.

= x(x + 2) + 3(x + 2)

3

Remove the brackets by multiplying each term in the brackets by the term outside the bracket.

= x2 + 2x + 3x + 6

4

Collect like terms.

= x2 + 5x + 6 Chapter 7 Quadratic expressions

221

number AND algebra • Patterns and algebra b

b (x - 7)(6 - x)

1

Write the expression.

2

Multiply the terms in the second bracket by the first term in the first bracket and then the second term in the first bracket. Notice that the minus sign stays with the second term in the first bracket (-7).

= x(6 - x) - 7(6 - x)

3

Remove the brackets by multiplying each term in the brackets by the term outside the bracket. Remember to change the sign when the term outside the bracket is negative.

= 6x - x2 - 42 + 7x

4

Collect like terms.

= -x2 + 13x - 42

FOIL method ■■

The word FOIL provides us with an acronym for the expansion of a binomial product.

■■

First: multiply the first terms in each bracket together

■■

Outer: multiply the two outer terms

■■

Inner: multiply the two inner terms

I (x + a)(x - b)

■■

Last: multiply the last terms in each bracket together

L (x + a)(x - b)

F (x + a)(x - b) O (x + a)(x - b)

Worked Example 2

Use FOIL to expand (x + 2)(x - 5). Think

Write

1

Write the expression.

(x + 2)(x - 5)

2

Multiply the first term in each bracket, then the outer terms, the inner terms and finally the last two terms.

= x ì x + x ì -5 + 2 ì x + 2 ì -5 = x2 - 5x + 2x - 10

3

Collect like terms.

= x2 - 3x - 10

■■

If there is a term outside the pair of brackets, expand the brackets and then multiply each term of the expansion by that term.

Worked Example 3

Expand 3(x + 8)(x + 2). Think

222

Write

1

Write the expression.

3(x + 8)(x + 2)

2

Use FOIL to expand the pair of brackets.

= 3(x2 + 2x + 8x + 16)

3

Collect like terms within the brackets.

= 3(x2 + 10x + 16)

4

Multiply each of the terms inside the brackets by the term outside the brackets.

= 3x2 + 30x + 48

Maths Quest 10 for the Australian Curriculum

number AND algebra • Patterns and algebra

■■

This method can be extended to include three or even more sets of brackets. In such examples, expand two brackets first and then multiply the result by the third bracket.

Expanding expressions that are perfect squares ■■ ■■

A special binomial expansion involves the expansion of a perfect square. The expansion of (a + b)2 can be represented visually by this area model. +

a

a

b

a ì a = a2

aìb = ab

a ì b = ab

bìb = b2

+ b

■■

■■ ■■

(a + b)2 = a2 + ab + ab + b2 = a2 + 2ab + b2

This result tells us that to expand a perfect square: 1. square the first term 2. multiply the two terms together and then double them 3. square the last term. Similarly (a - b)2 = a2 - 2ab + b2. (Try this expansion for yourself.) Any perfect square can also be expanded using FOIL; however, this method provides a quicker alternative for performing such expansions. (a + b)2 = a2 + 2ab + b2 or (a - b)2 = a2 - 2ab + b2

Worked Example 4

Expand and simplify each of the following. a (2x - 5)2 b -3(2x + 7)2 Think a

b

Write a (2x - 5)2

1

Write the expression.

2

Expand using the rule (a - b)2 = a2 - 2ab + b2.

1

Write the expression.

2

Expand the brackets using the rule (a + b)2 = a2 + 2ab + b2.

= -3[(2x)2 + 2 ì 2x ì 7 + (7)2] = -3(4x2 + 28x + 49)

3

Multiply every term inside the brackets by the term outside the brackets.

= -12x2 - 84x - 147

= (2x)2 - 2 ì 2x ì 5 + (5)2 = 4x2 - 20x + 25 b -3(2x + 7)2

Difference of two squares rule ■■ ■■

This rule results from the expansion of an expression of the form (a + b)(a - b). Note: The brackets can be in any order. The two brackets contain the same terms, but one has the terms added and the other subtracted. Chapter 7 Quadratic expressions

223

number AND algebra • Patterns and algebra

■■ ■■

The area model for the difference of two squares rule shows a large square with a smaller square removed from it. Consider the larger square has a side length of a, while the smaller square has a side length of b. a

a a

b

a2 - b

a

b2

a

a2 - b2

a-b

a-b =

= a-b

b

b

a-b

The final figure shows two rectangles with dimensions a by (a - b) and b by (a - b). So, a(a - b) + b(a - b) = a2 - b2 To factorise, take out a common factor of (a - b) on the left-hand side. (a - b)(a + b) = a2 - b2 Alternatively, the difference of two squares rule is usually written as (a + b)(a - b) = a2 - b2 Worked Example 5

Expand and simplify each of the following. a (3x + 1)(3x - 1) b 4(2 x - 7)(2 x + 7) Think a

b

Write a (3x + 1)(3x - 1)

1

Write the expression.

2

Expand using the rule (a + b)(a - b) =

1

Write the expression.

2

Expand using the difference of two squares rule.

= 4[(2x)2 - (7)2] = 4(4x2 - 49)

3

Multiply every term inside the brackets by the term outside the brackets.

= 16x2 - 196

a2

-

b2.

= (3x)2 - (1)2 = 9x2 - 1 b 4(2x - 7)(2x + 7)

remember

1. When expanding an algebraic expression with: (a) one bracket — multiply each term inside the bracket by the term outside the bracket (b) two brackets — multiply the terms in order: First terms, Outer terms, Inner terms and then Last terms (FOIL) (c) a term outside the two brackets — expand the pair of brackets first, then multiply each term of the expanded expression by the term outside the brackets (d) three brackets — expand any two of the brackets and then multiply the expanded expression by the third bracket. 2. Perfect squares rule: (a + b)2 = a2 + 2ab + b2 or (a - b)2 = a2 - 2ab + b2 3. Difference of two squares rule: (a + b)(a - b) = a2 - b2 224

Maths Quest 10 for the Australian Curriculum

number AnD AlgebrA • PAtterns AnD AlgebrA

exerCise

7A inDiViDuAl PAthWAys eBook plus

Activity 7-A-1

Review of expansion doc-5044 Activity 7-A-2

Expanding algebraic expressions doc-5045 Activity 7-A-3

Expanding more complex algebraic expressions doc-5046

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SkillSHEET 7.1 doc-5244

expanding algebraic expressions FluenCy 1 Expand■each■of■the■following. a 2(x■+■3) d -(x■+■3) g 3x(5x■-■2) j 2x2(2x■-■3)

c f i l

3(7■-■x) 2x(x■-■4) 2x(4x■+■1) 5x2(3x■+■4)

2 We1, 2 ■Expand■each■of■the■following. a (x■+■3)(x■-■4) b (x■+■1)(x■-■3) d (x■-■1)(x■-■5) e (2■-■x)(x■+■3) g (2x■-■3)(x■-■7) h (x■-■1)(3x■+■2) j (3■-■2x)(7■-■x) k (5■-■2x)(3■+■4x)

c f i l

(x■-■7)(x■+■2) (x■-■4)(x■-■2) (3x■-■1)(2x■-■5) (11■-■3x)(10■+■7x)

3 We3 ■Expand■each■of■the■following. a 2(x■+■1)(x■-■3) b 4(2x■+■1)(x■-■4) d 2x(x■-■1)(x■+■1) e 3x(x■-■5)(x■+■5) g -2x(3■-■x)(x■-■3) h -5x(2■-■x)(x■-■4)

c -2(x■+■1)(x■-■7) f 6x(x■-■3)(x■+■3) i 6x(x■+■5)(4■-■x)

4 Expand■each■of■the■following. a (x■-■1)(x■+■1)(x■+■2) d (x■-■1)(x■-■2)(x■-■3)

c (x■-■5)(x■+■1)(x■-■1) f (3x■+■1)(2x■-■1)(x■-■1)

b e h k

4(x■-■5) x(x■+■2) 5x(2■-■3x) 3x2(2x■-■1)

b (x■-■3)(x■-■1)(x■+■2) e (2x■-■1)(x■+■1)(x■-■4)

5 Expand■each■of■the■following■and■simplify. a (x■+■2)(x■-■1)■-■2x c (2x■-■3)(x■+■1)■+■(3x■+■1)(x■-■2) e (x■+■1)(x■-■7)■-■(x■+■2)(x■-■3) g (x■-■3)(x■+■1)■+■ 3x

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6 mC ■a■ ■(3x■-■1)(2x■+■4)■expands■to: A 6x2■+■10x■-■4 B D 6x2■-■10x■-■4 E b -2x(x■-■1)(x■+■3)■expands■to: A x2■+■2x■-■3 B D -2x3■+■4x2■-■6x E

b 3x■-■(2x■-■5)(x■+■2) d (3■-■2x)(2x■-■1)■+■(4x■-■5)(x■+■4) f (x■-■2)(x■-■5)■-■(x■-■1)(x■-■4) h ( 2■- 3x)( 3■+■2x)■-■ 5x

5x2■-■24x■+■3 6x2■-■4

C 3x2■+■2x■-■4

-2x2■-■4x■+■6 -2x3■-■3

C -2x3■-■4x2■+■6x

7 mC ■The■expression■(x■-■1)(x■-■3)(x■+■2)■is■not■the■same■as: A (x■–■3)(x■–■1)(x■+■2) B (x■+■3)(x■–■1)(x■–■2) C (x■–■1)(x■+■2)(x■–■3) D (x■+■2)(x■–■1)(x■–■3) E (x■–■3)(x■+■2)(x■–■1) 8 We4a ■Expand■and■simplify■each■of■the■following. a (x■-■1)2 b (x■+■2)2 2 d (4■+■x) e (7■-■x)2 2 g (3x■-■1) h (12x■-■3)2 2 j (2■-■3x) k (5■-■4x)2

c f i l

9 We4b ■Expand■and■simplify■each■of■the■following. a 2(x■-■3)2 b 4(x■-■7)2 2 d -(2x■+■3) e -(7x■-■1)2 2 g -3(2■-■9x) h -5(3■-■11x)2

c 3(x■+■1)2 f 2(2x■-■3)2 i -4(2x■+■1)2

10 We5 ■Expand■and■simplify■each■of■the■following. a (x■+■7)(x■-■7) b (x■+■9)(x■-■9) d (x■-■1)(x■+■1) e (2x■-■3)(2x■+■3) g (7■-■x)(7■+■x) h (8■+■x)(8■-■x)

c (x■-■5)(x■+■5) f (3x■-■1)(3x■+■1) i (3■-■2x)(3■+■2x)

(x■+■5)2 (12■-■x)2 (5x■+■2)2 (1■-■5x)2

Chapter 7 Quadratic expressions

225

number AnD AlgebrA • PAtterns AnD AlgebrA unDerstAnDing 11 The■length■of■the■side■of■a■rectangle■is■(x■+■1)■cm■and■the■width■is■(x■-■3)■■cm. a Find■an■expression■for■the■area■of■the■rectangle. b Simplify■the■expression■by■expanding. c If■x■=■5■■cm,■fi■nd■the■dimensions■of■the■rectangle■and,■hence,■its■area. 12 Chickens■are■kept■in■a■square■enclosure■with■sides■measuring■x■m.■The■number■of■chickens■is■

increasing■and■so■the■size■of■the■enclosure■is■to■have■1■metre■added■to■one■side■and■2■metres■to■ the■adjacent■side. a Draw■a■diagram■of■the■original■enclosure. b Add■to■the■fi■rst■diagram■or■draw■another■one■to■show■the■new■enclosure.■Mark■the■lengths■ on■each■side■on■your■diagram. c Find■an■expression■for■the■area■of■the■new■enclosure. d Simplify■the■expression■by■removing■the■brackets. e If■the■original■enclosure■had■sides■of■2■metres,■fi■nd■the■area■of■the■original■square■and■ then■the■area■of■the■new■enclosure. 13 A■jewellery■box■has■a■square■base■with■sides■measuring■(x■+■2)■■cm■and■is■5■■cm■high. a Write■an■expression■for■the■area■of■the■base■of■the■box. b Write■an■expression■for■the■volume■of■the■box.■ (V■=■area■of■base■ì■height) c Simplify■the■expression■by■expanding■the■brackets. d If■x■=■8■■cm,■fi■nd■the■volume■of■the■box■in■cm3. e Find■the■area■of■the■lid■of■the■box■and,■hence,■fi■nd■how■many■ 1-cm■square■tiles■could■be■inlaid■in■the■lid. 14 In■redesigning■their■courtyard,■Linda■and■Finn■removed■a■section■of■ paving■as■shown■in■the■diagram■below. x Section of paving removed

x+1

x+3 3x + 5 a Write■down■an■expression,■in■terms■of■x,■for■the■area■of■the■section■of■paving■that■is■ b c

removed.■Write■your■answer■in■expanded■form.■ Find■the■area■of■paving■that■remains■in■terms■of■x. A■circular■fountain■is■to■be■placed■in■the■section■created■from■removing■the■paving.■ i Write■down■the■largest■possible■radius,■in■terms■of■x,■for■the■circular■fountain. ii If■the■area■of■the■circular■fountain■is■1.77■m2,■determine■the■value■of■x.■Write■your■ answer■to■the■nearest■centimetre.

reAsoning 15 Find■the■values■a,■b,■c,■d and■e■that■make■the■following■identity■true.

4x2(x■−■2)(x■+■3)■+■3■=■ax4■+■bx3■+■cx2■+■dx■+ e 16 A■tissue■box■has■the■side■lengths■shown■at■right.■ a Write■an■expression■in■factorised■form■for■the■ volume■of■the■box. b Find■the■volume■when■x is■5■■cm. c What■is■the■volume■when x■=■1 226

maths Quest 10 for the Australian Curriculum

(x - 3) cm (x + 2) cm

(x - 1) cm

number AND algebra • Patterns and algebra

Is it possible for x to be zero? Justify your answer. What is the smallest number x can be so that all the measurements are feasible? Find the volume of the box when x is 4  cm. If the volume of the box is 120  cm3, find the value reflection    of x. (Hint: Try substituting values of x into the expression until you find the correct one.) Why does the Difference of Two h Write the expression for volume in expanded form. Squares rule have that name? Use this to calculate the volume when x = 6  cm. d e f g

7B

Factorising expressions with three terms ■■ ■■

An expression with three terms is called a trinomial. Quadratic trinomials can be written in the form ax2 + bx + c where the highest power is a squared term.

Factorising ax2 + bx + c when a = 1 ■■ ■■ ■■ ■■ ■■ ■■ ■■

■■

+ 3 x Factorising is the inverse of expanding, so the area model for expanding a binomial can be considered in reverse. 3x x2 x Looking in reverse, it can be seen that ■ x2 + 3x + 2x + 6 = (x + 3)(x + 2). + This means that the factorised form of x2 + 5x + 6 is (x + 3)(x + 2). 2x 6 2 The technique is to find two factors of 6 which add to 5 (3 and 2). Note that there are other factors of 6 (6 and 1, -3 and -2, -6 and -1), but there is only one pair of factors which adds to 5. Obviously, it is not necessary to draw an area model to factorise every trinomial. The following method works for every possible trinomial (when a = 1) that can be factorised. Step 1.  Place the trinomial in the correct order or standard form x2 + bx + c. Step 2.  Find all the factor pairs of c (the constant term). Step 3.  Identify the factor pair whose sum equals b. Step 4.  Express the trinomial x2 + bx + c in factor form; that is, (x + __)(x + __). Remember to first check for and take out any common factors. As always, you can check your answer by expanding the brackets to re-create the original expression.

Worked Example 6

Factorise each of the following. a x2 - x - 20 b -2x2 + 16x - 14 Think a

Write a x2 - x - 20

1

Write the expression.

2

Check for a common factor (none).

3

Identify the factors of x2 as x and x and find the factors of the last term (-20) which add to equal the coefficient of the middle term (-1).

-20: 5 + -4 = 1, -5 + 4 = -1

4

Write the expression and its factorised form.

x2 - x - 20 = (x - 5)(x + 4)

Chapter 7 Quadratic expressions

227

number AND algebra • Patterns and algebra b

b -2x2 + 16x - 14

1

Write the expression.

2

Check for a common factor. (-2 can be taken out.)

= -2(x2 - 8x + 7)

3

Identify the factors of x2 as x and x and find the factors of the last term (7) which add to equal the coefficient of the middle term (-8).

7: 1 + 7 = 8, -1 + -7 = -8

4

Write the expression and its factorised form.

-2(x2 - 8x + 7) = -2(x - 1)(x - 7)

Factorising ax2 + bx + c when a ò 1 ■■

■■ ■■

If the coefficient of x2 is not 1, and there is not a common factor, we factorise the expression by splitting up the x-term so that the expression can then be factorised by grouping. A quadratic trinomial of the form ax2 + bx + c is broken up into four terms by finding two numbers that multiply to give ac and add to give b. Alternatively, the cross-product method could be used. This is illustrated in the following worked example.

Worked Example 7

Factorise 10x2 - x - 2 by: a grouping b the cross-product method Think a

b

a 10x2 - x - 2

1

Write the expression and check for a common factor (none).

2

Find the factor pair of ac (-20) which gives a sum of b (-1).

-20: 2 + -10 = -8, -4 + 5 = 1, 4 + -5 = -1

3

Rewrite the expression by breaking the x-term into two terms using the factor pair from step 2.

10x2 - x - 2 = 10x2 + 4x - 5x - 2

4

Factorise by grouping terms.

= 2x(5x + 2) - (5x + 2) = (5x + 2)(2x - 1)

1

Write the expression.

2

List the factor pairs of the first term (10x2) and the last term (-2). Note: There are four possible factor pairs for the first term: x and 10x; 2x and 5x; -x and -10x; -2x and -5x. In some cases, all variations will have to be tested to obtain the required middle term.

3

228

Write

Calculate the sum of each crossproduct pair until you find the combination that produces the middle term from the original expression (shown in red).

Maths Quest 10 for the Australian Curriculum

b 10x2 - x - 2

Factors of Sum the 10x2 -2 cross-products 2x 5x

-2 1

2x 5x

Result

2x 5x

-2 1

2x - 10x = -8x

1 -2

2x 5x

1 -2

-4x + 5x = x

2x 5x

2 -1

2x 5x

2 -1

-2x + 10x = 8x

2x 5x

-1 2

2x 5x

-1 2

4x – 5x = -x

number AnD AlgebrA • PAtterns AnD AlgebrA

4

10x2■-■x■-■2■=■(2x■-■1)(5x■+■2)

Express■the■trinomial■in■factor■form. Note:■The■fi■rst■pair■of■brackets■contain■ the■fi■rst■row■entries■and■the■second■pair■ of■brackets■contain■the■second■row■ entries■that■produce■the■middle■term■ from■the■original■expression.

remember

1.■ When■factorising■any■expression,■look■for■a■common■factor■fi■rst. 2.■ To■factorise■a■quadratic■trinomial■when■the■coeffi■cient■of■x2■is■1■(that■is,■x2■+■bx■+■c): (a)■ identify■the■factor■pair■of■c■whose■sum■is■equal■to■b (b)■express■the■trinomial■in■factor■form,■x2■+■bx■+■c■=■(x■+■__)(x■+■__). 3.■ To■factorise■a■quadratic■trinomial■when■the■coeffi■cient■of■x2■is■not■1■(that■is,■ ax2■+■bx■+■c■where■a■ò■1): (a)■ identify■the■factor■pair■of■ac■whose■sum■is■equal■to■b (b)■rewrite■the■expression■by■breaking■the■x-term■into■two■terms■using■the■factor■pair■ from■the■previous■step (c)■ factorise■the■resulting■expression■by■grouping. Alternatively,■the■cross-product■method■could■be■used■to■solve■any■quadratic■trinomial. 4.■ All■factorisations■can■be■checked■by■expanding■to■re-create■the■original■expression. exerCise

7b inDiViDuAl PAthWAys eBook plus

Activity 7-B-1

Introducing quadratic factorisation doc-5050 Activity 7-B-2

Practising quadratic factorisation doc-5051 Activity 7-B-3

Tricky quadratic factorisation doc-5052

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SkillSHEET 7.5 doc-5250

Factorising expressions with three terms FluenCy 1 We6a ■Factorise■each■of■the■following. a x2■+■3x■+■2 b x2■+■4x■+■3 d x2■+■8x■+■16 e x2■-■2x■-■3 g x2■-■11x■-■12 h x2■-■4x■-■12 2 j x ■+■4x■-■5 k x2■+■6x■-■7 2 m x ■-■4x■+■3 n x2■-■9x■+■20

c f i l o

x2■+■10x■+■16 x2■-■3x■-■4 x2■+■3x■-■4 x2■+■3x■-■10 x2■+■9x■-■70

2 We6b ■Factorise■each■of■the■following. a -2x2■-■20x■-■18■ b -3x2■-■9x■-■6 2 d -x ■-■11x■-■10 e -x2■-■7x■-■10■ 2 g -x ■-■7x■-■12 h -x2■-■8x■-■12 2 j 3x ■+■33x■+■30 k 5x2■+■105x■+■100

c f i l

-x2■-■3x■-■2 -x2■-■13x■-■12 2x2■+■14x■+■20 5x2■+■45x■+■100

3 Factorise■each■of■the■following. a a2■-■6a■-■7 d m2■+■2m■-■15 g k2■+■22k■+■57 j v2■-■28v■+■75

c f i l

b2■+■5b■+■4 c2■+■13c■-■48 g2■-■g■-■72 x2■-■19x■+■60

b e h k

t2■-■6t■+■8 p2■-■13p■-■48 s2■-■16s■-■57 x2■+■14x■-■32

4 mC ■a■ To■factorise■-14x2■-■49x■+■21,■the■fi■rst■step■is■to: A fi■nd■factors■of■14■and■21■that■will■add■to■-49 B take■out■14■as■a■common■factor C take■out■-7■as■a■common■factor D fi■nd■factors■of■14■and■-49■that■will■add■to■make■21 E take■out■-14■as■a■common■factor Chapter 7 Quadratic expressions

229

number AND algebra • Patterns and algebra b The expression 42x2 - 9x - 6 can be completely factorised to: A (6x - 3)(7x + 2) B 3(2x - 1)(7x + 2) C (2x - 1)(21x + 6) D 3(2x + 1)(7x - 2) E 42(x - 3)(x + 2) 5   MC  When factorised, (x + 2)2 – (y + 3)2 equals: A (x + y – 2)(x + y + 2) B (x – y – 1)(x + y – 1) D (x – y + 1)(x + y + 5) E (x + y – 1)(x + y + 2)

C (x – y – 1)(x + y + 5)

6 Which method of factorising is the most appropriate for each of the following expressions? a Factorising using common factors b Factorising using the difference of two squares rule c Factorising by grouping d Factorising quadratic trinomials   i 5x2 + 3x − 2 ii 25a2 − b2 iii x2 + 6x + 9 – y2 2 2 2 iv 16x – 25x v 4x – 4y + x – y vi x2 + 14x - 32 7   WE 7  Factorise each of the following using an appropriate method. a 2x2 + 5x + 2 b 2x2 - 3x + 1 c 2 2 d 4x + 4x - 3 e 2x - 9x - 35 f g 6x2 - 17x + 7 h 12x2 - 13x - 14 i j 20x2 + 3x - 2 k 12x2 + 5x - 2 l

4x2 - 17x - 15 3x2 + 10x + 3 10x2 - 9x - 9 15x2 + x - 2 8 Factorise each of the following, remembering to look for a common factor first. a 4x2 + 2x - 6 b 9x2 - 60x - 21 2 c 72x + 12x - 12 d -18x2 + 3x + 3 2 e -60x + 150x + 90 f 24ax2 + 18ax - 105a 2 g -8x + 22x - 12 h -10x2 + 31x + 14 2 i -24x + 35x - 4 j -12x2 - 2xy + 2y2 2 2 k -30x + 85xy + 70y l -600x2 - 780xy - 252y2 Understanding 9 Consider the expression (x - 1)2 + 5(x - 1) - 6. a Substitute w = x - 1 in this expression. b Factorise the resulting quadratic. c Replace w with x - 1 and simplify each factor. This is the factorised form of the original

expression. 10 Use the method outlined in question 9 to factorise each of the following expressions. a (x + 1)2 + 3(x + 1) - 4 b (x + 2)2 + (x + 2) - 6 c (x - 3)2 + 4(x - 3) + 4 d (x + 3)2 + 8(x + 3) + 12 e (x - 7)2 - 7(x - 7) - 8 f (x - 5)2 - 3(x - 5) - 10 11 Factorise x2 + x - 0.75. 12 Students decide to make Valentine’s Day cards. The total area of each card is equal to (x2 - 4x - 5) cm2. a Factorise the expression to find the dimensions of the cards in terms of x. b Write down the length of the shorter side in terms of x. c If the shorter sides of a card are 10  cm in length and the longer sides are 16  cm in length, find the value of x. d Find the area of the card proposed in part c. Happy e If the students want to make 3000 Valentine’s Day cards, how Valentine's much cardboard will be required? Give your answer in terms Day of x. 230

Maths Quest 10 for the Australian Curriculum

number AnD AlgebrA • PAtterns AnD AlgebrA 13 The■area■of■a■rectangular■playground■is■given■by■the■general■expression■(6x2■+■11x■+■3)■m2■

where■x■is■a■positive■whole■number. a Find■the■length■and■width■of■the■play■ground■in■terms■of■x. b Write■an■expression■for■the■perimeter■of■the■playground. c If■the■perimeter■of■a■particular■playground■is■88■metres,■fi■nd x. reAsoning 14 Cameron■wants■to■build■an■in-ground■‘endless’■pool.■Basic■models■have■a■depth■of■2■metres■

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Digital doc

WorkSHEET 7.1 doc-5251

7C

and■a■length■triple■the■width.■A■spa■will■also■be■attached■to■the■end■of■the■pool. a The■pool■needs■to■be■tiled.■Write■an■expression■for■the■surface■area■of■the■empty■pool■ (that■is,■the■fl■oor■and■walls■only). b The■spa■needs■an■additional■16■m2■of■tiles.■Write■an■expression■for■the■total■area■of■tiles■ needed■for■both■the■pool■and■the■spa. c Factorise■this■expression. d Cameron■decides■to■use■ tiles■that■are■selling■at■a■ discount■price,■but■there■ are■only■280■m2■of■the■tile■ available.■Find■the■maximum■ dimensions■of■the■pool■if■the■ width■is■in■whole■metres. e What■area■of■tiles■is■actually■ needed■to■construct■the■ pool? f What■volume■of■water■can■the■ pool■hold? 15 A■quilt■is■made■by■repeating■the■patch■at■right. b y y The■letters■indicate■the■colours■of■fabric■that■make■up■the■patch■—■ yellow,■black■and■white.■The■yellow■and■white■pieces■are■square■and■ b b w the■black■pieces■are■rectangular.■Many■of■these■patches■are■sewn■ together■in■rows■and■columns■to■make■a■pattern.■The■fi■nished■quilt,■ b y y made■from■100■patches,■is■a■square■with■an■area■of■1.44■m2.■ ■ An■interesting■feature■is■created■when■the■blocks■are■sewn■together:■ each■colour■forms■a■shape.■The■shape■and■its■area■are■exactly■the■same■ for■each■colour.■(The■feature■appears■throughout■the■quilt,■except■at■the■edges.) a Determine■the■size■of■each■yellow,■black■and■white■ fabric■piece■in■a■patch. reFleCtion    b How■much■(in■m2)■of■each■of■the■different■colours■ In your own words, describe would■be■needed■to■construct■the■quilt?■(Ignore■ how you would factorise a seam■allowances.) quadratic trinomial. c Sketch■a■section■of■the■fi■nished■product.

Factorising expressions with two or four terms ■■ ■■ ■■

Factorising■to■the■inverse■or■opposite■of■expanding. The■factorised■form■shows■the■expression■as■a■product■of■factors,■while■the■expanded■form■ shows■the■expression■as■a■sum■or■difference■of■terms. The■most■straightforward■type■of■factorisation■is■where■a■common■factor■is■removed■from■ the■expression.■Once■this■has■been■done,■we■need■to■consider■the■number■of■terms■in■the■ expression■to■see■whether■other■types■of■factorisation■may■be■possible■to■further■simplify. Chapter 7 Quadratic expressions

231

number AND algebra • Patterns and algebra

Factorising expressions of the type a2 - b2 ■■ ■■

Recall the area model for the difference of two squares rule. When factorising an algebraic expression of the type a2 - b2, follow these steps. 1. Look for a common factor first. If there is one, factorise by taking it out. 2. Rewrite the expression showing the two squares and identifying the a and b parts of the expression. 3. Factorise, using the rule a2 - b2 = (a + b)(a - b).

Worked Example 8

Factorise each of the following. a 4x2 - 9 b 7x2 - 448 c x2 - 17 Think a

b

c

Write a 4x2 - 9

1

Write the expression.

2

Check for a common factor and write as two perfect squares.

= (2x)2 - 32

3

Factorise using a2 - b2 = (a + b)(a - b).

= (2x + 3)(2x - 3)

1

Write the expression.

2

Check for a common factor and take it out.

= 7(x2 - 64)

3

Write the terms in the bracket as two perfect squares.

= 7(x2 - 82)

4

Factorise using a2 - b2 = (a + b)(a - b).

= 7(x + 8)(x - 8)

1

Write the expression.

2

Check for a common factor and write as two perfect squares. In this case, a surd needs to be used to rewrite 17 as a perfect square.

= x2 - ( 17 )2

3

Factorise using a2 - b2 = (a + b)(a - b).

= (x + 17 )(x - 17 )

b 7x2 - 448

c

x2 - 17

Factorising expressions with four terms ■■ ■■

If there are four terms to be factorised, look for a common factor first. Then group the terms in pairs and look for a common factor in each pair. It may be that a new common factor emerges as a bracket (common binomial factor).

Worked Example 9

Factorise each of the following. a x - 4y + mx - 4my b x2 + 3x - y2 + 3y Think a

232

1

Write

Write the expression and look for a common factor. (There isn’t one.)

Maths Quest 10 for the Australian Curriculum

a x - 4y + mx - 4my

number AND algebra • Patterns and algebra

b

2

Group the terms so that those with common factors are next to each other.

= (x - 4y) + (mx - 4my)

3

Take out a common factor from each group (it may be 1).

= 1(x - 4y) + m(x - 4y)

4

Factorise by taking out a common binomial factor. The factor (x - 4y) is common to both groups.

= (x - 4y)(1 + m)

1

Write the expression and look for a common factor.

b x2 + 3x - y2 + 3y

2

Group the terms so that those with common factors are next to each other.

= (x2 - y2) + (3x + 3y)

3

Factorise each group.

= (x + y)(x - y) + 3(x + y)

4

Factorise by taking out a common binomial factor. The factor (x + y) is common to both groups.

= (x + y)(x - y + 3)

■■ ■■

In Worked example 9, grouping occurred in pairs. This is known as grouping ‘two and two’. Now we will look at grouping a different combination, known as grouping ‘three and one’.

Worked Example 10

Factorise the following expression: x2 + 12x + 36 - y2. Think

Write

1

Write the expression and look for a common factor.

x2 + 12x + 36 - y2

2

Group the terms so that those that can be factorised are next to each other.

= (x2 + 12x + 36) - y2

3

Factorise the quadratic trinomial. This is the form of a perfect square.

= (x + 6)(x + 6) - y2 = (x + 6)2 - y2

4

Factorise the expression using a2 - b2 = (a + b)(a - b).

= (x + 6 + y)(x + 6 - y)

remember

1. To factorise an expression with two terms: (a) take out any common factors (b) check whether the difference of two squares rule can be used. 2. To factorise an expression with four terms: (a) take out any common factors (b) check whether they can be grouped using the ‘two and two’ method or the ‘three and one’ method. Chapter 7 Quadratic expressions

233

number AnD AlgebrA • PAtterns AnD AlgebrA

exerCise

7C inDiViDuAl PAthWAys eBook plus

Activity 7-C-1

Factorising expressions with two or four terms doc-5047 Activity 7-C-2

More factorising expressions with two or four terms doc-5048 Activity 7-C-3

Advanced factorising expressions with two or four terms doc-5049

Factorising expressions with two or four terms FluenCy 1 Factorise■each■of■the■following■by■taking■out■a■common■factor. a x2■+■3x b x2■-■4x 2 d 4x ■+■16x e 9x2■-■3x 2 g 12x■-■3x h 8x■-■12x2

c 3x2■-■6x f 8x■-■8x2 i 8x2■-■11x

2 Factorise■each■of■the■following■by■taking■out■a■common■binomial■factor. a 3x(x■-■2)■+■2(x■-■2) b 5(x■+■3)■-■2x(x■+■3) c (x■-■1)2■+■6(x■-■1) d (x■+■1)2■-■2(x■+■1) e (x■+■4)(x■-■4)■+■2(x■+■4) f 7(x■-■3)■-■(x■+■3)(x■-■3) 3 We8a ■Factorise■each■of■the■following. a x2■-■1 b x2■-■9 2 d x ■-■100 e y2■-■k2 2 g 16a ■-■49 h 25p2■-■36q2

c x2■-■25 f 4x2■-■9y2 i 1■-■100d 2

4 We8b ■Factorise■each■of■the■following. a 4x2■-■4 b 5x2■-■80 2 2 d 2b ■-■8d e 100x2■-■1600 2 g 4px ■-■256p h 36x2■-■16

c ax2■-■9a f 3ax2■-■147a i 108■-■3x2

5 mC ■a■ ■If■the■factorised■expression■is■(x■+■7)(x■-■7),■then■the■original■expression■must■have■

been:

A x2■-■7 D x2■+■49

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Digital doc

been:

x2 3 A − 4 5

 x 3  x 3 ■then■the■original■expression■must■have■ + − 4 5   4 5 

6

Digital doc

SkillSHEET 7.4 doc-5247

7

8

9

x2 ( 3 ) − C 4 ( )2 5 2

x2 9 B − 16 25

x2 ( 3 ) E − 16 ( )2 5 2 2 c The■factorised■form■of■64x ■-■9y ■is: A (64x■+■9y)(64x■-■9y) B (8x■+■3y)(8x■-■3y) C (8x■-■3y)(8x■-■3y) D (8x■+■3y)(8x■+■3y) E (16x■+■3y)(16x■-■3y) mC ■Which■of■the■following■expressions■would■be■factorised■by■grouping■‘two■and■two’? A x2■–■a2■+■12a■−■36 B x2■−7x −10 2■ C 2x −■6x■−■xy +■3y D (s■–■5)2■–■25(s■+■3)2 E (r■+■5)■–■(r■+■3)(r■+■5) We8c ■Factorise■each■of■the■following. a x2■-■11 b x2■-■7 c x2■-■15 2 2 d 4x ■-■13 e 9x ■-■19 f 3x2■-■66 2 2 g 5x ■-■15 h 2x ■-■4 i 12x2■-■36 Factorise■each■of■the■following■expressions. a (x■-■1)2■-■4 b (x■+■1)2■-■25 c (x■-■2)2■-■9 d (x■+■3)2■-■16 e 49■-■(x■+■1)2 f 36■-■(x■-■4)2 g (x■-■1)2■-■(x■-■5)2 h 4(x■+■2)2■-■9(x■-■1)2 i 25(x■-■2)2■-■16(x■+■3)2 We9a ■Factorise■each■of■the■following. a x■-■2y■+■ax■-■2ay b 2x■+■ax■+■2y■+■ay c ax■-■ay■+■bx■-■by d 4x■+■4y■+■xz■+■yz e ef■-■2e■+■3f■-■6 f mn■-■7m■+■n■-■7 g 6rt■-■3st■+■6ru■-■3su h 7mn■-■21n■+■35m■-■105 x2 9 D − 4 25

234

C x2■-■49

b If■the■factorised■expression■is■  

SkillSHEET 7.3 doc-5246

eBook plus

B x2■+■7 E x2■-■14x■+■49

maths Quest 10 for the Australian Curriculum

2

number AND algebra • Patterns and algebra i 64 - 8j + 16k - 2jk k 5x2 + 10x + x2y + 2xy

j l

3a2 - a2b + 3ac - abc 2m2 - m2n + 2mn - mn2

10 Factorise each of the following. a xy + 7x - 2y - 14 c pq + 5p - 3q - 15 e a2b - cd - bc + a2d

b mn + 2n - 3m - 6 d s2 + 3s - 4st - 12t f xy - z - 5z2 + 5xyz

11   WE 9b  Factorise each of the following. a a2 - b2 + 4a - 4b c m2 - n2 + lm + ln e 5p - 10pq + 1 - 4q2

b p2 - q2 - 3p + 3q d 7x + 7y + x2 - y2 f 49g2 - 36h2 - 28g - 24h

12   WE 10  Factorise each of the following. a x2 + 14x + 49 - y2 c a2 - 22a + 121 - b2 e 25p2 - 40p + 16 - 9t2

b x2 + 20x + 100 - y2 d 9a2 + 12a + 4 - b2 f 36t2 - 12t + 1 - 5v

13   MC  a  In the expression 3(x - 2) + 4y(x - 2), the common binomial factor is: A 3 + 4y B 3 - 4y C x D -x + 2 E x - 2 b Which of the following terms is a perfect square? A 9 B (x + 1)(x - 1) C 3x2 2 D 5(a + b) E 25x c Which of the following expressions can be factorised using grouping? A x2 - y2 B 1 + 4y - 2xy + 4x2 C 3a2 + 8a + 4 2 2 D x + x + y - y E 2a + 4b - 6ab + 18 14   MC  When factorised, 6(a + b) − x(a + b) equals: A 6 – x(a + b) B (6 – x)(a + b) D (6 + x)(a – b) E (6 + x)(a + b)

C 6(a + b – x)

understanding 15 The area of a rectangle is (x2 - 25) cm2. a Factorise the expression. b Using the factors, find a possible length and width of the rectangle. c If x = 7  cm, find the dimensions of the rectangle. d Hence, find the area of the rectangle. e If x = 13  cm, how much bigger would the area of this rectangle be? Reasoning 16 A circular garden of diameter 2r m

is to have a gravel path laid around it. The path is to be 1  m wide. a Find the radius of the garden. b Find the radius of the circle that includes both the garden and the path. c Find the area of the garden in terms of r. d Find the area of the garden and path together in terms of r, using the formula for the area of a circle. e Write an equation to find the area of the path, then write your equation in fully factorised form. f If the radius of the garden is 5  m, use the answer to part e to find the area of the path, correct to 2 decimal places. Chapter 7 Quadratic expressions

235

number AnD AlgebrA • PAtterns AnD AlgebrA 17 A■roll■of■material■is■(x■+■2)■metres■wide.■Annie■buys■(x■+■3)■metres■and■Bronwyn■buys■

5■metres. a Write■an■expression,■in■terms■of■x,■for■the■area■of■ each■piece■of■material. b If■Annie■has■bought■more■material■than■Bronwyn,■ write■an■expression■for■how■much■more■she■has■than■ Bronwyn. c Factorise■and■simplify■this■expression. d Find■the■width■of■the■material■if■Annie■has■5■■m2■more■ than■Bronwyn. e How■much■material■does■each■person■have?■explain■ your■answer. reFleCtion 

 

What do you always check for first when factorising?

7D eBook plus

Interactivity Completing the square

int-2783

Factorising by completing the square A■quadratic■equation■can■be■written■in■general■form:■y■=■ax2■+■bx■+■c■and■in■turning■point■ form:■y■=■a(x■–■h)2■+■k. ■■ A■process■called■‘completing■the■square’■makes■it■possible■to■change■from■the■general■form■ to■the■turning■point■form.■This■process■is■useful■when■a■quadratic■equation■will■not■factorise■ easily■and■the■turning■point■and■the■x-intercepts■are■required. ■■ Completing■the■square■relies■on■the■knowledge■of■perfect■squares■and■the■difference■of■two■ squares.■Recall: ■■

Perfect■squares:■ ■ Difference■of■two■squares:■

(x■+■a)2■=■x2■+■2ax■+■a2 (x■-■a)2■=■x2■–■2ax■+■a2 x2■–■a2■=■(x■-■a)(x■+■a) x

Suppose■we■want■to■factorise■x2■+■4x■+■2.■There■are■no■factors■of■2■which■add■to■ give■4,■so■we■cannot■factorise■using■the■cross-product■method.■Instead,■the■ fi■rst■two■terms■can■be■used■to■‘build’■a■perfect■square.■ Start■with■a■square,■x2,■as■shown■at■right.

x2

x

x

2

x

x2

2x

2

2x

To■show■4x,■add■two■rectangles,■each■with■an■area■of■2x.

We■now■have■two■sides■of■a■square■with■side■length■(x■+■2). Two■units■with■a■value■of■1■are■added■to■make■x2■+■4x■+■2.■ There■are■2■units■missing. This■square,■then,■can■be■expressed■as:■x2■+■4x■+■2■=■(x■+■2)2■-■2 This■can■then■be■factorised■using■surds■and■the■difference■of■two■ squares■rule,■with■2■expressed■as■( 2 )2 .■ ■ ■ 236

(x■+■2)2■-■2■=■(x■+■2)2■-■( 2 )2 =■(x■+■2■-■ 2)(x■+■2■+■ 2)

maths Quest 10 for the Australian Curriculum

x

2

x

x2

2x

2

2x

1

1

number AND algebra • Patterns and algebra

■■

The factors include surds. This means that the expression has been factorised over the Real Number field. If the factors are not surds they have been factorised over the Rational Number field. To complete the square algebraically, we add the square of half the coefficient of x and, to compensate for adding this number, we need to subtract the same number to keep the equation equivalent. For example: x2 + 6x + 1 = x2 + 6x + (3)2 + 1 - (3)2 = (x + 3)2 + 1 - 9 = (x + 3)2 - 8



Using the difference of two squares rule, this factorises to: = (x + 3 - 8)(x + 3 + 8).

■■

Odd coefficients of x are more difficult to deal with; Fractions or decimals will need to be used.

Worked Example 11

Factorise each of the following by first completing the square. a  x2 - 8x + 5 b  x2 + 5x + 1 c  2x2 + 8x - 3 Think a

b

Write a x2 - 8x + 5

1

Write the expression.

2

Identify the coefficient of x, halve it and square the result.

3

Add the result of step 2 to the expression, placing it after the x-term. To balance the expression, we need to subtract the same amount as we have added.

= x2 - 8x + 16 + 5 - 16

4

Insert brackets around the first three terms to group them and then simplify the remaining terms.

= (x2 - 8x + 16) - 11

5

Factorise the first three terms to produce a perfect square.

= (x - 4)2 - 11

6

Rewrite the expression as the difference of two squares.

= (x - 4)2 - ( 11)2

7

Factorise using the difference of two squares rule.

= (x - 4 + 11)(x - 4 - 11)

1

Write the expression.

2

Identify the coefficient of x, halve it and square the result.

3

Add the result of step 2 to the expression, placing it after the x-term. To balance the expression, we need to subtract the same amount as we have added.

(

1 2

)

2

× −8 = (-4)2 = 16

b x2 + 5x + 1

(

1 2

) =( )

×5

2

5 2

= x 2 + 5x +

25 4

2

=

25 4

+1−

25 4

Chapter 7 Quadratic expressions

237

number AND algebra • Patterns and algebra

4

5

Insert brackets around the first three terms to group them and simplify the remaining terms. (Convert them to equivalent fractions.)

( = (x

Factorise the first three terms to produce a perfect square.

= x+ 2

) )−

25 = x 2 + 5 x + 25 + 44 − 4 4

+ 5x +

25 4

( )−

21 4

2

5

2

21 4

2

6

Rewrite the expression as the difference of two squares.

2 5   21   =x+  −  2   2 

7

Factorise using the difference of two squares rule.

 5 21   5 21  = x+ + x+ −   2 2  2 2    5 + 21   5 − 21  or  x + x+   2  2  

c

c

2x2 + 8x − 3

1

Write the expression.

2

Factorise the expression.

3

Identify the coefficient of x, halve it and square the result.

4

Add the result of step 3 to the expression, placing it after the x-term. To balance the expression, we need to subtract the same amount as we have added.

= 2 x 2 + 4 x + 4 − 32 − 4

5

Insert brackets around the first three terms to group them and simplify the remaining terms. (Convert them to equivalent fractions.)

= 2 ( x 2 + 4 x + 4) − 32 − 4

6

Factorise the first three terms to produce a perfect square.

7

Rewrite the expression as the difference of two squares.

8

Factorise using the difference of two squares rule. Leave the factor of 2 outside the difference of two squares expression.

9

Rationalise the surd denominator.

(

= 2 x 2 + 4 x − 32

(

1 2

×4

)

2

)

= (2)2 = 4

(

)

( ) = 2 ( ( x + 4 x + 4) − ) = 2 ( ( x + 2) − ) 11 2

2

2

 = 2  ( x + 2)2 − 

(

= 2 ( x + 2) −

11 2

( )  )((x + 2) + ) 11 2

11 2

2

11 2

 22   22  = 2  ( x + 2) −   ( x + 2) + 2  2     4 − 22   4 + 22  or 2  x + x+   2  2  

■■ ■■

238

Remember that you can expand the brackets to check your answer. If the coefficient of x2 ò 1, factorise the expression before completing the square.

Maths Quest 10 for the Australian Curriculum

number AnD AlgebrA • PAtterns AnD AlgebrA

remember

1.■ If■a■quadratic■trinomial■cannot■be■factorised■by■fi■nding■an■integer■factor■pair,■then■ factorise■using■the■completing■the■square■method: (a)■ if■possible,■take■out■a■common■factor■and■write■it■outside■the■brackets (b)■halve■the■value■of■the■coeffi■cient■of■the■x-term■and■square■the■result (c)■ add■this■number■to■the■expression,■writing■it■after■the■x-term.■Balance■the■ expression■by■also■inserting■the■necessary■subtraction. (d)■factorise■the■fi■rst■three■terms■as■a■perfect■square■and■then■simplify■the■remaining■ terms (e)■ rewrite■the■expression■as■the■difference■of■two■squares (f)■ factorise■using■the■difference■of■two■squares■rule. 2.■ All■factorisations■can■be■checked■by■expanding■to■the■original■expression.

exerCise

7D inDiViDuAl PAthWAys eBook plus

Activity 7-D-1

Introducing completing the square doc-5053 Activity 7-D-2

Practising completing the square doc-5054 Activity 7-D-3

Completing the square doc-5055

Factorising by completing the square FluenCy 1 Complete■the■square■for■each■of■the■following■expressions. a x2■+■10x b x2■+■6x 2 c x ■-■4x d x2■+■16x 2 e x ■-■20x f x2■+■8x 2 g x ■-■14x h x2■+■50x 2 i x ■-■2x 2 We11a ■Factorise■each■of■the■following■by■fi■rst■completing■the■square. a x2■-■4x■-■7 b x2■+■2x■-■2 2 c x ■-■10x■+■12 d x2■+■6x■-■10 2 e x ■+■16x■-■1 f x2■-■14x■+■43 2 g x ■+■8x■+■9 h x2■-■4x■-■13 2 i x ■-■12x■+■25 3 We11b ■Factorise■each■of■the■following■by■fi■rst■completing■the■square. a x2■-■x■-■1 b x2■-■3x■-■3 2 c x ■+■x■-■5 d x2■+■3x■-■1 2 e x ■+■5x■+■2 f x2■+■5x■-■2 2 g x ■-■7x■-■1 h x2■-■9x■+■13 2 i x ■-■x■-■3 4 We11c ■Factorise■each■of■the■following■by■fi■rst■looking■for■a■common■factor■and■then■

completing■the■square. a 2x2■+■4x■-■4 c 5x2■+■30x■+■5 e 5x2■-■30x■+■10 g 3x2■+■30x■+■39 i 6x2■+■36x■-■30

b d f h

4x2■-■8x■-■20 3x2■-■12x■-■39 6x2■+■24x■-■6 2x2■-■8x■-■14

unDerstAnDing 5 Which■method■of■factorising■is■the■most■appropriate■for■each■of■the■following■expressions? a Factorising■using■common■factors b Factorising■using■the■difference■of■two■squares■rule c Factorising■by■grouping Chapter 7 Quadratic expressions

239

number AnD AlgebrA • PAtterns AnD AlgebrA

Factorising■quadratic■trinomials Completing■the■square i 3x2■−■8x■−■3 ii 49m2■−■16n2 iii x2■+■8x■+■4■–■y2 iv 7x2■–■28x v 6a –■6b■+■a2■–■b2 vi x2■+■x■–■5 vii (x■–■3)2■+■3(x■–■3)■–■10 viii x2■–■7x■–■1 6 mC ■a■ ■To■complete■the■square,■the■term■which■should■be■added■to■x2■+■4x■is: A 16 B 4 D 2 C 4x E 2x b To■factorise■the■expression■x2■-■3x■+■1,■the■term■that■must■be■both■added■and■ subtracted■is: d e

A 9

B 3

C 3x

D

7 mC ■The■factorised■form■of■x2■–■6x■+■2■is:

3 2

E

A (x■+■3■-■ 7)(x■+■3■+■ 7)

B (x■+■3■-■ 7 )(x■-■3■+■ 7 )■

C (x■-■3■-■ 7)(x■-■3■-■ 7)

D (x■-■3■-■ 7 )(x■+■3■+■ 7 )

9 4

E (x -■3■+■ 7)(x■-3■-■ 7) reAsoning 8 A■square■measuring■x■cm■in■side■length■has■a■cm■added■to■its■

length■and■b■cm■added■to■its■width.■The■resulting■rectangle■ has■an■area■of■(x2■+■6x■+■3)■cm2.■Find■the■values■of■a■and■b,■ correct■to■2■decimal■places.

7e 7e inDiViDuAl PAthWAys eBook plus

Activity 7-E-1

Mixed factorisation doc-5056 Activity 7-E-2

Harder mixed factorisation doc-5057 Activity 7-E-3

Advanced mixed factorisation doc-5058

240

 

Why is this method called completing the square?

mixed factorisation ■■

exerCise

reFleCtion 

The■following■exercise■will■help■you■to■practise■recognising■the■appropriate■method■of■ factorising■needed■for■a■given■expression.

mixed factorisation FluenCy

Factorise■each■of■the■following■expressions■in■questions■1–45. 2 x2■+■4x■+■4■-■9y2

3 x2■-■36

4 x2■-■49

6 15x■-■20y

1 3x■+■9

5 5x2■-■9x■-■2

5x2■-■80

7 5c■+■de■+■dc■+■5e

8

10 x2■+■x■-■12

11 mn■+■1■+■m■+■n

9 -x2■-■6x■-■5 12 x2■-■7

13

16x2■-■4x

14

5x2■+■60x■+■100

16

x2■-■8x■+■16■-■y2

17

4x2■+■8

19

x2■-■5

20 10mn■-■5n■+■10m■-■5

21 x2■+■6x■+■5

23 x2■-■4

24 -5a■+■bc■+■ac■-■5b

22 x2■-■10x■-■11

15 18■+■9x■-■6y■-■3xy 18 fg■+■2h■+■2g■+■f h

25 xy■-■1■+■x■-■y

26

3x2■+■5x■+■2

27 7x2■-■28

28 -4x2■-■28x■-■24

29 2p■-■rs■+■pr■-■2s

30 3x2■-■27

31 -3u +■tv■+■ut■-■3v

32

34 (x■-■1)2■-■4

35 (x■+■2)2■-■16

36 (2x■+■3)2■-■25

37 3(x■+■5)2■-■27

38 25■-■(x■-■2)2

39 4(3■-■x)2■-■16y2

40 (x■+■2y)2■-■(2x■+■y)2

41 (x■+■3)2■-■(x■+■1)2

42 (2x■-■3y)2■-■(x■-■y)2

43 (x■+■3)2■+■5(x■+■3)■+■4

44 (x■-■3)2■+■3(x■-■3)■-■10

45 2(x■+■1)2■+■5(x■+■1)■+■2

maths Quest 10 for the Australian Curriculum

x2■-■11

33 12x2■-■7x■+■1

number AnD AlgebrA • PAtterns AnD AlgebrA

eBook plus

Digital doc

SkillSHEET 7.6 doc-5252

unDerstAnDing 46 Consider■the■following■product■of■algebraic■fractions.■

x 2 + 3 x − 10 x2 − 4

×

x2 + 4x + 4 x2 − 2x − 8

a Factorise■the■expression■in■each■numerator■and■denominator. b Cancel■factors■common■to■both■the■numerator■and■the■denominator. c Simplify■the■expression■as■a■single■fraction. eBook plus

Digital doc

SkillSHEET 7.7 doc-5248

47 Use■the■procedure■in■question■46■to■factorise■and■simplify■each■of■the■following. a c

e

eBook plus

g

Digital doc

WorkSHEET 7.2 doc-5254

i

x2 − 4x + 3 x 2 − 4 x − 12 6 x − 12 2

x −4

×

×

x 2 + 5x + 6

b

x2 − 9

3x + 6 x ( x − 5)

x2 + 4x − 5 x2 + x − 2

÷

d

x 2 + 10 x + 25 x2 + 4x + 4

4 ab + 8a 5ac + 5a ÷ 2 (c − 3) c − 2c − 3 m 2 + 4 m + 4 − n2 2

4 m − 4 m − 15

÷

f

h

2m 2 + 4 m − 2mn 2

10 m + 15m

j

3 x 2 − 17 x + 10 6 x 2 + 5x − 6 6x2 − x − 2

×

×

x2 − 1 x2 − 6x + 5

2x2 + x − 1

2 x 2 + 3 x + 1 3 x 2 + 10 x − 8 x2 − 7x + 6 x2 + x − 2 p2 − 7 p p2 − 49

÷

÷

x 2 − x − 12 x2 − 2x − 8

p2 + p − 6 p2 + 14 p + 49

d 2 − 6d + 9 − 25e 2 2

4 d − 5d − 6

reFleCtion 

÷

4 d − 12 − 20 e 15d − 10

 

When an expression is fully factorised, what should it look like?

Chapter 7 Quadratic expressions

241

number AND algebra • Patterns and algebra

Summary Expanding algebraic expressions ■■

■■ ■■

When expanding an algebraic expression with: (a) one bracket — multiply each term inside the bracket by the term outside the bracket (b) two brackets — multiply the terms in order: First terms, Outer terms, Inner terms and then Last terms (FOIL) (c) a term outside the two brackets — expand the pair of brackets first, then multiply each term of the expanded expression by the term outside the brackets (d) three brackets — expand any two of the brackets and then multiply the expanded expression by the third bracket. Perfect squares rule: (a + b)2 = a2 + 2ab + b2 or (a - b)2 = a2 - 2ab + b2 Difference of two squares rule: (a + b)(a - b) = a2 - b2 Factorising expressions with three terms

■■ ■■

■■

■■

When factorising any expression, look for a common factor first. To factorise a quadratic trinomial when the coefficient of x2 is 1 (that is, x2 + bx + c): (a) identify the factor pair of c whose sum is equal to b (b) express the trinomial in factor form, x2 + bx + c = (x + __)(x + __). To factorise a quadratic trinomial when the coefficient of x2 is not 1 (that is, ax2 + bx + c where a ò 1): (a) identify the factor pair of ac whose sum is equal to b (b) rewrite the expression by breaking the x-term into two terms using the factor pair from the previous step (c) factorise the resulting expression by grouping. Alternatively, the cross-product method could be used to solve any quadratic trinomial. All factorisations can be checked by expanding to re-create the original expression. Factorising expressions with two or four terms

■■

■■

To factorise an expression with two terms: (a) take out any common factors (b) check whether the difference of two squares rule can be used. To factorise an expression with four terms: (a) take out any common factors (b) check whether they can be grouped using the ‘two and two’ method or the ‘three and one’ method. Factorising by completing the square

■■

■■

242

If a quadratic trinomial cannot be factorised by finding an integer factor pair, then factorise using the completing the square method: (a) if possible, take out a common factor and write it outside the brackets (b) halve the value of the coefficient of the x-term and square the result (c) add this number to the expression, writing it after the x-term. Balance the expression by also inserting the necessary subtraction. (d) factorise the first three terms as a perfect square and then simplify the remaining terms (e) rewrite the expression as the difference of two squares (f) factorise using the difference of two squares rule. All factorisations can be checked by expanding to the original expression.

Maths Quest 10 for the Australian Curriculum

number AnD AlgebrA • PAtterns AnD AlgebrA

■■

■■

This■method■will■not■always■give■a■difference■of■two■squares.■A■sum■of■two■squares■will■ sometimes■be■obtained.■If■the■coeffi■cient■of■x2■ò■1,■factorise■the■expression■before■completing■ the■square. This■method■can■be■used■to■convert■an■expression■into■turning■point■form■to■fi■nd■the■turning■ point■of■a■quadratic■graph.

MAPPING YOUR UNDERSTANDING

Homework Book

Using■terms■from■the■summary,■and■other■terms■if■you■wish,■construct■a■concept■map■that■ illustrates■your■understanding■of■the■key■concepts■covered■in■this■chapter.■Compare■your■ concept■map■with■the■one■that■you■created■in■What do you know?■on■page■219. Have■you■completed■the■two■Homework sheets,■the■Rich task■and■two■Code puzzles■in■ your■Maths Quest 10 Homework Book?

Chapter 7 Quadratic expressions

243

number AND algebra • Patterns and algebra

Chapter review Fluency 1 When expanded, -3x(x + 4)(5 - x) becomes: A -3x3 - 3x2 – 27x B -3x3 + 3x2 – 27x C 3x3 + 3x2 – 60x D -3x3 + 3x2 – 60x 3 2 E 3x - 3x – 60x 2 When expanded, (3x + 7)2 becomes: A 9x2 + 49 B 3x2 + 49 C 3x2 + 21x + 49 D 9x2 + 42x + 49 E 9x2 + 21x + 49 3 The factorised form of -3d 2 - 9d + 30 is: A -3(d - 5)(d - 2) B -3(d + 5)(d - 6) C -(3d + 5)(d - 2) D -(3d + 5)(d - 6) E -3(d + 5)(d - 2) 4 If the factorised expression is (2x – 5)(2x + 5), then

the original expression must have been: A 2x2 – 5 B 4x2 – 5 2 C 4x – 25 D 4x2 – 20x + 25 2 E 2x + 25 5 To factorise -5x2 - 45x + 100, the first step is to: A find factors of 5 and 100 that sum to -45 B take out 5 as a common factor C take out -5 as a common factor D find factors of 5 and -45 that will add to

make 100 E take out -5x as a common factor 6 To complete the square, the term which should be

added to x2 – 12x is: A 36 B -12 D -6 E -6x

C -12x

7 Which of the following is equivalent to 5x2 – 20x – 5? A 5(x – 2)2 B 5(x – 2)2 – 3 2 C 5(x – 2) – 15 D 5(x – 2)2 – 20 2 E 5(x – 2) – 25

(2x - 5)(x - 3) (4x - 1)(3x - 5) 3(x - 4)(2x + 7) (2x - 5)(x + 3)(x + 7) (x + 5)(x + 7) + (2x - 5)(x - 6) (x + 3)(5x - 1) - 2x

10 Expand and simplify each of the following. a (x - 7)2 b (2 - x)2 2 c (3x + 1) d -2(3x - 2)2 2 e -7(2x + 5) f -10(4x - 5)2 g (x + 9)(x - 9) h (3x - 1)(3x + 1) i (5 + 2x)(5 - 2x) 11 Factorise each of the following. a 2x2 - 8x b -4x2 + 12x c 3ax - 2ax2 d (x + 1)2 + (x + 1) e 3(2x - 5) - (2x - 5)2 f (x - 4)(x + 2) - (x - 4) 12 Factorise each of the following. a x2 - 16 b x2 - 25 2 c 2x - 72 d 3x2 - 27y2 2 2 e 4ax - 16ay f (x - 4)2 - 9 13 Factorise each of the following by grouping. a ax - ay + bx - by b 7x + ay + ax + 7y c xy + 2y + 5x + 10 d mn - q - 2q2 + 2mnq 2 e pq - 5r - r + 5pqr f uv - u + 9v - 9 g a2 - b2 + 5a - 5b h d2 - 4c2 - 3d + 6c 2 i 2 + 2m + 1 - m 14 Factorise each of the following by grouping. a 4x2 + 12x + 9 - y2 b 49a2 - 28a + 4 - 4b2 c 64s2 - 16s + 1 - 3t

following is incorrect? A The value of the constant is -15. B The coefficient of the x term is 2. C The coefficient of the x term is -8. D The coefficient of the x2 term is 1. E The expansion shows this to be a trinomial expression.

15 Factorise each of the following. a x2 + 10x + 9 b x2 - 11x + 18 2 c x - 4x - 21 d x2 + 3x - 28 2 e -x + 6x - 9 f 3x2 + 33x - 78 2 g -2x + 8x + 10 h -3x2 + 24x - 36 2 i 8x + 2x - 1 j 6x2 + x - 1 2 k 8x + 4x - 12 l 105x2 - 10x - 15 m -12x2 + 62x - 70 n -45x2 - 3x + 6 o -60x2 - 270x - 270

9 Expand each of the following and simplify where

16 Factorise each of the following by completing the

8 In the expanded form of (x − 3)(x + 5), which of the

necessary. a 3x(x - 4) b -7x(3x + 1) c (x - 7)(x + 1) 244

d e f g h i

Maths Quest 10 for the Australian Curriculum

square. a x2 + 6x + 1 c x2 + 4x - 2 e x2 + 7x - 1

b x2 - 10x - 3 d x2 - 5x + 2 f 2x2 + 18x - 2

number AnD AlgebrA • PAtterns AnD AlgebrA 17 Factorise■each■of■the■following■using■the■most■

appropriate■method. a 3x2■-■12x b x2■+■6x■+■2 c 4x2■-■25 d 2x2■+■9x■+■10 e 2ax■+■4x■+■3a■+■6 f -3x2■-■3x■+■18 18 First■factorise■then■simplify■each■of■the■following. x+4 2 x − 12 3 x + 6 7 x − 42 b a × × 5 x − 30 x +1 4 x − 24 6 x + 12 x2 − 4 x2 + 4x − 5 c × x 2 + 5x x 2 − 2 x − 8 Problem solVing 1 A■large■storage■box■has■a■square■base■with■sides■

measuring■(x■+■2)■cm■and■is■32■■cm■high.

a Write■an■expression■for■the■area■of■the■base■of■

the■box. b Write■an■expression■for■the■volume■of■the■box■

(V■=■area■of■base■ì■height). c Simplify■the■expression■by■expanding■the■

brackets.

d If■x■=■30■■cm,■fi■nd■the■volume■of■the■box■in■cm3.

2 A■section■of■garden■is■to■have■a■circular■pond■of■

radius■2r■with■a■2■■m■path■around■its■edge.■ a State■the■diameter■of■the■pond.■ b State■the■radius■of■the■pond■and■path. c State■the■area■of■the■pond. d State■the■area■of■the■pond■and■path. e Write■an■expression■to■fi■nd■the■area■of■the■path■ only■and■write■it■in■factorised■form. f If■the■radius■of■the■pond■is■3■metres,■fi■nd■the■ area■of■the■path. 3 In■order■to■make■the■most■of■the■space■available■for■ headlines■and■stories,■the■front■page■of■a■newspaper■ is■given■an■area■of■x2■–■5x■–■14■■cm2.■■ a Factorise■the■expression■to■fi■nd■the■dimensions■ of■the■paper■in■terms■of■x. b Write■down■the■length■of■the■shorter■side■in■ terms■of x. c If■the■shorter■side■of■the■front■page■is■28■■cm,■ fi■nd■the■value■of■x. d Find■the■area■of■this■particular■paper.■ 4 Here■is■a■well-known■puzzle. Let■a■=■b■=■1 Step■1:■Write■a■=■b. a■=■b Step■2:■■Multiply■both■ a2 = ab sides■by■a. Step■3:■■Subtract■b2■ a2 - b2 = ab - b2 from■both■sides. Step■4:■■Factorise. (a■+■b)(a■-■b)■=■b(a■-■b) Step■5:■■Simplify■by■ (a + b)■=■b dividing■by■ (a - b). Step■6:■■Substitute■ 1■+■1■=■1 a = b =■1. Where■is■the■error?■ Show■your■thinking. eBook plus

Interactivities

Test yourself Chapter 7 int-2846 Word search Chapter 7 int-2844 Crossword Chapter 7 int-2845

Chapter 7 Quadratic expressions

245

eBook plus

ACtiVities

Chapter opener Digital doc

•■ Hungry■brain■activity■(doc-5243)■(page 219) Are you ready?

(page 220) •■ SkillSHEET■7.1■(doc-5244):■Expanding■brackets■ •■ SkillSHEET■7.2■(doc-5245):■Expanding■a■pair■of■ brackets■ •■ SkillSHEET■7.3■(doc-5246):■Factorising■by■taking■ out■the■highest■common■factor •■ SkillSHEET■7.4■(doc-5247):■Factorising■by■taking■ out■a■common■binomial■factor •■ SkillSHEET■7.7■(doc-5248):■Simplifying■algebraic■ fractions •■ SkillSHEET■7.8■(doc-5249):■Simplifying■surds Digital docs

7A Expanding algebraic expressions

(page 225) Activity■7-A-1■(doc-5044):■Review■of■expansion Activity■7-A-2■(doc-5045):■Expanding■algebraic■ expressions Activity■7-A-3■(doc-5046):■Expanding■more■ complex■algebraic■expressions SkillSHEET■7.1■(doc-5244):■Expanding■brackets SkillSHEET■7.2■(doc-5245):■Expanding■a■pair■of■ brackets

Digital docs

•■ •■ •■ •■ •■

7B Factorising expressions with three terms Digital docs

•■ Activity■7-B-1■(doc-5050):■Introducing■quadratic■ factorisation■(page 229) •■ Activity■7-B-2■(doc-5051):■Practising■quadratic■ factorisation■(page 229) •■ Activity■7-B-3■(doc-5052):■Tricky■quadratic■ factorisation■(page 229) •■ SkillSHEET■7.5■(doc-5250):■Finding■a■factor■pair■ that■adds■to■a■given■number■(page 229) •■ WorkSHEET■7.1■(doc-5251):■Factorising■and■ expanding■(page 231) 7C Factorising expressions with two or four terms Digital docs (page 234) •■ Activity■7-C-1■(doc-5047):■Factorising■expressions■ with■two■or■four■terms •■ Activity■7-C-2■(doc-5048):■More■factorising■ expressions■with■two■or■four■terms

246

maths Quest 10 for the Australian Curriculum

•■ Activity■7-C-3■(doc-5049):■Advanced■factorising■ expressions■with■two■or■four■terms •■ SkillSHEET■7.3■(doc-5246):■Factorising■by■taking■ out■the■highest■common■factor •■ SkillSHEET■7.4■(doc-5247):■Factorising■by■taking■ out■a■common■binomial■factor 7D Factorising by completing the square Digital docs (page 239) •■ Activity■7-D-1■(doc-5053):■Introducing■completing■ the■square •■ Activity■7-D-2■(doc-5054):■Practising■completing■ the■square •■ Activity■7-D-3■(doc-5055):■Completing■the■ square Interactivity

•■ Completing■the■square■(int-2783)■(page 236) 7E Mixed factorisation Digital docs

•■ Activity■7-E-1■(doc-5056):■Mixed■factorisation■ (page 240) •■ Activity■7-E-2■(doc-5057):■Harder■mixed■ factorisation■(page 240) •■ Activity■7-E-3■(doc-5058):■Advanced■mixed■ factorisation■(page 240) •■ SkillSHEET■7.6■(doc-5252):■Factorising■by■grouping■ three■and■one■(page 241) •■ SkillSHEET■7.7■(doc-5248):■Simplifying■algebraic■ fractions■(page 241) •■ WorkSHEET■7.2■(doc-5254):■Mixed■factorisation■ (page 241) Chapter review Interactivities (page 245) •■ Test■Yourself■Chapter■7■(int-2846):■Take■the■end-ofchapter■test■to■test■your■progress. •■ Word■search■Chapter■7■(int-2844):■an■interactive■ word■search■involving■words■associated■with■this■ chapter •■ Crossword■Chapter■7■(int-2845):■an■interactive■ crossword■using■the■defi■nitions■associated■with■the■ chapter

To access eBookPLUS activities, log on to www.jacplus.com.au

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships

8 Quadratic equations

8a Solving quadratic equations 8B The quadratic formula 8c Solving quadratic equations by inspecting graphs 8D Finding solutions to quadratic equations by interpolation and using the discriminant 8E Solving a quadratic equation and a linear equation simultaneously WhAt Do you knoW ? 1 List what you know about quadratic equations. Create a concept map to show your list. 2 Share what you know with a partner and then with a small group. 3 As a class, create a large concept map that shows your class’s knowledge of quadratic equations. eBook plus

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Hungry brain activity Chapter 8 doc-5255

opening Question

How can you tell that these images show shapes which are represented by quadratic equations?

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships

Are you ready?

Try■the■questions■below.■If■you■have■diffi■culty■with■any■of■them,■extra■help■can■be■obtained■by■ completing■the■matching■SkillSHEET.■Either■search■for■the■SkillSHEET■in■your■eBookPLUS■or■ ask■your■teacher■for■a■copy. eBook plus

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SkillSHEET 8.1 doc-5256

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SkillSHEET 8.2 doc-5257

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SkillSHEET 8.3 doc-5258

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SkillSHEET 8.4 doc-5259

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SkillSHEET 8.5 doc-5260

Factorising by taking out the highest common factor 1 Factorise■each■of■the■following■expressions. a x2■-■3x b 4x2■+■12x

Finding a factor pair that adds to a given number 2 For■each■of■the■following,■fi■nd■a■factor■pair■of■the■fi■rst■number■that■adds■to■the■second■number. a -6,■1 b 6,■-5 c -6,■-1

Simplifying surds 3 Simplify■each■of■the■following. a

b 2 50

24

c 3 288

Substituting into quadratic equations 4 Substitute■the■x-value■in■brackets■into■each■of■the■following■quadratic■equations■to■determine■

the■y-value.

a y■=■x2■-■4x■+■3■ (x■=■3) b y■=■-3x2■+■2x■-■8■ (x■=■2) c y■=■-8x2■-■3x■-■12■ (x■=■-2) Equation of a vertical line 5 Write■the■equation■for■each■of■the■lines■shown■below. y y a b

-4 -3 -2 -1 0 1 2 3

248

c 36x2■-■12x

x

maths Quest 10 for the Australian Curriculum

x -2-1 0 1 2 3 4

c

y

x -1 0 1 2 3 4 5

number AND algebra • Linear and non-linear relationships

8A

Solving quadratic equations ■■ ■■ ■■

The general form of a quadratic equation is ax2 + bx + c = 0. To solve an equation means to find the value of the pronumeral(s) or variables, which when substituted, will make the equation a true statement. This is done by using the Null Factor Law. If a ì b = 0 then a = 0 or b = 0 (or possibly both a and b equal 0).

■■

Equations that are not in factor form need to be factorised before the Null Factor Law can be applied.

Worked Example 1

Solve the equation (x - 7)(x + 11) = 0. Think

Write

1

Write the equation and check that the right-hand side equals zero.

(x - 7)(x + 11) = 0

2

The left-hand side is factorised so use the Null Factor Law to find two linear equations.

x - 7 = 0  or  x + 11 = 0

3

Solve for x.

x = 7

x = -11

Worked Example 2

Solve each of the following equations. a x2 - 3x = 0 c x2 - 13x + 42 = 0 Think a

b

b 3x2 - 27 = 0 d 36x2 - 21x = 2 Write a x2 - 3x = 0

1

Write the equation. Check that the righthand side equals zero.

2

Factorise by taking out any common factors (x).

x(x - 3) = 0

3

Use the Null Factor Law to write two linear equations.

x = 0  or  x - 3 = 0

4

Solve for x.

x = 0 b

x=3 3x2

- 27 = 0

1

Write the equation. Check that the righthand side equals zero.

2

Take out any common factors (3).



3

Look at the number of terms to factorise and identify the appropriate method. Factorise using the difference of two squares rule.

3(x2 - 32) = 0 3(x + 3)(x - 3) = 0

4

Use the Null Factor Law to write two linear equations.

x + 3 = 0  or  x - 3 = 0

5

Solve for x.

   x = -3 x=3   (Alternatively, x = ê3)

3(x2 - 9) = 0

Chapter 8 Quadratic equations

249

number AND algebra • Linear and non-linear relationships c

d

c

x2 - 13x + 42 = 0

1

Write the equation. Check that the right-hand side equals zero. Check for any common factors (none).

2

Look at the number of terms to factorise and identify the appropriate method. Factorise by finding a factor pair of ■ 42 that adds to -13.

   42: -6 + -7 = -13 (x - 6)(x - 7) = 0

3

Use the Null Factor Law to write two linear equations.

x - 6 = 0  or  x - 7 = 0

4

Solve for x.

   x = 6 36x2

d   

x=7

- 21x = 2

1

Write the equation. Check that the righthand side equals zero. (It does not.)

2

Rearrange the equation so the right-hand side of the equation equals zero. Check for any common factors (none).

36x2 - 21x - 2 = 0

3

Recognise that the expression to factorise is a quadratic trinomial. Identify a factor pair of ac (-72) which adds to the coefficient of x (-21).

   -72: 3 + -24 = -21

4

Rewrite the expression by breaking the x-term into two terms using the factor pair from step 3.

36x2 - 24x + 3x - 2 = 0

5

Factorise the expression by grouping.

12x(3x - 2) + (3x - 2) = 0     (3x - 2)(12x + 1) = 0

6

Use the Null Factor Law to write two linear equations.

3x - 2 = 0  or  12x + 1 = 0     3x = 2 12x = -1

7

Solve for x.



2

x = 3

1

x = - 12

Solving quadratic equations by completing the square ■■

■■

If it is not possible to find an integer factor pair when factorising a quadratic trinomial, the completing the square method can be used before applying the Null Factor Law to the equation. This method allows us to find irrational solutions. In other words, the solutions will be surds.

Worked Example 3

Find the solutions to the equation x2 + 2x - 4 = 0. Give exact answers. Think

250

1

Write the equation. Check that the right-hand side equals zero.

2

Identify the coefficient of x, halve it and square the result.

Maths Quest 10 for the Australian Curriculum

Write

x2 + 2x - 4 = 0

( × 2) 1 2

2

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships

3

Add■the■result■of■step■2■to■the■equation,■placing■ it■after■the■x-term.■To■balance■the■equation,■we■ need■to■subtract■the■same■amount■as■we■have■ added.

4

Insert■brackets■around■the■fi■rst■three■terms■to■ group■them■and■then■simplify■the■remaining■ terms.

5

Factorise■the■fi■rst■three■terms■to■produce■a■ perfect■square.

6

Express■as■the■difference■of■two■squares■and■ then■factorise.

x2 + 2x +

( × 2) − 4 − ( × 2) ■=■0 2

1 2

1 2

2

x2■+■2x■+■(1)2■-■4■−■(1)2■=■0 x2■+■2x■+■1■-■4■-■1■=■0 (x2■+■2x■+■1)■-■5■=■0

(x■+■1)2■-■5■=■0 (x■+■1)2■-■( 5)2■=■0 (x■+■1■+■ 5)(x■+■1■-■ 5)■=■0

7

Use■the■Null■Factor■Law■to■fi■nd■linear■ equations.

x■+■1■+■ 5■=■0■or■x■+■1■-■ 5■=■0

8

Solve■for■x.■Keep■the■answer■in■surd■form■to■ provide■an■exact■answer.

x■=■-1■-■ 5 ■ x■=■-1■+■ 5

■■ ■■

(Alternatively,■x■=■-1■ê■ 5 .)

There■are■many■problems■that■can■be■modelled■by■a■quadratic■equation.■You■should■fi■rst■form■ the■quadratic■equation■that■represents■the■situation■before■attempting■to■solve■such■problems. Recall■that■worded■problems■should■always■be■answered■with■a■sentence.

WorkeD exAmple 4

eBook plus

When two consecutive numbers are multiplied together, the result is 20. Determine the numbers. think

eLesson Completing the square

eles-0174

Write

1

Defi■ne■the■terms■by■using■a■pronumeral■for■one■ of■the■numbers■and■adding■1■to■it■to■give■the■ second■number.

2

Write■an■equation■multiplying■the■numbers■to■ give■the■answer.

3

Rearrange■the■equation■so■that■the■right-hand■ side■equals■zero.

4

Expand■to■remove■the■brackets.

5

Factorise.

(x■+■5)(x■-■4)■=■0

6

Use■the■Null■Factor■Law■to■solve■for■x.

x■+■5■=■0■ or■ x■-■4■=■0 x■=■-5■ x■=■4

7

Use■the■answer■to■determine■the■second■ number.

If■x■=■-5,■x■+■1■=■-4. If■x■=■4,■x■+■1■=■5.

8

Answer■the■question■in■a■sentence.

The■numbers■are■4■and■5■or■-5■and■-4.

9

Check■the■solutions.

Check:■4■ì■5■=■20■ -5■ì■-4■=■20

Let■the■two■numbers■be■x■and■(x■+■1).

x(x■+■1)■=■20 x(x■+■1)■-■20■=■0 x2■+■x■-■20■=■0

Chapter 8 Quadratic equations

251

number AND algebra • Linear and non-linear relationships

Worked Example 5

The height of a football after being kicked is determined by the formula h = -0.1d 2 + 3d, where d is the horizontal distance from the player. a How far away is the ball from the player when it hits the ground? b What horizontal distance does the ball cover when the height of the ball first reaches 20  m? Think a

b

Write a h = -0.1d 2 + 3d

1

Write the formula.

2

The ball hits the ground when h = 0. Write the formula with h = 0 on the right-hand side.

-0.1d 2 + 3d = 0

3

Factorise the expression.

-0.1d 2 + 3d = 0 d(-0.1d + 3) = 0

4

Use the Null Factor Law and then simplify the expression.

d = 0  or  -0.1d + 3 = 0 -0.1d = -3 −3 d = −0.1 = 30

5

Interpret the solutions.

d = 0 is the origin of the kick. d = 30 is the distance from the origin the ball has travelled when it lands.

6

Answer the question in a sentence.

The ball is 30  m from the player when it hits the ground.

1

To find the height of the ball at 20 m, substitute 20 for h.

2

Rearrange the expression.

0.1d2 – 3d + 20 = 0

3

Multiply both sides of the equation by 10 to remove the decimal from the coefficient.

d2 – 30d + 200 = 0

4

Factorise the expression.

    (d – 20)(d – 10) = 0

5

Apply the Null Factor Law.

d − 20 = 0  or  d – 10 = 0

6

Solve.

   d = 20

7

Interpret the solution. The first time the ball reaches a height of 20  m is the smaller value of d. Answer in a sentence.

The ball first reaches a height of 20  m after it has travelled a distance of 10 m.

b

h = -0.1d 2 + 3d 20 = -0.1d 2 + 3d

d = 10

remember

1. The general form of a quadratic equation is ax2 + bx + c = 0. 2. To solve a quadratic equation: (a) make sure the right-hand side of the equation equals zero (b) take out any common factors (c) factorise the left-hand side if applicable (d) use the Null Factor Law to solve for x. 3. An exact answer is a surd or an answer that has not been rounded or approximated. 252

Maths Quest 10 for the Australian Curriculum

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships

exerCise

8A inDiViDuAl pAthWAys eBook plus

Activity 8-A-1

Solving simple quadratics doc-5059 Activity 8-A-2

Solving quadratic equations doc-5060 Activity 8-A-3

Solving more complex quadratics doc-5061

solving quadratic equations fluenCy 1 We 1 ■Solve■each■of■the■following■equations. a (x■+■7)(x■-■9)■=■0 b (x■-■3)(x■+■2)■=■0 d x(x■-■3)■=■0 e x(x■-■1)■=■0 g 2x(x■-■3)■=■0

c (x■-■2)(x■-■3)■=■0 f x(x■+■5)■=■0

h 9x(x■+■2)■=■0

j -(x■+■1.2)(x■+■0.5)■=■0 k 2(x■-■0.1)(2x■-■1.5)■=■0 2 Solve■each■of■the■following■equations. a (2x■-■1)(x■-■1)■=■0 b (3x■+■2)(x■+■2)■=■0 d (7x■+■6)(2x■-■3)■=■0 e (5x■-■3)(3x■-■2)■=■0 g x(x■-■3)(2x■-■1)■=■0 h x(2x■-■1)(5x■+■2)■=■0 3 We 2a ■Solve■each■of■the■following■equations. a x2■-■2x■=■0 b x2■+■5x■=■0 2 d 3x ■=■-2x e 4x2■-■6x■=■0 g 4x2■-■2 7x■=■0

h 3x2■+■ 3x■=■0

4 We 2b ■Solve■each■of■the■following■equations. a x2■-■4■=■0 b x2■-■25■=■0 2 d 4x ■-■196■=■0 e 9x2■-■16■=■0 g 9x2■=■4 j

1 2 x ■-■ 4 ■=■0 36 9

1

1

i

(x■-■ 2 )(x■+■ 2 )■=■0

l

(x■+■ 2)(x■-■ 3)■=■0

c (4x■-■1)(x■-■7)■=■0 f (8x■+■5)(3x■-■2)■=■0 i x(x■+■3)(5x■-■2)■=■0 c x2■=■7x f 6x2■-■2x■=■0 i 15x■-■12x2■=■0 c 3x2■-■12■=■0 f 4x2■-■25■=■0 1

h 36x2■=■9

i

x2■-■ 25 ■=■0

k x2■-■5■=■0

l

9x2■-■11■=■0

c f i l

x2■-■6x■-■7■=■0 x2■-■3x■-■4■=■0 x2■-■8x■+■12■=■0 x2■-■7x■+■12■=■0

5 We 2c ■Solve■each■of■the■following■equations. a x2■-■x■-■6■=■0 b x2■+■6x■+■8■=■0 2 d x ■-■8x■+■15■=■0 e x2■-■2x■+■1■=■0 2 g x ■-■10x■+■25■=■0 h x2■-■3x■-■10■=■0 2 j x ■-■4x■-■21■=■0 k x2■-■x■-■30■=■0 6 mC ■The■solutions■to■the■equation■x2■+■9x■-■10■=■0■are: a x■=■1■and■x■=■10 B x =■1■and■x =■-10 D x■=■-1■and■x■=■-10 E x■=■1■and■x■=■9 7 mC ■The■solutions■to■the■equation■x2■–■100■=■0■are: a x■=■0■and■x■=■10 B x■=■0■and■x■=■-10 D x■=■0■and■x■=■100 E x■=■-100■and■x■=■100 8 We 2d ■Solve■each■of■the■following■equations. a 2x2■-■5x■=■3 b 3x2■+■x■-■2■=■0 2 d 6x ■-■11x■+■3■=■0 e 14x2■-■11x■=■3 2 g 6x ■-■7x■=■20 h 12x2■+■37x■+■28■=■0 2 j 6x ■-■25x■+■24■=■0 k 30x2■+■7x■-■2■=■0

c x■=■-1■and■x■=■10

c x■=■-10■and■x■=■10

5x2■+■9x■=■2 12x2■-■7x■+■1■=■0 10x2■-■x■=■2 3x2■-■21x■=■-36 9 We 3 ■Find■the■solutions■for■each■of■the■following■equations.■Give■exact■answers. a x2■-■4x■+■2■=■0 b x2■+■2x■-■2■=■0 c x2■+■6x■-■1■=■0 2 2 d x ■-■8x■+■4■=■0 e x ■-■10x■+■1■=■0 f x2■-■2x■-■2■=■0 2 2 g x ■+■2x■-■5■=■0 h x ■+■4x■-■6■=■0 i x2■+■4x■-■11■=■0 10 Find■the■solutions■for■each■of■the■following■equations.■Give■exact■answers. a x2■-■3x■+■1■=■0 b x2■+■5x■-■1■=■0 c x2■-■7x■+■4■=■0 2 2 d x ■-■5■=■x e x ■-■11x■+■1■=■0 f x2■+■x■=■1 g x2■+■3x■-■7■=■0 h x2■-■3■=■5x i x2■-■9x■+■4■=■0 11 Solve■each■of■the■following■equations,■rounding■answers■to■2■decimal■places. a 2x2■+■4x■-■6■=■0 b 3x2■+■12x■-■3■=■0 c 5x2■-■10x■-■15■=■0 2 2 d 4x ■-■8x■-■8■=■0 e 2x ■-■6x■+■2■=■0 f 3x2■-■9x■-■3■=■0 2 2 g 5x ■-■15x■-■25■=■0 h 7x ■+■7x■-■21■=■0 i 4x2■+■8x■-■2■=■0 c f i l

Chapter 8 Quadratic equations

253

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships unDerstAnDing 12 Are■there■real■solutions■to■the■equation■x2■+■4x■+■10■=■0?■Give■reasons■for■your■answer. 13 We 4 ■When■two■consecutive■numbers■are■multiplied,■the■result■is■72.■Find■the■numbers. 14 When■two■consecutive■even■numbers■are■multiplied,■the■result■is■48.■Find■the■numbers. 15 When■a■number■is■added■to■its■square■the■result■is■90.■Find■the■number. 16 Twice■a■number■is■added■to■three■times■its■square.■If■the■result■is■16,■fi■nd■the■number. 17 Five■times■a■number■is■added■to■two■times■its■square.■If■the■result■is■168,■fi■nd■the■number. 18 We 5 ■A■soccer■ball■is■kicked.■The■height,■h,■in■metres,■of■the■soccer■ball■

t■seconds■after■it■is■kicked■can■be■represented■by■the■equation■h■=■-t(t■-■6).■ Find■how■long■it■takes■for■the■soccer■ball■to■hit■the■ground■again. 19 The■length■of■an■Australian■fl■ag■is■twice■its■width■and■the■diagonal■length■ is■45■■cm. a If■x■cm■is■the■width■of■the■fl■ag,■fi■nd■the■length■in■terms■of■x. b Draw■a■diagram■of■the■fl■ag■marking■in■the■diagonal.■Mark■the■length■and■the■width■in■ terms■of■x. c Use■Pythagoras’■theorem■to■write■an■equation■relating■the■lengths■of■the■sides■to■the■ length■of■the■diagonal. d Solve■the■equation■to■fi■nd■the■dimensions■of■the■Australian■fl■ag.■Round■your■answer■to■ the■nearest■cm. 20 If■the■length■of■a■paddock■is■2■■m■more■than■its■width■and■the■area■is■48■■m2,■fi■nd■the■length■and■ width■of■the■paddock. 21 Solve■for■x. a

b

c

x2 + 4x + 3 2

x + 4x + 4 x2 − 2x − 3 x2 − 6x + 8

= =

x2 + 2x + 1 x 2 + 5x + 6 x 2 + x − 12 ( x 2 − 16)( x − 2)

( x − 1)( x 2 + 2 x + 1) 2x2 + 6x + 4

=

3x 2 + 4 x + 1 3 x 2 + 15 x + 18

reAsoning 22 H enrietta■is■a■pet■rabbit■who■lives■in■an■

enclosure■that■is■2■■m■wide■and■4■■m■ long.■Her■human■family■has■decided■to■ purchase■some■more■rabbits■to■keep■ her■company■and■so■the■size■of■the■ enclosure■must■be■increased. a Draw■a■diagram■of■Henrietta’s■ enclosure,■clearly■marking■the■ lengths■of■the■sides. b If■the■length■and■width■of■the■ enclosure■are■increased■by■x■m,■ fi■nd■the■new■dimensions. c If■the■new■area■is■to■be■24■■m2,■ write■an■equation■relating■the■ sides■and■the■area■of■the■enclosure■ (Area■=■length■ì■width). d Use■the■equation■to■fi■nd■the■value■of■x■and,■hence,■the■length■of■the■sides■of■the■new■ enclosure. 254

maths Quest 10 for the Australian Curriculum

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships 23 A■student■is■required■to■cover■an■area■of■620■■cm2■with■mosaic■tiles.■The■tile■pattern■is■to■be■

eBook plus

Digital doc

WorkSHEET 8.1 doc-5261

8b

surrounded■by■a■border■2■■cm■wide■to■complete■the■display■page.■The■length■of■the■display■ page■is■l■cm■and■its■width■is■4■■cm■less■than■its■length. a Find■the■width■of■the■display■page■in■terms■of■l. b Find■the■width■and■length■of■the■tile■pattern■in■terms■of■l. c Using■the■answers■from■b,■write■an■equation■relating■the■area■of■the■tile■pattern■to■its■ dimensions. d Use■the■completing■the■square■method■to■solve■the■equation■and,■hence,■fi■nd■the■length,■ l■cm,■of■the■display■page.■Round■your■answer■to■the■nearest■cm. e Find■the■area■of■the■display■page.■Round■your■answer■to■the■nearest■cm2. 24 The■cost■per■hour,■C(s),■in■thousands■of■dollars■of■running■two■cruise■ships,■Annabel■and■Betty,■ travelling■at■a■speed■of■s■knots■is■given■by■the■following■relationships. CAnnabel(s)■=■0.3s2■+■4.2s■+■12■and■CBetty(s)■=■0.4s2■+■3.6s■+■8 a Determine■the■cost■per■hour■for■each■ship■if■they■are■both■ travelling■at■28■knots. b Find■the■speed■in■knots■at■which■both■ships■must■travel■ refleCtion    for■them■to■have■the■same■cost. What does the Null c Explain■why■only■one■of■the■solutions■obtained■in■your■ Factor Law mean? working■for■part■b■is■valid.■

the quadratic formula ■■ ■■

The■method■of■solving■quadratic■equations■by■completing■the■square■can■be■generalised■to■ produce■what■is■called■the■quadratic formula. Consider■solving■the■general■equation■ax2■+■bx■+■c■=■0.■We■will■fi■rst■follow■the■steps■involved■ in■completing■the■square. b c x2 + x + = 0 a a

1.■ Divide■both■sides■of■the■equation■by■a.■ 2

x2 +

2.■ Complete■the■square.■

2

b c  b  b x+  −  + =0  2a   2a  a a 2

3.■ Factorise■the■fi■rst■three■terms■as■a■perfect■square.■

b b2 c  x + − + =0  2a  4a 2 a

4.■ Add■the■fi■nal■two■terms.■

b b 2 − 4 ac  =0  x + 2a  − 4a2

2

2

2  b 2 − 4 ac  b  x + −   =0    2a  2a

5.■ Write■as■the■difference■of■two■squares.■

6.■ ■Factorise■using■the■difference■of■two■■ squares■rule. 7.■ Solve■the■two■linear■factors.■ ■

x+

 b b 2 − 4 ac   b b 2 − 4 ac  + − x+ x+  =0     2a 2a 2a 2a b b 2 − 4 ac b b 2 − 4 ac + = 0 ■ or■ x + − =0 2a 2a 2a 2a x=

−b b 2 − 4 ac −b b 2 − 4 ac ■ ■■ x= + − 2a 2a 2a 2a Chapter 8 Quadratic equations

255

number AND algebra • Linear and non-linear relationships

■■

■■

− b ± b 2 − 4 ac where a is the coefficient of x2, b is 2a the coefficient of x and c is the constant or the term without an x. This formula can be used to solve any quadratic equation. The solution can be summarised as x =

Worked Example 6

Use the quadratic formula to solve each of the following equations. a 3x2 + 4x + 1 = 0 (exact answer) b -3x2 - 6x - 1 = 0 (round to 2 decimal places) Think a

b

Write a 3x2 + 4x + 1 = 0

1

Write the equation.

2

Write the quadratic formula.

x=

3

State the values for a, b and c.

where a = 3, b = 4, c = 1

4

Substitute the values into the formula.

x=

5

Simplify and solve for x.

1

Write the equation.

2

Write the quadratic formula.

x=

3

State the values for a, b and c.

where a = -3, b = -6, c = -1

4

Substitute the values into the formula.

x=

5

Simplify the fraction.

=

6

Solve for x.

x ö -1.82   or  x ö -0.18

− b ± b 2 − 4 ac 2a − 4 ± (4)2 − (4 × 3 × 1) 2×3

−4 ± 4 6 −4 ± 2 = 6 −4 − 2 −4 + 2 x=   or x = 6 6 1 x = - 3 x = -1 =

b -3x2 - 6x - 1 = 0

− b ± b 2 − 4 ac 2a −( −6) ± 36 − 4 × −3 × −1 2 × −3

6 ± 24 −6 6±2 6 = −6 3± 6 = −3 3+ 6 3− 6 x=   or  −3 −3

Note: When asked to give an answer in exact form, you should simplify any surds as necessary. ■■

256

If the value inside the square root sign is negative, then there are no solutions to the equation.

Maths Quest 10 for the Australian Curriculum

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships

remember

The■quadratic■formula■x = form■ax2■+■bx■+■c■=■0. exerCise

8b inDiViDuAl pAthWAys

− b ± b 2 − 4 ac ■can■be■used■to■solve■quadratic■equations■of■the■ 2a

the quadratic formula fluenCy

Introducing the quadratic formula doc-5062

1 State■the■values■for■a,■b■and■c■in■each■of■the■following■equations■of■the■form■ax2■+■bx■+■c■=■0. a 3x2■-■4x■+■1■=■0 b 7x2■-■12x■+■2■=■0 2 c 8x ■-■x■-■3■=■0 d x2■-■5x■+■7■=■0 2 e 5x ■-■5x■-■1■=■0 f 4x2■-■9x■-■3■=■0 2 g 12x ■-■29x■+■103■=■0 h 43x2■-■81x■-■24■=■0 i 6x2■-■15x■+■1■=■0

Activity 8-B-2

2 We 6a ■Use■the■quadratic■formula■to■solve■each■of■the■following■equations.■Give■exact■

eBook plus

Activity 8-B-1

Practice using the quadratic formula doc-5063 Activity 8-B-3

Using the quadratic formula doc-5064

eBook plus

3

Digital doc

SkillSHEET 8.6 doc-5262

eBook plus

4

Digital doc

SkillSHEET 8.7 doc-5263

5

6

7

answers. a x2■+■5x■+■1■=■0 b x2■+■3x■-■1■=■0 2 c x ■-■5x■+■2■=■0 d x2■-■4x■-■9■=■0 2 e x ■+■2x■-■11■=■0 f x2■-■7x■+■1■=■0 2 g x ■-■9x■+■2■=■0 h x2■-■6x■-■3■=■0 2 i x ■+■8x■-■15■=■0 j -x2■+■x■+■5■=■0 2 k -x ■+■5x■+■2■=■0 l -x2■-■2x■+■7■=■0 We 6b ■Use■the■quadratic■formula■to■solve■each■of■the■following■equations.■Give■approximate■ answers■rounded■to■2■decimal■places. a 3x2■-■4x■-■3■=■0 b 4x2■-■x■-■7■=■0 2 c 2x ■+■7x■-■5■=■0 d 7x2■+■x■-■2■=■0 2 e 5x ■-■8x■+■1■=■0 f 2x2■-■13x■+■2■=■0 2 g -3x ■+■2x■+■7■=■0 h -7x2■+■x■+■8■=■0 2 i -12x ■+■x■+■9■=■0 j -6x2■+■4x■+■5■=■0 2 k -11x ■-■x■+■1■=■0 l -4x2■-■x■+■7■=■0 2 m -2x ■+■12x■-■1■=■0 n -5x2■+■x■+■3■=■0 2 0 -3x ■+■5x■+■2■=■0 mC ■The■solutions■of■the■equation■3x2■-■7x■-■2■=■0■are: a 1,■2 B 1,■-2 c -0.257,■2.59 D -0.772,■7.772 E -1.544,■15.544 mC ■In■the■expansion■of■(6x■–■5)(3x■+■4),■the■coeffi■cient■of■x■is: a 18 B -15 c 9 D 6 E -2 mC ■In■the■expanded■form■of■(x■−■2)(x■+■4),■which■of■the■following■is■incorrect? a The■value■of■the■constant■is■-8. B The■coeffi■cient■of■the■x■term■is■-6. c The■coeffi■cient■of■the■x■term■is■2. D The■coeffi■cient■of■the■x2■term■is■1. E The■expansion■shows■this■to■be■a■trinomial■expression. mC ■An■exact■solution■to■the■equation■x2■+■2x■–■5■=■0■is: a -3.449 D

2 + −16 2

B -1■+■ 24 E

c -1■+■ 6

2 + 24 2 Chapter 8 Quadratic equations

257

number AND algebra • Linear and non-linear relationships Understanding 8 Solve each of the following equations using any suitable method. Round to 3 decimal places

where appropriate. a 2x2 - 7x + 3 = 0 d x2 - 3x + 1 = 0 g x2 - 5x + 8 = 0 j 3x2 + 3x - 6 = 0 m -x2 + 9x - 14 = 0

b e h k n

x2 - 5x = 0 x2 - 7x + 2 = 0 x2 - 7x - 8 = 0 2x2 + 11x - 21 = 0 -6x2 - x + 1 = 0

c f i l o

x2 - 2x - 3 = 0 x2 - 6x + 8 = 0 x2 + 2x - 9 = 0 7x2 - 2x + 1 = 0 -6x2 + x - 5 = 0

Reasoning 9 The surface area of a closed cylinder is given by the formula SA = 2πr(r + h), where r cm is

the radius of the can and h cm is the height.   The height of a can of wood finish is 7  cm and its surface area is 231  cm2. a Substitute values into the formula to form a quadratic equation using the pronumeral, r. b Use the quadratic formula to solve the equation and, hence, find the radius of the can. Round the answer to 1 decimal place. c Calculate the area of the paper label on the can. Round the answer to the nearest square centimetre. x 10 To satisfy lighting requirements, a window must have an area of 1500  cm2. 30 cm a Find an expression for the area of the window in terms of x. b Write an equation so that the window satisfies the lighting requirements. x c Use the quadratic formula to solve the equation and find x to the nearest mm. 11 Two competitive neighbours build rectangular pools that cover the same area but are different

shapes. Pool A has a width of (x + 3) m and a length that it 3 m longer than its width. Pool B has a length that is double the width of Pool A. The width of Pool B is 4 m shorter than its length. a Find the exact dimensions of each pool if their areas are the same. b Verify mathematically that the areas are the same. reflection    12 A block of land is in the shape of a rightWhat kind of answer will you get if angled triangle with a perimeter of 150 m and a the value inside the square root sign hypotenuse of 65 m. Determine the lengths of the in the quadratic formula is zero? other two sides.

8c

Solving quadratic equations by inspecting graphs y

■■ ■■

■■

■■

258

The graph of a quadratic equation is called a parabola. To solve quadratic equations graphically means to find the values of x where y = 0 or where the parabola intercepts the x-axis. A quadratic equation written in standard form has solutions when the graph of y = ax2 + bx + c is equal to zero. In this section, we will find solutions (also called roots or zeros) of quadratic equations by inspecting their corresponding graphs.

Maths Quest 10 for the Australian Curriculum

y = ax2 + bx + c

(0, c) 0

x Solutions/roots/zeros to ax2 + bx + c = 0

number AND algebra • Linear and non-linear relationships

Worked Example 7

Determine the solution (or roots) of each of the following quadratic equations by inspecting their corresponding graphs. Round answers to 1 decimal place where appropriate. a x2 + x - 2 = 0 b 2x2 - 4x - 5 = 0 Think

Write/Draw

a The graph of y = x2 + x - 2 is equal to zero

a

when y = 0. Look at the graph to find where y = 0; that is, where it intersects the x-axis.

y 3 2 1 0 -3 -2 -1 -1 -2 -3

1 2 3x y = x2 + x - 2

x2 + x - 2 = 0 From the graph, the solutions are x = 1 and x = -2. b The graph of y = 2x2 - 4x - 5 is equal to zero

b

when y = 0. Look at the graph to see where y = 0; that is, where it intersects the x-axis. By sight, we can only give estimates of the solutions.

y 6 4 2 0 -3 -2 -1 -2

1 2 3x

-4 -6 -8

2x2 - 4x – 5 = 0 From the graph, the solutions are x ö -0.9 and x ö 2.9.

■■

Some quadratic equations have only one solution. For example, the graph of x2 - 4x + 4 = 0 has the one solution of x = 2. That is, the graph of equation touches the x-axis only at x = 2. y y = x2 - 4x + 4 5

-2 ■■

0

2

4

x y

There are also quadratic equations that have no real solutions. For example, the graph of y = 3x2 - 4x + 4 does not intersect the x-axis and so 3x2 - 4x + 4 = 0 has no real solutions (that is, no solutions that are real numbers).

10 5

-2

y = 3x2 - 4x + 4 x 0 2

Chapter 8 Quadratic equations

259

number AND algebra • Linear and non-linear relationships

Confirming solutions ■■

It is possible to confirm the solutions obtained by sight. As we saw with linear equations, this is achieved by substituting the solution or solutions into the original quadratic equation. If both sides of the equation are equal, the solution is correct.

Worked Example 8

Confirm, by substitution, the solutions obtained in Worked example 7. a x2 + x - 2 = 0; solutions: x = 1 and x = -2 b 2x2 - 4x - 5 = 0; solutions: x ö -0.9 and x ö 2.9 Think a

b

Write a

When x = 1, x2 + x - 2 = 12 + 1 - 2

1

Write the left-hand side of the equation and substitute x = 1 into the expression.

2

Simplify to check that the expression is equal to zero.

3

Write the expression and substitute x = -2.

  When x = -2, x2 + x - 2 = (-2)2 + -2 - 2

4

Simplify to check that the expression is equal to zero.



1

Write the left-hand side of the equation and substitute x = -0.9 into the expression.

2

Simplify. As the x-values are only estimates, the results should be reasonably close to zero.

= 1.62 + 3.6 - 5 = -0.22 As -0.9 is only an estimate, the left-hand side expression can be said to be close to zero.

3

Write the expression and substitute x = 2.9 into the expression.

When x = 2.9, 2x2 - 4x - 5 = 2 ì (2.9)2 - 4 ì 2.9 - 5

4

Simplify to check that the expression is reasonably close to zero.

= 16.82 - 11.6 - 5 = -0.22 As 2.9 is only an estimate, the left-hand side expression can be said to be close to zero.

  = 0   Solution is confirmed.

b

=4-2-2 = 0   Solution is confirmed.

When x = -0.9, 2x2 - 4x - 5 = 2 ì (-0.9)2 - 4 ì -0.9 - 5

Worked Example 9

A golf ball hit along a fairway follows the path shown in the graph. The height, h metres after it has 1 travelled x metres horizontally, follows the rule h = - 270 (x2 - 180x). Use the graph to find how far the ball landed from the golfer. h 30

1 2 h = - ––– 270 (x - 180x)

20 10 0 260

Maths Quest 10 for the Australian Curriculum

90

180 x

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships

think

Write

On■the■graph,■the■ground■is■represented■by■the■x-axis■ since■this■is■where h =■0.■The■golf■ball■lands■when■ the■graph■intersects■the■x-axis.

The■golf■ball■lands■180■■m■from■the■golfer.

remember

1.■ The■solution(s)■(also■known■as■roots■or■zeros)■of■a■quadratic■equation■can■be■found■by■ inspecting■the■graph■of■the■equation.■You■may■need■to■draw■the■graph■of■the■equation■ fi■rst■using■a■calculator■or■graphing■software. 2.■ The■root■of■any■graph■is■the■x-intercept■or■the■x-coordinate■of■the■point■where■the■graph■ crosses■the■x-axis. 3.■ The■roots■or■intercepts■of■the■quadratic■graph y =■ax2■+■bx■+ c are■the■solutions■to■the■ equation■ax2■+■bx■+ c =■0. exerCise

8C inDiViDuAl pAthWAys eBook plus

Activity 8-C-1

solving quadratic equations by inspecting graphs fluenCy 1 We 7 ■Determine■the■roots■of■each■of■the■following■quadratic■equations■by■inspecting■the■

corresponding■graphs.■Round■answers■to■1■decimal■place■where■appropriate. b x2■-■11x■+■10■=■0

a x2■- x -■6■=■0

Finding solutions to quadratic equations by inspecting graphs doc-5065 Activity 8-C-2

Solving quadratic equations by inspecting graphs doc-5066 Activity 8-C-3

Harder solutions to quadratic equations by inspecting graphs doc-5067



y

y

12 8 4

8

-6 -4 -2 0 2 4 6 x -4 2 -8 y = x - x - 6

c -x2■+■25■=■0





2 4 6 x

e x2■-■3x■-■4■=■0 y 2 15 y = x - 3x - 4 10 5

2 4 6 8 10 12 x y = x2 - 11x + 10

d 2x2■-■8x■+■8■=■0 y y = 2x2 - 8x + 8 20 10

y y = -x2 + 25 30 20 10 -6 -4 -2 0 -10

-2 0 -8 -16 -24

-1 0 1 2 3 4 5 6 x -10



-1 0 -10

1 2 3 4 5 x

f x2■-■3x■-■6■=■0 y 2 15 y = x - 3x - 6 10 5 -1 0 1 2 3 4 5 6 x -10 Chapter 8 Quadratic equations

261

number AND algebra • Linear and non-linear relationships g x2 + 15x - 250 = 0 y

h -x2 = 0 y 5 0

200 100 -30 -20 -10 0 -100 -200 -300 -400 y = x2 + 15x - 250

i

10

-5

x

j

2x2 + x - 3 = 0

5

-5



5 x

-5 -10

x2 + x - 3 = 0

-4 -2 0

y = -x2

5

2 4

x

-2 -1 0

y = x2 + x - 3

-5

1 2

x

y = 2x2 + x - 3

Understanding 2   WE 8  Confirm, by substitution, the solutions obtained in question 1. 3   WE9  A golf ball hit along a fairway follows the path shown in the graph. h 28

1 2 h = - ––– 200 (x - 150x)

0

150 x

75

The height, h metres after it has travelled x metres horizontally, follows the rule 1

h = - 200 (x2 - 150x). Use the graph to find how far the ball lands from the golfer. 4 A ball is thrown upwards from a building and follows the path shown in the graph until it lands on the ground. h 25

h = -x2 + 4x + 21

21

0

2

7

x

The ball is h metres above the ground when it is a horizontal distance of x metres from the building. The path of the ball follows the rule h = -x2 + 4x + 21. Use the graph to find how far from the building the ball lands. 262

Maths Quest 10 for the Australian Curriculum

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships reAsoning 5 a■ ■Use■a■graphics■calculator■to■sketch■the■two■functions■

y■=■2x2■-■5x■-■3■and■y■=■-x2■–■3x.

refleCtion 

b Use■the■calculator■to■fi■nd■the■solution■to■

What does ‘the root of a graph’ mean?

2x2■-■5x■-■3■=■-x2■-■3x,■correct■to■2■decimal■places. c Comment■on■your■answers■to■parts■a■and■b.

8D

 

finding solutions to quadratic equations by interpolation and using the discriminant interpolation

eBook plus

■■

Interactivity Solving by interpolation

Consider■the■quadratic■equation■x2■-■3x■–■6■=■0. Using■a■graphing■calculator■or■graphing■software■to■sketch■the■graph■of■the■equation y =■x2■-■3x■–■6,■we■can■see■there■is■a■solution■between x =■4■and x =■5. y 20

int-1147

y = x2 - 3x - 6

10 -2 -1 0 -10

1 2 3 4 5 6

x

This■can■be■confi■rmed■using■the■following■logic: Step■1.■ The■value■of y =■x2■-■3x■-■6■when x =■4■can■be■expressed■as y(4)■=■42■-■3■ì■4■-■6■ ■ ■ =■-2 ■ The■value■of■y■=■x2■-■3x■-■6■when■x■=■5■can■be■expressed■as y(5)■=■52■-■3■ì■5■-■6■ ■ ■ =■4 Since■the■graph■moves■from■below■the x-axis■at x =■4,■to■above■the x-axis■at x =■5,■it■is■ reasonable■to■assume■that■there■is■a■solution■somewhere■between x =■4■and x =■5. Step■2.■ Choose■a■value■between x =■4■and x =■5;■for■example,■4.5. ■■

■ ■

y(4)■=■42■-■3■ì■4■-■6■ =■-2 y(4.5)■=■4.52■-■3■ì■4.5■-■6■ =■0.75 Since■the■graph■moves■from■below■the x-axis■at x =■4,■to■above■the x-axis■at x =■4.5,■it■is■ reasonable■to■assume■there■is■a■solution■somewhere■between x =■4■and x =■4.5.

■■

Step■3.■ ■Repeat■step■2,■checking■that■your■equations■are■approaching■zero.■The■solution■is■ approximately x =■4.372. Repeat■the■process■to■fi■nd■the■other■root,■somewhere■between■-2■and■0. Notes 1.■ ■This■process■can■also■be■done■on■a■spreadsheet. 2.■ ■A■CAS■calculator■can■also■help■you■to■fi■nd■the■roots.■Rather■than■using■the■trace■function,■ try■using■the■table■function■after■drawing■the■graph.■Step■up■in■increments■of■0.1■then■0.01■ to■pinpoint■the■solution. Chapter 8 Quadratic equations

263

number AND algebra • Linear and non-linear relationships

Using the discriminant ■■

■■ ■■

−b ± b 2 − 4 ac gives the solutions to the general quadratic equation 2a 2 ax + bx + c = 0. By examining the expression under the square root sign, b2 - 4ac, we can determine the number and type of solutions produced and, hence, the number of x-intercepts to expect when the quadratic equation is graphed. The expression b2 - 4ac is known as the discriminant and is denoted by the symbol D (delta). The formula x =

Case 1: D < 0 If x2 + 2x + 3 = 0, then a = 1, b = 2 and c = 3. x=

D = b2 - 4ac = 22 - (4 ì 1 ì 3) = -8

−b ± b 2 − 4 ac 2a

− 2 ± −8 2 If the discriminant is less than zero, there are no real solutions because the expression under the square root sign is negative. It is not possible to find a real number that is the square root of a negative number. Hence, the graph of y = x2 + 2x + 3 will not intersect the x-axis; i.e., there will be no x-axis intercepts. ■■

=

Case 2: D = 0 If 4x2 + 12x + 9 = 0, then a = 4, b = 12 and c = 9. D = b2 - 4ac = 122 - (4 ì 4 ì 9) = 144 - 144 =0

−b ± b 2 − 4 ac 2a −12 ± 0 = 2×4

x=

12

= - 8 3

= - 2 If the discriminant is equal to zero then the two solutions are the same. That is, if b2 - 4ac = 0, −b −b + 0 −b − 0 then x = and x = . This may be regarded as one rational solution that is equal to . 2a 2a 2a One solution indicates that the quadratic trinomial is a perfect square that can be factorised easily using the perfect squares rule; that is, 4x2 + 12x + 9 = (2x + 3)2. Hence, the graph of y = 4x2 + 12x + 9 will touch the x-axis once. ■■

Case 3: D > 0 If the discriminant is positive, there are two distinct solutions. We can determine more information than this by checking whether the discriminant is also a perfect square. (a)  If 2x2 - 7x - 4 = 0, then a = 2, b = -7 and c = -4. D = b2 - 4ac = (-7)2 - (4 ì 2 ì -4) = 49 + 32 = 81

2 x = − b ± b − 4 ac 2a

7 ± 81 2×2 7±9 = 4 1 x = 4 or x = - 2 =

If the discriminant is positive and a perfect square, the quadratic trinomial will have ■ two rational solutions. This means the quadratic trinomial can be factorised easily; that is,■ 2x2 - 7x - 4 = (2x + 1)(x - 4). 264

Maths Quest 10 for the Australian Curriculum

number AND algebra • Linear and non-linear relationships

(b)  If x2 - 5x - 1 = 0 then a = 1, b = -5 and c = -1. D = b2 - 4ac = (-5)2 - (4 ì 1 ì -1) = 25 + 4 = 29

x =

−b ± b 2 − 4 ac 2a

x = 5 ± 29 2 ×1

5 ± 29 2 If the discriminant is positive but not a perfect square, the factors are irrational and the quadratic formula must be used to find the two irrational (surd) solutions. Hence, the graphs of both equations shown in (a) and (b) will each have two x-intercepts. The table below summarises the three cases. x =

■■

D > 0 (positive) D < 0 (negative)

D = 0 (zero)

Not a perfect square

Perfect square

Number of solutions

No solutions

1 rational solution

2 rational solutions

Graph

Graph does not cross or touch the x-axis

Graph touches the x-axis

Graph crosses the x-axis twice

y

y

2 irrational (surd) solutions

y

y

x x

a

b

x

-b -a

a

b

x

 

Worked Example 10

By using the discriminant, determine whether the following equations have:   i two rational solutions ii two irrational solutions iii one rational solution (two equal solutions) iv no real solutions. a x2 - 9x - 10 = 0 b x2 - 2x - 14 = 0 2 c x - 2x + 14 = 0 d x2 + 14x = -49 Think a

Write a x2 - 9x - 10 = 0

1

Write the equation.

2

Identify the coefficients a, b and c.

a = 1, b = -9, c = -10

3

Find the discriminant.

D = b2 - 4ac = (-9)2 - (4 ì 1 ì -10) = 121

4

Identify the number and type of solutions when D > 0 and is a perfect square.

The equation has two rational solutions.

Chapter 8 Quadratic equations

265

number AND algebra • Linear and non-linear relationships b

c

d

b x2 - 2x - 14 = 0

1

Write the equation.

2

Identify the coefficients a, b and c.

a = 1, b = -2, c = -14

3

Find the discriminant.

D = b2 - 4ac = (-2)2 - 4 ì 1 ì -14 = 60

4

Identify the number and type of solutions when D > 0 but not a perfect square.

The equation has two irrational solutions.

1

Write the equation.

2

Identify the coefficients a, b and c.

a = 1, b = -2, c = 14

3

Find the discriminant.

D = b2 - 4ac = (-2)2 - (4 ì 1 ì 14) = -52

4

Identify the number and type of solutions when D < 0.

The equation has no real solutions.

1

Write the equation, then rewrite it so the right side equals zero.

c

x2 - 2x + 14 = 0

d x2 + 14x = -49

x2 + 14x + 49 = 0

2

Identify the coefficients a, b and c.

a = 1, b = 14, c = 49

3

Find the discriminant.

D = b2 - 4ac = 142 - (4 ì 1 ì 49) =0

4

Identify the number and types of solutions when D = 0.

The equation has 1 rational solution.

Remember, the number of solutions of a quadratic equation is the same as the number of x-intercepts obtained when the equation is graphed.

remember

1. Interpolation can be used to find approximate solutions to quadratic equations. 2. The discriminant of a quadratic equation is given by D = b2 - 4ac. 3. If D < 0, there are no real solutions to the equation. 4. If D = 0, there is only one rational solution (or two equal solutions) to the equation. The equation can be factorised easily. 5. If D > 0, there are two distinct solutions to the equation. (a) If the discriminant is a perfect square, the solutions are rational and the equation can be factorised easily. (b) If the discriminant is not a perfect square, the solutions are irrational and the equation can be solved using the quadratic formula or the completing the square method. 6. The number of solutions of a quadratic equation corresponds to the number of x-intercepts obtained when the equation is graphed. 266

Maths Quest 10 for the Australian Curriculum

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships

7.■ This■information■can■be■summarised■in■the■following■table: D > 0 (positive) D < 0 (negative)

exerCise

8D inDiViDuAl pAthWAys eBook plus

Activity 8-D-1

Finding solutions to quadratic equations by interpolation doc-5068 Activity 8-D-2

Harder solutions to quadratic equations by interpolation doc-5069 Activity 8-D-3

Difficult solutions to quadratic equations by interpolation doc-5070

D = 0 (zero)

Perfect square

Not a perfect square

Number of solutions

No■solutions

1■rational■ solution

2■rational■ solutions

Graph

Graph■does■not■ cross■or■touch■ the■x-axis

Graph■touches■ the■x-axis

Graph■crosses■the■x-axis■twice

2■irrational■ (surd)■solutions

finding solutions to quadratic equations by interpolation and using the discriminant fluenCy 1 Use■a■graphing■calculator■or■graphing■software■to■sketch■the■graph■of■each■of■the■following■

equations■and■then■use■the■process■of■interpolation■to■fi■nd■approximate■solutions. a x2■+■3x■-■7■=■0 b 3x2■-■2x■-■4■=■0 c 2x2■+■7x■-■10■=■0 2 Determine■the■discriminant■for■each■of■the■following■equations. a x2■-■3x■+■5 b 4x2■-■20x■+■25■=■0 c x2■+■9x■-■22■=■0 2 2 d 9x ■+■12x■+■4 e x ■+■3x■-■7■=■0 f 25x2■-■10x■+■1■=■0 2 2 g 3x ■-■2x■-■4■=■0 h 2x ■-■5x■+■4■=■0 i x2■-■10x■+■26■=■0 2 2 j 3x ■+■5x■-■7■=■0 k 2x ■+■7x■-■10■=■0 l x2■-■11x■+■30■=■0 3 We10 ■By■using■the■discriminant,■determine■whether■the■equations■in■question■2■have: i two■rational■solutions ii two■irrational■solutions iii one■rational■solution■(two■equal■solutions) iv no■real■solutions. 4 With■the■information■gained■from■the■discriminant,■use■the■most■effi■cient■method■to■solve■each■ equation■in■question■2.■Where■appropriate,■round■answers■to■3■decimal■places. unDerstAnDing 5 Consider■the■equation■3x2■+■2x■+■7■=■0. a What■are■the■values■of■a,■b■and■c? b What■is■the■value■of■b2■-■4ac? c How■many■real■solutions,■and■hence■x-intercepts,■are■there■for■this■equation? 6 Consider■the■equation■-6x2■+■x■+■3■=■0. a What■are■the■values■of■a,■b■and■c? b What■is■the■value■of■b2■-■4ac? c How■many■real■solutions,■and■hence■x-intercepts,■are■there■for■this■equation? d With■the■information■gained■from■the■discriminant,■use■the■most■effi■cient■method■to■solve■

the■equation.■Give■an■exact■answer. 7 mC ■The■discriminant■of■the■equation■x2■-■4x■-■5■=■0■is: a 36 B 11 c 4 D 0 E -4 8 mC ■Which■of■the■following■quadratic■equations■has■two■irrational■solutions? a x2■-■8x■+■16■=■0 B 2x2■-■7x■=■0 c x2■+■8x■+■9■=■0 2■ 2■ D x -■4■=■0 E x -■6x■+■15■=■0 Chapter 8 Quadratic equations

267

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships 9 mC ■The■equation■x2■=■2x■-■3■has: a two■rational■solutions c no■solutions E one■rational■and■one■irrational■solution

B exactly■one■solution D two■irrational■solutions

reAsoning 10 Find■the■value■of■k if■x2■-■2x - k■=■0■has■one■solution. 11 Find■the■values■of■m■for■which■mx2■-■6x■+■5■=■0■has■one■solution. 12 Find■the■values■of■n when■x2■-■3x■-■n■=■0■has■two■solutions. 13 Show■that■3x2■+■px■-■2■=■0■will■have■real■solutions■for■all■values■of■p. 14 The■path■of■a■dolphin■as■it■leaps■out■of■the■water■can■be■modelled■by■the■equation■

h■=■-0.4d2■+■d,■where■h■is■the■dolphin’s■height■above■water■and■d■is■the■horizontal■distance■ from■its■starting■point.■Both■h■and■d■are■in■metres.

a How■high■above■the■water■is■the■dolphin■when■it■has■travelled■2■■m■horizontally■from■its■

starting■point? b What■horizontal■distance■has■the■dolphin■covered■when■it■fi■rst■reaches■a■height■of■25■■cm? c What■horizontal■distance■has■the■dolphin■covered■when■it■next■reaches■a■height■of■25■■cm?■

Explain■your■answer. d What■horizontal■distance■does■the■dolphin■cover■in■one■leap?■(Hint:■What■is■the■value■of■

h■when■the■dolphin■has■completed■its■leap?)

eBook plus

Digital doc

WorkSHEET 8.2 doc-5264

268

e Can■this■dolphin■reach■a■height■of: i 0.5■■m ii 1■■m■during■a■leap?

■ ow■can■you■determine■this■without■actually■solving■ H the■equation? f Find■the■greatest■height■the■dolphin■reaches■during■ a■leap.

maths Quest 10 for the Australian Curriculum

refleCtion 

 

What does the discriminant tell us?

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships

8e

solving a quadratic equation and a linear equation simultaneously There■are■occasions■when■it■is■necessary■to■fi■nd■the■intersections■points■(if■any)■of■a■quadratic■ and■a■linear■equation. ■■ There■are■three■possible■outcomes■in■this■situation. 1.■ The■straight■line■can■intersect■twice■with■the■parabola. y ■■

eBook plus

Interactivity Simultaneous quadratic equations

int-2784

0

2.■ ■The■straight■line■can■be■a■tangent■to■the■parabola.■In■ this■case■the■straight■line■touches■the■parabola■at■ one■point.

y

0

3.■ ■The■straight■line■may■not■intersect■at■all■with the■parabola.

■■ ■■ ■■ ■■

x

y

0

■■

x

x

A■quadratic■equation■can■be■solved■simultaneously■with■a■linear■equation■using■the■ substitution■method. The■x2■term■in■the■quadratic■equation■makes■it■impractical■to■use■the■elimination■method. Each■equation■is■best■written■with■y■as■the■subject■so■that■the■right-hand■side■of■each■equation■ can■then■be■set■equal■to■one■another. The■result■can■then■be■simplifi■ed■to■produce■a■new■quadratic■equation. The■quadratic■equation■will■have■two■solutions■if■the■straight■line■cuts■the■parabola■twice,■ one■solution■if■the■straight■line■is■a■tangent■to■the■parabola,■and■no■solution■if■the■line■does■ not■intersect■the■parabola.

Chapter 8 Quadratic equations

269

number AND algebra • Linear and non-linear relationships

Worked Example 11

Solve the simultaneous equation pair y = x2 + 2x + 2 and y = 7 - 2x. Think

Write

1

Write the equations, one under the other, and number them.

y = x2 + 2x + 2 y = 7 - 2x

2

Both equations are written with y as the subject, so equate them.

x2 + 2x + 2 = 7 - 2x

3

Move every term to the left-hand side to create a new quadratic equation.

x2 + 4x - 5 = 0

4

Factorise and use the Null Factor Law to solve the quadratic equation.

(x + 5)(x - 1) = 0 x + 5 = 0   or  x - 1 = 0 x = -5  or x=1

5

Substitute each answer for x into equation [2] to find the corresponding values of y.

Substituting x = -5 into [2]:   y = 7 - 2(-5)   y = 17 Substituting x = 1 into [2]:   y = 7 - 2(1)   y=5

6

Answer the question.

Solution: x = -5, y = 17  (-5, 17)  or x = 1, y = 5  (1, 5)

7

Check the answers by substituting the point of intersection into equation [1].

Check: Substitute (-5, 17) into y = x2 + 2x + 2 RHS = (-5)2 + 2(-5) + 2 = 25 - 10 + 2 = 17 = LHS Substitute (1, 5) into y = x2 + 2x + 2 RHS = (1)2 + 2(1) + 2 =1+2+2 =5 = LHS Both solutions are correct.

■■ ■■

[1] [2]

When there is only one solution to the simultaneous equation pair, the straight line is a tangent to the parabola. When this situation arises, the new quadratic equation formed will be a perfect square.

Worked Example 12

Solve the simultaneous equation pair y = x2 - 5x + 2 and y = x - 7. Think

270

Write

1

Write the equations, one under the other, and number them.

y = x2 - 5x + 2   [1] y = x - 7 [2]

2

Both equations are written with y as the subject, so equate them.

x2 - 5x + 2 = x - 7

Maths Quest 10 for the Australian Curriculum

number AND algebra • Linear and non-linear relationships

3

Move every term to the left-hand side to create a new quadratic equation.

x2 - 6x + 9 = 0

4

Factorise and use the Null Factor Law to solve the quadratic equation.



5

Substitute for x into equation [2] to find the corresponding value of y.

Substituting x = 3 into [2]: y=3-7 = -4

6

Answer the question.

Solution: x = 3, y = -4  (3, -4)

7

Check the answer by substituting the point of intersection into equation [1].

Check: Substitute into y = x2 - 5x + 2 RHS = (3)2 -5(3) + 2 = 9 - 15 + 2 = -4 = LHS The solution is correct.

■■

(x - 3)2 = 0 x-3=0 x=3

When there is no intersection of the parabola and the straight line, the discriminant has a value less than zero.

Worked Example 13

Show that the equations y = x2 + x + 4 and y = 2x - 1 have no solution when solved simultaneously. Think

Write

1

Write the equations, one under the other, and number them.

y = x2 + x + 4 y = 2x - 1

2

Both equations are written with y as the subject, so equate them.

x2 + x + 4 = 2x - 1

3

Move every term to the left-hand side to create a new quadratic equation.

x2 - x + 5 = 0

4

If the equations are to have no solution, ■ then D < 0.

D = b2 - 4ac = (-1)2 - 4 ì 1 ì 5 = 1 - 20 = -19 0 (positive) Perfect square Not a perfect square 2 rational ■ solutions

2 irrational (surd) solutions

Graph does not cross■ Graph touches Graph crosses the x-axis twice or touch the x-axis the x-axis

Solving a quadratic equation and a linear equation simultaneously ■■ ■■

274

To solve a quadratic equation with a linear equation we use the substitution method. Make y the subject of both the quadratic equation and the linear equation, and then equate the right-hand side expressions formed. This will leave a new quadratic equation to solve for x.

Maths Quest 10 for the Australian Curriculum

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships

■■ ■■ ■■

If■the■new■quadratic■equation■has■two■solutions,■both■must■be■substituted■back■into■one■of■the■ original■equations■to■fi■nd■the■corresponding■values■of■y. In■some■cases■there■will■only■be■one■value■of■x.■This■occurs■when■the■linear■equation■is■a■ tangent■to■the■parabola. In■some■cases■there■will■be■no■solution■to■a■straight■line■and■a■parabola.■If■this■is■the■case,■the■ discriminant■of■the■equation■formed■when■solving■will■be■less■than■0,■that■is,■b2■-■4ac■■1,■so■the■graph■is■narrower■than■that■ of■y■=■x2.

3

The■dilation■doesn’t■change■the■turning■point.

The■turning■point■is■(0,■0).

eBook plus

Interactivity Vertical translation of y = x2 + c

1 ■ 1, the graph is narrower than that of y = x2.

y

y = x2

(0, 0)

y = -2x2

290

Maths Quest 10 for the Australian Curriculum

x

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships

remember

1.■ If■the■graph■of y =■x2■is■translated■k■units■vertically,■the■equation■becomes■y =■x2■+■k. 2.■ If■the■graph■of y =■x2■is■translated■h■units■horizontally,■the■equation■becomes■ y =■(x■-■h)2. 3.■ If■the■graph■of y =■x2■is■dilated■by■factor■a,■the■graph■becomes■narrower■if■a■>■1■and■ wider■if■0■ 1.

y = -2(0 + 1)2 + 6 = -2 ì 1 + 6 =4 The y-intercept is 4.

-2(x + 1)2 + 6 = 0 2(x + 1)2 = 6 (x + 1)2 = 3 x + 1 = 3  or  x + 1 = - 3 x = -1 + 3      x = -1 - 3 The x-intercepts are -1 - 3 and -1 + 3 (or approximately -2.73 and 0.73).

vi

y (–1, 6) 4

Label the graph. –1 – 3

0

–1 + 3

x

y = –2(x + 1)2 + 6

Chapter 9 Functions

295

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships

Unless■otherwise■stated,■exact■values■for■the■intercepts■should■be■shown■on■sketch■graphs.■ remember

1.■ If■the■equation■of■a■parabola■is■in■turning■point■form, y =■a(x■-■h)2■+■k,■then■the■turning■ point■is■(h,■k). 2.■ If■a■is■positive,■the■graph■is■upright■with■a■minimum■turning■point. 3.■ If■a■is■negative,■the■graph■is■inverted■with■a■maximum■turning■point. 4.■ If■the■magnitude■of■a■is■greater■than■1,■the■graph■is■narrower■than■the■graph■of y =■x2. 5.■ If■the■magnitude■of■a■is■between■0■and■1,■the■graph■is■wider■than■the■graph■of y =■x2. 6.■ To■fi■nd■the■y-intercept,■substitute x =■0■into■the■equation. 7.■ To■fi■nd■the■x-intercepts,■substitute y =■0■into■the■equation■and■solve■for■x.■ exerCise

9C inDiviDuAl pAthWAys eBook plus

Activity 9-C-1

Reviewing turning point form doc-5080 Activity 9-C-2

Turning point form doc-5081 Activity 9-C-3

Interpreting turning point form trends doc-5082

sketching parabolas in turning point form FluenCy 1 We8 ■For■each■of■the■following■equations,■state■the■coordinates■of■the■turning■point■of■the■

graph■and■whether■it■is■a■maximum■or■a■minimum. b y =■(x■+■2)2■-■1 e y =■-(x■-■5)2■+■3

c y =■(x■+■1)2■+■1 f y =■(x■+■2)2■-■6

g y =■(x■+ 2 )2■-

i

a y =■(x■-■1)2■+■2 d y =■-(x■-■2)2■+■3 1

3 4

1

h y =■(x■- 3)2■+

2 3

y =■(x■+■0.3)2■-■0.4

2 For■each■of■the■following■state: i the■coordinates■of■the■turning■point ii whether■the■graph■has■a■maximum■or■a■minimum■turning■point iii whether■the■graph■is■wider,■narrower■or■the■same■width■as■that■of y =■x2. a y =■2(x■+■3)2■-■5 b y =■-(x■-■1)2■+■1 c y =■-5(x■+■2)2■-■4 1 1 1 1 f y =■0.2(x■+ )2■-■ 2 d y =■ (x■-■3)2■+■2 e y =■- (x■+■1)2■+■7 4

5

2

3 Select■the■equation■that■best■suits■each■of■the■following■graphs. i ii y y

iii

y

3 1 0

x

-1 0

x

0

x

2

-3 iv

v

y

vi

y

y

3 0 1 -2

0

x

a y =■(x■-■1)2■-■3 d y =■-(x■+■2)2■+■3 296

maths Quest 10 for the Australian Curriculum

0 -1

b y =■-(x■-■2)2■+■3 e y =■-x2■+■1

x -3 c y =■x2■-■1 f y =■(x■+■1)2■-■3

x

number AND algebra • Linear and non-linear relationships

1

1 3

4   MC  a The translations required to change y = x2 into y = (x - 2 )2 + are: 1 2 1 , 2 1 , 3

a right , up c right e right

1 3 1

down 3 up

1 2 1 , 2

b left , down d left

1 3

1

up 3

1 2 1

1 2

1 3

b For the graph 4 (x - )2 + , the effect of the

1 4

on the graph is:

no effect to make the graph narrower to make the graph wider to invert the graph 1 e to translate the graph up of a unit 4 a b c d

c Compared to the graph of y = x2, y = -2(x + 1)2 - 4 is: a inverted and wider b inverted and narrower c upright and wider d upright and narrower e inverted and the same width d A graph with minimum turning point (1, 5) and which is narrower than the graph of

y = x2 is: a y = (x - 1)2 + 5

b y = (x + 1)2 + 5

c y = 2(x - 1)2 + 5

d y = 2(x + 1)2 + 5

1 2

1 2

e y = (x - 1)2 + 5 e Compared to the graph of y = x2, the graph of y = -3(x - 1)2 - 2 has the following

features. Maximum TP at (-1, -2), narrower Maximum TP at (1, -2), narrower Maximum TP at (1, 2), wider Minimum TP at (1, -2), narrower Minimum TP at (-1, -2), wider 5   WE9  Determine i the y-intercept and ii the x-intercepts (where they exist) for the parabolas with equations: a y = (x + 1)2 - 4 b y = 3(x - 2)2 c y = -(x + 4)2 - 2 2 2 d y = (x - 2) - 9 e y = 2x + 4 f y = (x + 3)2 - 5 a b c d e

Understanding 6   WE10  For each of the following:   i write the coordinates of the turning point  ii state whether the graph has a maximum or a minimum turning point iii state whether the graph is wider, narrower or the same width as the graph of y = x2 iv find the y-intercept  v find the x-intercepts vi sketch the graph. a y = (x - 4)2 + 2 b y = (x - 3)2 - 4 c y = (x + 1)2 + 2 2 2 d y = (x + 5) - 3 e y = -(x - 1) + 2 f y = -(x + 2)2 - 3 2 2 g y = -(x + 3) - 2 h y = 2(x - 1) + 3 i y = -3(x + 2)2 + 1 7 Consider the equation 2x2 - 3x - 8 = 0. a Complete the square. b Use the result to determine the exact solutions to the original equation. c Determine the turning point of y = 2x2 - 3x - 8 and indicate its type. Chapter 9 Functions

297

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships 8 a■ ■Find■the■equation■of■a■quadratic■which■has■a■turning■point■of■(-4,■6)■and■has■an■x-intercept■

at■(-1,■0).■ b State■the■other■x-intercept■(if■any).■ reAsoning 9 The■price■of■shares■in■fl■edgling■company■‘Lollies’r’us’■plunged■dramatically■one■afternoon,■

following■the■breakout■of■a■small■fi■re■on■the■premises.■However,■Ms■Sarah■Sayva■of■Lollies■ Anonymous■agreed■to■back■the■company,■and■share■prices■began■to■rise. ■ Sarah■noted■at■the■close■of■trade■that■afternoon■that■the■company’s■share■price■followed■ the■curve:■P■=■0.1(t■-■3)2■+■1■where■$P■is■the■price■of■shares■t■hours■after■noon. a Sketch■a■graph■of■the■relationship■between■time■and■share■price■to■represent■the■ situation. b What■was■the■initial■share■price? c What■was■the■lowest■price■of■shares■that■afternoon? d At■what■time■was■the■price■at■its■lowest? e What■was■the■fi■nal■price■of■‘Lollies’r’us’■shares■as■trade■closed■at■5■pm? 10 Rocky■is■practising■for■a■football■kicking■competition.■After■being■kicked,■the■path■that■the■ball■ follows■can■be■modelled■by■the■quadratic■relationship: h=

−1 (d − 15)2 + 8, 30

where■h■is■the■vertical■distance■the■ball■reaches■(in■metres),■and■d■is■the■horizontal■distance■ (in■metres).■ a Determine■the■initial■vertical■height■of■

the■ball. b Determine■the■exact■maximum■horizontal■

distance■the■ball■travels.■ c Write■down■both■the■maximum■height■and■ the■horizontal■distance■when■the■maximum■ height■is■reached.■

9D eBook plus

Interactivity Sketching parabolas

int-2785

reFleCtion 

 

Does a in the equation y = a(x - h)2 + k have any impact on the turning point?

sketching parabolas of the form y = ax 2 + bx + c ■■ ■■

The■standard■form■of■a■quadratic■equation■is y =■ax2■+■bx■+■c■where■a,■b■and■c■are■ constants.■ As■seen■in■the■previous■section,■to■sketch■a■parabola■we■need■to■know: 1.■y-intercept 2.■x-intercepts 3.■the■nature■of■the■turning■point;■that■is,■whether■it■is■a■maximum■or■a■minimum■ turning■point 4.■the■coordinates■of■the■turning■point.

Finding the turning point of a parabola when the equation is not in turning point form 1. Changing to turning point form ■■ As■seen■previously,■when■an■equation■is■written■in■turning■point■form■the■coordinates■of■the■ turning■point■can■be■read■from■the■equation.■That■is,■the■coordinates■of■the■turning■point■for y =■a(x■-■h)2■+■k■are■(h,■k). ■■ To■change■to■turning■point■form■we■use■the■completing■the■square■method.■ 298

maths Quest 10 for the Australian Curriculum

number AND algebra • Linear and non-linear relationships

Worked Example 11

Change each of the following equations into turning point form and hence state the coordinates of the turning point for each one. a  y = x2 + 6x + 2 b  y = -2x2 + 7x - 3 Think a

b

Write

1

Write the equation.

2

Complete the square:  (i) Halve the coefficient of x and square it. (ii) Add this new term to the equation, placing it after the x-term. (iii) Balance the equation by subtracting the same term from the right-hand side of the equation.

3

Factorise the perfect square and simplify the remaining terms.

4

State the coordinates of the turning point.

1

Write the equation.

2

Take out a common factor.

3

Complete the square:  (i) Halve the coefficient of x and square it. (ii) Add this new term to the equation, placing it after the x-term. (iii) Balance the equation by subtracting the same term from the right-hand side of the equation.

a y = x2 + 6x + 2  6

2

 6

2

= x2 + 6x +  2  −  2  + 2 = x2 + 6x + (3)2 - (3)2 + 2 = (x2 + 6x + 9) - 9 + 2

= (x + 3)2 - 9 + 2 = (x + 3)2 - 7 The turning point is (-3, -7). b y = -2x2 + 7x - 3

= -2  x 2 − 7 x + 2



= −2  x 2 − 72 x + = = =

3  2 1  2

2

1

7 2

× − 72  −  2 × − 2  +

 2 2 3  7 −2  x 2 − 2 x +  − 47  −  − 47  + 2     2 7 49 3 49  8 −2  x − x + 16  − 16 + 2 × 8  2    2 7  −2  x − x + 49  − 49 + 24  2 16 16 16   2   −2  x − 7  − 25  2 16  

4

Factorise the perfect square and simplify the remaining terms.

=

5

Multiply the common factor by each term in the square brackets so that the equation is in turning point form.

 = −2  x − 47  +

6

State the coordinates of the turning point.

2

3  2 

25 8

The turning point is

 7 25  ,  4 8 

or 13 , 31  . 4 8

Worked Example 12

Sketch the graph of y = 2x2 - 4x - 2 using the completing the square method to find the coordinates of the turning point. Show all relevant points. Think

Write/draw

1

Write the equation.

y = 2x2 - 4x - 2

2

Find the y-intercept by substituting x = 0.

y-intercept: when x = 0, y=0-0-2 = -2 The y-intercept is -2. Chapter 9 Functions

299

number AND algebra • Linear and non-linear relationships

3

Find the x-intercepts by substituting y = 0. Factors cannot be easily found, so use the quadratic formula to solve for x.

x-intercepts: when y = 0, 0 = 2x2 - 4x - 2 2x2 - 4x - 2 = 0 2(x2 - 2x - 1) = 0 x2 - 2x - 1 = 0 − b ± b 2 − 4 ac 2a where a = 1, b = -2, c = -1 x=

2 ± (−2)2 − 4(1)(−1) 2 2± 8 x= 2 x=

2±2 2 2 The x-intercepts are 1 - 2 and 1 + 2 (approx. -0.41 and 2.41). x=

4

Find the turning point by taking out a common factor from the original equation.

y = 2x2 - 4x - 2 = 2(x2 - 2x - 1)

5

Complete the square:  (i)  Halve the coefficient of the x-term and square it. (ii) Add this new term to the equation, placing it after the x-term. (iii) Balance the equation by subtracting the same term from the right-hand side of the equation.

  2 2 2 2 = 2  x 2 − 2 x + − 2 − − 2 − 1   = 2[(x2 - 2x + (-1)2) - (-1)2 - 1] = 2[(x2 - 2x + 1) - 1- 1]

6

Factorise the perfect square and simplify the remaining terms.

= 2[(x - 1)2 - 2]

7

Multiply the common factor by each term in the square brackets so that the equation is in turning point form. State the coordinates of the turning point.

= 2(x - 1)2 - 4

State the nature of the turning point. As the sign of a is positive, the parabola has a minimum turning point.

The parabola has a minimum turning point.

8 9

10

Sketch the graph.

( ) ( )

Turning point is (1, -4).

y y = 2x2 - 4x - 2

11

Label the graph. 1- 2 0 1 -2 -4

1+ 2 x

(1, -4)

2. Using the x-intercepts to find the x-coordinate of the turning point ■■ A parabola is symmetrical, so the x-intercepts are the same distance from the axis of symmetry (the line which divides the graph exactly in half ). This means that the x-coordinate of the turning point is halfway between the x-intercepts. 300

Maths Quest 10 for the Australian Curriculum

number AND algebra • Linear and non-linear relationships

■■

y

In this graph, the x-intercepts are exactly 1 unit from the y-axis which is the axis of symmetry. One way to find the x-coordinate of the turning point is to calculate the average of the two x-intercepts. In this case, 1 + −1 = 0 is the x-coordinate of the

y = x2 - 1

2

turning point. The y-coordinate of the turning point can then be found by substituting the x-coordinate into the equation.

-1 0 -1

1

x

Worked Example 13

Sketch the graph of y = x2 - 10x + 21, using the x-intercepts to find the coordinates of the turning point. Think

Write/draw

1

Write the equation.

y = x2 - 10x + 21

2

Find the y-intercept by substituting x = 0.

y-intercept: when x = 0, y = 0 - 0 + 21 = 21 The y-intercept is 21.

3

Find the x-intercepts by substituting y = 0.

x-intercepts: when y = 0, x2 - 10x + 21 = 0

4

Factorise and solve for x by using the Null Factor Law.

5

Find the x-coordinate of the turning point by averaging the x-intercepts, x1 + x 2 . (This is halfway between the 2 x-intercepts.)

(x - 7)(x - 3) = 0 x - 7 = 0  or  x - 3 = 0 x = 7      x = 3 The x-intercepts are 3 and 7. 3+7 x-value of the turning point = 2 10 = 2 =5

6

Find the y-coordinate of the turning point by substituting the x-coordinate into the equation and solving for y.

When x = 5, y = 52 - 10(5) + 21 = -4

7

State the coordinates of the turning point.

The turning point is (5, -4).

8

State the nature of the turning point. As the sign of a is positive, the parabola has a minimum turning point.

Parabola has a minimum turning point.

9

Sketch the graph.

10

Label the graph.

y

y = x2 - 10x + 21

21

0 -4

3 5 7 (5, -4)

x

Chapter 9 Functions

301

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships

remember

1.■ If■the■equation■is■in■the■form y =■ax2■+■bx■+■c,■the■coordinates■of■the■turning■point■can■ be■found■by: (a)■ ■using■the■completing■the■square■method■to■change■the■equation■into■turning■point■form (b)■■fi■nding■the■x-coordinate■of■the■point■exactly■halfway■between■the■two■x-intercepts.■ This■is■the■x-coordinate■of■the■turning■point.■Then■substitute■the x-value■into■the■ equation■to■fi■nd■the■y-coordinate. −b .■Then■substitute■the■x-value■into■the■equation■to■fi■nd■the■y-coordinate. (c)■ ■using■ x = 2a 2.■ The■graph■should■also■show■both■the■y-intercept■and■the■x-intercepts■of■the■parabola■if■ they■exist. exerCise

9D inDiviDuAl pAthWAys

sketching parabolas of the form y = ax 2 + bx + c FluenCy 1 We11 ■Change■each■of■the■following■equations■into■turning■point■form■and■write■the■

eBook plus

Activity 9-D-1

Understanding parabola sketching doc-5083

2

Activity 9-D-2

Parabola sketching doc-5084 Activity 9-D-3

Sketching tricky parabolas doc-5085

3 eBook plus

Digital doc

SkillSHEET 9.3 doc-5268

4 eBook plus

Digital doc

SkillSHEET 9.4 doc-5269

5

coordinates■of■the■turning■point■for■each■one. a y =■x2■+■4x■-■2 b y =■x2■+■12x■-■4 c y =■x2■-■8x■+■6 2 2 d y =■x ■-■2x■+■12 e y =■x ■+■3x■+■1 f y =■x2■+ x■-■2 2 2 g y =■x ■+■7x■+■2 h y =■2x ■+■4x■+■8 i y =■3x2■-■12x■+■6 We12 ■Sketch■the■graph■of■each■of■the■following■using■the■completing■the■square■method■to■ fi■nd■the■coordinates■of■the■turning■point.■Show■all■relevant■points. a y =■x2■+■2x■-■5 b y =■x2■-■4x■+■7 c y =■x2■+ 6x■-■3 d y =■x2■-■5x■+■1 e y =■-x2■-■5x■+■1 f y =■-x2■+ x■-■3 g y =■3x2■+■3x■-■12 h y =■-5x2■+■10x■-■35 2 i y =■-7x ■-■7x■+■49 We13 ■Sketch■the■graph■of■each■of■the■following,■using■the■x-intercepts■to■fi■nd■the■coordinates■ of■the■turning■point.■ a y =■x2■+ x -■12 b y =■x2■-■12x■+■32 2 c y =■x ■-■8x■-■9■ d y =■-x2■-■6x■-■8 2 e y =■-x ■-■6x■+■27 f y =■-x2■+■2x■+■35 2 g y =■x ■+■4x■- 5 Sketch■the■graphs■of■each■of■the■following. a y =■2x2■-■17x■-■9 b y =■3x2■-■23x■+■14 2 c y =■5x ■+■27x■+■10 d y =■6x2■+■7x■-■3 2 e y =■-2x ■+■7x■+■4 f y =■-2x2■+■11x■+■21 2 g y =■-6x ■+■5x■+■6 h y =■-18x2■+■67x■-■14 2 i y =■2x ■-■7x■+■8 y mC ■a■ The■equation■that■best■suits■the■graph■shown■is: a y =■x2■+■2x■-■24 b y =■3x2■+■6x■-■72 c y =■x2■-■2x■-■24 d y =■3x2■-■6x■-■72 0 (-6, 0) (4, 0) x e y =■2x2■+■4x■-■72 -72

302

maths Quest 10 for the Australian Curriculum

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships

eBook plus

Digital doc

b The■equation y =■x2■+■5x■-■24■is■best■represented■by: a b y

SkillSHEET 9.5 doc-5270

y 24

0

-3

eBook plus

Digital doc

8

x

-3 0

x

8

-24

SkillSHEET 9.6 doc-5271

c

d

y

y 24

0

-8

3

x

3

x

-8

0

3

x

2

3

x

-24

e

y

0

-8

-12

6 Match■each■of■the■following■graphs■with■the■appropriate■equation. a

b

y

y 0

0

c

4

x

d

y

1

y 9

-2 0

6 x

-3

0

3

x

Chapter 9 Functions

303

number AND algebra • Linear and non-linear relationships

e

f

y

y

5 0 1

g

x

5

0

h

y

-1

0

x

5– 2

y

x

1

3– 2

0

1

x

-1

i   y = x2 - 6x + 5 iii   y = -x2 + 9   v   y = 2x - 2x2 vii   y = -x2 + 5x - 6

 ii   y = x2 - 1     iv   y = 3x2 - 12x     vi   y = -x2 + 4x + 12 viii   y = -4x2 + 16x - 15

Understanding 7 a Find the equations for parabolas A and B in both turning point form and standard form. b For the two parabolas, A and B, state the transformations on A to create B. y 8 A

6

(3, 4)

4 2 -6

0 -4 -2 (-1, -1) -2 B

2

4

6x

-4 -6

8 Use a graphical method on your calculator to find the points of intersection of these parabolas,

correct to 2 decimal places. y = x2 + 6x + 5 y = -x2 - 4x + 2 9 Use simultaneous equations to show that the parabola y = x2 + 6x + 5 and the straight line y = 2x + 1 intersect at one point only. a Find the coordinates of this point of intersection. b Verify this by graphing the simultaneous equations on your calculator and finding the point of intersection. 304

Maths Quest 10 for the Australian Curriculum

number AND algebra • Linear and non-linear relationships Reasoning 10 Consider the family of parabolas y = x2 + px + 5. a Sketch, on the same set of axes, the parabolas for p = -4, -2, 2, 4. b Discuss the effect of p on the graph. 11 The height, h metres, of a model rocket above the ground t seconds after launching is given by

the equation h = 4t(50 - t) for 0 Ç t Ç 50. a Sketch the graph of the rocket’s flight. b Determine the height of the rocket when it is launched. c What is the greatest height the rocket reaches? d After how long does the rocket reach the greatest height? e How long is the rocket in the air? 12 A farmer decides to fence a new rectangular paddock with the greatest possible area, using an existing fence for one side and 40 metres of fencing to make the other three sides. Let the area of the paddock be defined as A  m2. a Write an equation using x and y to describe the area of the paddock. b Write an equation relating x and y and the length of fencing Existing fence available. c Rearrange the equation so that y is the subject. d Substitute this value of y into the equation for the area. xm e Using the intercept method find the coordinates of the New paddock turning point. xm f Sketch the graph. ym g Use the graph to find the maximum area of the paddock and its dimensions. 13 A daring feat performed in Acapulco, Mexico, is for a person to dive from a cliff into the ocean. Starting from about 17 metres above the water, ■ the height, h (in metres), of a diver t seconds after he jumps can be represented by the equation ■ h = -4.9t2 + 1.5t + 17. a Sketch a graph to represent the diver’s height after jumping. b How long does it take for the diver to reach the water (to the nearest second)? c When does the diver reach his maximum height above the water? Give your answer correct to 2 decimal places. d What is the diver’s greatest height above the water? Give your answer correct to the nearest cm. 14 A farmer has 300 metres of fencing with which to fence 3 sides of a rectangular paddock. a Using the method described in question 8, find an equation relating the area and the width of the paddock. b Sketch the graph. c Use the graph to find the greatest possible area for the paddock and its dimensions. Chapter 9 Functions

305

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships 15 In■preparation■for■an■Archery■Games■opening■ceremony,■an■archer■shoots■a■fi■ery■arrow■that■

follows■a■parabolic■path■into■a■cauldron.■The■cauldron■is■15■metres■high■with■its■centre■a■ distance■of■10■metres■from■a■wall. The■archer■releases■the■arrow■at■a■distance■of■34■metres■horizontally■from■the■wall■as■ represented■by■the■diagram■below. P

Wall

Cauldron

15 m 10 m

34 m

This■event■can■be■represented■on■a■set■of■axes■and■the■path■of■the■arrow■can■be■modelled■by■ a■quadratic■equation■of■the■form■y■=■ax2■+■bx■+■c. a Given■that■if■the■cauldron■was■not■in■the■way,■the■arrow■would■land■2■metres■from■the■ wall.■Show■that■the■path■of■the■arrow■can■be■represented■by■the■equation −5 2 45 85 y= x + x− . 64 16 16 b Use■your■calculator■to■graph■the■equation■ reFleCtion    and■hence■fi■nd■the■exact■coordinates■ Which feature is most clearly displayed in where■the■arrow■reaches■its■maximum■ an equation of the type y = ax 2 + bx + c: height,■P. the x-intercept(s), the y-intercept or the c Convert■the■equation■in■part■a■to■turning■ turning point? point■form■to■show■that■your■answer■to■b■ is■correct.

eBook plus

Digital doc

WorkSHEET 9.2 doc-5273

9e

exponential functions and their graphs ■■ ■■

Relationships■of■the■form y =■ax■are■called■exponential functions■with■base■a,■where■a■is■a■ real■number■not■equal■to■1,■and x is■the■index■power■or■exponent. The■term■‘exponential’■is■used,■as■the■independent■variable x is■the■exponent■(or■index).

WorkeD exAmple 14

Complete the table of values below and use it to plot the graph of y = 2x. x

-4

-3

-2

-1

0

1

2

3

4

y think 1

2

306

Substitute■each■value■of x into■the■function y =■2x■to■ obtain■the■corresponding y-value. Plot■each■point■generated■on■a■set■of■axes.

maths Quest 10 for the Australian Curriculum

Write/DrAW

x

-4 -3 -2 -1

0■

1

2

3

4

y

1 16

1

2

4

8

16

1 8

1 4

1 2

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships

3

y

Join■with■a■smooth■curve.

y = 2x

18 16 14 12 10 8 6 4 2 -4 -3 -2 -1 0

4

x

1 2 3 4

Label■the■graph.

eBook plus

Interactivity Exponential graphs

int-1149

The■graph■in■Worked■example■14■has■several■important■features. ■■ The■graph■passes■through■(0,■1).■That■is,■the■y-intercept■is■1.■The■graph■of■any■equation■in■the■ form y =■ax■will■pass■through■this■point. ■■ The■graph■passes■through■the■point■(1,■2).■All■graphs■of■the■form y =■ax■will■pass■through■the■ point■(1,■a). ■■ y >■0■for■all■values■of■x.■You■will■notice■that■for■negative■values■of x, the■graph■gets■very■close■ to■but■will■never■touch■the■x-axis.■When■this■occurs,■the■line■that■the■graph■approaches■is■ called■an■asymptote.■The■equation■of■the■asymptote■for y =■ax■is y =■0;■i.e.■the■x-axis.

WorkeD exAmple 15 a Plot the graph of y = 3 ì 2x for -3 Ç x Ç 3. c Write the equation of the horizontal asymptote.

b State the y-intercept.

think a

1

2

Write/DrAW

Prepare■a■table■of■values■taking x-values■ from■-3■to■3.■Fill■in■the■table■by■substituting■ each■value■of x into■the■given■equation.■

a

Draw■a■set■of■axes■on■graph■paper■to■plot■ the■points■from■the■table■and■join■them■with■ a■smooth■curve.

x

-3

-2

-1

0■

1

2

3

y

3 8

3 4

1 12

3

6

12

24

24 22 20 18 16 14 12 10 8 6 4 2 -3 -2 -1 0

3

y y = 3 ì 2x

y=0 1 2 3

x

Label■the■graph.

b

Locate■where■the■curve■cuts■the■y-axis.■ Alternatively,■fi■nd■the■y-value■for x =■0■in■the■table.

b The■y-intercept■is■3.

c

Find■an■imaginary■line■to■which■the■curve■gets■ closer■and■closer■but■does■not■cross.■As■it■is■a■ horizontal■asymptote,■the■equation■will■be■of■the■ form y =■constant.

c

The■equation■of■the■asymptote■is y =■0.

Chapter 9 Functions

307

number AND algebra • Linear and non-linear relationships

■■

■■

Compare the graphs drawn in Worked examples 14 and 15. When 2x was multiplied by a constant, the graph was dilated; that is, its width changed. Since the constant was a positive number greater than 1, the graph became narrower. If the constant had been a fraction or decimal between 0 and 1, the graph would have become wider. The following worked example considers the effect of a negative exponent.

Worked Example 16

Plot the graph of y = 3-x for -3 Ç x Ç 3, clearly showing the y-intercept and the horizontal asymptote. Think 1

Write/draw

Draw up a table of values.

2

Substitute the values of x into the equation to find the corresponding y-values.

3

Draw a set of axes, plot the points generated from the table and join with a smooth curve.

x

-3

-2

-1

0

1

2

3

y

27

 9

 3

1

1 3

1 9

1 27

y y = 3–x

28 26 24 22 20 18 16 14 12 10 8 6 4 2

-3 -2 -1 0

4

y=1 1 2 3

y=0 x

Label the graph.

Further exponential graphs ■■ ■■ ■■

Recall the dilation, reflection and translation rules for quadratic graphs. These rules also apply to exponential graphs. Adding to or subtracting from the basic function shifts the graph up or down the y-axis. y = ax + k  or  y = ax - k Adding to or subtracting from x shifts the graph left or right along the x-axis. y = ax - h  or  y = ax + h remember

1. Relationships of the form y = ax, where a ò 1 are called exponential functions with base a. 2. To obtain the graph of an exponential function, construct a table of values first and then plot the points from the table and join them with a smooth curve. Alternatively use a graphics calculator, CAS calculator or graphing software. 3. An asymptote is a line which the graph approaches but never cuts or touches. 4. Multiplying by a constant dilates the basic graph — that is, makes it narrower or wider. 5. If x is a negative number, the graph is reflected across the y-axis. 308

Maths Quest 10 for the Australian Curriculum

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships

exerCise

9e inDiviDuAl pAthWAys eBook plus

exponential functions and their graphs You■may■use■a■graphing■calculator■or■graphing■software■to■assist■you■in■this■exercise. FluenCy 1 a■ ■ We14 ■Complete■the■table■below■and■use■the■table■to■plot■the■graph■of y =■10x■for■

-2 Ç■x Ç■2.

Activity 9-E-1

Exploring exponential graphs doc-5086

Activity 9-E-3

Tricky exponential graphs doc-5088

eBook plus

Digital doc

SkillSHEET 9.7 doc-5274

-3

-2

-1

0

1

2

3

4

y

Activity 9-E-2

Features of exponential graphs doc-5087

-4

x

b Why■would■it■be■diffi■cult■to■draw■the■graph■for■-4 Ç■x Ç■4? 2 Plot■the■graph■of■each■of■the■following■exponential■functions. a y =■4x b y =■5x

c y =■6x

3 On■the■one■set■of■axes,■draw■the■graphs■of y =■2x, y =■2-x,■y =■3x, y =■3-x,■y =■4x and■y =■4-x. 4 Use■your■answer■to■question■3■to■describe■the■effect■of■increasing■the■value■of■a■on■the■graph■

of y =■ax. 5 We15 ■a■ Plot■the■graph■of y =■2■ì■3x■for■-3 Ç■x Ç■3. b State■the■y-intercept. c Write■the■equation■of■the■horizontal■asymptote. 6 Complete■the■following■table■of■values■and■then■plot■the■graph■of■y =■2x,■y =■3■ì■2x,■and■ y =■ 15■ì■2x on■the■same■set■of■axes. -3

x

-2

-1

0

1

2

3

2x 3■ì■2x 1 ■ì■2x 5

7 Study■the■graphs■in■question■6■and■state■the■effect■that■the■value■of■k■has■on■graphs■with■ 8 9 10 11

equation y =■k■ì■ax. We16 ■Plot■the■graph■of y =■2-x for■-3■Ç x Ç■3,■clearly■showing■the■y-intercept■and■the■ horizontal■asymptote. On■the■one■set■of■axes,■sketch■the■graphs■of y =■3x■and■y =■3-x. Use■your■answer■to■question■9■to■describe■the■effect■of■a■negative■index■on■the■graph■of y =■ax. a■ ■■Complete■the■table■of■values■below■and■use■the■points■generated■to■sketch■the■graph■of

( ).

y =■

1 2

x

-3

x

-2

-1

0

1

2

3

y 1 2

( ) ■and

b By■writing■ ■with■a■negative■index,■show■algebraically■that■the■functions y =■

y =■2-x■are■identical. 12 Draw■the■graphs■of y =■(1.2)x, y =■(1.5)x■and y =■(1.8)x. 13 a■ Draw■the■graph■of y =■10■ì■(1.3)x. b State■the■y-intercept. c Write■the■equation■of■the■horizontal■asymptote.

1 2

x

Chapter 9 Functions

309

number AND algebra • Linear and non-linear relationships 14 Use a calculator to draw the graphs of each of the following on the one set of axes. a y = 3x b y = 3x + 2 c y = 3x - 3 15 For the graphs drawn in question 14, state the equation of the horizontal asymptote. 16 Use your answers to questions 14 and 15 to state the effect that changing the value of c has on

the graph of y = 3x + c.

Understanding 17 Match each of the graphs (a–d) with the correct equation below (i–iv). y 10 8 6 4 2 (1, 1.6)

a

-10

0

-5

y 20 16 12 8 4

c

-4

0

-2

5

y 12 10 8 6 4 2

b

10

x -4

y 10 8 6 4 2

(1, 16)

2

4

x

0

-2

d

-4

(1, 12)

2

(1, 0.75)

0

-2

x

4

2

x

4

     i  y = 3 ì 4x   ii  y = 2 ì 8x iii  y = 3 ì 0.25x iv  y = 2 ì 0.8x 18   MC  The equation for the graph at right is: A y = 2x B y = 3x C y = 2 ì 3x D y = 3 ì 2x -x E y = 2

y 10 8 6 4 2 -4

-2

(1, 3)

0

2

19   MC  The graph of y = -3 ì 4x is best represented by: A

y

B

y x

0 (0, 1) 0

310

x

Maths Quest 10 for the Australian Curriculum

(0, -3)

4

x

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships

c

y

d

y

(0, 3) x

0 (0, -1) x

0

e

y

0

(0, -1)

x

20 On■the■same■set■of■axes■draw■the■graphs■of: a y =■4x b y =■4x + 1

c y =■4x -■3.

21 Use■your■answer■to■question■20■to■state■the■effect■that■changing■the■value■of■b■has■on■the■graph■

of■y =■4x - b. 22 Sketch■graphs■of■each■of■the■following■on■the■one■set■of■axes,■showing■the■y-intercept■and■the■ horizontal■asymptote.■(Remember■that■a■sketch■graph■shows■the■basic■shape■of■the■relationship■ and■its■key■features.■A■sketch■graph■is■not■drawn■by■plotting■points■from■a■table■of■values.) a y =■2x■and y =■2x■+■4 b y =■2x■and y =■2x■-■1 c y =■2x■and y =■2x■+■4 d y =■2x■and y =■2x■-■1 Check■your■answers■with■a■graphing■calculator■or■graphing■software. reAsoning 23 Myung-Hye■invests■$1000■at■10%■p.a.■interest■compounded■annually.■This■investment■can■be■

represented■by■the■function■A■=■1000■ì■(1.1)n,■where■A■is■the■amount■to■which■the■investment■ grows■and n is■the■number■of■years■of■the■investment. a Prepare■a■table■of■values■for■0■Ç n Ç■6.■Substitute■integer■values■of n into■the■equation■and■ use■a■calculator■to■determine■corresponding■values■of■A. b Plot■the■points■generated■by■the■table,■clearly■labelling■the■axes.■Join■the■points■with■a■ smooth■curve. c Use■the■table■of■values■or■the■graph■to■fi■nd■the■value■of■the■investment■after■3■years. 24 Kevin■buys■a■car■for■$40■■000.■The■car■depreciates■at■the■rate■of■15%■p.a.■The■value,■$V,■of■the■ car■after n years■can■be■given■by■the■equation■V■=■40■■000■ì■(0.85)n. a Prepare■a■table■of■values■for■ 0■Ç n Ç■5.■Substitute■integer■ values■of n into■the■equation■ and■use■a■calculator■to■fi■nd■ corresponding■values■of■V.■ Round■answers■to■the■nearest■ whole■number■as■required. b Plot■the■points■generated■by■the■table,■clearly■labelling■the■axes.■Join■the■points■with■a■ smooth■curve. c Describe■what■is■happening■to■the■value■of■the■car■as n increases. d Find■the■value■of■the■car■after■5■years.■Give■the■answer■to■the■nearest■dollar.■ Chapter 9 Functions

311

25 The graph shows the growth rate of two different bacteria. a Determine when each bacteria reaches a

population of 500  000. b Estimate the starting population of each bacteria. c Which bacteria grows at a faster rate? d When are the populations equal? 26 The rat population in Hamlin is very prolific; the rats double their population every 2 days. An initial count of rats in the town shows 2048 rats. a What is the rat population 10 days after the initial count? b Predict the population after 100 days. c Write an equation that enables you to predict the rat population. d Predict when the rat population will reach:   i 1 million ii 10 million iii 1 billion (1000 million).

9F

Bacteria population ( ì 1000)

number AND algebra • Linear and non-linear relationships

1400 1200 1000

B A

800 600 400 200 0

50 100 150 200 250 300 Time (s)

reflection 

 

What are two different transformations that can be done to exponential graphs?

The hyperbola ■■

k A hyperbola is a function of the form xy = k or y = . x

Worked Example 17

1 Complete the table of values below and use it to plot the graph of y = . x x

-3

-2

-1

1

-2

1 2

0

1

2

3

y Think 1

2

Write/draw

Substitute each x value into the function 1 y = to obtain the corresponding y value. x

1

x

-3 -2 -1 - 2

y

- 3 - 2 -1 -2 Undefined

1

-3 -2 -1

312

1

1 2

2

1

2

3

1

1 2

1 3

y

Draw a set of axes and plot the points from the table. Join them with a smooth curve.

■■

0

2 1 0

y = —1x 1 2 3 -1 -2

x

The graph in Worked example 17 has several important features. 1. There is no function value (y value) when x = 0. At this point the hyperbola is undefined. When this occurs, the line that the graph approaches (x = 0) is called a vertical asymptote.

Maths Quest 10 for the Australian Curriculum

number AND algebra • Linear and non-linear relationships

■■

2. As x becomes larger and larger, the graph gets very close to but will never touch the x-axis. The same is true as x becomes smaller and smaller. The hyperbola also has a horizontal asymptote at y = 0. 3. The hyperbola has two separate branches. It cannot be drawn without lifting your pen from the page and is an example of a discontinuous graph. k 1 As with exponential functions, graphs of the form y = are the same basic shape as y = x x with y values dilated by a factor of k.

Worked Example 18 a Plot the graph of y =

4 for -2 Ç x Ç 2. x

b  Write down the equation of each asymptote.

Think a

1

2

Write/draw

Prepare a table of values taking x values from -2 to 2. Fill in the table by substituting each x value into the given equation to find the corresponding y value.

a

1

x

-2 -1 - 2

y

-2 -4 -8 Undefined

Consider any lines that the curve approaches but does not cross.

■■

1 2

1

2

8

4

2

y

Draw a set of axes and plot the points from the table. Join them with a smooth curve. -2 -1

b

0

8 4 0

y = —4x x

1 2 -4 -8

b Vertical asymptote is x = 0.

Horizontal asymptote is y = 0.

Consider the effect of negative values of k.

Worked Example 19

Plot the graph of y =

−3 for -3 Ç x Ç 3. x

Think 1

2

Draw a table of values and substitute each x value into the given equation to find the corresponding y value. Draw a set of axes and plot the points from the table. Join them with a smooth curve.

Write/draw 1

x -3 -2 -1 - 2

0

1

  2   1      2   3

y   1 1.5   3   6 Undefined -6 -3 -1.5 -1 y 6 3 -3 -2 -1 0 -3

— y = –3 x

1 2 3 x

-6

Chapter 9 Functions

313

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships

remember

k 1.■ A■hyperbola■is■a■function■of■the■form■y = . x 2.■ To■obtain■the■graph■of■a■hyperbola,■construct■a■table■of■values.■Plot■the■points■and■ join■them■with■a■smooth■curve.■Alternatively,■use■a■graphics■calculator■or■a■computer■ graphing■package. 3.■ A■hyperbola■will■often■have■both■a■horizontal■and■a■vertical■asymptote.

exerCise

9F inDiviDuAl pAthWAys

the hyperbola You■may■use■a■graphics■calculator■or■computer■graphing■package■to■assist■you■in■this■exercise. FluenCy

eBook plus

Activity 9-F-1

1 We17 ■Complete■the■table■of■values■below■and■use■it■to■plot■the■graph■of■y =

Hyperbola graphs doc-5089 Exploring the hyperbola doc-5090 Activity 9-F-3

More hyperbola graphs doc-5091

-5

x

Activity 9-F-2

-4

-3

-2

-1

0

1

2

4

5

y 2 We18 ■a■ Plot■the■graph■of■each■hyperbola. b Write■down■the■equation■of■each■asymptote.

100 x 2 3 4 On■the■same■set■of■axes,■draw■the■graphs■of■y = , y = ■and■y = . x x x Use■your■answer■to■question■3■to■describe■the■effect■of■increasing■the■value■of■k■on■the■ k graph■of■y = . x −10 We19 ■Plot■the■graph■of■y = ■for■-5■Ç■x■Ç■5. x 6 −6 On■the■same■set■of■axes,■draw■the■graphs■of■y = ■and■y = . x x −k Use■your■answer■to■question■6■to■describe■the■effect■of■the■negative■in■y = . x 1 Complete■the■table■of■values■below■and■use■the■points■to■plot■y = .■State■the■equation■of■ x −1 the■vertical■asymptote. i y=

3 4

5 6 7 8

5 ■ x

x

ii■ y =

-3

-2

20 ■ x

-1

iii y =

0

1

2

y 9 Plot■the■graph■of■each■hyperbola■and■label■the■vertical■asymptote.

1 x−2 1 b y= x−3 1 c y= x +1 a y=

314

3

10 . x

maths Quest 10 for the Australian Curriculum

3

4

number AND algebra • Linear and non-linear relationships Understanding

1 . x−a 11 Sketch each of the following, showing the position of the vertical asymptote. −4 a y = x +1 2 b y = x −1 5 reflection    c y = x+2 How could you summarise the effect of the 12 Give an example of the equation of a transformations dealt with in this exercise on 1 hyperbola that has a vertical asymptote of: the shape of the basic hyperbola y = ? a x = 3 x b x = -10. 10 Use your answers to question 9 to describe the effect of a in y =

9G

The circle ■■ ■■

A circle is the path traced out by a point at a constant distance (the radius) from a fixed point (the centre). Consider the circles shown below right. The first circle has its centre at the origin and radius r.   Let P (x, y) be a point on the circle. y   By Pythagoras: x 2 + y2 = r 2. P(x, y)   This relationship is true for all points, P, on the circle. r y

■■

If the circle is translated h units to the right, parallel to the x-axis, and k units upwards, parallel to the y-axis, then: The equation of a circle, with centre (h, k) and radius r, is: (x - h)2 + (y - k)2 = r 2

x

x

The equation of a circle, with centre (0, 0) and radius r, is: x2 + y2 = r 2 y y k

P(x, y) (y - k) (x - h) h

x x

Worked Example 20

Sketch the graph of 4x2 + 4y2 = 25, stating the centre and radius. Think 1

Express the equation in standard form by dividing both sides by 4.

Write/draw

x2 + y2 = r2 + 4y2 = 25

4x2

x2 + y2 =

25 4

2

State the coordinates of the centre.

Centre (0, 0)

3

Find the length of the radius by taking the square root of both sides. (Ignore the negative results.)

r2 =

25 4

r=

5 2

Radius = 2.5 units Chapter 9 Functions

315

number AND algebra • Linear and non-linear relationships

4

y 2.5

Sketch the graph. -2.5

2.5

x

-2.5

Worked Example 21

Sketch the graph of (x - 2)2 + ( y + 3)2 = 16, clearly showing the centre and radius. Think

Write/draw

1

Express the equation in standard form.

(x - h)2 + ( y - k)2 = r 2 (x - 2)2 + ( y + 3)2 = 16

2

State the coordinates of the centre.

Centre (2, -3)

3

State the length of the radius.

r 2 = 16 r=4 Radius = 4 units

4

Sketch the graph.

y 1 -2 -3

2

6 x

4

-7

Worked Example 22

Sketch the graph of the circle x2 + 2x + y2 - 6y + 6 = 0. Think 1

Express the equation in standard form by completing the square on the x terms and again on the y terms.

3

State the coordinates of the centre. State the length of the radius.

4

Sketch the graph.

2

Write/draw

(x - h)2 + ( y - k)2 = r 2 + 2x + y2 - 6y + 6 = 0 2 (x + 2x + 1) - 1 + ( y2 - 6y + 9) - 9 + 6 = 0 (x + 1)2 + ( y - 3)2 - 4 = 0 (x + 1)2 + ( y - 3)2 = 4 Centre (-1, 3) r2 = 4 r=2 Radius = 2 units x2

y 5 3 1 -3 -1 1

316

Maths Quest 10 for the Australian Curriculum

x

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships

remember

Circle■graphs: x2■+■y 2■=■r 2■ (x■-■h)2■+■(■y■-■k)2■=■r 2■ exerCise

9g inDiviDuAl pAthWAys eBook plus

Activity 9-G-1

Circle graphs doc-5092 Activity 9-G-2

Exploring the circle doc-5093 Activity 9-G-3

More circle graphs doc-5094

centre■(0,■0)■ centre■(h,■k)■

radius■r radius■r

the circle FluenCy 1 We20 ■Sketch■the■graphs■of■the■following,■stating■the■centre■and■radius■of■each. a x2■+■y2■=■49 b x2■+■y2■=■42 c x2■+■y2■=■36 d x2■+■y2■=■81■ e 2x2■+■2y2■=■50■ f 9x2■+■9y2■=■100 2 We21 ■Sketch■the■graphs■of■the■following,■clearly■showing■the■centre■and■the■radius. a (x■-■1)2■+■(■y■-■2)2■=■52 b (x■+■2)2■+■(■y■+■3)2■=■62 c (x■+■3)2■+■(■y■-■1)2■=■49 d (x■-■4)2■+■(■y■+■5)2■=■64 e x2■+■(■y■+■3)2■=■4■ f (x■-■5)2■+■y2■=■100 3 We22 ■Sketch■the■graphs■of■the■following■circles. a x2■+■4x■+■y2■+■8y■+■16■= 0 b x2■-■10x■+■y2■-■2y■+■10■= 0 c x2■-■14x■+■y2■+■6y■+■9■= 0 d x2■+■8x■+■y2■-■12y■-■12■= 0 e x2■+■y2■-■18y■-■19■= 0 f 2x2■-■4x■+■2y2■+■8y■-■8■= 0 unDerstAnDing 4 mC ■The■graph■of■(x■-■2)2■+■(■y■+■5)2■=■4■is: a

y

y

b

5

5 x

–2

x

-2 d

y c

2 -5

y

x

2

x

-5

Chapter 9 Functions

317

number AND algebra • Linear and non-linear relationships 5   MC  The centre and radius of the circle (x + 1)2 + ( y - 3)2 = 4 is: a (1, -3), 4 b (-1, 3), 2 c (3, -1), 4 d (1, -3), 2 y 6 Find the equation representing

the outer edge of the galaxy as shown in the photo at right, using the astronomical units provided.

3

5

reflection 

9

 

How could you write equations representing a set of concentric circles (circles with the same centre, but different radii)?

318

Maths Quest 10 for the Australian Curriculum

x

number AND algebra • Linear and non-linear relationships

Summary Plotting parabolas ■■ ■■ ■■ ■■ ■■ ■■ ■■

Produce a table of values by substituting each integer value of x into the equation. Plot a graph by drawing and labelling a set of axes, plotting the points from the table and joining the points to form a smooth curve. The axis of symmetry is the line that divides the parabola exactly in half. The turning point is the point where the graph changes direction or turns. The turning point is a maximum if it is the highest point on the graph and a minimum if it is the lowest point on the graph. The x-intercepts are the x-coordinates of the points where the graph crosses the x-axis. The y-intercept is the y-coordinate of the point where the graph crosses the y-axis. Sketching parabolas using the basic graph of y = x 2

■■ ■■ ■■ ■■ ■■

If the graph of y = x2 is translated c units vertically, the equation becomes y = x2 + c. If the graph of y = x2 is translated h units horizontally, the equation becomes y = (x − h)2. If the graph of y = x2 is dilated by factor a, the graph becomes narrower if a > 1 and wider if 0 < a < 1. If the x2 term is positive, the graph is upright. If there is a negative sign in front of the x2 term, the graph is inverted. Invariant points are points that do not change under a transformation. Sketching parabolas in turning point form

■■ ■■ ■■ ■■ ■■ ■■ ■■

If the equation of a parabola is in turning point form, y = a(x - h)2 + k, then the turning point is (h, k). If a is positive, the graph is upright with a minimum turning point. If a is negative, the graph is inverted with a maximum turning point. If the magnitude of a is greater than 1, the graph is narrower than the graph of y = x2. If the magnitude of a is between 0 and 1, the graph is wider than the graph of y = x2. To find the y-intercept, substitute x = 0 into the equation. To find the x-intercepts, substitute y = 0 into the equation and solve for x. Sketching parabolas of the form y = ax2 + bx + c

■■

■■

If the equation is in the form y = ax2 + bx + c, the coordinates of the turning point can be found by: (a) using the completing the square method to change the equation into turning point form (b) finding the x-coordinate of the point exactly halfway between the two x-intercepts. This is the x-coordinate of the turning point. Then substitute the x-value into the equation to find the y-coordinate. −b . Then substitute the x-value into the equation to find the y-coordinate. (c) using x = 2a The graph should also show both the y-intercept and the x-intercepts of the parabola if they exist. Exponential functions and their graphs

■■ ■■

■■ ■■ ■■

Relationships of the form y = ax, where a ò 1 are called exponential functions with base a. To obtain the graph of an exponential function, construct a table of values first and then plot the points from the table and join them with a smooth curve. Alternatively use a graphics calculator, CAS calculator or graphing software. An asymptote is a line which the graph approaches but never cuts or touches. Multiplying by a constant dilates the basic graph — that is, makes it narrower or wider. If x is a negative number, the graph is reflected across the y-axis. Chapter 9 Functions

319

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships The hyperbola ■■ ■■ ■■

k A■hyperbola■is■a■function■of■the■form■y = . x To■obtain■the■graph■of■a■hyperbola,■construct■a■table■of■values.■Plot■the■points■and■join■them■ with■a■smooth■curve.■Alternatively,■use■a■graphics■calculator■or■a■computer■graphing■package. A■hyperbola■will■often■have■both■a■horizontal■and■a■vertical■asymptote.

The circle ■■

Circle■graphs: x2■+■y 2■=■r 2■ (x■-■h)2■+■(■y■-■k)2■=■r 2■

centre■(0,■0)■ centre■(h,■k)■

radius■r radius■r

MaPPING YOUR UNdeRSTaNdING

Homework book

320

Using■terms■from■the■summary,■and■other■terms■if■you■wish,■construct■a■concept■map■that■ illustrates■your■understanding■of■the■key■concepts■covered■in■this■chapter.■Compare■your■ concept■map■with■the■one■that■you■created■in■What do you know?■on■page■279. Have■you■completed■the■two■Homework sheets,■the■Rich task■and■two■Code puzzles■in■ your■Maths Quest 10 Homework Book?

maths Quest 10 for the Australian Curriculum

number AND algebra • Linear and non-linear relationships

Chapter review y 10

b

Fluency 1 The turning point for the graph y = 3x2 − 4x + 9 is: A

 1 2  , 1  3 3

B

 1 2 ,  3 3 

C

 1 1  6 , 16 

D

2  3 ,

E

2 , 2  3 6 3 

5 -4

-2

2

4x

2

4x

2

4x

2

4x

-10

7 2  3

C

y 10

2 Which graph of the following equations has the

x-intercepts closest together? A y = x2 + 3x + 2 B y = x2 + x - 2 C y = 2x2 + x - 15 D y = 4x2 + 27x - 7 E y = x2 - 2x - 8 3 Which graph of the equations below has the largest y-intercept? A y = 3(x - 2)2 + 9 B y = 5(x - 1)2 + 8 C y = 2(x - 1)2 + 19 D y = 2(x - 5)2 + 4 E y = 12(x - 1)2 + 10 4 The translation required to change y = x2 into 1 y = (x - 3)2 + 4 is: A right 3, up

1 4

C left 3, down D left 3, up

-2 -5 -10

D

y 10 5 -4

-2

1 0 -5 -10 y 10 5 3

1 4

1 4

-4

1 4

-2

0 -5

5 The graph of y = -3 ì 2x is best represented by:

turning point for each of the following graphs. a y = x2 - 8x + 1 b y = x2 + 4x - 5

5 3 0 -5 -10

-10 6 Use the completing the square method to find the

y 10

-2

-4

1 4

E right , up 3

A

5 3

E

B right 3, down

-4

0 -3 -5

2

4x

7 For the graph of the equation y = x2 + 8x + 7,

produce a table of values for the x-values between -9 and 1, and then plot the graph. Show the y-intercept and turning point. From your graph, state the x-intercepts.

Chapter 9 Functions

321

number AND algebra • Linear and non-linear relationships 8 For each of the following, find the coordinates of

the turning point and the x- and y-intercepts and sketch the graph. a y = (x - 3)2 + 1 b y = 2(x + 1)2 - 5

21 Find the equation of this circle. y 6

9 For the equation y = -x2 - 2x + 15, sketch the

graph and determine the x- and y-intercepts and the coordinates of the turning point.

10 For the exponential function y = 5x: a complete the table of values below

x -3 -2 -1  0  1  2  3

y

6 x

-6

1 Consider the quadratic equation: y = x2 - 4x + 7. a Determine the equation of the quadratic which

just touches the one above at the turning point. b Confirm your result graphically. 2 The height, h, in metres of a golf ball t seconds

11 Draw the graph of y = 10 ì 3x for -4 Ç x Ç 4. 12 Draw the graph of y = 10-x for -4 Ç x Ç 4. 13 a On the same axes draw the graphs of y = (1.2)x

and y = (1.5)x. b Use your answer to part a to explain the effect of changing the value of a in the equation of y = a x.

14 a On the one set of axes draw the graphs of 1

y = 2 ì 3x, y = 5 ì 3x and y = 2 ì 3x. b Use your answer to part a to explain the effect of changing the value of k in the equation y = kax. 15 a On the same set of axes sketch the graphs of

y = (2.5)x and y = (2.5)-x. b Use your answer to part a to explain the effect of a negative index on the equation y = ax.

16 Sketch each of the following.

4 x

0

problem solving

b plot the graph.

a y =

-6

b y = −

after it is hit is given by the formula h = 4t - t2. a Sketch the graph of the path of the ball. b What is the maximum height the golf ball reaches? c How long does it take for the ball to reach the maximum height? d How long is it before the ball lands on the ground after it has been hit? 3 A ball is thrown upwards from a building and follows the path given by the formula h = -x2 + 4x + 21. The ball is h metres above the ground when it is a horizontal distance of x metres from the building. a Sketch the graph of the path of the ball. b What is the maximum height the ball reaches? c How far is the ball from the wall when it reaches the maximum height? d How far from the building does the ball land? 4 A soccer ball is kicked upwards in the air. The height, h, in metres, t seconds after the kick is modelled by the quadratic equation h = -5t2 + 20t.

2 x

−3 x−2 18 Give an example of an equation of a hyperbola that has a vertical asymptote at x = -3.

17 Sketch y =

19 Sketch each of these circles. Clearly show the

centre and the radius. a x2 + y2 = 16

b (x - 5)2 + (y + 3)2 = 64

20 Sketch the following circles. Remember to first

complete the square. a x2 + 4x + y2 - 2y = 4 b x2 + 8x + y2 + 8y = 32 322

Maths Quest 10 for the Australian Curriculum

a Sketch the graph of this relationship. b For how many seconds is the ball in the air? c For how many seconds is the ball above a

height of 15 m? That is, solve the quadratic inequation -5t2 + 20t > 15. d For how many seconds is the ball above a height of 20  m?

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships 5 The■height■of■the■water■level■in■a■cave■is■determined■

by■the■tides.■At■any■time,■t,■in■hours■after■9■am,■ the■height,■h(t),■in■metres,■can■be■modelled■by■the■ function■h(t)■=■t2■-■12t■+■32,■0■Ç■t■Ç■12.■ a What■values■of■t■is■the■model■valid■for?■Write■ your■answer■in■interval■notation.■ b Determine■the■initial■height■of■the■water. c Bertha■has■dropped■her■keys■onto■a■ledge■ which■is■7■metres■from■the■bottom■of■the■cave.■ ■ By■using■a■graphics■calculator,■determine■the■ times■in■which■she■would■be■able■to■climb■ down■to■retrieve■her■keys.■Write■your■answers■ correct■to■the■nearest■minute. 6 A■grassed■area■is■planted■in■a■courtyard■that■has■ a■width■of■5■metres■and■length■of■7■metres.■The■ perimeter■of■the■grassed■area■is■described■by■the■ function■P■=■-x2■+■5x,■where■P■is■the■distance,■ in■metres,■from■the■house■and■x■is■the■distance,■ in■metres■from■the■side■wall.■The■diagram■below■ represents■this■information■on■a■Cartesian■plane. 7m Wall

5m House

x

a In■terms■of■P,■write■down■an■inequality■that■

describes■the■region■where■the■grass■has■been■ planted. b Determine■the■maximum■distance■the■grass■ area■will■be■from■the■house.■ c The■owners■of■the■house■have■decided■that■ they■would■prefer■the■grassed■area■to■be■in■ a■maximum■distance■of■3.5■metres■from■the■ house.■The■perimeter■of■the■lawn■following■this■ design■can■be■described■by■the■equation N(x)■=■ax2■+■bx■+■c■ i Using■algebra,■show■that■this■new■ design■can■be■described■by■the■function■ N(x)■=■-0.48x(x■-■5) ii Describe■the■transformation■that■maps■P(x)■ to■N(x)■

d If■the■owners■decide■on■the■fi■rst■design,■P(x),■

the■percentage■of■area■within■the■courtyard■ without■grass■is■40.5%.■By■using■any■method,■ fi■nd■the■approximate■percentage■of■area■of■ courtyard■without■lawn■with■the■new■design,■ N(x)■ 7 A■stone■arch■bridge■has■a■span■of■50■metres.■The■ shape■of■the■curve■AB■can■be■modelled■using■a■ quadratic■equation.■ b(x)

4.5 m A (0, 0)

50 m

B

x

a Taking■A■as■the■origin■(0,■0)■and■given■that■the■

maximum■height■of■the■arch■above■the■water■ level■is■4.5■metres,■show■using■algebra,■that■the■ shape■of■the■arch■can■be■modelled■using■the■ equation■b(x)■=■−0.0072x2■+■0.36x,■where■b(x)■ is■the■vertical■height■of■the■bridge,■in■metres,■ and■x■is■the■horizontal■distance,■in■metres.■ b A■fl■oating■platform■20■metres■wide■and■ p■metres■high■is■towed■under■the■bridge.■Given■ that■the■platform■needs■to■have■a■clearance■of■ at■least■30■centimetres■on■each■side,■explain■ why■the■maximum■value■of■p■is■less■than■ 10.7■centimetres. eBook plus

Interactivities

Test yourself Chapter 9 int-2852 Word search Chapter 9 int-2850 Crossword Chapter 9 int-2851

Chapter 9 Functions

323

eBook plus

ACtivities

chapter opener Digital doc

•■ Hungry■brain■activity■Chapter■9■(doc-5265)■(page 279) are you ready? Digital docs

(page 280)

•■ SkillSHEET■9.1■(doc-5266):■Substitution■into■quadratic■ equations •■ SkillSHEET■9.2■(doc-5267):■Equation■of■a■vertical■line •■ SkillSHEET■9.3■(doc-5268):■Completing■the■square •■ SkillSHEET■9.4■(doc-5269):■Solving■quadratic■ equations■using■the■quadratic■formula •■ SkillSHEET■9.5■(doc-5270):■Solving■quadratic■ equations■of■the■type■ax2■+■bx■+■c■=■0■where■a■=■1 •■ SkillSHEET■9.6■(doc-5271):■Solving■quadratic■ equations■of■the■type■ax2■+■bx■+■c■=■0■where■a■ò■1 9a

Plotting parabolas

Digital docs

•■ Activity■9-A-1■(doc-5074):■Review■of■plotting■ parabolas■(page 284) •■ Activity■9-A-2■(doc-5075):■Plotting■parabolas■(page 284) •■ Activity■9-A-3■(doc-5076):■Trends■in■plotting■parabolas■ (page 285) •■ SkillSHEET■9.1■(doc-5266):■Substitution■into■quadratic■ equations■(page 285) •■ SkillSHEET■9.2■(doc-5267):■Equation■of■a■vertical■line■ (page 285) 9b

Sketching parabolas using the basic graph of y = x 2

Interactivities

•■ Dilation■of■y■=■x2■(int-1148)■(page 287) •■ Vertical■translation■of■y■=■x2■+■c■(int-1192)■(page 287) •■ Horizontal■translation■of■y■=■(x■-■h)2■(int-1193)■(page 288) Digital docs

•■ Activity■9-B-1■(doc-5077):■Review■of■sketching■basic■ parabolas■(page 291) •■ Activity■9-B-2■(doc-5078):■Sketching■basic■parabolas■ (page 291) •■ Activity■9-B-3■(doc-5079):■Trends■in■sketching■basic■ parabolas■(page 291) •■ WorkSHEET■9.1■(doc-5272):■Quadratic■graphs■ (page 292) 9c

Sketching parabolas in turning point form

Digital docs

(page 296)

•■ Activity■9-C-1■(doc-5080):■Reviewing■turning■point■ form■ •■ Activity■9-C-2■(doc-5081):■Turning■point■form■ •■ Activity■9-C-3■(doc-5082):■Interpreting■turning■point■ form■trends■ 9d

Sketching parabolas of the form y = ax 2 + bx + c

Interactivity

•■ Sketching■parabolas■(int-2785)■(page 298) 324

maths Quest 10 for the Australian Curriculum

Digital docs

•■ Activity■9-D-1■(doc-5083):■Understanding■parabola■ sketching■(page 302) •■ Activity■9-D-2■(doc-5084):■Parabola■sketching■ (page 302) •■ Activity■9-D-3■(doc-5085):■Sketching■tricky■parabolas■ (page 302) •■ SkillSHEET■9.3■(doc-5268):■Completing■the■square■ (page 302) •■ SkillSHEET■9.4■(doc-5269):■Solving■quadratic■ equations■using■the■quadratic■formula■(page 302) •■ SkillSHEET■9.5■(doc-5270):■Solving■quadratic■ equations■of■the■type■ax2■+■bx■+■c■=■0■where■a■=■1■ (page 303) •■ SkillSHEET■9.6■(doc-5271):■Solving■quadratic■ equations■of■the■type■ax2■+■bx■+■c■=■0■where■a■ò■1■ (page 303) •■ WorkSHEET■9.2■(doc-5273):■y■=■ax2■+■bx■+■c■ (page 306) 9e

exponential functions and their graphs

Interactivity

•■ Exponential■graphs■(int-1149)■(page 307) Digital docs

(page 309)

•■ Activity■9-E-1■(doc-5086):■Exploring■exponential■ graphs■ •■ Activity■9-E-2■(doc-5087):■Features■of■exponential■ graphs■ •■ Activity■9-E-3■(doc-5088):■Tricky■exponential■ graphs■ •■ SkillSHEET■9.7■(doc-5274):■Substitution■into■index■ expressions 9F

The hyperbola

Digital docs

(page 314)

•■ Activity■9-F-1■(doc-5089):■Hyperbola■graphs •■ Activity■9-F-2■(doc-5090):■Exploring■the■ hyperbola■ •■ Activity■9-F-3■(doc-5091):■More■hyperbola■graphs■ 9G

The circle

Digital docs

(page 317)

•■ Activity■9-G-1■(doc-5092):■Circle■graphs■ •■ Activity■9-G-2■(doc-5093):■Exploring■the■circle■ •■ Activity■9-G-3■(doc-5094):■More■circle■graphs■ chapter review Interactivities

(page 323)

•■ Test■yourself■Chapter■9■(int-2852):■Take■the■end-ofchapter■test■to■test■your■progress •■ Word■search■Chapter■9■(int-2850):■an■interactive■ word■search■involving■words■associated■with■this■ chapter •■ Crossword■Chapter■9■(int-2851):■an■interactive■ crossword■using■the■defi■nitions■associated■with■the■ chapter To access eBookPLUS activities, log on to www.jacplus.com.au

measurement anD geometry • geometric reasoning

10

10a 10B 10C 10d

Congruence review Similarity review Congruence and proof Quadrilaterals: definitions and properties 10e Quadrilaterals and proof What Do you knoW ?

Deductive geometry

1 List what you know about geometry. Create a concept map to show your list. 2 Share what you know with a partner and then with a small group. eBook plus 3 As a class, create a large concept map Digital doc that shows your Hungry brain activity class’s knowledge Chapter 10 of geometry. doc-5275

opening Question

Can you predict the path of a ball after it has been hit by the cue? Will the ball land in a pocket?

measurement anD geometry • geometric reasoning

are you ready?

Try■the■questions■below.■If■you■have■diffi■culty■with■any■of■them,■extra■help■can■be■obtained■by■ completing■the■matching■SkillSHEET.■Either■search■for■the■SkillSHEET■in■your eBookPLUS■or■ ask■your■teacher■for■a■copy. eBook plus

Digital doc

SkillSHEET 10.1 doc-5276

eBook plus

Digital doc

SkillSHEET 10.2 doc-5277

Naming angles, lines and figures 1 For■the■triangle■shown■at■right,■use■correct■mathematical■

notation■to■name: a the■triangle b the■angle■marked■ñ c the■line■opposite■the■right■angle.

C A B

Corresponding sides and angles of congruent triangles 2 The■two■triangles■below■are■congruent. A

Q P

C a b c d eBook plus

Digital doc

SkillSHEET 10.3 doc-5278

B Triangle 1

Triangle 2

R

Which■side■in■Triangle■1■corresponds■to■side■QR■in■Triangle■2? Which■side■in■Triangle■2■corresponds■to■side■BC■in■Triangle■1? Name■the■angle■in■Triangle■1■that■corresponds■to■±PQR■in■Triangle■2. Name■the■angle■in■Triangle■2■that■corresponds■to■±ABC■in■Triangle■1.

Writing similarity statements 3 For■the■pair■of■similar■triangles■shown■at■right,■write■the■similarity■

A

statement■and■list■the■ratios■of■the■corresponding■sides.■ D

B eBook plus

Digital doc

SkillSHEET 10.4 doc-5279

326

Identifying quadrilaterals 4 Name■the■following■shapes. a

maths Quest 10 for the australian curriculum

b

c

E

C

measurement AND geometry • geometric reasoning

10A

Congruence review ■■ ■■

Congruent figures have the same size and the same shape; that is, they are identical in all respects. It can be said that congruent figures are similar figures with a scale factor of 1. The symbol used for congruency is @. For example, DABC in the diagram below is congruent to DPQR. This is written as DABC @ DPQR. C

A

■■

Q

B

R

Note that the vertices of the two triangles are written in corre­sponding order. Of all the shapes that are being tested for congruency, we are par­ticularly interested in triangles. There are four tests designed to check whether triangles are congruent. Three of these tests are for any type of triangle and one is specifically designed for right-angled triangles. The tests are summarised in the table below. Test

■■

P

Diagram

Abbreviation

All three sides in one triangle are equal in length to the corresponding sides in the other triangle.

SSS

Two corresponding sides and the included angle are the same in both triangles.

SAS

Two corresponding angles and a pair of corresponding sides are the same.

ASA

The hypotenuse and one pair of the other corresponding sides in rightangled triangles are the same.

RHS

In each of the tests we need to show three equal measurements about a pair of ­triangles in order to show they are congruent. Chapter 10 Deductive geometry

327

measurement AND geometry • geometric reasoning

Worked Example 1

Select a pair of congruent triangles from the diagram below, giving a reason for your answer. A

18 cm

Q 50è

15 cm

95è C

95è

P

B

L

35è 15 cm

Think

N

35è

95è

R

M Write

1

In each triangle the length of the side opposite the 95è angle is given. If triangles are to be congruent, the sides opposite the angles of equal size must be equal in length. Draw your conclusion.

AC = PR = 15 cm, LN = 18 cm Since LN ò AC and LN ò PR, DLMN is not congruent to DABC and DPQR.

2

We have a pair of equal sides. For DABC and DPQR to be congruent, 2 pairs of corresponding angles must be shown to be equal.

DABC: ±A = 50è, ±B = 95è, ±C = 180è - 50è - 95è = 35è ±B = ±Q ±C = ±R

3

Triangles ABC and PQR have a pair of corresponding sides equal in length and 2 pairs of angles the same, so draw your conclusion.

DABC @ DPQR (ASA)

■■

■■

Note that in the above worked example the congruent triangles were identified by using the ASA test, which implies that two angles and one pair of corresponding sides must be the same. Note that if two pairs of corresponding angles are equal, the third pair must also be equal.

Worked Example 2

Given that DABD @ DCBD, find the values of the pronumerals in the figure at right.

B

A Think

328

40è

x

z

y D

Write

1

In congruent triangles corresponding sides are equal in length. Side AD (marked x) corresponds to side DC, so state the value of x.

DABD @ DCBD AD = CD, AD = x, CD = 3 So x = 3 cm.

2

Since triangles are congruent, corresponding angles are equal. State the angles corresponding to y and z and hence find the values of these pronumerals.

±A = ±C ±A = 40è, ±C = y So y = 40è. ±BDA = ±BDC ±BDA = z, ±BDC = 90è So z = 90è.

Maths Quest 10 for the Australian Curriculum

3 cm

C

measurement anD geometry • geometric reasoning

WorkeD example 3

Prove that DPQS is congruent to DRSQ.

think

P

Q

S

R

Write

1

Study■the■diagram■and■state■which■sides,■and/or■ angles■are■equal.■

QP■=■SR■(given) ±SPQ■= ±SRQ■=■90°■(given) QS■is■common.

2

Select■the■appropriate■congruency■test.■(In■this■ case■it■is■RHS■because■the■triangles■have■an■equal■ side,■a■right■angle■and■a■common■hypotenuse.)

So■DPQS■@■DRSQ■(RHS).

rememBer

1.■ Congruent■fi■gures■are■identical■in■all■respects;■that■is,■they■have■the■same■shape■and■the■ same■size. 2.■ Triangles■are■congruent■if■any■one■of■the■following■applies: (a)■ corresponding■sides■are■the■same■(SSS) (b)■two■corresponding■sides■and■the■included■angle■are■the■same■(SAS) (c)■ two■angles■and■a■pair■of■corresponding■sides■are■the■same■(ASA) (d)■the■hypotenuse■and■one■pair■of■the■other■corresponding■sides■are■the■same■in■a■ right-angled■triangle■(RHS). 3.■ The■symbol■used■for■congruency■is■@.

exercise

10a inDiViDual pathWays eBook plus

Activity 10-A-1

Review of congruent shapes doc-5095 Activity 10-A-2

Practice with congruent figures doc-5096

congruence review fluency 1 We1 ■Select■a■pair■of■congruent■triangles■in■each■of■the■following,■giving■a■reason■for■your■

answer.■All■side■lengths■are■in■cm. a

65è

65è 4

3 I

65è

II

3 70è

70è

4

III

4

3 45è

chapter 10 Deductive geometry

329

measurement anD geometry • geometric reasoning

inDiViDual pathWays

b

110è

eBook plus

6 cm

I

40è

6 cm III

II

Activity 10-A-3

Tricky congruent figures doc-5097

110è

110è

40è 6 cm

40è eBook plus

c

3

Digital doc

SkillSHEET 10.1 doc-5276

4

3

5

II

III

4

I

3

eBook plus

Digital doc

SkillSHEET 10.2 doc-5277

d

3.5

2

3.5

2

I III

II

4.8

2.5

4.8

3.5

4.8

unDerstanDing 2 We2 ■Find■the■value■of■the■pronumeral■in■each■of■the■following■pairs■of■con■gruent■triangles.■

All■side■lengths■are■in■cm. b

a 4

c

80è

3

30è

85è z x

4

e

d

7 30è

330

y x

maths Quest 10 for the australian curriculum

x

x

x

y

40è

n m

z

y

measurement anD geometry • geometric reasoning reasoning eBook plus

3 We3 ■Prove■that■each■of■the■following■pairs■of■triangles■are■congruent. b a P P

Q

Digital doc

SkillSHEET 10.5 doc-5280

c

R

S

Q

P

S

Q

d

S

R

A

B

D

C

R e

Q

P

R

S 4 mc ■Note:■There■may■be■more■than■one■correct■answer.■

Which■of■the■following■is■congruent■to■the■triangle■ shown■at■right?

3 cm

5 cm

35è a

3 cm

B

5 cm

3 cm 35è

35è

C

d 3 cm 35è

5 cm

3 cm 35è

5 cm 5 cm

chapter 10 Deductive geometry

331

measurement AND geometry • geometric reasoning 5 Prove that DABC @ DADC and hence find the values of the pronumerals in each of the

following.

b   B

A

a

30è 30è

D

w

7 cm

x

x

B

70è

A B

c

30è y 65è

A

C

x

C

D

y D

4 cm 40è 40è z C 6 Explain why the triangles

shown at right are not ■ necessarily congruent.

40è

5 cm

5 cm 7 cm

40è 7 cm 7 Explain why the triangles

8 cm

shown at right are not ■ congruent.

8 cm

30è

30è 70è

70è

8 Show that DABO @ DACO, if O is the centre of the circle.

reflection 

B

 

How can you be certain that two figures are congruent?

10B

A

O C

Similarity review ■■ ■■ ■■

■■

Similar figures have identical shape but different size. The corresponding angles in similar figures are equal in size and the corresponding sides are in the same ratio, called a scale factor. The sign used to denote similarity is ~ which is read as ‘is similar to’. •• Similar figures can be obtained as a result of an enlargement or reduction. •• If an enlargement (or a reduction) took place, the original figure can be called the object and the enlarged (or reduced) figure called the image. It can also be said that the object maps to the image. For any two similar figures, the scale factor can be obtained using the following formula: length of the image length of the object Note: The size of the scale factor indicates whether the original object has been enlarged or reduced. That is, if the scale factor is greater than 1, an enlargement has occurred. If it is less than 1, a reduction in size has occurred. Scale factor =

332

Maths Quest 10 for the Australian Curriculum

measurement AND geometry • geometric reasoning

■■

Consider the pair of similar triangles below. U A 3 B

10

6

5 C

4

V

8

W

The following can be said about these ­triangles. – Triangle UVW is similar to ­triangle ABC or, using symbols, DUVW ~ DABC. – The corresponding angles of the two triangles are equal in size: ±CAB = ±WUV, ±ABC = ±UVW and ±ACB = ±UWV. UV VW UW = = = 2; – The corresponding sides of the two triangles are in the same ratio. AB BC AC that is, DUVW has each of its sides twice as long as the ­corresponding sides in DABC. – The scale factor is 2.

Testing triangles for similarity ■■

Triangles can be checked for similarity using one of the tests described in the table below. Test

■■

Abbreviation

All corresponding angles are equal in size.

AAA or equiangular (angle–angle–angle)

All corresponding sides are in the same ratio.

SSS (side–side–side)

Two pairs of corresponding sides are in the same ratio and the included angles are equal in size.

SAS (side–angle–side)

Both are right-angled triangles; the hypotenuses and one other pair of corresponding sides are in the same ratio.

RHS (right angle–hypotenuse–side)

Note: When using the equiangular test, only two corresponding angles have to be checked. Since the sum of the interior angles in any tri­angle is a constant number (180è), the third pair of corresponding angles will automatically be equal, provided that the first two pairs match exactly.

Worked Example 4

Find a pair of similar triangles among those shown. Give a reason for your answer. a  b  c  3 cm

140è 2 cm

6 cm

3 cm

140è

5 cm

140è 4 cm

Chapter 10 Deductive geometry

333

measurement AND geometry • geometric reasoning

Think 1

Write

In each triangle we know the size of two sides and the included angle, so the SAS test can be applied. Since all included angles are equal (140è), we need to find ratios of corresponding sides, taking two triangles at a time.

For triangles a and b: 6 4 =2=2 3 For triangles a and c: 5 3 = 1.6 , 2 = 1.5 3 For triangles b and c: 5 3 = 0.83 , 4 = 0.75 6

2

Triangle a ~ triangle b (SAS)

Only triangles a and b have corresponding sides in the same ratio (and included angle of equal size). State your conclusion, specifying the similarity test that has been used.

Worked Example 5

Prove that DABC is similar to DEDC.

A

D C

B Think

E Write

1

AB is parallel to DE. Transversal BD forms two alternate angles: ±ABC and ±EDC.

±ABC = ±EDC (alternate angles)

2

Transversal AE forms two alternate angles: ±BAC and ±DEC.

±BAC = ±DEC (alternate angles)

3

While crossing each other, the two transversals form vertically opposite angles at C.

±BCA = ±DCE (vertically opposite angles)

4

Triangles ABC and EDC have three pairs of corresponding angles of equal size and therefore are similar. State this using appropriate mathematical symbols and specify the similarity test being used.

\ DABC ~ DEDC (equian­gular, AAA)

remember

1. Similar figures have the same shape but different size. 2. Corresponding angles of similar figures are equal in size. 3. Corresponding sides of similar figures are in the same ratio, called the scale factor. 4. Triangles can be tested for similarity using the following requirements: (a) corresponding angles are equal in size (AAA or equiangular) (b) corresponding sides are in the same ratio (SSS) (c) two pairs of corresponding sides are in the same ratio, and angles included between those sides are equal in size (SAS) (d) one angle in each triangle is right (90­è); the hypotenuses and one pair of corresponding sides are in the same ratio (RHS). 5. The symbol for similarity is ~. 334

Maths Quest 10 for the Australian Curriculum

measurement anD geometry • geometric reasoning

exercise

10B inDiViDual pathWays eBook plus

Activity 10-B-1

similarity review fluency 1 We4 ■Find■a■pair■of■similar■triangles■among■those■shown■in■each■part.■Give■a■reason■for■your■

answer. a i■

ii

Review of similar shapes doc-5098

iii

5

10

5

Activity 10-B-2

Similarity practice doc-5099 Activity 10-B-3

Tricky similarity problems doc-5100

3

6

4

b i■

iii

ii 4 20è

c i■

20è

5 ii

2 4

8

2

12

iii

2 5

3

20è

2.5

6

4.5

4

3 d i■

iii

ii

40è

50è

60è

e i■

40è

60è

ii

iii

4 6

3

60è

2

8

7

5 4

4

unDerstanDing 2 Name■two■similar■triangles■in■each■of■the■following■fi■gures. b a Q A

c

Q

P

B B A P d

C

D

R

C R

A

B

e

B D

D

S

T

E A

C

E

C chapter 10 Deductive geometry

335

measurement anD geometry • geometric reasoning reasoning eBook plus

Digital doc

3 We5 ■Prove■that■DABC■is■similar■to■DEDC■in■each■of■the■following. a b C D A

SkillSHEET 10.3 doc-5278

E A

eBook plus

c

SkillSHEET 10.6 doc-5281

C

B

B

E d

E

A

Digital doc

C

D

D B

B

D A C

E

AB BC = = . AD AE b Find■the■value■of■the■pronumerals.

4 a■ Complete■this■statement:■

2

A

D

4

B 3

f

4 C

g E

5 Find■the■value■of■the■pronumeral■in■the■diagram■at■right.■

Q A

x

2 P

4

R

4

B

6 The■triangles■shown■at■right■are■similar.■Find■the■

value■of■x■and■y.■

45è

4 45è 1

20è

7 Find■the■values■of■x■and■y■in■the■diagram■at■right. eBook plus

reflection 

Digital doc WorkSHEET 10.1 doc-5282

 

How can you be certain that two figures are similar?

P

9

y S

1.5

3 y Q

x

R

8 6 x T

10c

congruence and proof ■■ ■■

336

In■geometry,■congruence■is■one■of■the■main■tools■we■can■use■to■prove■both■familiar■and■ unfamiliar■properties■of■shapes. In■this■section,■we■will■aim■to■prove■key■deductive■geometry■theorems■about■triangles.■In■a■ proof,■it■is■important■to■give■reasons■for■all■steps.

maths Quest 10 for the australian curriculum

measurement AND geometry • geometric reasoning

Worked Example 6 A

Prove that if two sides of a triangle are equal, then the angles opposite those sides are equal, that is, ±ABC = ±ACB.

C

B Think 1

Write/draw

Draw a diagram. Construct AD so that ■ ±BAD = ±CAD.

A

B

C

D

2

State the known facts about the sides and angles.

AB = AC (given) ±BAD = ±CAD (by construction) AD is common.

3

Summarise the given information.

Two sides and the included angle in DBAD and DCAD are equal.

4

State which congruency test applies.

DBAD @ DCAD (SAS)

5

State the conclusion.

±ABC = ±ACB (corresponding angles in congruent triangles are equal)

Worked Example 7

PQR is an isosceles triangle with PQ = PR. Also, MQ = NR. Prove that MR = NQ.

P

M

N R

Q Think 1

Write/draw P

Draw a diagram, labelling all given information.

M Q

N R Chapter 10 Deductive geometry

337

measurement anD geometry • geometric reasoning

2

State■the■known■facts■about■the■sides■and■angles.

MQ■=■NR■(given) and■PQ■=■PR■(given) \■PM■=■PN also■±QPR■is■common.

3

Summarise■the■given■information.

Two■sides■and■the■included■angle■in■DMPR■and■ DNPQ■are■equal.

4

State■which■congruency■test■applies.

DMPR■@■DNPQ■(SAS)

5

State■the■conclusion.

\■MR■=■NQ■(corresponding■sides■in■congruent■ triangles■are■equal)

rememBer

1.■ Many■deductive■geometry■proofs■can■be■completed■using■congruent■triangle■tests. 2.■ In■a■proof,■it■is■important■to■give■reasons■for■all■steps. exercise

10c inDiViDual pathWays eBook plus

Activity 10-C-1

Congruent triangles doc-5101

congruence and proof reasoning 1 We6 ■Prove■that■if■two■angles■of■a■triangle■are■equal,■the■sides■

A

opposite■those■angles■are■equal.■(See■the■fi■gure■shown■at■right.) ■ ■ Hint:■Construct■a■line■perpendicular■to■BC■through■A■ and■prove■that■DABD■@■DACD.

Activity 10-C-2

Matching congruent triangles doc-5102 Activity 10-C-3

Harder congruent triangles doc-5103

B

C

D

2 Prove■that■each■angle■of■an■equilateral■triangle■is■60è. M

3 Prove■that■the■bisector■of■the■vertical■angle■of■an■isosceles■

triangle■bisects■the■base.■(See■the■fi■gure■shown■lower■right.) 4 Prove■that■the■intervals■joining■the■midpoints■of■ the■three■sides■of■a■triangle■cut■the■original■triangle■ into■four■congruent■triangles. N 5 We7 ■Use■congruence■to■prove■that■AB■||■CD.

P

O

C

A E D

B 6 Prove■that■DWXY■@■DYZW.

W

X

Z 338

maths Quest 10 for the australian curriculum

Y

measurement AND geometry • geometric reasoning 7 Prove that DADO @ DABO.

B

A

O C

D 8 Prove that DPTS @ DQTR.

Q

P

T

S

R

9 DABC is isosceles with AB = AC. D lies on BC so that AD ^ BC. Prove that AD bisects

±BAC and D is the midpoint of BC. 10 PQR is a triangle. M lies on PR so that QM ^ PR. N lies on PQ so that PQ ^ RN. Also RN = QM. Prove that ±PRQ = ±PQR and hence that DPQR is an isosceles triangle. (Hint: Prove DQNR @ DRMQ.) B 11 Can we prove Pythagoras’ theorem using our knowledge of similar triangles? Consider the triangle ■ shown. DABC is a right-angled triangle with ±ABC = 90°. We can construct a perpendicular from B to AC, meeting AC at D. Let ±BAD = x. x A C D Our aim is to show that AB2 + BC2 = AC2. a Copy the diagram and label the size of all other angles in the triangle. b Using the equiangular test, prove that DBAD ~ DCAB. Hint: Show that ±ABD = 90° - x and ±ACB = 90° - x. c Copy and complete: AD = AC \ AB2 = d Using the equiangular test, prove that DBCD ~ DACB. e Copy and complete: CD = AC \ BC2 = f Combining step c and step e, copy and complete: AB2 + BC2 = __________ = __________  (by factorising) = __________  (by simplifying) \ AB2 + BC2 = AC2 g Challenge: Can you prove the converse of Pythagoras’ theorem? That is, if the square on one side of a triangle equals the sum of the squares on the other two sides, then the angle between these other two sides is a right angle. reflection    What is the most important thing to include in a proof?

Chapter 10 Deductive geometry

339

measurement anD geometry • geometric reasoning

10D eBook plus

Interactivity Quadrilateral definitions

int-2786

Quadrilaterals: definitions and properties ■■

There■are■many■important■properties■of■quadrilaterals■that■can■be■shown■using■deductive■ geometry. Shape

Definition

Properties

Trapezium

A■trapezium■is■a■ quadrilateral■with■two■pairs■ of■equal■adjacent■angles.

One■pair■of■opposite■sides■is■ parallel.

Parallelogram

A■parallelogram■is■a■ quadrilateral■with■both■pairs■ of■opposite■sides■parallel.

■■

A■rhombus■is■a■ parallelogram■with■four■ equal■sides.

■■

A■rectangle■is■a■ parallelogram■whose■ interior■angles■are■right■ angles.

Diagonals■are■equal■and■ bisect■each■other.

A■square■is■a■parallelogram■ whose■interior■angles■are■ right■angles■with■four■equal■ sides.

■■

Rhombus

Rectangle

Square

■■ ■■

■■

■■ ■■

■■

340

maths Quest 10 for the australian curriculum

Opposite■angles■are■equal. Opposite■sides■are■equal. Diagonals■bisect■each■ other.

Diagonals■bisect■each■ other■at■right■angles. Diagonals■bisect■the■ angles■at■the■vertex■ through■which■they■pass.

All■angles■are■right■angles. All■side■lengths■are■equal. Diagonals■are■equal■in■ length■and■bisect■each■ other■at■right■angles. Diagonals■bisect■the■ vertex■through■which■they■ pass■(45è).

measurement anD geometry • geometric reasoning

WorkeD example 8

Use the definitions and properties of the five special quadrilaterals to answer the following statements as true or false. a A parallelogram is a trapezium. b A trapezium is a rectangle. c A square is a parallelogram. think a

b

Write a A■trapezium■has■one■pair■of■parallel■sides.

1

Consider■the■properties■of■a■trapezium.

2

Consider■the■properties■of■a■ parallelogram.

A■parallelogram■is■a■quadrilateral■with■both■ pairs■of■opposite■sides■parallel.

3

Decide■if■a■parallelogram■fi■ts■the■ defi■nition■of■a■trapezium.

Statement■is■false.

1

Consider■the■defi■nition■of■a■rectangle.

b A■rectangle■is■a■parallelogram■whose■interior■

angles■are■right■angles.

c

2

Consider■the■defi■nition■of■a■trapezium.

A■trapezium■is■a■quadrilateral■two■pairs■of■equal■ adjacent■angles.

3

Decide■if■a■trapezium■fi■ts■the■defi■nition■of■ a■rectangle.

A■trapezium■does■not■necessarily■have■a■right■ angle,■therefore,■the■statement■is■false.

1

Consider■the■defi■nition■of■a■ parallelogram.

2

Consider■the■defi■nition■of■a■square.

A■square■is■a■parallelogram■whose■interior■ angles■are■right■angles■with■four■equal■sides.

3

Decide■if■the■square■fi■ts■the■defi■nition■of■ a■parallelogram.

A■square■is■a■■parallelogram;■therefore,■the■ statement■is■true.

c

A■parallelogram■is■a■quadrilateral■whose■ appropriate■sides■are■parallel.

rememBer

1.■ A■trapezium■is■a■quadrilateral■with■two■pairs■of■equal■adjacent■angles. 2.■ A■parallelogram■is■a■quadrilateral■with■both■pairs■of■opposite■sides■parallel. 3.■ A■rhombus■is■a■parallelogram■with■four■equal■sides. 4.■ A■rectangle■is■a■parallelogram■whose■interior■angles■are■right■angles. 5.■ A■square■is■a■parallelogram■whose■interior■angles■are■right■angles■with■four■equal■sides. exercise

10D inDiViDual pathWays eBook plus

Activity 10-D-1

Quadrilaterals doc-5104

Quadrilaterals: definitions and properties fluency 1 We8 ■Use■the■defi■nitions■of■the■fi■ve■special■quadrilaterals■to■decide■if■the■following■■statements■

are■true■or■false. A■square■is■a■rectangle. A■square■is■a■rhombus. A■square■is■a■trapezium. A■trapezium■is■a■rhombus.

a c e g

b d f h

A■rhombus■is■a■parallelogram. A■rhombus■is■a■square. A■parallelogram■is■a■rectangle. A■rectangle■is■a■square. chapter 10 Deductive geometry

341

measurement anD geometry • geometric reasoning

inDiViDual pathWays eBook plus

Activity 10-D-2

Harder quadrilaterals doc-5105 Activity 10-D-3

Tricky quadrilaterals doc-5106

eBook plus

Digital doc SkillSHEET 10.4 doc-5279

unDerstanDing 2 Draw■three■different■trapeziums.■Using■your■ruler■and■protractor,■decide■which■of■the■

following■properties■are■true■in■a■trapezium. a Opposite■sides■are■equal. b All■sides■are■equal. c Opposite■angles■are■equal. d All■angles■are■equal. e Diagonals■are■equal■in■length. f Diagonals■bisect■each■other. g Diagonals■are■perpendicular. h Diagonals■bisect■the■angles■they■pass■through. 3 Draw■three■different■parallelograms.■Using■your■ruler■and■protractor■to■measure,■decide■which■

of■the■following■properties■are■true■in■a■parallelogram. a Opposite■sides■are■equal. b All■sides■are■equal. c Opposite■angles■are■equal. d All■angles■are■equal. e Diagonals■are■equal■in■length. f Diagonals■bisect■each■other. g Diagonals■are■perpendicular. h Diagonals■bisect■the■angles■they■pass■through. 4 Draw■three■different■rhombuses.■Using■your■ruler■and■protractor■to■measure,■decide■which■of■

the■following■properties■are■true■in■a■rhombus. a Opposite■sides■are■equal. b All■sides■are■equal. c Opposite■angles■are■equal. d All■angles■are■equal. e Diagonals■are■equal■in■length. f Diagonals■bisect■each■other. g Diagonals■are■perpendicular. h Diagonals■bisect■the■angles■they■pass■through. 5 Draw■three■different■rectangles.■Using■your■ruler■and■protractor■to■measure,■decide■which■of■

the■following■properties■are■true■in■a■rectangle. a Opposite■sides■are■equal. b All■sides■are■equal. c Opposite■angles■are■equal. d All■angles■are■equal. e Diagonals■are■equal■in■length. f Diagonals■bisect■each■other. g Diagonals■are■perpendicular. h Diagonals■bisect■the■angles■they■pass■through. 6 Draw■three■different■squares.■Using■your■ruler■and■protractor■to■measure,■decide■which■of■the■

following■properties■are■true■in■a■square. a Opposite■sides■are■equal. b All■sides■are■equal. c Opposite■angles■are■equal. d All■angles■are■equal. e Diagonals■are■equal■in■length. f Diagonals■bisect■each■other. g Diagonals■are■perpendicular. h Diagonals■bisect■the■angles■they■pass■through. 7 Name■two■quadrilaterals■that■have■diagonals■that■bisect■each■other■at■right■angles. 8 Name■two■quadrilaterals■with■all■angles■equal. 9 Name■four■quadrilaterals■that■have■at■least■one■pair■of■opposite■sides■that■are■parallel■and■equal. 10 Name■a■quadrilateral■that■has■equal■diagonals■that■bisect■each■other■and■bisect■the■angles■they■

pass■through. reasoning 11 Pool■is■played■on■a■rectangular■table.■Balls■are■hit■with■a■cue■

and■bounce■off■the■sides■of■the■table■until■they■land■in■one■of■ the■holes■or■pockets. a Draw■a■rectangular■pool■table■measuring■5■cm■ by■3■cm■on■graph■paper. ■ Mark■the■four■holes,■one■in■each■corner. 342

maths Quest 10 for the australian curriculum

measurement anD geometry • geometric reasoning b A■ball■starts■at■A.■It■is■hit■so■that■it■travels■at■

a■45è■diagonal■across■the■grid.■When■it■hits■ the■side■of■the■table,■it■bounces■off■at■a■45è■ diagonal■as■well.■How■many■sides■does■the■ ball■bounce■off■before■it■goes■in■a■hole? c A■different■size■table■is■7■cm■by■2■cm.■How■ many■sides■does■a■ball■bounce■off■before■it■ goes■in■a■hole■when■hit■from■A? d Complete■the■following■table. Table size

A

Number of sides hit

5■cm■ì■3■cm 7■cm■ì■2■cm 4■cm■ì■3■cm 4■cm■ì■2■cm 6■cm■ì■3■cm 9■cm■ì■3■cm 12■cm■ì■4■cm e Can■you■see■a■pattern?■How■many■sides■would■a■

f

■

■ g

■

h

i

j

ball■bounce■off■before■going■in■a■hole■when■hit■ from■A■on■an■m■ì■n■table? The■ball■is■now■hit■from■B■on■a■5■cm■ì■3■cm■ pool■table. How■many■different■paths■can■a■ball■take■when■ hit■along■45è■diagonals?■Do■these■paths■all■hit■ the■same■number■of■sides■before■going■in■a■hole?■ B Does■the■ball■end■up■in■the■same■hole■each■time? Justify■your■answer. The■ball■is■now■hit■from■C■along■the■path■shown. What■type■of■triangles■and■quadrilaterals■ are■formed■by■the■path■of■the■ball■with■ itself■and■the■sides■of■the■table?■Are■any■ of■the■triangles■congruent? A■ball■is■hit■from■C■on■a■6■cm■by■3■cm■ table.■What■shapes■are■formed■by■the■path■ of■the■ball■with■itself■and■the■sides■of■the■table?■ C Is■there■only■one■path■possible? Challenge:■A■ball■is■hit■from■A■along■45è■ diagonals.■The■table■is■m■ì■n.■Can■you■fi■nd■a■ formula■to■predict■which■hole■the■ball■will■go■in? Challenge: What■would■happen■if■the■game■was■played■on■a■trapezoidal■table? reflection 

eBook plus

Digital doc WorkSHEET 10.2 doc-5283

 

What is the difference between the definitions and properties of shapes?

chapter 10 Deductive geometry

343

measurement AND geometry • geometric reasoning

10E

Quadrilaterals and proof ■■ ■■

In the previous exercise, we investigated some of the properties of quadrilaterals by construction and measurement. It is also important to be able to prove these properties from the definitions of the shapes.

Worked Example 9

Use the definition of a parallelogram to prove that the opposite sides of a parallelogram are equal. Think 1

Write/draw

Draw a diagram.

A

B

D 2

State the definition of a parallelogram.

3

Construct diagonal BD on the diagram.

4

Prove that the two triangles formed are congruent.

In DBAD and DDCB, BD is common. ±ADB = ±DBC (alternate angles equal as AD || BC) ±ABD = ±BDC (alternate angles equal as AB || DC) \ DBAD @ DDCB (AAS)

5

State conclusions.

AD = BC and AB = DC (corresponding sides in congruent triangles are equal) \ Opposite sides in a parallelogram are equal.

■■

A parallelogram is a quadrilateral with both pairs of opposite sides par­allel.

It is also often useful to prove that a particular quadrilateral is a parallelogram, for instance. If we know that the opposite sides are parallel, then we can use the definition and show the quadri­lateral is a parallelogram. However, there are also other tests for some of the quadrilaterals as outlined in the table below. Shape

344

C

Tests

Parallelogram

A quadrilateral is a parallelogram if: (a) opposite sides are parallel or (b) opposite sides are equal or (c) opposite angles are equal or (d) one pair of sides is both equal and parallel or (e) the diagonals bisect each other.

Rhombus

A quadrilateral is a rhombus if: (a) all sides are equal or (b) the diagonals bisect each other at right angles or (c) the diagonals bisect the angles they pass through.

Rectangle

A quadrilateral is a rectangle if: (a) all angles are equal or (b) the diagonals are equal and bisect each other.

Maths Quest 10 for the Australian Curriculum

measurement anD geometry • geometric reasoning

WorkeD example 10

Prove that PQRS is a parallelogram.

P

Q

S think

R

Write

1

State■the■given■information.

PS■=■QR ±PSQ■=■±RQS

2

Draw■conclusions■from■the■given■facts.

\■PS || QR■(alternate■angles■are■equal)

3

State■reasons■why■PQRS■is■a■parallelogram.

\■PQRS■is■a■parallelogram■since■PS■and■QR■are■ both■equal■and■parallel.

rememBer

1.■ A■quadrilateral■is■a■parallelogram■if: (a)■ opposite■sides■are■parallel■or (b)■opposite■sides■are■equal■or (c)■ opposite■angles■are■equal■or (d)■one■pair■of■sides■is■both■equal■and■parallel■or (e)■ the■diagonals■bisect■each■other. 2.■ A■quadrilateral■is■a■rhombus■if: (a)■ all■sides■are■equal■or (b)■the■diagonals■bisect■each■other■at■right■angles■or (c)■ the■diagonals■bisect■the■angles■they■pass■through. 3.■ A■quadrilateral■is■a■rectangle■if: (a)■ all■angles■are■equal (b)■the■diagonals■are■equal■and■bisect■each■other. exercise

10e inDiViDual pathWays eBook plus

Quadrilaterals and proof reasoning W

1 We9 ■Use■congruence■to■prove■that■the■opposite■angles■

X

(±ZWX■and■±XYZ)■are■equal■in■a■parallelogram.

Activity 10-E-1

Quadrilateral proofs doc-5107

Z

Y

Activity 10-E-2

Harder quadrilateral proofs doc-5108

2 Use■congruence■on■DADE■and■DCBE■to■prove■that■

A

B

the■diagonals■of■a■parallelogram■bisect■each■other.

E

Activity 10-E-3

Tricky quadrilateral proofs doc-5109

D chapter 10 Deductive geometry

C 345

measurement anD geometry • geometric reasoning 3 a■ Prove■that■DAED■@■DCED. b Hence,■show■that■±AED■=■±CED■=■90è.■(That■is,■

B

A E

the■diagonals■of■a■rhombus■are■perpendicular.) c Show■that■BD■bisects■±ADC.■(That■is,■the■diagonals■ of■a■rhombus■bisect■the■angles■they■pass■through.) 4 Prove■that■the■diagonals■of■a■rhombus■bisect■each■other. 5 Prove■that■all■angles■in■a■rectangle■are■right■angles.

D

6 Use■congruence■on■DADC■and■DBCD■to■show■that■

C

P

Q

S

R

A

B

D

C

the■diagonals■in■a■rectangle■are■the■same■length.

7 We10 ■ABCD■is■a■parallelogram.■X■is■the■midpoint■of■AB■

X

A

and■Y■is■the■midpoint■of■DC.■Prove■that■AXYD■is■ also■a■parallelogram. D

midpoints■of■their■respective■sides■of■ABCD. a Prove■DPAS■@■DRCQ. b Prove■DSDR■@■DPBQ. c Hence,■prove■that■PQRS■is■also■a■parallelogram.

C

Y P

A

8 ABCD■is■a■parallelogram.■P,■Q,■R■and■S■are■all■

B

B

S

Q

D

C

R

9 AC■and■BD■are■diameters■of■a■circle■with■centre■O.■

A

Prove■that■ABCD■is■a■rectangle. D

O

B

10 The■diagonals■of■a■parallelogram■meet■at■right■angles.■

C

Prove■that■the■parallelogram■is■a■rhombus. 11 Two■congruent■right-angled■triangles■are■arranged■

as■shown.■Show■that■PQRS■is■a■parallelogram.

Q

P

R

S eBook plus

Digital doc WorkSHEET 10.3 doc-5284

12 Two■circles,■centred■at■M■and■N,■have■equal■radii■and■

reflection 

 

How do you know if a quadrilateral is a rhombus?

346

P

intersect■at■P■and■Q.■Prove■that■PNQM■is■a■rhombus.

maths Quest 10 for the australian curriculum

M

N Q

measurement AND geometry • geometric reasoning

Summary Congruence review ■■

■■

■■

Congruent figures are identical in all respects; that is, they have the same shape and the same size. Triangles are congruent if any one of the following applies: (a) corresponding sides are the same (SSS) (b) two corresponding sides and the included angle are the same (SAS) (c) two angles and a pair of corresponding sides are the same (ASA) (d) the hypotenuse and one pair of the other corresponding sides are the same in a rightangled triangle (RHS). The symbol used for congruency is @. Similarity review

■■ ■■ ■■ ■■

■■

Similar figures have the same shape but different size. Corresponding angles of similar figures are equal in size. Corresponding sides of similar figures are in the same ratio, called the scale factor. Triangles can be tested for similarity using the following requirements: (a) corresponding angles are equal in size (AAA or equiangular) (b) corresponding sides are in the same ratio (SSS) (c) two pairs of corresponding sides are in the same ratio, and angles included between those sides are equal in size (SAS) (d) one angle in each triangle is right (90­è); the hypotenuses and one pair of corresponding sides are in the same ratio (RHS). The symbol for similarity is ~. Congruence and proof

■■ ■■

Many deductive geometry proofs can be completed using congruent triangle tests. In a proof, it is important to give reasons for all steps. Quadrilaterals: definitions and properties

■■ ■■ ■■ ■■ ■■

A trapezium is a quadrilateral with two pairs of equal adjacent angles. A parallelogram is a quadrilateral with both pairs of opposite sides parallel. A rhombus is a parallelogram with four equal sides. A rectangle is a parallelogram whose interior angles are right angles. A square is a parallelogram whose interior angles are right angles with four equal sides. tQuadrilaterals and proof

■■

■■

A quadrilateral is a parallelogram if: (a) opposite sides are parallel or (b) opposite sides are equal or (c) opposite angles are equal or (d) one pair of sides is both equal and parallel or (e) the diagonals bisect each other. A quadrilateral is a rhombus if: (a) all sides are equal or (b) the diagonals bisect each other at right angles or (c) the diagonals bisect the angles they pass through. Chapter 10 Deductive geometry

347

measurement anD geometry • geometric reasoning

■■

A■quadrilateral■is■a■rectangle■if: (a)■ all■angles■are■equal (b)■the■diagonals■are■equal■and■bisect■each■other.

MaPPINg Your uNderStaNdINg

Homework Book

348

Using■terms■from■the■summary,■and■other■terms■if■you■wish,■construct■a■concept■map■that■ illustrates■your■understanding■of■the■key■concepts■covered■in■this■chapter.■Compare■your■ concept■map■with■the■one■that■you■created■in■What do you know?■on■page■325. Have■you■completed■the■two■Homework sheets,■the■Rich task■and■two■Code puzzles■in■ your■Maths Quest 10 Homework Book?

maths Quest 10 for the australian curriculum

measurement AND geometry • geometric reasoning

Chapter review Fluency

4 Test whether the following pairs of triangles are

1 Select a pair of congruent triangles in each of the

following sets of triangles, giving a reason for your answer. All angles are in degrees and side lengths in cm. (The figures are not drawn to scale.) a

4

75è

4

40è II

6

65è

6

a

47è

6

47è

2 110è

75è

III

4

similar. For similar triangles find the scale factor. All angles are in degrees and side lengths in cm. 3

110è 5

7.5

b

I 75è

3

5

b I

6

10

6

8

6

II

50è 1

III

8

50è 2

c

2 Find the value of the pronumeral in each pair of

congruent triangles. All angles are given in degrees and side lengths in cm. b a

2

2 70è

8

4

x 2

5 Find the value of the pronumeral in each pair of

similar triangles. All angles are given in degrees and side lengths in cm.

x c

y

z

x

a

60è

A

5 y

30è D x A

3 a Prove that the two

B

E 3

C b ••

A 1 50è

D S

2

C

triangles shown in ■ the diagram at right ■ are congruent.

b Prove that DPQR

B

48è

R

B

C 1.5

44è

z

E

x

is congruent to ■ DQPS.

8 y P

Q

D Chapter 10 Deductive geometry

349

measurement anD geometry • geometric reasoning c

d What■does■this■mean■about■AG,■BG■and■CG? e A■circle■centred■at■G■is■drawn■through■A.■What■

P x

other■points■must■it■pass■through? y

9

Q

A

z

3

30è

C

B

4

6 Prove■that■

D

A

DABC ~ DEDC.

DPQS■is■isosceles.

P

5

R

12 PR■is■the■perpendicular■bisector■of■QS.■Prove■that■

Q

S

R

C

B

13 Name■any■quadrilaterals■that■have■diagonals■that■

E

bisect■the■angles■they■pass■through.

Q

7 Prove■that■

DPST ~ DPRQ.

14 State■three■tests■that■can■be■used■to■show■that■a■

S P

T

R

quadrilateral■is■a■rhombus. 15 Prove■that■WXYZ■is■a■parallelogram.■ W

8 Prove■that■the■angles■opposite■the■equal■sides■in■an■

isosceles■triangle■are■equal. 9 mc ■Note:■There■may■be■more■than■one■correct■ answer. ■ ■ A■quadrilateral■with■two■adjacent■sides■equal■could■ be■a: a rhombus■ B■ square C rectangle■ d■ parallelogram 10 True■or■false?■ a A■rhombus■is■a■square. b A■square■is■a■rectangle. c A■rectangle■is■a■trapezium. proBlem solVing 11 ABC■is■a■triangle.■D■is■the■midpoint■of■AB,■E■is■

130è

50è

Z

Y

16 Prove■that■the■diagonals■in■a■rhombus■bisect■the■

angles■they■pass■through. 17 Explain■why■the■triangles■shown■below■are■not■

congruent. 5 cm 80è

80è

25è

5 cm

E

F O

A H

20 Name■any■quadrilaterals■

G F

G

19 State■the■defi■nition■of■a■rhombus.

E

D

that■have■equal■diagonals. C

a Prove■that■DGDA■@■DGDB. b Prove■that■DGAE■@■DGCE. c Prove■that■DGBF■@■DGCF. 350

25è

18 Prove■that■DEFO ~ DGHO.

the■midpoint■of■AC■and■F■is■the■midpoint■of■BC.■ DG■^■AB,■EG■^■AC■and■FG■^■BC.

B

X

maths Quest 10 for the australian curriculum

eBook plus

Interactivities

Test yourself Chapter 10 int-2855 Word search Chapter 10 int-2853 Crossword Chapter 10 int-2854

eBook plus

actiVities

Chapter opener Digital doc

•■ Hungry■brain■activity■Chapter■10■(doc-5275)■ (page 325) are you ready? Digital docs (page 326) •■ SkillSHEET■10.1■(doc-5276):■Naming■angles,■lines■ and■fi■gures •■ SkillSHEET■10.2■(doc-5277):■Corresponding■sides■ and■angles■of■congruent■triangles •■ SkillSHEET■10.3■(doc-5278):■Writing■similarity■ statements •■ SkillSHEET■10.4■(doc-5279):■Identifying■ quadrilaterals

10a Congruence review Digital docs

•■ Activity■10-A-1■(doc-5095):■Review■of■congruent■ shapes■(page 329) •■ Activity■10-A-2■(doc-5096):■Practice■with■congruent■ fi■gures■(page 329) •■ Activity■10-A-3■(doc-5097):■Tricky■congruent■ fi■gures■(page 330) •■ SkillSHEET■10.1■(doc-5276):■Naming■angles,■lines■ and■fi■gures■(page 330) •■ SkillSHEET■10.2■(doc-5277):■Corresponding■sides■ and■angles■of■congruent■triangles■(page 330) •■ SkillSHEET■10.5■(doc-5280):■Angles■and■parallel■ lines■(page 331) 10B Similarity review Digital docs

•■ Activity■10-B-1■(doc-5098):■Review■of■similar■ shapes■(page 335) •■ Activity■10-B-2■(doc-5099):■Similarity■practice■ (page 335) •■ Activity■10-B-3■(doc-5100):■Tricky■similarity■ problems■(page 335) •■ SkillSHEET■10.3■(doc-5278):■Writing■similarity■ statements■(page 336) •■ SkillSHEET■10.6■(doc-5281):■Calculating■unknown■ side■lengths■in■a■pair■of■similar■triangles■(page 336) •■ WorkSHEET■10.1■(doc-5282):■Deductive■geometry■I■ (page 336)

10C Congruence and proof

(page 338) •■ Activity■10-C-1■(doc-5101):■Congruent■triangles •■ Activity■10-C-2■(doc-5102):■Matching■congruent■ triangles •■ Activity■10-C-3■(doc-5103):■Harder■congruent■triangles Digital docs

10d Quadrilaterals: definitions and properties Interactivity

•■ Quadrilateral■defi■nitions■(int-2786)■(page 340) Digital docs

•■ Activity■10-D-1■(doc-5104):■Quadrilaterals■(page 341) •■ Activity■10-D-2■(doc-5105):■Harder■quadrilaterals■ (page 342) •■ Activity■10-D-3■(doc-5106):■Tricky■quadrilaterals■ (page 342) •■ SkillSHEET■10.4■(doc-5279):■Identifying■ quadrilaterals■(page 342) •■ WorkSHEET■10.2■(doc-5283):■Deductive■ geometry■II■(page 343) 10e Quadrilaterals and proof Digital docs

•■ Activity■10-E-1■(doc-5107):■Quadrilateral■proofs■ (page 345) •■ Activity■10-E-2■(doc-5108):■Harder■quadrilateral■ proofs■(page 345) •■ Activity■10-E-3■(doc-5109):■Tricky■quadrilateral■ proofs■(page 345) •■ WorkSHEET■10.3■(doc-5284):■Deductive■ geometry■III■(page 346) Chapter review

(page 350) •■ Test■yourself■Chapter■10■(int-2855):■Take■the■end-ofchapter■test■to■test■your■progress •■ Word■search■Chapter■10■(int-2853):■an■interactive■word■ search■involving■words■associated■with■this■chapter •■ Crossword■Chapter■10■(int-2854):■an■interactive■ crossword■using■the■defi■nitions■associated■with■the■ chapter Interactivities

To access eBookPLUS activities, log on to www.jacplus.com.au

chapter 10 Deductive geometry

351

eBook plus

ict actiVity Process ■■

Backyard flood SearCHLIgHt Id: Pro-0099

Scenario Mrs■Effi■ciency■is■at■home■busy■preparing■for■ Christmas■and■needs■to■water■the■fruit■trees■in■the■ back■yard.■When■she■goes■to■connect■the■hose■to■the■ tap■she■discovers■that■Mr■Effi■ciency■has■taken■the■one■ she■usually■uses.■Her■only■option■is■to■use■one■twice■ the■diameter.■She■assumes■this■will■deliver■water■at■ twice■the■rate■of■the■smaller■hose■and■so■leaves■it■for■ half■her■usual■time■of■20■minutes.■She■returns■to■a■ fl■ood■in■the■backyard.■ At■the■evening■meal■she■discusses■the■event■with■ the■family.■She■is■really■distressed■because■she■ realises■that■her■action■has■led■to■a■waste■of■water■(a■ very■scarce■resource).■Her■Year■10■daughter■decides■ to■research■the■situation■and■present■a■summary■to■her■ mother■to■explain■what■has■happened.

■■ ■■

■■

■■

task ■■

■■ ■■ ■■

352

You■will■need■to■look■at■the■Interactive■websites■ found■in■your■eBookPLUS■to■summarise■how■ doubling■and■halving■dimensions■affects■length,■ area■and■volume. You■will■use■Google■Sketchup■to■design■Babushka■ solids■to■be■used■as■measures■for■cooking. You■will■also■investigate■if■the■weight■gain■of■a■ baby■fi■ts■this■same■model. You■will■then■write■up■your■procedure■and■ conclusions■in■Word■and■present■it■as■a■book■in■ Calameo.■You■should■assume■the■role■of■Mrs■ Effi■ciency’s■daughter■who■is■preparing■a■report■for■ her■mother.■Remember■to■provide■some■advice■to■ Mrs■Effi■ciency■regarding■conservation■of■water.

maths Quest 10 for the australian curriculum

■■

■■

Open■the■ProjectsPLUS■application■in■your■ eBookPLUS.■Watch■the■introductory■video■lesson,■ click■the■‘Start■Project’■button■and■then■set■up■ your■project■group.■You■can■complete■this■project■ individually■or■invite■other■members■of■your■class■ to■form■a■group.■After■saving■your■settings,■the■ project■will■be■launched. Navigate■to■your■Research Forum.■Here■you■will■fi■nd■ a■series■of■topics■to■help■you■complete■your■task. Research.■Keep■a■journal■of■your■discoveries■as■ you■proceed■with■the■task.■The■journal■entries■ should■be■short■phrases■to■explain■what■you■are■ doing/thinking,■similar■to■using■twitter■or■sending■a■ text■message■to■a■friend■—■do■not■use■abbreviations■ for■your■words.■Start■each■message■with■the■title■ of■the■topic■you■are■reporting■on.■Each■person■in■ your■group■should■report■on■at■least■3■different■ sources■of■information.■You■can■view■and■comment■ on■the■other■group■members’■entries■and■rate■the■ comments■they■have■made.■When■your■research■is■ complete,■print■your■Research■Report■to■hand■to■ your■teacher. Go■to■the■Media Centre■in■your■eBookPLUS■and■ open■the■Enlargements■Teaching■Tool.■Complete■ the■table.■Answer■the■questions■provided■in■the■ Enlargements■Teaching■Tool■fi■le■in■the■Media Centre.■Summarise■your■discoveries■in■200■words■ or■fewer.■Include■screenprints■of■each■enlargement. Wordle■is■a■site■that■creates■an■image■of■the■words■ in■an■article■according■to■the■frequency■of■their■ usage.■Open■the■Wordle■site.■Select■the■create■tab.■ Copy■the■summary■of■your■article■into■the■text■box.■ Keep■selecting■the■Randomise button■until■you■are■ happy with■the■result.■Print■your■fi■nal■choice.■Take■ a■screenprint■of■your■fi■nal■choice.■Take■it■into■Paint for■use■as■an■image■in■your■presentation.■Save■your■ Paint■fi■le■as■a■jpeg■fi■le. Go■to■the■Media Centre■in■your■eBookPLUS■ and■open■the■Area■Calculator■site.■Make■sure■the■ calculations■are■done■in■metres.■For■each■of■the■ shapes,■choose■a■number■to■enter■into■the■box■to■ calculate■the■area.■Recalculate■the■area■for■half■ this■dimension■and■double■it.■How■can■this■be■ summarised?■Repeat■this■process■for■the■Volume■ Calculator■site.■Summarise■your■fi■ndings.■Include■ this■summary■in■your■report. Download■Google Sketchup. Watch■the■ introductory■video■to■get■started.■Use■the■measuring■

tool to help you create a set of nesting solids to be used instead of babushka (matryoshka) measuring cups — look at the photo to see the proportions of the cups. Take a screenprint of your masterpiece to include in your report.

area, surface area and suggested volume in terms of the software scale factor. Discuss • ProjectsPLUS whether or not the • Microsoft Word weight gain of a baby • Internet Explorer fits this same model. • Google Sketchup Write a paragraph • Wordle • Calameo explaining the error of • Enlargements Mrs Efficiency’s Teaching Tool calculations, and why her backyard flooded. Assume the role of Mrs Efficiency’s daughter who is presenting her research findings to her mother.

Remember: 1 cup is equal to 250 mL.

■■

■■

Use the Australian Icon weblink in your eBookPLUS and choose 3 icons to compare with the real object. Calculate the scale factor (range of numbers). Give measurements and calculations to support your conclusion. Calculate the enlargement factor of the icon from the real object. Calculate how many times larger the surface area and volume of the icon is compared with real object. Go to the Media Centre in your eBookPLUS and open the Sydney Harbour Bridge link. Download the model. Use the measuring tool in Google Sketchup to find the length and height of the model. Compare this with the actual measurements of the Sydney Harbour Bridge. Comment on its accuracy in your report. Take a screen capture, with your measurements visible on the model, to include in your report.

■■

■■

■■

Use Microsoft Word when preparing your presentation. Your Media Centre includes images that can help to liven up your presentation. As you arrange your images on your Word page make them appear as they would in the pages of a book. Each page of your Word document will be a page in your Calameo production of your report. Summarise the findings of your research. Give formulas for length,

Use Word to develop your presentation. Remember that you are trying to deliver a comprehensive summary of similarity and clarify why there was a flood in the backyard. You are also showing applications of similarity. Make sure you include all the results of your research, and that your presentation will grab the attention of those listening. Try to limit your report to either 4 or 8 pages for the most professional delivery. Use Word to type up your dialogue to present your report (200–500 words). Before you present your findings, use the Calameo weblink in your eBookPLUS and sign up for a free account. You will need an email address so ensure you have your teacher’s or your parents’ permission to do this. Take the guided tour to learn how to prepare a Calameo presentation. Create a publication and upload your Word report. Choose the radio button for a Private Publication Mode. Click the Start Uploading button. To show your report, click the button Read the publication and choose full screen view before you commence your delivery. (If the pagination alters using Word — and this is a problem — delete your Calameo publication and repeat the process by uploading a pdf of your report.)

ICT Activity — projectsplus

353

probleM solving

11 problem solving I

60 cm

m

40 c

opening QUesTion

How far does the centre of the racquet travel if the girl swings through an angle of 300è?

problem solving 1 A cuboid has dimensions 10 cm by 12 cm by 18 cm. Find the length of the diagonal space. 2 Expand (3x - 2y)4. 3 Sketch a possible graph of y = 2x2 -3x + g. Determine axial intercepts and the coordinates of 4 5 6 7



the turning point, if any. Find the volume of a right cone with a base diameter of 14 cm and slant height of 25 cm. Consider a right-angled triangle, such that the two shorter sides are 6.4 mm and 8.9 mm in length. Find the angle between the shortest side and the hypotenuse. The perimeter of a rectangle is 20 cm and its area is 14 cm2. Calculate the dimensions of the rectangle, correct to 1 decimal place. Solve for x and y. y = ax - 3b dx + ey = f

8 When two algebraic fractions are equal, a method known as ‘cross-multiplying’ makes

finding the value of x a lot quicker. a c =   so, a(x + d) = c(x + b) x+b x+d Expand and solve normally. Use the above method to find x in each of the following. a

3 5 = x−2 x+2

b

3x + 4 5x − 4 = 2 6

2x + 1 4 = 1 + 3x 5 9 Mary has baked a birthday cake in the shape below. c

16 cm 14 cm 18 cm

10 11 12

13

She has 60 cm of ribbon which she wants to wrap around the sides of the cake. Does she have enough ribbon? Explain your answer. Find the angle of elevation to the top of a 27.3-m high Norfolk Pine tree that is 83.6 m from the observer. Assume that the observer’s eye is 1.667 m above ground level. Solve a(x - p)(x + q) í 0 for x if a < 0 and p > -q. A cylinder of length, l m, has both circular ends removed and replaced with hemispheres. The container now has a length, L m. Determine the volume of the container now in terms of L and l. Marlon substituted numbers into the equation below until he had a true statement. x(x - 3) = 10 Marlon’s answer for the problem was 5. a Is Marlon’s answer right or wrong? Explain.

356

Maths Quest 10 for the Australian Curriculum

problem solving b A friend of Marlon’s showed him another way to solve the problem.

x(x - 3) = 10 x = 10 x - 3 = 10 x = 13 (10, 13)

14 15

16

17

Is Marlon’s friend correct? If the solution is correct explain why, and if the solution is incorrect, provide a correct solution. Sketch the graph of y = 2(4)x - 8, showing axial intercepts and asymptotes, if any. A piece of flat pastry is cut in the shape of a right-angled triangle. The longest side is 6b cm and the shortest is 2b cm a Find the length of the third side. Give your answer in exact form. b Find the sizes of the angles in the triangle. c Prove that the area of the triangle is equal to 4 2b 2 cm2. Cameron purchased 500 tickets for an AFL game so that all of the teachers and students in the school would be able to attend the match. Teachers AFL tickets: $15/ticket Students AFL tickets: $5/ticket The total cost for the game was $3500. How many students attended the game? Determine the length of the diagonal x.

8 cm

x

7 cm

5 cm

18 Parallel lines on a Cartesian plain have the same gradient but different y-intercepts. Find the

19

20

21 22

pairs of parallel lines from the following list and state the gradient and y-intercept for each. a 3y + 6x = -36 b 4y = -4x + 20 c 3y + 1 = 9x d 12 = 2x + 2y 1 e 10 y = − x − 8 2 f 12x = -6y + 12 g 2y - 6x - 7 = 0 h 20y = -x + 5 The formula that can be used to find the surface area, A cm2, of a solid cylinder with radius r cm and height h cm is A = 2p r(r + h). a Find an exact value for A when r = 4, h = 6 b Describe in words the changes that will occur to A if r and h are both halved. Justify your reasoning mathematically. A rocket is fired from ground level (from an underground concealed bunker) and lands h kilometres away, across horizontal terrain. If the maximum height the rocket reaches is k kilometres, find the equation of its path in terms of h and k. The perimeter, P, of a square lies in the range e to f, i.e. e  Ç  P  Ç  f. In terms of e and f what is the range of values for its area, A? Find the equation of the straight line going through (-1, 5), parallel to the line which passes through (0, 4) and (5, -3). Chapter 11 Problem solving I

357

problem solving 23 One method of measuring the height of a building (h) is shown in the figure below. This

method is often used when you are not able to measure the distance along the ground (x) because of a lake or some other obstacle.

h q

a D

x

Step 1: From a location on ‘your’ side of the lake, measure the angle a, using an angle measuring device called a transit or an inclinometer. Write an equation involving x, h and a. Express this equation with x on the LHS and all other terms on the RHS. Step 2: Move a further distance, D, away from the building, where D is the known distance between the first location and the second location. Measure the angle, q, to the top of the building from the second location.

24 25

26 27

28

Write an equation involving x, h, D and q. Express this formula with x + D on the LHS.

Step 3: Solve the equations from Steps 1 and 2 for x in terms of all other variables. Then since each of the expressions equals x, equate the expressions. Step 4:  Solve this equation for h, in terms of D, a and q. Step 5: Use this equation to determine the height of a building where D = 50 m, a = 34.3è and q = 30.7è. P2 The perimeter of a rectangle is P cm and its area is cm2. Determine the dimensions of the 16 rectangle in terms of P. Warwick was solving a pair of simultaneous equations using the elimination method and reached the result that 0 = -5. Suggest a solution to the problem, giving a reason for your answer. A piece of wire 1 m long is cut into two pieces. One piece is used to make a circle, the other a square. Determine where to cut the wire so that the areas of the square and circle are equal. Translate each of these parabolas by the given amounts, then write the new equation in standard form. a y = x2 - 4x + 1 translated 3 units left, 2 units up. b y = -4x2 + 6x - 2 translated 3 units down and 1 unit right. c y = (x - 4)2 - 5 translated 2 units left and 5 units up. A cone has a radius (r) of 8 cm and a height (h) of 16 cm, as shown in the figure below. The ‘top’ is sliced off to leave a ‘frustum’ (shaded area).

Cone Frustum

h r

a Calculate the total volume of the cone. b If the volume of the sliced off top is 20% of the total volume, determine the height of the

frustum. 358

Maths Quest 10 for the Australian Curriculum

problem solving 29 A movie projector uses 35 mm film (35 mm wide and 24 mm high) with a light source 60 mm

from the film’s surface. a How far away is the projector’s light source from the screen if the width of the image on the screen is 16.5 m? b If the distance between the film and the light source is halved, what happens to the width of the image on the screen? 30 A landscape gardener wishes to put a fence around a rectangular lawn. The lawn’s width is 3 m shorter than its length and there is to be allowance for a 2-m wide gate. a Develop a formula for the total length of the fence in terms of the length of the lawn. b The cost of the fence is $23 per metre plus a $100 additional fixed fee. Modify your formula to provide the cost of the fence. c Use your formula to determine the cost of a fence for a lawn whose width is 12 m. 31 The equation of a quadratic can be determined directly from a table of values and differences. Consider the table below. x

-3

-2

-1

 0

1

 2

 3

y

 3

-3

-5

-3

3

13

27

Now determine the 1st differences x

-3

-2

-1

 0

1

 2

 3

y

 3

-3

-5

-3

3

13

27

-6

1st difference

-2

 2

6

10

14

Add the 2nd differences to the table x

-3

-2

-1

 0

1

 2

 3

y

 3

-3

-5

-3

3

13

27

1st difference 2nd difference

-6

-2  4

 2  4

6 4

10  4

14  4

Note how the 2nd differences are constant. The theory of differences states that this constant is equal to 2a in the equation y = ax2 + bx + c. a Determine the value of a. b Re-do the table to subtract the term ax2 from each y-value, leaving a table for bx + c. Use this table to determine b. Note that this equation is now linear and the first difference gives the value of b. c Use a similar method to determine c. d What is the equation of the quadratic? Confirm your result by plotting the graph. 32 A box with a lid is to be constructed with a total surface area of 260 cm2. The box is to have a square base of side length x and height L. a Write the equation for total surface area and make L the subject of the equation. b Write the formula for the volume in terms of L and x. Substitute in your result from part a  to derive a formula for the volume in terms of x. c By trial-and-error, or another method, determine the values of L and x that make the volume as large as possible. Chapter 11 Problem solving I

359

problem solving 33 A right-angled triangle is inscribed in a circle of diameter length d cm as shown in the

diagram below.

d

ø8 4

a Show that d = 4 6 cm b Show that the proportion of the area of the triangle to the area of the circle is

2 6π

34 Develop a formula for the volume of a cone with a radius r and a slant height s. 35 Given the quadratic equation y = x2 - 4x + 7 and a ‘general’ quadratic y = ax2 + bx + c,

determine the conditions on a, b and c such that: a the two quadratics never intersect b the two quadratics intersect (or touch) once c the two quadratics intersect twice. 36 A set of wine bottles is stacked lying down as shown in the diagram below, making a triangular effect. The radius of each bottle is r cm.

a b c d

How many bottles are required if there are 10 bottles on the base and 1 bottle on the top? What is the length of the base of the 10-bottle ‘triangle’? What is the total height of the triangle? Generalise your result for the height of a ‘triangle’ of n bottles on the base.

37 The cost of a return airline ticket to Perth from Sydney varies between airlines. If a ticket

travelling with Virgin Green Airline costs $458 and a ticket travelling with Qintas costs $506, determine the number of people who travelled by air to Perth from Sydney if there were 20% more passengers that flew with Virgin Green and the total price for all airline tickets was $63  336. 38 The diagram below represents the safety ratio for placing ladders against vertical structures.

4 units q 1 unit a Using the values shown in the diagram, determine the value of the angle, q. Write your

answer correct to the nearest minute.

b A 3-metre ladder is placed against a vertical wall. Determine the horizontal distance,

to the nearest centimetre, that the ladder should be placed so that it satisfies the safety regulations. 360

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problem solving 39 Concentric circles are circles that share a common centre. A circle is drawn with a radius of

x cm. The next circle drawn has a radius of (x + 1) cm. This pattern continues until five circles are drawn. The diagram below shows the 5 concentric circles.

x+1 x

a Write down the radii of the 3rd, 4th and 5th circles, in terms of x. b Write down an expression, in terms of x, that can be used to determine the ratio of the

area of the 2nd circle to the area of the 5th circle. If this ratio is 94 , determine the value of x. c What is the percentage increase in circumference, in terms of x, in moving from the 3rd circle to the 4th circle? 40 Carol is celebrating her 16th birthday. In her excitement she cuts the cake into unequal

sections. Her brother takes the largest piece which is twice as much as her mother’s piece. Her sister takes one of the smaller pieces which is 1 the size of her mother’s piece of cake. 3

Carol and her father each take a piece of cake which is 1 12 times as large as Carol’s mother’s

piece. If Carol had cut the cake into 8 equal slices, each slice would have been the same size as her mother’s piece of cake. a What fraction of the cake was eaten by Carol and her family? b What fraction of the cake remains? c If the exact amount of cake, in cm2, remaining after Carol and her family have eaten their one piece of cake is 160p, determine the exact diameter of the cake.

41 Calculate the x-coordinate of the intersection point(s), if any, of:

y = 2x2 - 5x - 3 and y = -x2 - 3x Give your answer in exact form. 42 A large advertising banner is to be placed on the side of a building. The banner has a diagonal

of length 4 17 m and a height of 5 5m. a Determine the exact width, in metres, of the banner. b Determine the exact area, in m2, of the banner. c To attach the banner to the side of the building, anchor points are attached around the border. There is an anchor point attached to each corner and anchor points across the width and the height. i There are 7 anchor points across the width at the top and 7 at the bottom of the banner. Determine the exact length, in metres, between these points. ii Anchor points are placed

5 5 3

metres apart along both sides of the height of the

banner. Determine the total number of anchor points needed for the banner. 43 Shane is a coach driver who conducts tours in outback Australia. All tours are based on twin share at a cost of $x per passenger. For passengers wishing to have their own room, an additional cost of $385 is added to the overall tour price. Usually on any tour, an average of 75% of passengers choose twin share. a If there are n passengers, write down an equation, in terms of n and x, that can be used to determine the total amount, A, in dollars, collected in tour money. Chapter 11 Problem solving I

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problem solving b Shane conducts a tour with 50 passengers. On the next tour there are 45 passengers.

The difference in the total amount between the first and second tour is $17  981.25. Determine the value of x. 44 A section of a stained glass window is shown below. y 2 3

1

D

A

0

B

C

x

The pattern formed can be modelled using three intersecting parabolas labelled 1, 2 and 3 with equations: 2 10 Parabola 1: y = − x 2 + x [1] 15 3 Parabola 2:  y = -0.192x2 + 9.6x - 90 Parabola 3: y = −

[2]

2 2 500 x + 10 x − 15 3

[3]

a By solving the equation [2] for y = 0 using any method, find the coordinates of points

B and C.

b      i Set up an equation, in terms of x, that will determine the point of intersection

between parabolas 1 and 2. Write your values in exact form. ii Using your calculator, find the coordinates of point A. c      i By finding the turning point format of equations [1] and [3], state the transformation

made to equation [1] so that it maps onto equation [3]. ii Hence, write down the coordinates of point D. 45 Chilly Treats want to remodel their ice snacks packages. Currently frozen fruit juice is sold

in containers in the shape of an equilateral triangular-based prism. The sides of the triangular base are x cm and the height of the container is 10 cm.

10 cm

x

x x

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The total capacity of the container is 250 mL. Chilly Treats’ new design will be in the shape of a cylinder. a Find the exact value of x2. b If the capacity and vertical height of the container is to remain the same, determine the exact diameter, in centimetres, of the cylinder. c If the area of one of the side faces of the triangular based prism is approximately 40 cm2, determine the difference in the total surface area, in cm2, between the two different containers. Write your answer correct to 2 decimal places. Clearly indicate which container has the larger surface area. 46 Jacques is test driving a new model Rocket Roadster. The speed of the car can be modelled using the equation S(t) = -3t2 + 12t + 27, where S is the speed in metres per second (m/s) and t is time, in seconds. a What was the initial speed, in km/h, of the car when the testing began? Write your answer correct to 1 decimal place. b By converting S(t) into turning point format, determine the maximum speed, in km/h correct to 1 decimal place, that the Rocket Roadster can reach in this road test and the time taken, in seconds, to reach that speed. 47 A circular piece of paper has a 90è sector removed as shown in the diagram below. r 90è

The remaining area is carefully folded to make the shape of a cone. a What is the slant height of this cone? b Develop a formula for the radius of the base of this cone. 3 c Show that the volume of the cone is 7 π r 3. 64 48 A ball is thrown upwards in the air. The maximum height of the ball is given by the rule h = h0 + V0t - 16t2, where h = height (in metres) and t = time (in seconds). a What is the height at t = 0? b The ball is thrown up in the air from the edge of a roof 20 m high. It goes up, reaches a maximum height and returns just past the edge of the roof and hits the ground. The initial speed of the ball, V0, is 12 m/s (about 43 km/h). Sketch a graph of this situation. c What is the maximum height of the ball? Estimate your answer from the graph then calculate the exact answer. d How long does it take to reach the maximum height? Estimate your answer from the graph then calculate the real answer. e How long is the ball in the air before it hits the ground? Estimate your answer from the graph then calculate the real answer. 49 Farmer Gordon has two paddocks ready for sowing. One paddock has dimensions (2x - 1) metres by (3x + 1) metres, and the second paddock has an area of (36x2 - 6x - 6) m2. a Show that the area ratio of paddock 1 to paddock 2 is 1 : 6. b The area of paddock 2 is 97  026 m2. Show that the value of x is 52. Hence find a set of possible dimensions for paddock 2 using your answer from part a. 50 Greg is building a garden shed. He measured out the width and length of the shed to be (3 m ê 5 cm) by (4.5 m ê 5 cm). a i Determine the possible range of values for both the width and length. ii D  etermine the percentage error in Greg’s measurement for the width if the actual width was 2.98 m. Write your answer correct to 2 decimal places. Chapter 11 Problem solving I

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probleM solving b The walls of the shed will be 2 metres high and the roof will be a height of 3 metres

from the base. The shed will be constructed entirely from corrugated iron. The diagram below shows the shed and its dimensions. The shed will be built on a concrete slab. Using Greg’s measurements of length of 4.5 m and width of 3 m, determine the minimum amount of iron, in m2, required. Write your answer correct to 2 decimal places.

1m

2m

4.5 m 3m c Corrugated iron is sold in lineal metres at $11.80 per metre. The effective width of the

corrugated iron sheet is 762 mm (this is allowing for overlapping of sheets). Determine the minimum cost for the corrugated iron. Write your answer to the nearest $10. 51 To estimate the length along the side of an inaccessible bushland, a surveyor marks out a circular path around the bushland so that the four corners of the bushland lie on the circumference of the circle. She is able to measure three of the four sides of the bushland as shown in the sketch below. (The diagram is not drawn to scale.) D 790 m A

pè

2sè 3

xm

980 m

3 pè 2 sè

C

850 m

B a Using one of the theorems of circle geometry, determine the exact values of p and s, in

degrees. b Describe the shape of the figure ABCD. 52 Explain how you could develop a pattern to determine two consecutive natural numbers

whose squares differ by 75. 53 A certain type of carpet has a width of (x + 2) metres. Customers can purchase the carpet in any length. Mr Barnes buys (x + 5) m of this carpet for his rumpus room and Mr Snowdon buys a 4 m length for his family room. a Write an expression for the area of carpet that each man buys. b Write an expression for the difference in area if Mr Barnes has the longer piece of carpet. c Factorise and simplify this expression. d If Mr Barnes has bought 6 m2 more than Mr Snowdon, find the width of the carpet. e What area of carpet did each man buy? 364

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problem solving 54 A square is drawn within a semicircle as shown in this diagram. The area of the square is 32 cm2.

A square is drawn inside a circle with the same radius as the semicircle.

Determine the length of the side of the square in the circle. 55 Rebecca and Bethany are participating in a fund raising charity door knock. In the first hour, each girl collected x number of gold coins. In the second hour Rebecca collected (x - 1) gold coins and in the third hour she collected (3x - 4) gold coins. In the fourth hour, the number of gold coins each girl collected was the product of the number of coins collected during the second and third hours. Bethany’s number of gold coins collected in the fourth hour can be expressed as 2x2 + 5x - 7. a Show that the number of coins Rebecca collected during the fourth hour was 3x2 - 7x + 4. b Show that the expression, in terms of x, for the combined number of coins both Rebecca and Bethany collected in the fourth hour is (x - 1)(5x + 3). 56 The shape of this vase be approximated using a truncated cone, as shown in the diagram below. 20 cm

30 cm 15 cm

Show that the total amount of water, in litres, that can be poured into the vase is 2.3p. 57 Consider the diagram below. If the length of AO is one-third the length of AD and the length

of AC is 2 units, explain why the length of BD is 4 units. C

D

O A

B

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problem solving

58 In an election for Year 10 representatives on the school council,



1 4

1

2 5

of the votes went to James,

1

to Jennifer, 6 to Raoul, 10 to Amy and the remaining 20 votes went to Diana. How many

students were there in year 10? 59 An Xbox games package comes in a box that has a length 10 cm longer than its width and a height that is 7 cm greater than its width. a Using the variable x to represent the width of the box, write an equation for the box’s volume. b Find the volume of the box if its width is 30 cm. c The manufacturer wants to include another controller and decides that the box should have a volume of 94  000 cm3 but retain the same shape. Using trial and error, find the width for this volume, if x must be an integer.

60 Find values of k for which the simultaneous equations 2x + y = 6 and 2y = -4x + k have: a an infinite number of solutions b no solutions.

Explain how you arrived at your answers. 61 A yacht sails for 25 km on a bearing of 163è20ÅT and then for a further 19 km on a bearing of 121è40ÅT. a Draw a fully labelled diagram to show the yacht’s course. b Determine how far east and south the yacht is from its starting point. c Calculate the true bearing of the yacht from its starting point. Give your answer correct to the nearest minute. 62 Martha decides to redesign the front cover of her diary which has an area equal to (x2 - 3x - 10) cm2. a Factorise this expression to find the dimensions of the diary cover in terms of x. b Write down the shorter length in terms of x. c If the shorter sides of the diary cover are 12 cm in length, find the value of x. d What is the area of the front cover of Martha’s diary? 63 A rectangular hallway rug is five times as long as it is wide. Its diagonal length is 410 cm. How wide and long is the rug? 366

Maths Quest 10 for the Australian Curriculum

probleM solving 64 A circular dining table made of cedar timber is inlaid with glass as shown in the diagram

below. The radius of the glass top is 2r cm with a 20 cm ring of cedar around it.

State the diameter of the glass (in terms of r). Give the radius of the glass and wood (the table top). Calculate the area of the glass. Determine the area of the top of the table (glass and wood) Write an equation to find the area of the wood section only and write it in factorised form. f If the radius of the glass is 40 cm, find the area of the wood needed to surround the glass. Give your answer in m2. g The manufacturers want to make a slightly larger table in the same design using the same width cedar ring. If the area of the table top is to be 2 m2, find the size of r (to the nearest cm). 65 How many solutions can you find for the equation (x - 6)(y - 4) = 24 if x and y are positive integers? 66 A cross brace (shown in red) has been placed to support the roof of a garage as shown. Find the length (in mm) of this supporting beam. a b c d e

1800 mm

5200 mm 3000 mm 67 When the movie The Fellowship of the Ring was shown in the cinema, every seat (550) was

taken. The price of admission for adults was $9.50 and for children $4.50. The takings for one night were $4275. How many adults and children were present at the movie? 68 A coffee table rectangular cloth is to be decorated by sewing lace onto the edge of the material. Its length is four times its width. a If the width of the material is x cm, express the dimensions of the cloth in terms of x. b Give an equation for the perimeter and the area of the cloth in terms of x. c Find the length and width of the cloth if its perimeter is 3 m. d If the width of the lace is 6 cm, what is the outside perimeter of the cloth now, and how much area does it cover? (Answer in terms of x) e Given the original perimeter was 3 m, what increase in area of the cloth was achieved by adding the lace? 69 Solve each of the following. a Find the value of r if x2 - 4x - r = 0 has one solution. b Find the value of s if 2x2 - 5x + s = 0 has two solutions. c Find the value of t when tx2 - 3x - 8 = 0 has one solution. Chapter 11 problem solving I

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probleM solving 70 Five years ago Dan was twice as old as he was 15 years ago. What is Dan’s age now? 71 A surveyor measures the angle of elevation to the top of a lighthouse from a point on the

ground 130 m from its base as 37è. When he looks further down the lighthouse, he sees a large balcony. The angle of elevation to the balcony from the same point is 31è. What is the distance from the balcony to the top of the lighthouse? 72 Robyn keeps guinea pigs in a small square enclosure with sides measuring x m. The number of guinea pigs is increasing so she wants to increase the size of the enclosure by 1 m on one side and 3 m on the adjacent side. a Draw a labelled diagram of the original square and show the additions to it. b Write an expression for the area of the new enclosure. c To satisfy animal safety requirements, the area of the enclosure must be at least 15 m2. Find the dimensions of the enclosure. d To make sure the enclosure is big enough, Robyn decides to make the area 17 m2. Determine the new dimensions of the enclosure (to the nearest cm). 73 During an 8-hour period, an experiment is done in which the temperature of a room follows the relationship T = h2 - 8h + 21, where T is the temperature in degrees Celsius h hours after starting the experiment. a Change the equation into turning point form and hence sketch the graph of this quadratic. b What is the initial temperature? c After three hours, is the temperature increasing or decreasing? d After five hours is the temperature increasing or decreasing? e State the minimum temperature and when it occurred. f What is the temperature after 8 hours? 74 The height of a playground swing above the ground of is 2 3 m. The base of the swing’s pipe supports must be 5 m apart so that the structure is stable once children start to swing.

5m 2 3m

a Find the length of the struts (use exact values) that make up the supports for the swing. b If the base of the swing seat is to be 1 m off the ground, how much chain is required for

2 swings? (Use exact values.) c How much pipe is required to build the swing if the length at the top of the swing is to

be 30 m long? 75 A bike chain is wrapped around 3 gear wheels that are the same size. The radius of each wheel is 8 cm. How long is the chain?

368

Maths Quest 10 for the Australian Curriculum

problem solving 76 The parabola y = x2 + bx + c has x intercepts (2, 0) and (6, 0). a Find the values of b and c. b State the equation. c Complete the square and find the turning point. d Sketch the parabola. 77 Two guide wires are used to support a flagpole as shown. One reaches the top of the flagpole

and the other part way down the pole.

Wire 9.5 m 3m 6m a What is the height of the flagpole (to the nearest metre)? b What angle (to the nearest degree) does the longer guide wire make with the ground? c The shorter wire is attached to the flagpole 1 m from the top. How long is this wire? 78 When a drop of water hits the flat surface of a pool, circular ripples are made. One ripple is

represented by the equation x2 + y2 = 9 and 5 seconds later, the ripple is represented by the equation x2 + y2 = 225, where the lengths of the radii are in cm. a State the radius of each of the ripples. b Sketch these equations. c How fast is the ripple moving outwards? d If the ripple continues to move at the same rate, when will it hit the edge of the pool which is 2 m from its centre? 79 Stephie, a tennis player, serves the ball in a tournament. She throws the ball in the air and hits it over the net. Her arm length is 60 cm and it is 40 cm from her grip on the tennis racquet to the centre of the racquet. How far does the centre of the racquet travel if she swings through an angle of 300è? 80 This 8 cm by 12 cm rectangle is cut into two sections as shown. 6 cm

8 cm

6 cm

10 cm

12 cm a Draw labelled diagrams to show how the two sections can be rearranged to form a:    i parallelogram ii right-angled triangle iii trapezium. b Show that these figures, as well as the original rectangle, all have the same area. c Comment on the perimeters of the figures. 81 Bridgette is practising her golf drives. The path the golf ball takes is defined by the quadratic

1 equation h = − (d − 6)2 + 6, where h is the height of the ball above the ground for a 6 horizontal distance of d. Both h and d are in metres. a Find the value of h when d = 0. 1 b State the turning point of the equation h = − (d − 6)2 + 6 6 c Sketch the graph of this relationship. d What horizontal distance does the golf ball cover in its flight? Chapter 11 Problem solving I

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problem solving e What is the maximum height the golf ball reaches? f At what horizontal distance was the golf ball at its maximum height? 82 A yacht is anchored off an island. It is 2.3 km from the yacht club and 4.6 km from a weather

station. The three points form a right angled triangle at the yacht club. Weather station

Yacht club

2.3 km 4.6 km Yacht a Calculate the angle at the yacht between the yacht club and the weather station. b Calculate the distance between the yacht club and the weather station.

The next day the yacht travels directly towards the yacht club, but is prevented from reaching the club because of dense fog. The weather station notifies the yacht that it is now 4.2 km from the station. c Calculate the new angle at the yacht between the yacht club and the weather station. d Determine how far the yacht is now from the yacht club. 83 The minute hand in Penny’s watch is 1 cm long. Someone told her that the tip of the hand travels more than 30 m in 8 hours. Is this true? Show full working to justify your answer. 84 This cable drum has the measurements shown. 10 cm 25 cm

30 cm

10 cm 50 cm a What volume of wood was used in its construction? b Determine its surface area. 85 Jan is practising for the World Diving Championships. The path she takes from the diving board

into the water is given by the quadratic equation y = -0.75x2 + 3x + 8, where y metres is the height above the water level and x metres is the horizontal distance from the edge of the board. a Using a graphics calculator, sketch the graph of y = -0.75x2 + 3x + 8. b What is the height of the diving board above the water? c What was the maximum height Jan reached during her dive? d What was the horizontal distance Jan covered before she hit the water? 86 A skip bin for waste has been delivered to a building site. Its shape is in the form of a trapezoidal prism with dimensions as shown in the diagram. 4m 2m

3.5 m 3m 370

Maths Quest 10 for the Australian Curriculum

problem solving a Calculate the volume of material this skip can hold. (Assume that it is not loaded beyond

the top rim.) b A smaller skip has a volume one-eighth the size of the larger one. If its shape is similar

to that of the larger one, what would its dimensions be? 87 In a children’s play gym, two cylindrical foam shapes are placed on the ground and a board covered in foam rests on it. The cylinders have radii of 50 cm and 40 cm and their distance apart on the ground is 1.5 m. Calculate the angle the board makes with the ground.

50 cm 40 cm 1.5 m 88 The small and large triangles in this figure are similar.

25

19

x 15

y 22

Determine the lengths of the pronumerals. 89 The equation (x - 1)2 + (y - 2)2 = 4 describes a circle. a State the centre of the circle. b State the radius of the circle. c Find the x- and y-intercepts of this circle. d Sketch the circle, clearly marking the centre. 90 The parabola y = ax2 + bx + c has a turning point (-3, 4) and passes through the point

(1, -28). a Determine its equation. b State the values of a, b and c. c Sketch the parabola. 91 Find the sum of the angles at the tips of this regular star.

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problem solving 92 Tina is re-covering a footstool in the shape of a cylinder with diameter 50 cm and height

30 cm. She also intends to cover the base of the cushion.

She has 1 m2 of fabric to make this footstool. When calculating the area of fabric required, allow an extra 20% of the total surface area to cater for seams and pattern placings. Explain whether Tina has enough material to cover the footstool. 93 The Gold Coast City Council has decided to construct a ceremonial arch at the entrance to Surfers Paradise beach. The arch is to be in the shape of a parabola. The maximum height is to be 15 metres and the width at the base is to be 20 metres. A path, 16 metres wide will pass beneath the arch with its centre immediately beneath the highest point of the arch.

15 m 16 m 20 m

The designers decide to use a mathematical model of the arch in the design process. They place an origin of coordinates at ground level immediately beneath the highest point of the arch. y 14 12 10 8 6 4 2 -10

-5

0

5

10 x

a Prove that the equation of this curve is y = 15 - 0.15x2 b The designers are concerned that there will not be sufficient clearance for vehicles up to

6 metres in height to pass along the path beneath the arch. Show that the vertical height 372

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problem solving

of this arch above the road level at the edge of the road would not be sufficient to allow a 6 m high vehicle to pass through. c How wide must the road be to allow a 6-m high vehicle to pass through the arch? a−b 94 a Q b is defined as . a b 1 1 Θ What is the value of 2 3 in simplest terms? 1 1 Θ 3 2 95 David has calculated the time, in minutes, it takes him to drive to work in the morning as m2 - 10m + 50, where m is the number of minutes after 8 am that he leaves home. a How long does it take David to reach work if he leaves at 8 am. b At what time can he leave so that the trip takes 30 minutes? c The trip will take one hour if he leaves at what time? d When should he leave to take the smallest amount of time. e How long will it take him if he leaves home at this time? f If he decides that he cannot take longer than 30 minutes to get to work, between what times would he have to leave home? 96 There is a theorem which says: If two distinct numbers are exactly divisible by the difference of the two numbers, the difference is the HCF of the two numbers. Explain what this means, illustrating with an example. 97 Matt and his brother Steve start from home in their cars. Matt travels directly east, while

Steve travels directly north at a speed 15 km/h faster than Matt’s speed. After travelling for 1 hour 20 minutes, the two cars are 100 km apart. At what speeds are the two cars travelling? 98 A regular octagon is inscribed inside a circle with all its vertices lying on the circumference of the circle. The circle has a radius of 10 cm. Determine the perimeter of the octagon. 99 An obtuse-angled isosceles triangle has equal angles of xè and equal side lengths of y cm. xè y cm xè y cm

There is not enough information about the triangle to use a traditional formula 1 A = ( 2 base ì height) to find its area. Show how you could use trigonometry to develop the following formula to calculate its area. 1 Area = y2sin (2xè) 2 100 Farmer Max has a rectangular field 150 m by 100 m. His son offers to help him mow the field, but says he will do only half. The ride-on mower cuts a strip 1 m wide. Max starts mowing at a corner, and mows around the field towards the centre. He stops and hands over to his son when he has done n circuits of the field. Construct an algebraic equation involving n, and solve to find its value. Give your answer correct to 1 decimal place. Chapter 11 Problem solving I

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problem solving 101 Pulsars are rapidly spinning stars. They spin at incredible rates when they are first formed —

about 30 times every second. As they age, they slow down. Astronomers have represented the spinning rates of two pulsars (Crab Nebula and AP 2016 + 28) by two simple equations. Crab Nebula: P = 0.033 + 0.000 013T AP 2016 + 28:  P = 0.558 + 0.000 000 004 7T P is the time (in seconds) it takes for the pulsar to spin once on its axis, and T is the number of years since today. In approximately how many years from now will the two pulsars be spinning at the same rate? 102 The two arches of the Sydney Harbour Bridge can both be modelled as parabolas.

Using the reference point (0, 0) as the bottom of the left side of the lower arch, the two equations are: Lower arch: y = -0.001  92x2 + 0.96x Upper arch: y = -0.001  28x2 + 0.64x + 60 Note: The measurements are in metres. Write a description of the two arches, giving as much information as you can. Remember to support your information with mathematical evidence. 1 1 1 1 103 The unit fraction can be expressed as = + where a and b are natural 12 12 12 + a 12 +b numbers. Use this equation to find the product of a and b, then list the 8 different representations for 1 the unit fraction . 12 104 Explain why all perfect squares have an odd number of factors. Give an example to support your explanation. 105 A cone is formed from a sector of a circle with a central angle of 72è. The radius of the base of the cone is 3.18 cm. What is the radius of the circle from which the sector was taken? 106 This tile pattern is made using congruent triangular tiles. There is 1 tile in row 1, 3 in row 2, and so on.

Develop a formula to determine the number of tiles needed to complete a pattern of this type with r rows. 374

Maths Quest 10 for the Australian Curriculum

problem solving 107 The school cafeteria sells apples at one price, and bananas at a different price. Six apples and

4 bananas cost $7, while 1 apple and 9 bananas cost $4.50. a Show how you can determine how much more an apple costs, compared with a banana, without actually finding the cost of each. b What is this difference in price? 108 When children are sick, it’s important they’re given the correct dosage of medicine. If adult medicines are the only ones available, you need to convert the adult dosage into a safe dose for young children. One rule which can be used is: Age in years + 1 ì adult dose Child’s dose = 24 a For an adult dose of 5 mL, how many mL would you give a 5-year old? Another rule commonly used is: Age in years ì adult dose Child’s dose = age + 12 b How many mL of a 5-mL adult dose would you administer to a 5-year old using this formula? c Your two answers should be different. Comment on this difference. d Is there any age for which these two formulae give the same dosage? 109 Ben’s teacher shows him a graph of a quadratic equation. It is not labelled, except for three points (-2, 19), (0, 1) and (3, 4) on the curve. His task is to find the equation of the quadratic. What is the equation of the curve? 110 Draw two straight lines across the face of this clock, so that the sum of the numbers in each region formed is the same. 11 12

1 2

10 9

3

8

4 7

6

5

111 A group of four people out bushwalking comes across a suspension bridge as the last obstacle

they need to cross to reach their campsite. They can’t all cross at once, because the bridge can only support a maximum of 2 people at a time. Unfortunately it is approaching dark, and they only have 1 torch among the 4 of them.

When walking alone, the four people would take 1, 2, 5 and 10 minutes to cross one way. With 2 people walking together, because they need the flashlight for safety reasons, they must travel at the speed of the slower person. One person must then travel back across the bridge each time to bring the flashlight back. How can the group arrange themselves for the bridge-crossing to take the minimum time? What is this minimum time? Chapter 11 Problem solving I

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problem solving 112 Tennis balls are stacked in the shape of a triangular pyramid, with 5 balls on each side of the

base. How many balls are in the whole stack? 113 Timber railings are manufactured to be 100 cm long, with a possible error of 4%. Dana’s deck is 55 m long, and she plans to place the railings end-to-end. How many of these railings should she order to ensure she can cover the whole length? 114 Suppose you had some ‘blue-tack’ in the shape of a cylinder. How could you cut through the cylinder to expose a surface in the shape of a parabola? Explain any restrictions. 115 Prove that the set of numbers represented by: 2n, (n2 - 1) and (n2 + 1) produces Pythagorean triples for all values of n greater than 1. Explain the restriction on the value of n. 116 The Jackson family wants to put an L-shaped deck on one corner of their house. It is to be symmetrical around the corner, as shown. Deck

House

  They have 30 m of a specially designed hand railing to use around the perimeter of the deck. What dimensions of the deck will give them the maximum area, using the whole 30 m of the handrail? 117 A palimage of a number is the number that has the same digits as the given number, but in the reverse order. For example, the palimage of 476 is 674. If the sum of a number and its palimage is 968, what could the original number have been? There is more than one answer. See how many you can find. 118 Consider this question on a test paper. Solve the equation x = −3. Here is Kelly’s solution.

( x )2 = (−3)2 x=9

Comment on the solution. 119 A domino is a 2 dimensional figure formed by two congruent squares that share a common

side.

A tetromino is a 2 dimensional figure formed by four congruent squares that share common sides. a Draw the shape of all the different tetrominoes. b Compare their perimeters. 120 A motor boat leaves a ship at sea and travels north at 80 km/h. The ship precedes 30è south of east at 32 km/h. The motor boat only has fuel for 4 hours. How far north can the motor boat travel, so it can safely return to the ship in time? 376

Maths Quest 10 for the Australian Curriculum

problem solving

121 You are familiar with the quadratic formula x =

− b ± b 2 − 4 ac 2a

An alternative form of the quadratic formula is x =

2c − b ± b 2 − 4 ac

Choose a quadratic equation and show that the two formulae give the same answers. 122 Consider the number 234 written in words.

TWO HUNDRED AND THIRTY FOUR The letters of each word are cycled separately as shown below, and placed in a numbered vertical list. 1 TWO HUNDRED AND THIRTY FOUR 2 WOT UNDREDH NDA HIRTYT OURF 3 OTW NDREDHU DAN IRTYTH URFO .. .. .. n TWO HUNDRED AND THIRTY FOUR If n > 500, what is its smallest value? 123 You will be familiar with the following unit fraction additions.

1 1 1 = + 2 3 6 1 1 1 = + 3 4 12 1 1 1 = + 5 6 30 Write a general equation involving n of the type 1 = . . . . . . . . n to represent unit fraction additions of this type. 124 A puzzle company prides itself on its unique designs. The following design of a map when folded flat is printed on one piece of paper. 3 6

4 5

2 1

7 8

The numbers represent the page numbers of the map. How can the map be folded so that its pages are in the correct order? 125 Trains travel along two straight parallel tracks between Allensville and Bentley, with the journey taking 4 h 15 min each way. The trains leave both towns on the hour every hour. If I leave Allensville at 12 noon and travel towards Bentley, how many trains will pass by me coming in the opposite direction? 126 It has been said that if you multiply the y-coordinates for a particular x-value of two straight lines, then plot this y2-value against the particular x-value, a parabola will result. To investigate this claim, consider the two straight lines y = -2x + 4 and y = x - 3. Complete the following tables. y = -2x + 4 x

-2

-1

0

1

2

3

4

5

6

y Chapter 11 Problem solving I

377

problem solving

y=x-3 x

-2

-1

0

1

2

3

4

5

6

-2

-1

0

1

2

3

4

5

6

y x 2

y 127

128

129 130

378

What is your conclusion regarding the claim? A small company manufactures and sells muesli bars. The set-up cost to make the bars themselves was $5400, and the cost for ingredients is 45c per bar. The set-up cost to package them was $7500, and the cost of materials for each bar is 35c. The bars are sold in a multipack at 5 for $6. How many multi-packs would the manufacturer need to sell in order to break even? Consider the following puzzle. ■■ Take a calendar displaying the dates row by row of any month in the year. ■■ Choose 4 days that form a square on the calendar. ■■ Find the total of those 4 dates. Explain how, from the total, you can determine which 4 dates have been selected. The hour and the minute hand of a clock each trace a circular path as they sweep around the clock’s face. How many times will they lie on top of each other in a 12-hour period? You are familiar with the equation of the parabola with its vertical axis of symmetry. It is also possible to get a horizontal parabola. Its shape is similar to that of a vertical parabola, but its axis of symmetry is horizontal. A horizontal parabola can be represented by the following general equation. x - h = a(y - k)2 Where (h, k) is the vertex of the parabola, and a is a constant. Investigate this general equation with particular values for h, k and a, then complete the following. a How can you distinguish a vertical parabola from a horizontal parabola, simply from its equation? b What is the effect of the value of a? c A horizontal parabola has the equation x + 4 = (y + 1)2. Draw a sketch of its shape, labelling its vertex and all x- and y-intercepts.

Maths Quest 10 for the Australian Curriculum

stAtistiCs AND probAbility • ChANCe

12

12A Review of probability 12B Complementary and mutually exclusive events 12C Two-way tables and tree diagrams 12D Independent and dependent events 12E Conditional probability 12F Subjective probability WhAt Do yoU kNoW ?

probability

1 List what you know about chance. Create a concept map to show your list 2 Share what you know with a partner and then with a small group 3 As a class, create a large concept map that shows your class’s knowledge of chance. eBook plus

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Blackjack (or 21) is a popular card game. What is the chance of being dealt two cards whose total is 21?

stAtistiCs AND probAbility • ChANCe

Are you ready?

Try the questions below. If you have difficulty with any of them, extra help can be obtained by completing the matching SkillSHEET. Either search for the SkillSHEET in your eBookPLUS or ask your teacher for a copy. eBook plus

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380

Set notation 1 Three different sets are identified as follows:

A = {1, 4, 9} B = {2, 4, 6, 8} C = {2, 3, 5, 7}. Use these sets to answer the following. a How many numbers appear in each set? b Which number is common to set A and set B? c Which number is common to set B and set C?

Simplifying fractions 2 Write each of the following fractions in simplest form. a

13 52

b

4 36

c

8 12

Determining complementary events 3 Two events which have nothing in common, but when added together form the sample

space, are called complementary events. Determine the complementary event for each of the following. a Drawing an ace from a standard deck of playing cards. b Drawing a black card from a standard deck of playing cards. c Obtaining a factor of 6 when a six-sided die is rolled. Addition and subtraction of fractions 4 Simplify each of the following. a

1 2

+

1 6

b

1 52

+

5 6

c

1 13

c

1 2

+

1 2

−

1 26

Multiplying fractions for calculating probabilities 5 Simplify each of the following. a

1 2

×

1 6

Maths Quest 10 for the Australian Curriculum

b

1 52

×

5 6

×

1 2

×

1 2

stAtistiCs AND probAbility • ChANCe

12A eBook plus

review of probability ■

Interactivity Random number generator

■

int-0089

■

■

■

■

This chapter, investigates such things as the probability of selecting 3 hearts from a deck of cards and the probability of tossing two heads when a coin has been tossed twice. Probability deals with the likelihood or chance of an event occurring. The probability of an event is represented by a number in the range 0 to 1 inclusive, which can be expressed as a fraction, decimal or percentage. There are times when it is certain that an event will not occur; for example, it is certain that an athlete will not complete a 100 m race in less than 5 seconds. Therefore, the probability for the event ‘an athlete completes 100 m in less than 5 seconds’ is 0. Alternatively, it can be certain that an event will occur. For instance, it is certain that the day following Saturday is Sunday. The probability of such an event is 1. In the range 0 to 1 inclusive there is an infinite set of numbers giving the probabilities of various events, where the chance of an event occurring increases as the probability gets closer to 1. The probability scale shown below displays the range of probabilities in the range 0 to 1 inclusive. Chances decrease Highly unlikely Impossible

0

0.1

Unlikely Very unlikely

0.2

Less than even chance

0.3

Likely

Even chance

0.4

0%

Better than even chance

0.5

0.6

Very likely

0.7

0.8

Highly likely Certain

0.9

50%

1 100%

Chances increase ■

■

■

A probability of 0.5 indicates that there is an equal chance of an event occurring as there is for the event not occurring. The probability of an event can also be described by words and phrases, such as impossible, highly unlikely, very unlikely, less than even chance, even chance, better than even chance, very likely, highly likely, certain and so on. Some terms that are used in the study of probability are defined below.

Definitions ■ ■

■ ■

■

■

Trial: the number of times a probability experiment is conducted. Outcome: the result of an experiment. For example, if a die is rolled, the outcome is a number in the range 1 to 6 inclusive. Event: a desired or favourable outcome. Equally-likely outcomes: outcomes that have the same chance of occurring. For example, if a coin is tossed, then the chance of tossing a Head is equal to the chance of tossing a Tail. Hence, they are equally-likely outcomes. Sample space, S: the set of all possible outcomes for an experiment. For example, in rolling a die, the sample space, S, is S = {1, 2, 3, 4, 5, 6}. Frequency: the number of times an outcome occurs.

experimental probability ■

The experimental probability of an event is based on past experience. Experimental probability =

number of times an event has occurred total number of trials Chapter 12 probability

381

statistics AND probability • Chance

Worked Example 1

A discus thrower has won 7 of her last 10 competitions. a  What is the probability that she will win the next competition? b  What is the probability that she will not win the next competition? Think a

b

1

Write a Number of wins = 7

Write the number of wins and the total number of competitions.

Total number of competitions = 10 number of times this event occurred total number of trials

2

Write the rule for probability.

P(event) =

3

Substitute the known values into the rule.

P(she wins) = 10

4

Write your answer.

The probability she will win the next competition 7 is 10 .

1

Write the number of losses and the total number of competitions.

7

b Number of losses = 3

Total number of competitions = 10 number of times this event occurred total number of trials

2

Write the rule for probability.

P(event) =

3

Substitute the known values into the rule.

P(she loses) = 10

4

Write your answer.

The probability she will lose the next competition 3 is 10 .

■■

3

The event ‘she will win the next competition’ and the event ‘she will not win the next competition’ are called complementary events. Complementary events will be discussed in more detail in the next section.

Relative frequency frequency of the score f or total sum of frequencies S f

■■

Relative frequency of a score =

■■

The symbol S (sigma) means ‘the sum of’. The relative frequency of a score is the same as the experimental probability of that score and is useful when analysing tabulated results.

■■

Worked Example 2

A Year 10 class has the following composition. a Calculate the relative frequency of 16-year-old girls in the class. b If a student is selected at random, determine the probability that the student is a boy. Frequency (  f  )

382

15-year-olds

16-year-olds

Total (S f  )

Boys

 7

 9

16

Girls

 6

 8

14

Total (S f )

13

17

30

Maths Quest 10 for the Australian Curriculum

statistics AND probability • Chance

Think a

1

Write

Write the number of 16-year-old girls and the total number of students in the class.

a Number of 16-year-old girls (  f  ) = 8

Total number in the class (S f  ) = 30 f S f

2

Write the rule for relative frequency.

Relative frequency =

3

Substitute the known values into the rule.

Relative frequency = 30

4

Simplify and evaluate.

5

Write your answer.

8

4

= 15 The relative frequency of 16-year-old girls in the class is 4 . 15

b

1

Write the number of boys in the class and the total number of students in the class.

b Number of boys = 16

Total number in the class = 30 number of times this event occurred total number of trials

2

Write the rule for probability.

P(event) =

3

Substitute the known values into the rule.

P(boy selected) = 30

4

Simplify and evaluate.

5

Write your answer.

16 8

= 15 The probability of a boy being chosen is 8 . 15

Theoretical probability ■■ ■■

■■

The theoretical probability of an event, P(E), depends on the number of favourable outcomes and the total number of possible outcomes (that is, the sample space). The theoretical probability of an event is given by the rule: number of favourable outcomes P(event) = number of possible outcomes This may be simplified to: n(E) P(E) = n(S) where n(E) = number of times or ways an event, E, can occur and n(S) = number of elements in the sample space or number of ways all outcomes can occur, given all the outcomes are possible.

Worked Example 3

A card is drawn from a shuffled pack of 52 cards. Determine the probability that the card chosen is: a  a heart b  a king. Think a

1

Write

Define the events and write the number of favourable outcomes and the total number of possible outcomes. Note: There are 13 cards in each of the 4 suits.

a H is the event that a heart is chosen.

S is the sample space. n(H ) = 13 n(S ) = 52 Chapter 12 Probability

383

stAtistiCs AND probAbility • ChANCe

2

b

Write the rule for probability.

3

Substitute the known values into the rule.

4

Simplify and evaluate.

5

Write your answer.

1

Define the event and write the number of favourable outcomes and the total number of possible outcomes.

2

Write the rule for probability.

3

Substitute the known values into the rule.

4

Simplify and evaluate.

5

Write your answer.

Using P(E) =

n(E ) n(S)

P(H) =

n(H) n(S)

13

P(H) =  52 =

1 4

The probability of choosing a heart is 1 . 4

b K is the event that a king is chosen.

S is the sample space. n(K) = 4 n(S) = 52 Using P(E ) =

n(E ) n(S)

P(K) =

n(K) n(S)

P(K) =

4 52 1

= 13 The probability of choosing a king is 1 . 13

venn diagrams ■ ■

■

■

eBook plus

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SkillSHEET 12.1 doc-5286

384

Venn diagrams provide a means of representing outcomes diagrammatically. A common way of drawing Venn diagrams is to use a rectangle which represents the sample space and a series of circles representing other smaller, sorted sets. In Venn diagrams, overlapping circles represent the intersection of, or common elements in, those sets. The sample space is also known as the universal set, x .

Definitions Terminology associated with Venn diagrams is defined below. 1. A set is a collection of similar elements. 2. The universal set, x , is the largest set that contains all the possible outcomes for that experiment and is represented by the rectangle of the Venn diagram. Consider all the outcomes from an experiment where a die is rolled. The sample space, S, for this experiment is also known as the universal set, x   = {1, 2, 3, 4, 5, 6}. 3. The intersection of sets (symbol ¶) is x = Universal set represented by the common elements in two (or more) sets. A B The shaded region is A ¶ B.

Maths Quest 10 for the Australian Curriculum

statistics AND probability • Chance

4. The union (symbol ß) of the sets A and B is given by the combined set of A and B. It is the set of elements that are in set A or set B or in both.   The shaded region is A ß B.   Note: Common elements are written only once.

x = Universal set

A

5. The complement of a set, A (written AÅ), is the set of elements that are in x   but not in A. The shaded region represents the complement of A.

B

x = Universal set

A

B

The definitions of set, universal set, intersection, union and complement are illustrated in the following example. Example 1 Consider when rolling a die the two events: event A: rolling an even number event B: rolling a multiple of 3. The universal set is written as, x  = {1, 2, 3, 4, 5, 6} and sets A and B are, A = {2, 4, 6} and B = {3, 6}. These are represented in the Venn diagram below. x = {1, 2, 3, 4, 5, 6}

A

B 2

6

3

4 1

5

Also, Intersection of sets A and B

Union of sets A and B

x = {1, 2, 3, 4, 5, 6}

A

A

B 2

6

x = {1, 2, 3, 4, 5, 6}

3

B 2

4

6

3

4 1

A ¶ B = {6}

5

1

5

A ß B = {2, 3, 4, 6} Chapter 12 Probability

385

statistics AND probability • Chance

Complement of set A x = {1, 2, 3, 4, 5, 6}

A

B 2

3

6

4 1

5

AÅ = {1, 3, 5} 6. The subset (symbol ´) of a set is a smaller set from within the set. The shaded region in the diagram shows that A is a subset of x ; that is, A ´ x . x

A

The definition, subset, is illustrated in the example below. If M = {2, 3, 4} and N = {2, 3}, then N is a subset of M, written as N ´ M. x

M N 2

3 4

7. Disjoint sets are sets that have nothing in common with each other. That is, A ¶ B = { } = f It can be argued that the intersection of disjoint sets has nothing in it. The set {} or f  is known as the empty set, or null set. x

A B

386

Maths Quest 10 for the Australian Curriculum

statistics AND probability • Chance

Worked Example 4 a Draw a Venn diagram representing the relationship between the following sets. Show the position

of all the elements in the Venn diagram. x  = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20} A = {3, 6, 9, 12, 15, 18} B = {2, 4, 6, 8, 10, 12, 14, 16, 18, 20} b Determine: i  P(A)    ii  P(B)    iii  P(A ¶ B)    iv  P(A ß B)    v  P(AÅ ¶ BÅ). Think a

Draw a rectangle with two partly intersecting circles labelled A and B.

1 2

b

Write/draw

A

Analyse sets A and B and place any common elements in the central overlap.

3

Place the remaining elements of set A in circle A.

4

Place the remaining elements of set B in circle B.

5

Place the remaining elements of the universal set x  in the rectangle. i

ii

n(x ) = 20

a

B 3 15

11 13 1 5 7 17 19 b

i n(A) = 6, n(x  ) = 20

1

Write the number of elements that belong to set A and the total number of elements.

2

Write the rule for probability.

P(A) =

n(A) n(x  )

3

Substitute the known values into the rule.

P(A) =

6 20

4

Evaluate and simplify.

1

Write the number of elements that belong to set B and the total number of elements.

2

Repeat steps 2 to 4 of part b i.

3

= 10 ii n(B) = 10, n(x  ) = 20

P(B) =

n(B) n(x  )

P(B) =

10 20

= iii

iv

1

Write the number of elements that belong to set A ¶ B and the total number of elements.

2

Repeat steps 2 to 4 of part b i.

1

Write the number of elements that belong to set A ß B and the total number of elements.

2

Repeat steps 2 to 4 of part b i.

2 4 8 10 14 16 20

6 12 18

9

1 2

iii n(A ¶ B) = 3, n(x  ) = 20

P(A ¶ B) =

n(A ¶ B) n(x  )

P(A ¶ B) =

3 20

i v n(A ß B) = 13, n(x  ) = 20

P(A ß B) = P(A ß B) =

n(A ß B) n(x  )

13 20

Chapter 12 Probability

387

statistics AND probability • Chance v

1

Write the number of elements that belong to set AÅ ¶ BÅ and the total number of elements.

2

Repeat steps 2 to 4 of part b i.

v n(AÅ ¶ BÅ) = 7, n(x  ) = 20

P(AÅ ¶ BÅ) =

n(AÅ ¶ BÅ) n(x  ) 7

P(AÅ ¶ BÅ) = 20

Worked Example 5

In a class of 35 students, 6 students like all three subjects: PE, Science and Music. Eight of the students like PE and Science, 10 students like PE and Music, and 12 students like Science and Music. Also, 22 students like PE only, 18 students like Science only and 17 like Music only. Two students don’t like any of the subjects. a Display this information on a Venn diagram. b Determine the probability of selecting a student who: i  likes PE only           ii  does not like Music. c Find P[(Science ß Music) ¶ PEÅ]. Think a

1

Write/draw

Draw a rectangle with three partly intersecting circles, labelled PE, Science and Music.

n(x ) = 35

a PE

Science

Music 2

Extract the information relating to students liking all three subjects. Note: The central overlap is the key to solving these problems. Six students like all three subjects, so place the number 6 into the section corresponding to the intersection of the three circles.

n(x ) = 35 PE

Science

6

Music 3

388

Extract the relevant information from the second sentence and place it into the appropriate position. Note: Eight students like PE and Science; however, 6 of these students have already been accounted for in step 2. Therefore, 2 will fill the intersection of only PE and Science. Similarly, 4 of the 10 who like PE and Music will fill the intersection of only PE and Music, and 6 of the 12 students will fill the intersection of only Science and Music.

Maths Quest 10 for the Australian Curriculum

n(x ) = 35 PE

Science 2 4

6

6

Music

statistics AND probability • Chance

4

5

6

b

i

ii

1

n(x ) = 35

Extract the relevant information from the third sentence and place it into the appropriate position. Note: Twenty-two students like PE and 12 have already been accounted for in the set. Therefore, 10 students are needed to fill the circle corresponding to PE only. Similarly, 4 students are needed to fill the circle corresponding to Science only to make a total of 18 for Science. One student is needed to fill the circle corresponding to Music only to make a total of 17 for Music.

PE 2

10 4

4

6

6

1 Music

n(x ) = 35

Extract the relevant information from the final sentence and place it into the appropriate position. Note: Two students do not like any of the subjects, so they are placed in the rectangle outside the three circles.

PE

Science 2

10 4

Check that the total number in all positions is equal to the number in the universal set. 10 + 2 + 4 + 4 + 6 + 6 + 1 + 2 = 35 Write the number of students who like PE only and the total number of students in the class.

Science

4

6

6

1 Music b

2

i n(students who like PE only) = 10

n(x  ) = 35

2

Write the rule for probability.

P(likes PE only) =

n(likes PE only) n(x  )

3

Substitute the known values into the rule.

P(likes PE only) =

10 35

4

Evaluate and simplify.

5

Write your answer.

1

Write the number of students who do not like Music and the total number of students in the class. Note: Add all the values that do not appear in the Music circle as well as the two that sit in the rectangle outside the circles.

=

2 7

The probability of selecting a student who 2 likes PE only is 7 . ii n(students who do not like Music) = 18

n(x  ) = 35

2

Write the rule for probability.

P(does not like Music) n(does not like Music) = n(x  )

3

Substitute the known values into the rule.

P(does not like Music) = 35

4

Write your answer.

The probability of selecting a student who does not like Music is 18 . 35

18

Chapter 12 Probability

389

statistics AND probability • Chance c

1

Write the number of students who like Science and Music but not PE. Note: Add the values that appear in the Science and Music circles but do not overlap with the PE circle.

2

Repeat steps 2 to 4 of part b ii.

n[(Science ß Music) ¶ PEÅ] = 11 n(x  ) = 35

c

P[(Science ß Music) ¶ PEÅ] n[(Science ß Music) ¶ PEÅ] = n(x  ) 11

P[(Science ß Music) ¶ PEÅ] = 35 The probability of selecting a student who likes Science or Music but not PE is 11 . 35

Odds ■■ ■■ ■■

Probabilities in gambling can be expressed as odds. 5 This is very common in racing, where odds are given as ratios; for example 5–1 (or 1 or 5  :  1). In the odds of a–b, a–b a is the chance against the event

■■

If the odds for a horse to win are given as 5–1, then from 6 races the horse is expected to lose 5 and win 1. The probability that the horse wins or loses can be calculated from the odds given. These calculations are shown below. P(win) =

■■

b is the chance for the event

n(expected wins) n(races) 1

P(lose) = =

= 6

n(expected losses) n(races) 5 6

If given odds of a–b, then: n(E) b = n(x  ) a + b ñ  P(the event does not occur), P(EÅ) = n(EÅ) = a n(x  ) a+b

ñ  P(the event occurs), P(E) =

Payouts ■■ ■■

■■

390

The payout in races is based on the odds given. a If the odds given are a–b, you can win $ for every $1 bet, and alternatively stated, $a for b every $b bet. The bookmaker will pay out your win plus the initial bet. The TAB quotes a whole payout figure for a horse, made up of the winnings and the initial bet. For example: Odds

Bet

Winnings

5–1

$10

$5 for every $1 bet: 1 ì $10 = $50

7–2

$14

$7 for every $2 bet: 2 ì $14 = $49

Maths Quest 10 for the Australian Curriculum

Payout figure

5

$60 ($50 + $10)

7

$63 ($49 + $14)

statistics AND probability • Chance

Worked Example 6

The odds given for the horse Gunnawin to win the Melbourne Cup are 9–4. a Determine the probability of Gunnawin winning the Melbourne Cup. b Tony decides to bet $12 on Gunnawin to win. If the horse does win, what is Tony’s payout? c In the same race, the probability that the horse ‘Can’t Lose’ wins is given as 5 . What are the odds 17 that this horse will win? Think a

b

c

1

Write

Write the number of ways Gunnawin can win (4) and the total number of outcomes (9 + 4 = 13).

a n(Gunnawin wins) = 4

n(x  ) = 13

2

Write the rule for probability.

P(Gunnawin wins) n(Gunnawin wins) = n(x  )

3

Substitute the known values into the rule.

P(Gunnawin wins) = 13

4

Write your answer.

The probability of Gunnawin winning the 4 Melbourne Cup is 13 .

1

Explain what the ratio means and relate it to the bet.

4

9

b In the odds 9–4 the punter can win $ 4 for

every $1 that is bet (or for every $4 bet the punter will win $9). Therefore, if Tony bets 9 $12 he will win 4 ì $12 = $27.

2

Add the original amount invested to the amount returned.

Payout = $27 + $12 = $39

3

Write your answer.

Tony’s payout will be $39.

1

Look at the given fraction. The numerator corresponds to the ‘win’ component (second number) of the ratio.

2

The lose component of the ratio is always the first number.

Therefore the lose–win ratio is 12–5.

3

Write your answer.

The odds of Can’t Lose winning the Melbourne Cup are 12–5.

c

This horse has been given the chance of 5 winning as 17 . Therefore its chance of losing 12 is 17 .

remember

1. Probabilities can be expressed as a percentage, fraction or decimal in the range 0 to 1, inclusive. number of times an event has occurred 2. Experimental probability = total number of trials frequency of the score f or 3. Relative frequency of a score = total sum of frequencies S f n(E) 4. Theoretical probability that an event, E, will occur is P(E) = where n(x  ) n(E) = number of times or ways an event, E, can occur and n(x  ) = the total number of ways all outcomes can occur. Chapter 12 Probability

391

stAtistiCs AND probAbility • ChANCe 5. P(x  ) = 1

6. Venn diagrams provide a diagrammatic representation of sample spaces. 7. If given odds of a – b, then: b n(E) ■ P(the event occurs) P(E) = = a + b n(x  ) n(EÅ) a ■ P(the event does not occur) P(EÅ) = . = n(x  ) a + b exerCise

12A iNDiviDUAl pAthWAys

review of probability flUeNCy 1 Complete the relative frequency column in the given table.

eBook plus

Activity 12-A-1

Review of probability doc-5110 Activity 12-A-2

General probability problems doc-5111

f

1

2

2

5

3

6

4

3

5

Activity 12-A-3

Tricky probability problems doc-5112

x

Relative frequency

4 S f =

2 For the table of values in question 1, what is the probability of selecting the following numbers

if a number is chosen at random? eBook plus

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SkillSHEET 12.2 doc-5287

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SkillSHEET 12.6 doc-5291

a 5 b 1 c 3 3 a We4 Draw a Venn diagram representing the relationship between the following sets.

Show the position of all the elements in the Venn diagram. x  = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20} A = {1, 3, 5, 7, 9, 11, 13, 15, 17, 19} B = {1, 4, 9, 16} b Calculate: i P(A) ii P(B) iii P(A ¶ B) iv P(A ß B) v P(AÅ ¶ BÅ). 4 Using the given Venn diagrams, indicate the set each of the following shaded areas represents. a

c

A

B x

A

B x

b X

d

Y

A

B

C 392

Maths Quest 10 for the Australian Curriculum

x

x

statistics AND probability • Chance 5 Write the following odds as probabilities. a 5–1 b 13–4

c 7–1

2 9

6   MC  The probability of written as odds is: A 7–2

B 2–7

C 2–9

D 9–2

E  11–9

Understanding 7   WE 1  Terry has kicked 9 goals of the 10 attempts he made during a football match. a What is the probability that he will kick a goal on his next attempt? b What is the probability that he will not kick a goal on his next attempt? 8 Rachel attended 12 meetings in December. She was elected as the chairperson in 10 of those

meetings. What is the probability that she will be elected as the chairperson in the next meeting she attends? 9 For a survey, a student counted the vehicles driving out of a sports complex at the end of day 1 of a sports carnival. She recorded the results in a table as shown below. Vehicle type Number

Bus

Car

Motorbike

4-wheel drive

3

17

4

6

Assuming that on day 2 there is a similar traffic movement, what is the probability that a randomly selected vehicle will be: a a car b a bus c not a 4-wheel drive? 10   WE 2  Visitors to the Queen Victoria market were interviewed. The composition of this survey group is given by the following table. Females

Males

Total

NSW

 7

 9

16

Qld

 5

 7

12

Tasmania

 3

 2

 5

Europe

16

17

33

Asia

10

 4

14

Total

41

39

80

a b

Calculate the relative frequency of:    i visitors from Queensland   ii European female visitors iii male visitors from New South Wales iv Asian visitors.

If a person is selected at random from this group, find the probability that the person is:    i a Tasmanian visitor   ii a European male visitor iii a female visitor from Queensland.

11   MC  Which statement is true for the information given in the table in question 10? a The probability of selecting a European visitor from this group is higher than that of

selecting a person from any other visitor group. b The probability of selecting a European visitor from this group is the same as that of

selecting a person from any other visitor group. c The probability of selecting a European male visitor from this group is the same as that of

selecting a European female visitor. d The probability of selecting an Asian visitor is the lowest. e The probability of selecting a visitor from New South Wales is the same as the probability

of selecting a visitor from Queensland. Chapter 12 Probability

393

stAtistiCs AND probAbility • ChANCe 12 We3 A card is drawn from a shuffled pack of 52 cards. Find the probability that the card

drawn is: a an ace b a club c a red card d not a jack e a green card f not a red card. 13 A bag contains 4 blue marbles, 7 red marbles and 9 yellow marbles. All marbles are of the same size. A marble is selected at random. What is the probability that the marble is: a blue b red c not yellow d black? 14 MC Fifty Year 10 students on an excursion were asked to indicate their preference for an evening activity. It was concluded that, if a student is selected at random, the probability that he or she has chosen ice-skating is 15. a The number of students who chose ice-skating is: A 5 B 1 C 10 D 40 E 8 b The probability that a randomly selected student did not choose ice-skating is: A D

1 5 4 5

B

2 5

C

3 5

E 1

c The probability that a randomly selected student

chose tenpin bowling is: A

1 5

D 1

B

4 5

C 0

E not able to be determined

15 A sporting club has members who play different sports, as shown by the given Venn diagram. Volleyball

Walking 10

15

x

38

8 2

17 6

Tennis

a b c d

394

Copy the given Venn diagram and shade the areas that represent: i members playing tennis only ii members walking only iii members playing both tennis and walking but not playing volleyball.

How many members belong to the sporting club? Determine the probability of members who: i play volleyball ii are involved in all three activities. Determine the probability of members who do not: i play tennis ii walk.

Maths Quest 10 for the Australian Curriculum

stAtistiCs AND probAbility • ChANCe 16 We5 Thirty students were asked which lunchtime sports they enjoyed — volleyball, soccer or

tennis. Five students chose all three sports. Six students chose volleyball and soccer, 7 students chose volleyball and tennis while 9 chose soccer and tennis. Fifteen students chose volleyball, 14 students chose soccer and 18 students chose tennis. a Copy the Venn diagram shown and enter the given information. n(x ) = 30 Volleyball

Soccer

Tennis b c

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If a student is selected at random, determine the probability of selecting a student who: chose volleyball chose all three sports chose both volleyball and soccer but not tennis did not choose tennis chose soccer. Determine: i P[(Soccer ß Tennis) ¶ VolleyballÅ] ii P[(Volleyball ß Tennis) ¶ SoccerÅ]. 17 Thirty-five Year 10 students were required to bring a calculator and a graph book to a maths lesson. On checking, it was found that 18 students had brought both, 7 students had the calculator only and 5 students had the graph book only. Five students had neither the calculator nor the graph book. a Show this information on a Venn diagram. b How many students had: i a calculator ii a graph book? c If a student is selected at random, determine the probability that the student: i had both the calculator and the graph book ii had a calculator iii had neither iv did not have a graph book. d Calculate: i P(calculator only) ii P(calculator or graph book or both) iii P(graph book only). 18 We6 The odds given for the greyhound ‘Dog’s Breakfast’ to win its race are 7–3. a Determine the probability of Dog’s Breakfast winning its race. b Maria decides to bet $15 on Dog’s Breakfast to win the race. If Dog’s Breakfast wins, calculate Maria’s payout. 4 c The dog ‘Zoom Top’ is also in the race. If the probability of Zoom Top winning is 13 , what odds should be given for Zoom Top? 3 19 The probability of a horse winning a race is given as . What are the horse’s chances, given as 7 odds? i ii iii iv v

Chapter 12 probability

395

statistics AND probability • Chance Reasoning 20 Azi and Robyn are playing a dice game. Azi has an eight-sided die (faces numbered 1 to 8

21 22

23

24 25

12B

inclusive) and Robyn has a six-sided die (faces numbered 1 to 6 inclusive). They both roll their die. a The person who rolls the number 5 wins. Is this game fair? b The person who rolls an even number wins. Is this game fair? A six-sided die has three faces numbered 1 and the other three faces numbered 2. Are the events ‘rolling a 1’ and ‘rolling a 2’ equally likely? Using a six-sided die, an eight-sided die, a twelve-sided die and a sixteen-sided die (all faces numbered consecutively beginning with 1): a analyse and comment on the fairness of a game that constitutes a win by rolling a multiple of 4. b devise rolling games where:   i the game is fair regardless of the die used.  ii it is more probable to win using a die with a smaller number of faces. iii it is more probable to win using a die with a larger number of faces. Alex places a $5 bet on a horse to win at 4–1 and Rene bets $10 on another horse. The pay-out figure for both bets reflection    is $25. What is the probability that Rene’s horse wins? What basic formula must Are the odds 10–6 the same as 5–3? Explain. be remembered in order to With the use of diagrams, show that calculate simple probabilities? P(AÅ ¶ BÅ) = P(A ß B)Å.

Complementary and mutually exclusive events Complementary events ■■ ■■

The complement of a given set is made up of all the elements that belong to the universal set, x , but not to the particular given set. This is illustrated in the Venn diagram below, where the complement of set M, denoted as MÅ, will constitute all the elements outside set M; that is {5, 6, 7, 8}. n(x ) = 8 M

N

1

5 3

2

4

7

6 8

■■

■■

396

Consider sets M and MÅ They have no elements in common and make up the universal set, x . Using set notation ñ  M ¶ MÅ = f , the null or empty set (indicating no elements) ñ  M ß MÅ = x . If A and AÅ are complementary events then P(A) + P(AÅ) = 1. This may be rearranged to P(AÅ) = 1 - P(A) or P(A) = 1 - P(AÅ).

Maths Quest 10 for the Australian Curriculum

statistics AND probability • Chance

Worked Example 7

A card is drawn from a pack of 52 playing cards. Determine: a  the probability of drawing a spade b  the probability of not drawing a spade. Think a

b

1

Write

Write the number of favourable outcomes; that is, the number of ways a spade may be drawn and the total number of possible outcomes.

a n(drawing a spade) = 13

n(x  ) = 52

n(drawing a spade) n(x  )

2

Write the rule for probability.

P(spade) =

3

Substitute the known values into the rule.

P(spade) = 52

4

Simplify and evaluate.

5

Write your answer.

1

Write the rule for obtaining the complement of drawing a spade.

2

Substitute the known values into the given rule.

13

=

1 4 1

The probability of drawing a spade is 4 . b P(AÅ) = 1 - P(A)

P(not a spade) = 1 - P(spade) =1-

3

Evaluate.

4

Write your answer.

=

1 4

3 4

The probability of not drawing a spade is 43 .

Worked Example 8

A player is chosen from a cricket team. Are the events ‘selecting a batsman’ and ‘selecting a bowler’ complementary events if a player can have more than one role? Give a reason for your answer. Think

Write

Explain the composition of a cricket team. Players who can bat and bowl are not necessarily the only players in a cricket team. There is a wicket-keeper as well. Some players (all rounders) can bat and bowl.

No, the events ‘selecting a batsman’ and ‘selecting a bowler’ are not complementary events. These events may have common elements; that is, the all rounders in the team who can bat and bowl. The cricket team also includes a wicket-keeper.

Mutually exclusive events ■■ ■■

Two events that have no common elements and that cannot occur simultaneously are defined as mutually exclusive events. That is A ¶ B = { } or f . Using set notation, if two events A and B are mutually exclusive then P(A ¶ B) = 0 since 0 n(A ¶ B) = P(A ¶ B) = =0 n(x  ) n(x  ) Chapter 12 Probability

397

statistics AND probability • Chance

■■

■■ ■■

Some examples of mutually exclusive events are listed below. 1. A die is rolled. Let event A = obtaining an even number = {2, 4, 6} and event B = obtaining a factor of 3 = {1, 3}. A 2. In the given Venn diagram, event A = {1, 7, 10} and 1 event B = {2, 3}. 3. Two coins are tossed. Let event A = obtaining 7 2 Heads and event B = obtaining 2 Tails. 10 Complementary events are mutually exclusive; their union forms the universal set. The union of mutually exclusive events do not always form the universal set and as such, mutually exclusive events are not necessarily complementary events.

n(x ) = 5 B 2 3

Complementary events, A and AÅ

Mutually exclusive events, A and B

Common elements

A ¶ AÅ = •

A¶B=•

Union of sets

A ß AÅ = x

AßBòx

The Addition Law of probability ■■

■■

Consider the shaded region in the Venn diagram shown. 1. Counting the number of elements in the shaded region gives, n(A ß B) = 12. 2. For sets A and B, n(A) = 7 and n(B) = 8. 3. Sets A and B have three common elements, n(A ¶ B) = 3. 4. Adding together the number of elements in sets A and B, counts the common elements twice and hence does not give the number of elements in the shaded region, n(A) + n(B) = 15 ò 12 = n(A ß B). 5. Adding together the number of elements in sets A and B and subtracting the number of common elements gives the number of elements in the shaded region, n(A) + n(B) - n(A ¶ B) = 7 + 8 - 3 = 12 = n(A ß B) Therefore the number of elements in A ß B is n(A ß B) = n(A) + n(B) - n(A ¶ B) So, n(A ß B) x P(A ß B) = n(x  ) A B n(A) + n(B) - n(A ¶ B) = n(x  ) n(A) + n(B) - n(A ¶ B) = n(x  ) n(x  ) n(x  ) = P(A) + P(B) - P(A ¶ B)

■■

■■

398

This is known as the Addition Law of probability. If events A and B are not mutually exclusive, the Addition Law of probability states that: P(A or B) = P(A) + P(B) - P(A and B)  or  P(A ß B) = P(A) + P(B) - P(A ¶ B) If events A and B are mutually exclusive, the Addition Law of probability states that: P(A or B) = P(A) + P(B)  or  P(A ß B) = P(A) + P(B) since P(A ¶ B) = 0

Maths Quest 10 for the Australian Curriculum

statistics AND probability • Chance

Worked Example 9

A card is drawn from a pack of 52 playing cards. What is the probability that the card is a heart or a club? Think

Write

1

Determine whether the given events are mutually exclusive.

The two events are mutually exclusive as they have no common elements.

2

Determine the probability of drawing a heart and of drawing a club.

P(heart) =

3

Write the Addition Law for two mutually exclusive events.

P(A or B) = P(A) + P(B) where A = drawing a heart and B = drawing a club

4

Substitute the known values into the rule.

P(heart or club) = P(heart) + P(club) 1 = 4 + 14

=

13 52 1 4

P(club) = =

= 5

Evaluate and simplify.

6

Write your answer.

Note: Alternatively, we can use the formula for theoretical probability.

13 52 1 4

2 4

= 12 1

The probability of drawing a heart or a club is 2 . P(heart or club) =

n(heart or club) n(x  ) 26

= 52 =

1 2

Worked Example 10

A die is rolled. Determine: a P(an odd number) b P(a number less than 4) c P(an odd number or a number less than 4). Think a

b

1

Write

Determine the probability of obtaining an odd number; that is, {1, 3, 5}.

2

Write your answer.

1

Determine the probability of obtaining a number less than 4; that is, {1, 2, 3}.

2

Write your answer.

a

3

P(odd) = 6 =

1 2 1

The probability of obtaining an odd number is 2 . b P(less than 4) =

=

3 6 1 2

The probability of obtaining a number less 1 than 4 is 2 .

Chapter 12 Probability

399

statistics AND probability • Chance c

c

1

Determine whether the given events are mutually exclusive.

2

Write the Addition Law for two non-mutually exclusive events.

P(A or B) = P(A) + P(B) - P(A and B) where A = selecting an odd number and B = selecting a number less than 4.

3

Substitute the known values into the rule. Note: P(A and B) = 62 = 13 since the

P[odd number ß (number < 4)] = P(odd number) + P[(number < 4)] - P[odd number ¶ (number < 4)] 1 1 1 =2+2−3

( )

events have two elements in common.

The two events are not mutually exclusive as they have common elements; that is, 1 and 3.

2

4

Evaluate and simplify.

=3

5

Write your answer.

The probability of obtaining an odd number or a 2 number less than 4 is 3 .

Note: Alternatively, we can use the formula for theoretical probability.

The set that has elements that are odd numbers or numbers less than 4 is {1, 2, 3, 5}. P[odd number ß (number < 4)] n[odd number ß (number less than 4)] = n(x  ) 4 =6 =

2 3

remember

1. Complementary events have no common elements and together make up the universal set. 2. If A and AÅ are complementary events then P(A) + P(AÅ) = 1. This may be rearranged to: P(AÅ) = 1 - P(A) or P(A) = 1 - P(AÅ). 3. Mutually exclusive events have no common elements and cannot occur simultaneously. 4. If events A and B are not mutually exclusive then: P(A or B) = P(A) + P(B) - P(A and B) or P(A ß B) = P(A) + P(B) - P(A ¶ B) where P(A ¶ B) is the probability of the intersection of sets A and B or the common elements in sets A and B. 5. If events A and B are mutually exclusive then: P(A or B) = P(A) + P(B) or P(A ß B) = P(A) + P(B) since P(A ¶ B) = 0. 6. Mutually exclusive events may or may not be complementary events. 7. Complementary events are always mutually exclusive. Exercise

12B

Complementary and mutually exclusive events fluency 1 Consider complementary events, A and B. a 1 - P(B) = ? c If P(B) = 0.65, then P(A) = ?

400

Maths Quest 10 for the Australian Curriculum

3

b If P(A) = 7 , then P(B) = ? d P(A ß B) = ?

stAtistiCs AND probAbility • ChANCe

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2 Events M and N are not mutually exclusive events. a Draw a Venn diagram to illustrate events M and N. b P(M ß N) = ? c State if the following statements are true, false or cannot be determined? i P(M ¶ N) = • ii P(M ¶ N) > 0 iii P(M ß N) = 1 UNDerstANDiNg 3 We7 A card is drawn from a pack of 52 playing cards. Determine: a the probability of obtaining an ace b the probability of not obtaining an ace. 4 The weather bureau announced that there is an 80% probability of having a rain shower on 5

6

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7

8

Friday. What is the probability of not having a rain shower on that day? A number is selected from the set {1, 2, 3 . . . 20}. Let E1 be the event of selecting an even number and E2 be the event of selecting an odd number. a Determine: i P(E1) ii P(E2). b Are E1 and E2 complementary? A bag contains 50 balls, of which there are 10 blue balls, 5 red balls and 3 yellow balls. What is the probability of picking a ball that is not blue, red or yellow? Questions 7 and 8 refer to the following information. A number is selected from 1 to 100, inclusive. Let: E1 = a multiple of 10 is picked E2 = a factor of 20 is picked E3 = a multiple of 2 is picked E4 = an odd number is picked. MC Which of the following represents a pair of complementary events? A E1 and E2 B E2 and E3 C E3 and E4 D E2 and E4 E E1 and E3 Calculate: a P(multiple of 10) b P(not a multiple of 10) c P(not a factor of 20). Questions 9 and 10 refer to the following information. The ages of 50 Year 10 students are shown in the following table. Age (years) 15

16

17

Total

Girls

7

10

9

26

Boys

9

8

7

24

Total

16

18

16

50

9 MC Which of the following represent a pair of complementary events? A Selecting a 15-year-old boy and selecting a 15-year-old girl B Selecting a 15-year-old student and selecting a 16-year-old student C Selecting a 17-year-old student and selecting a 15-year-old student D Selecting a 15- or 16-year-old student and selecting a 17-year-old student E Selecting a 17-year-old student and selecting a 15- or 16-year-old girl Chapter 12 probability

401

stAtistiCs AND probAbility • ChANCe 10 Calculate: a P(selecting a 15-year-old boy) b P(not selecting a 15-year-old boy) c P(selecting a boy) d P(selecting a girl). 1

11 a When a coin is tossed 4 times, the probability of getting 4 Heads is 16 . What is the

probability of not getting 4 Heads?

2 9

b The probability that a horse will win a race is . What is the probability that one of the

other horses will win the race?

12 Are the events ‘getting 2 Tails’ and ‘getting 0 Tails’ complementary when a coin is tossed

twice? 13 In a school raffle, 200 tickets were sold. Margaret and Julie bought 25 tickets between them. a What is the probability that Margaret or Julie will win? b What is the probability that neither of them will win? eBook plus

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14 We8 A die is rolled. What is the probability that the outcome is an even number or a 5? 15 A number is chosen from the set {1, 2, 3 . . . 25}. What is the probability that the number is: a a multiple of 4 or a multiple of 7 b a multiple of 4 or an odd number c less than 5 or more than 20? 16 A card is drawn from a well-shuffled pack of 52 playing cards. Calculate: a P(a spade or ace of hearts is drawn) b P(a king or a queen is drawn) c P(a jack or a king or an ace is drawn). 17 MC Which of the following represents a pair of mutually exclusive events when a die is

rolled? Obtaining an even number or obtaining a 4 Obtaining an odd number or obtaining a 3 Obtaining a number less than 3 or obtaining a number more than 5 Obtaining a multiple of 2 or obtaining a multiple of 3 Obtaining a factor of 6 or obtaining a multiple of 6

A B C D E

18 In a 3-horse race, the probability for each of the horses to win is given as:

Our Lady: 37

Shaka:

4 9

Speedy:

8 . 63

Determine the probability that: a either Our Lady or Speedy wins b either Shaka or Speedy wins. 19 Christine’s teaching timetable for Monday and Tuesday is given below.

Period and class 1

2

Monday

10B

Tuesday

8B

3

8A

4

5

8B

8A

10A

6

7 9A

9B

She is organising a music tuition class for a lesson when she is not teaching, but she cannot use the first lesson on any day because of her responsibility as a senior teacher. Determine the probability that: a she cannot take music tuition because she is teaching b she cannot take music tuition because it is the first lesson c she cannot take music tuition. 402

Maths Quest 10 for the Australian Curriculum

stAtistiCs AND probAbility • ChANCe 20 We9 A card is drawn from a well-shuffled pack of 52 playing cards. Calculate: a P(a king is drawn) b P(a heart is drawn) c P(a king or a heart is drawn). 21 Two coins are tossed. Event 1 is obtaining 2 Heads and event 2 is obtaining 2 Tails. a Are events 1 and 2 mutually exclusive? b What is P(event 1 or event 2)? reAsoNiNg 22 We10 For each of the following pairs of events: i state, giving justification, if the pair are complementary events ii alter the statements, where applicable, so that the events become complementary events. a Having Weet Bix or having Strawberry Pops for breakfast b Walking to a friend’s place or driving there c Watching TV or reading as a leisure activity d Rolling a number less than 5 or rolling a number greater than 5 with a ten-sided die

with faces numbered 1 to 10 e Passing a maths test or failing a maths test 23 Pat suggests that for a single roll of a die, getting a factor of 4 and getting a factor of 6 are eBook plus

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12C

mutually exclusive. Is he right? Why or why not? 24 Two tetrahedral dice (4-sided) are rolled and the sum of the outcome on each is taken. (Note: The outcome is the number on the bottom face.) Let: Event 1 = the sum is 6 Event 2 = the sum is 3 Event 3 = the sum is more than 4 Event 4 = the sum is less than 4. a Decide whether the following statements are true or false. i Events 1 and 2 are mutually exclusive. ii Events 2 and 4 are mutually exclusive. iii Events 2 and 3 are mutually exclusive. iv Events 1 and 2 are complementary events. v Events 2 and 4 are complementary events. vi Events 2 and 3 are complementary events. b Determine: i P(event 1) ii P(event 2) iii P(event 3) refleCtioN    iv P(event 4). How are the differences between c Determine: mutually exclusive events and i P(event 1 or event 2) complementary events reflected ii P(event 2 or event 4) in the addition law of probability? iii P(event 2 or event 3).

two-way tables and tree diagrams ■

■

When more than one event has to be considered, a diagrammatic representation of the sample space is helpful in calculating the probabilities of various events. Two-way tables and tree diagrams may be used.

two-way tables ■

A two-way table (sometimes referred to as a lattice diagram) is able to represent two events in a 2-dimensional table. Chapter 12 probability

403

statistics AND probability • Chance

■■ ■■

With the help of the information in each row and each column, all the pairs of outcomes are listed and the diagram ensures that none of the pairs is omitted. A two-way table for the experiment of tossing a coin and rolling a die simultaneously is shown in the following table. Die outcomes

Coin outcomes

1

■■

2

H (H, 1) (H, 2) T

(T, 1)

(T, 2)

3

4

5

6

(H, 3)

(H, 4)

(H, 5)

(H, 6)

(T, 3)

(T, 4)

(T, 5)

(T, 6)

Two-way tables can be used to display the combined outcomes of only two events.

Worked Example 11

Two dice are rolled. The outcome is the pair of numbers shown uppermost. a Show the results on a two-way table. b Calculate the probability of obtaining an identical ordered pair; that is, P[(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)]. Think

Rule a two-way table.

2

Label the first row as ‘Die 2’ and write all the outcomes.

3

Label the first column as ‘Die 1’ and write all the outcomes.

4

Write each ordered pair in its respective position.

1

Look at the two-way table from part a and highlight the identical pairs.

a

Die 2 outcomes 1

404

2

3

5

6

(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)

2

(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)

3

(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)

4

(4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)

5

(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)

6

(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)

Die 2 outcomes 1

2

3

4

5

6

1

(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)

2

(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)

3

(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)

4

(4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)

5

(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)

6

(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)

2

Write the number of identical pairs and the total number of possible pairs.

n(identical pairs) = 6 n(x  ) = 36

3

Since each outcome is equally likely, write the rule for probability.

P(identical pairs) =

Maths Quest 10 for the Australian Curriculum

4

1

b

Die 1 outcomes

b

1

Die 1 outcomes

a

Write

n(identical pairs) n(E) = n(x  ) n(S)

stAtistiCs AND probAbility • ChANCe

4

Substitute the known values into the rule.

5

Simplify and evaluate.

6

Write your answer.

■

6

P(identical ordered pairs) = 36 1

=6 The probability of obtaining an identical ordered pair is 16 .

Two-way tables are limited to displaying two events occurring simultaneously. Alternative representations are used to display more than two events.

tree diagrams ■

■

eBook plus

eLesson Games at Wimbledon

■

eles-1032 ■

■

■

■

Another way of representing the sample space is to construct a tree diagram. This is a branching diagram that helps list all the outcomes. Coin 1 Coin 2 Tree diagrams are very helpful when there are multiple 1– H events; for example, when a coin is tossed twice. Each stage 2 of a multiple event experiment produces a part of a tree. 1– H 2 1– The first stage of the experiment is tossing coin 1. The two T 2 possible results that can be obtained are Heads or Tails and 1– H these are listed at the end of each branch. The probability of 2 1– 2 T obtaining the result listed is written along the branches. 1– The second stage of the experiment is tossing coin 2, for T 2 which the possible results are also Heads or Tails. A pair of branches is attached to each of the ends of the existing branches. Again, the branches are labelled with the appropriate outcomes and probabilities. After the diagram has been completed, the outcomes are listed at the right-hand side of the tree diagram. This is done by beginning at the starting point and following along each set of branches, then listing the combinations. The possible results or outcomes obtained by following along the combined branches are (H, H), (H, T), (T, H) and (T, T). The probability for each outcome is calculated by taking the product of the probabilities associated with the respective branches. For example, the probability of (H, H) is obtained by multiplying the individual probabilities of the two H branches; that is P(H, H) = P(H) ì P(H) =2ì2

 

=

1 4

The completed tree diagram is illustrated below. Coin 1

Coin 2 Outcomes HH H

1– 2 1– 2

1– 2

■

1

1

 

H

T

Probability 1– 1– 1– 2 ì 2 = 4

1– 2

T

HT

1– 2

ì

1– 2

=

1– 4

1– 2

H

TH

1– 2

ì

1– 2

=

1– 4

1– 2

T

TT

1– 2

ì

1– 2

=

1– 4

— 1

When added together, all the probabilities should sum to 1. If more than one outcome is included in a particular event, then the respective probabilities are added. For example: P(1 Head) = P(H, T) + P(T, H) 1

 

=4+

 

=

1 4

1 2

Chapter 12 probability

405

statistics AND probability • Chance

■■

Tree diagrams may be extended to display three or more events occurring simultaneously.

■■

Tree diagrams are useful in working out the sample space and calculating probabilities of various events. On each branch of a tree diagram, the probability associated with the branch is listed. The products of the probabilities given on the branches are taken to calculate the probability for a particular outcome.

Worked Example 12

Three coins are tossed simultaneously. Draw a tree diagram for the experiment. Calculate the following probabilities. a  P(3 Heads) b  P(2 Heads) c  P(at least 1 Head) Think

Write/draw

1

Use branches to show the individual outcomes for the first part of the experiment (tossing the first coin). Place a 1 above the coin toss outcomes. Label the ends of the branches H and T and place the probabilities along the branches.

2

Link each outcome of the first toss with the outcomes of the second part of the experiment (tossing the second coin). Place a 2 above the second toss outcomes. Label the ends of the branches H and T and place the probabilities along the branches.

3

Link each outcome from the second toss with the outcomes of the third part of the experiment (tossing the third coin). Place a 3 above the third toss outcomes. Label the ends of the branches H and T and place the probabilities along the branches.

4

List each of the possible outcomes on the right-hand side of the tree diagram.

5

Determine the probability of each result. Note: The probability of each result is found by multiplying along the branches and in each case this will be 1 1 1 1 ì 2 ì 2 = 8. 2

1

2 1– 2

1– 2

1– 2

H

T

1– 2 1– 2 1– 2

H

1– 2 1– 2

T 1– 2

H

1– 2

T 1– 2

406

Maths Quest 10 for the Australian Curriculum

3 H

HHT

1– 2

H

HTH

T

HTT

H

THH

T

THT

H

TTH

T

TTT

1– 2 1– 2 1– 2 1– 2 1– 2 1– 2

T 1– 2

1– 2

1– 2

Outcomes Probability 1– 1– 1– HHH 2 ì 2 ì 2 = ì

1– 2

ì

1– 2 1– 2 1– 2 1– 2 1– 2 1– 2

ì ì ì ì ì

ì

1– 2

=

ì

1– 2 1– 2 1– 2 1– 2 1– 2 1– 2

=

ì ì ì ì ì

= = = = =

1– 8 1– 8 1– 8 1– 8 1– 8 1– 8 1– 8 1– 8

— 1

statistics AND probability • Chance

6

a

b

c

Place calculations and results next to the respective outcomes. Note that in this example, each of the outcomes has the same probability, therefore each outcome is equally likely.

Refer to the listed outcomes and calculations next to the tree diagram and write your answer. 1

Refer to the listed outcomes and calculations next to the tree diagram. Note: This combination occurs three times.

2

Write your answer.

1

Refer to the listed outcomes and calculations next to the tree diagram. Note: At least 1 Head means any outcome that contains one or more Head. This is every outcome except 3 Tails. That is, it is the complementary event to obtaining 3 Tails.

2

Write your answer.

■■

a The probability of obtaining 3 Heads is 1 . 8

b P(2 Heads)

= P(H, H, T) + P(H, T, H) + P(T, H, H) 1 1 1 =8+8+8 3

=8

The probability of obtaining exactly 2 Heads is 83 . c

P(at least 1 Head) = 1 - P(T, T, T) 1 =1-8 7

=8

7

The probability of obtaining at least 1 Head is 8 .

As can be seen from the tree diagram in Worked example 12, the probabilities of all outcomes add up to 1.

Worked Example 13

Two dice are rolled simultaneously. Draw a tree diagram for the experiment and find: a  P(two 6s)            b  P(one 6) c  P(no 6s)             d  P(at least one 6). Think

Write/draw

1

Draw the tree diagram using two outcomes — S (getting a 6) and SÅ (not getting a 6). Note: Recall for complementary events, P(S) + P(SÅ) = 1.

2

List each of the possible outcomes on the right-hand side of the tree diagram.

1

2 1– 6

1– 6

5– 6

S

5– 6 1– 6

Probability 1– 1– 1 — 6 ì 6 = 36

S

Outcomes SS

S'

SS'

1– 6

ì

5– 6

5 =— 36

S

S'S

5– 6

ì

1– 6

5 =— 36

S'

S'S'

5– 6

ì

5– 6

25 =— 36 — 1

S' 5– 6

Chapter 12 Probability

407

statistics AND probability • Chance

a

b

3

Determine the probability of each possible result by multiplying along the branches.

4

Place the calculations and results next to the respective outcomes.

1

Refer to the listed outcomes and calculations next to the tree diagram.

2

Write your answer.

1

Refer to the listed outcomes and calculations next to the tree diagram. Note: This combination occurs twice.

a P(two 6s) = P(S, S) 1 = 36

1 . The probability of obtaining two 6s is 36

b

P(one 6) = P(S, SÅ) + P(SÅ, S) 5 5 = 36 + 36 10

= 36 = 185

c

d

5

2

Write your answer.

1

Refer to the listed outcomes and calculations next to the tree diagram.

2

Write your answer. Refer to the listed outcomes and calculations next to the tree diagram. Note: P(at least one 6) means any outcome that contains one or more six. This includes every outcome except for the (SÅ, SÅ) combination. That is, it is the complementary event to obtaining the (SÅ, SÅ) combination.

1

2

The probability of obtaining one 6 is 18 . c

P(no 6s) = P(SÅ, SÅ) 25 = 36 25

The probability of obtaining no 6s is 36 . d P(at least one 6) = 1 - P(SÅ, SÅ). 25 = 1 - 36 = 11 36

11

Write your answer.

The probability of obtaining at least one 6 is 36.

Alternatively, part d of Worked example 13 could have been a calculated in the following manner. P(at least one 6) = P(one or more 6s) = P(S, S) + P(S, SÅ) + P(SÅ, S)

=



=

1 36 11 36

+

5 36

5

+ 36

Worked Example 14

The letters A, B, C and D are written on identical pieces of card and placed in a box. A letter is drawn at random from the box. Without replacing the first card, a second one is drawn. Use a tree diagram to find: a P(first letter is A)    b P(second letter is B)    c P(both letters are the same).

408

Maths Quest 10 for the Australian Curriculum

statistics AND probability • Chance

Think

Write/draw

1

Draw a tree diagram of the situation. There are four letters to choose as the first letter of the pair of letters. Each has a probability of 14 of being chosen.

2

For each letter chosen as the first letter, there are three letters remaining to choose from. Each has a probability 1 of 3 of being chosen.

3

Write the sample space and calculate the probability of each outcome.

1– 3

A

1– 3

1– 4 1– 4

B 1– 4

1– 3

C

1– 4

1– 3

D

a

1

Refer to the listed outcomes and calculations next to the tree diagram.

1– 3

1– 3

1– 3

1– 3

1– 3

1– 3

1– 3

1– 3

B C D A C D A B D A B C

Outcomes AB AC AD BA BC BD CA CB CD DA DB DC

Probability 1– 1 1 4 ì –3 = — 12 1– 1– 1 — ì = 4 3 12 1– 1– 1 — 4 ì 3 = 12 1– 1– 1 — 4 ì 3 = 12 1– 1– 1 — ì = 4 3 12 1– 1– 1 — 4 ì 3 = 12 1– 1– 1 4 ì 3 = — 12 1– 1– 1 — ì = 4 3 12 1– 1 1 4 ì –3 = — 12 1– 1– 1 4 ì 3 = — 12 1– 1– 1 — 4 ì 3 = 12 1– 1 1 4 ì –3 = — 12 — 1

a P(first letter A) = P(A, B) + P(A, C) + P(A, D) 1

1

1

P(first letter A) = 12 + 12 + 12 3

= 12 = 14

2

b

1

1

The probability that the first letter is A is 4 .

Write your answer. Refer to the listed outcomes and calculations next to the tree diagram.

b

P(second letter B) = P(A, B) + P(C, B) + P(D, B) 1 1 1 P(second letter B) = 12 + 12 + 12 3

= 12 =

c

2

Write your answer.

1

Refer to the listed outcomes and calculations next to the tree diagram.

2

Write your answer.

1 4

The probability that the second letter is B is 14 . c

P(both letters are the same) = 0

As the first card is not replaced before the second is drawn, the probability that both letters are the same is 0.

Chapter 12 Probability

409

stAtistiCs AND probAbility • ChANCe

reMeMber

1. Two-way tables give a clear diagrammatic representation of the sample space; however, they are limited to displaying two events. 2. Tree diagrams are useful in working out the sample space and calculating probabilities of various events, especially if there is more than one event. On each branch of a tree diagram, the probability associated with the branch is listed. The products of the probabilities given on the branches are taken to calculate the probability for an outcome. 3. The probabilities of all outcomes add to 1. exerCise

12C iNDiviDUAl pAthWAys

two-way tables and tree diagrams flUeNCy 1 For the tree diagram below, calculate the following probabilities:

eBook plus

Activity 12-C-1

Review of two-way tables and tree diagrams doc-5116

0.4 0.4

Activity 12-C-2

Practice with two-way tables and tree diagrams doc-5117

H

0.5

T

0.5

H

0.5

T

0.5

H

0.5

T

G

0.2 B

Activity 12-C-3

Tricky two-way tables and tree diagrams doc-5118

0.5 R

a P(R, H) b P(B, H) d P(H) e P(R, H or G, T) 2 i Copy and complete the two-way table below.

c P(B) f P(BÅ)

Card outcomes

Coin outcomes

Club, Ü H

H, Ü

T

T, Ü

Spade, â

ii Calculate the following probabilities. a P(T, Ü) b P(T)

Diamond, á

Heart, à

H, á T, à c P(red card)

UNDerstANDiNg eBook plus

Digital doc

SkillSHEET 12.5 doc-5290

410

3 We11 Two dice are rolled. The outcome is the pair of numbers shown on each die. a Show the results on a two-way table. b Calculate the probability of obtaining an ordered pair where the second digit is half the

value of the first. 4 A 10-sided die is rolled at the same time that a coin is tossed. a Show the outcomes on a two-way table. b Calculate the probability of event (H, n), where n is a factor of 10. c Calculate P(T, even number).

Maths Quest 10 for the Australian Curriculum

statistics AND probability • Chance 5 A green octahedron (a 3-dimensional shape with 8 regular faces) is rolled simultaneously with

6

7

8

9

10

11

12

a yellow octahedron. Both figures have the faces numbered 1, 2, 3  .  .  .  8. a Show the sample space on a two-way table. b On the diagram, highlight the event (n, n), where n is a number in the range 1 to 8 inclusive. c What is the probability of getting (n, n) as described in part b above?   WE 12  A circular spinner is divided into two equal halves, coloured red and blue, and spun 3 times. Draw a tree diagram for the experiment. Calculate the following probabilities. a P(3 red sectors) b P(2 red sectors) c P(1 red sector) d P(0 red sectors) e P(at least 1 red sector) f P(at least 2 red sectors) A bag contains 6 identical marbles, 2 of which are red, 1 green and 3 blue. A marble is drawn, the colour is noted, the marble is replaced and another marble is drawn. a Show the possible outcomes on a tree diagram. b List the outcomes of the event ‘the first marble is red’. c Calculate P(the first marble is red). d Calculate P(2 marbles of the same colour are drawn). Assuming that it is equally likely that a boy or a girl will be born, answer the following. a Show the possibilities of a 3-child family on a tree diagram. b In how many ways is it possible to have exactly 2 boys in the family? c What is the probability of getting exactly 2 boys in the family? d Which is more likely, 3 boys or 3 girls in the family? e What is the probability of having at least 1 girl in the family? A tetrahedron (prism with 4 identical triangular faces) is numbered 1, 1, 2, 3. It is rolled twice. The outcome is the number facing downwards. a Show the results on a tree diagram. b Are the outcomes 1, 2 and 3 equally likely? c Find the following probabilities:   i P(1, 1)  ii P(1 is first number) iii P(both numbers equal) iv P(both numbers are odd).   WE 13  A die is rolled twice to check whether a 3 occurs. Draw a tree diagram for the experiment and calculate: a P(two 3s) b P(one 3) c P(no 3s) d P(at least one 3). A card is drawn from a pack of 52 playing cards and checked to see whether a spade has been selected. The card is replaced, the pack reshuffled and another card is selected. a Draw a tree diagram for the activity and list the sample space. b What is the probability that both cards are spades? c What is the probability that neither of the cards is a spade? d What is the probability that one of the cards is a spade?   WE 14  The letters X, Y, W and Z are written on identical pieces of card and placed in a box. A letter is drawn at random from the box. Without replacing the first card, a second one is drawn. Use a tree diagram to find: a P(first letter is W) b P(second letter is Z) c P(both letters are different). Chapter 12 Probability

411

stAtistiCs AND probAbility • ChANCe 13 A group of students is made up of 6 girls and 4 boys. Two students are to be selected to

represent the group on the student representative council. They decide to write all names on identical pieces of paper, put them in a hat and choose two names randomly. They want to check the composition (boys or girls) of the two-person team. a Show the selections on a tree diagram (note that the probabilities for the second selection change). b Determine the probability of 2 boys being selected. c Determine the probability of 2 girls being selected. d Determine the probability of selection of 1 boy and 1 girl. e Are the events ‘0 boys’, ‘1 boy’ and ‘2 boys’ equally likely? reAsoNiNg 14 Robyn is planning to watch 3 footy

eBook plus

Digital doc

WorkSHEET 12.2 doc-5296

412

games on one weekend. She has a choice of two games on Friday night; (A) Carlton vs West Coast and (B) Collingwood vs Western Bulldogs. On Saturday, she can watch one of the three games; (C) Geelong vs Brisbane Lions, (D) Melbourne vs Fremantle and (E) Kangaroos vs Adelaide. On Sunday, she also has a choice of three games; (F) St Kilda vs Sydney, (G) Essendon vs Port Adelaide and (H) Richmond vs Hawthorn. a To determine the different combinations of games Robyn can watch, she draws a tree diagram using codes A, B, . . . H. Suggest a sample space for Robyn’s selections. b Robyn’s favourite team is Carlton. What is the probability that one of the games Robyn watches involves Carlton? c Robyn has a good friend that plays for St Kilda. What is the probability that Robyn watches both the matches involving Carlton and St Kilda? 15 A small hospital awaits the arrival of four children showing symptoms of a particular virus. The virus has two different strands, strand A and strand B. A child suffering from strand A cannot share a room with a child suffering from strand B. Given a room can fit at most two children and that there is an equal chance that a child has strand A or B, decide if 2 or 3 rooms need to be made up in order to house the children. 16 To pass an exam Susan must answer two of the last three multiple choice questions correctly. As Susan is running out of time she decides to guess the answers to these three questions. Susan notes that two of the questions give six possible answers rather than the usual standard four choices. Analyse and comment on how the inclusion of six possible answers (as opposed to 4) for two of these questions will affect her chances of passing the exam? 17 i Four identical counters, 2 red and 2 green, are placed in a bag. One counter is drawn, its colour refleCtioN    recorded, it is replaced in the bag and a second one is drawn. When calculating probabilities, what is the significance of an item a Show the sample space on a tree diagram. being replaced, or not replaced, b Calculate P(2 counters of the same colour). before a second event occurs? c Calculate P(2 counters of different colours). ii Suppose the first counter is not replaced. Analyse and explain how this will affect the probabilities?

Maths Quest 10 for the Australian Curriculum

stAtistiCs AND probAbility • ChANCe

12D eBook plus

Interactivity A pack of cards

int-2787

Independent and dependent events ■

■ ■

■

If a coin is tossed the outcome is a Head or a Tail. The outcome of the first toss does not affect the outcome of the next toss of the coin. The second toss will still yield a Head or a Tail irrespective of the outcome of the first toss. Similarly, the outcome on the roll of a die will not affect the outcome of the next roll. If successive events have no effect on each other, they are called independent events. If events A and B are independent then the Multiplication Law of probability states that: P(A and B) = P(A) ì P(B) or P(A ¶ B) = P(A) ì P(B) The reverse is also true. If: P(A and B) = P(A) ì P(B) or P(A ¶ B) = P(A) ì P(B) is true then event A and event B are independent events.

WorkeD exAMple 15

Adam is one of the 10 young golfers to represent his state. Paz is one of the 12 netball players to represent her state. All the players in their respective teams have an equal chance of being nominated as captains. a Are the events ‘Adam is nominated as captain’ and ‘Paz is nominated as captain’ independent? b Determine: i P(Adam is nominated as captain) ii P(Paz is nominated as captain). c What is the probability that both Adam and Paz are nominated as captains of their respective teams? thiNk a

b

Write

Determine whether the given events are independent and write your answer.

a Adam’s nomination has nothing to do with Paz’s

i

b

1

Determine the probability of Adam being nominated as captain. He is one of 10 players.

nomination and vice versa. Therefore, the events are independent. i P(Adam is nominated) = P(A)

=

n(Adam is nominated) n(x  ) 1

P(Adam is nominated) = 10

ii

2

Write your answer.

1

Determine the probability of Paz being nominated as captain. She is one of 12 players.

The probability that Adam is nominated as 1 captain is 10 . ii P(Paz is nominated) = P(P)

=

n(Paz is nominated) n(x  ) 1

P(Paz is nominated) = 12

c

2

Write your answer.

1

Write the Multiplication Law of probability for independent events.

The probability that Paz is nominated as 1 captain is 12 . c

P(A and P) = P(A ¶ P) = P(A) ì P(P) P(Adam and Paz are nominated) = P(Adam is nominated) ì P(Paz is nominated)

2

Substitute the known values into the rule.

1 1 = 10 × 12

Chapter 12 probability

413

statistics AND probability • Chance 1

3

Evaluate.

= 120

4

Write your answer.

The probability that both Adam and Paz are 1 nominated as captains is 120 .

■■

■■ ■■

Sometimes one event affects the outcome of another. For example, if a card is drawn from a 13 1 pack of playing cards, the probability that its suit is hearts, P(hearts), is 52 (or 4 ). If this card is not replaced, then this will affect the probability of subsequent draws. The probability that 12 the second card drawn is a heart will be 51 while the probability that the second card is not a 39 heart will be 51 . When one event affects the occurrence of another, the events are called dependent events. If two events are dependent, then the probability of occurrence of one event affects that of the other.

Worked Example 16

A bag contains 5 blue, 6 green and 4 yellow marbles. The marbles are identical in all respects except in their colours. Two marbles are picked in succession without replacement. Determine the probability of picking 2 blue marbles. Think 1

Determine the probability of picking the first blue marble.

Write/draw

P(picking a blue marble) =

n(B) n(x  ) 5

P(picking a blue marble) = 15 1

=3 2

3

Determine the probability of picking the second blue marble. Note: The two events are dependent since marbles are not being replaced. Since we have picked a blue marble this leaves 4 blue marbles remaining out of a total of 14 marbles. Calculate the probability of obtaining 2 blue marbles.

P(picking second blue marble) =

n(B) n(x  ) 4

P(picking second blue marble) = 14 2

=7 P(2 blue marbles) = P(1st blue) ì P(2nd blue) 1 2 = × 3

7

2 = 21

4

Write your answer.

Note: Alternatively, a tree diagram could be used to solve this question.   The probability of selecting 2 blue marbles successively can be read directly from the first branch of the tree diagram.

2

The probability of obtaining 2 blue marbles is 21.

5 — 15

10 — 15

4 — 14

Blue

10 — 14

Not blue

5 — 14

Blue

Blue

Not blue 9 — 14

P(2 blue marbles) = = = 414

Maths Quest 10 for the Australian Curriculum

5 × 4 15 14 1 ×2 3 7 2 21

Not blue

stAtistiCs AND probAbility • ChANCe

reMeMber

1. Events are independent if the occurrence of one event does not affect the occurrence of the other. 2. If events A and B are independent, then P(A ¶ B) = P(A) ì P(B). This is the Multiplication Law of probability. Conversely, if P(A ¶ B) = P(A) ì P(B) then events A and B are independent. 3. Dependent events affect the probability of occurrence of one another. exerCise

12D iNDiviDUAl pAthWAys eBook plus

Activity 12-D-1

Simple independent and dependent events doc-5119 Activity 12-D-2

Independent and dependent events doc-5120 Activity 12-D-3

Tricky independent and dependent events doc-5121

eBook plus

Interactivity Random numbers

Independent and dependent events flUeNCy 1 If A and B are independent events and P(A) = 0.7 and P(B) = 0.4, calculate: a P(A and B) b P(AÅ and B) where AÅ is the complement of A c P(A and BÅ) where BÅ is the complement of B d P(AÅ and BÅ). UNDerstANDiNg 2 We15 A die is rolled and a coin is tossed. a Are the outcomes independent? b Determine: i P(Head) on the coin ii P(6) on the die. c Determine P(6 on the die and Head on the coin). 3 A tetrahedron (4-faced) die and a 10-sided die are rolled simultaneously. What is the

probability of getting a 3 on the tetrahedral die and an 8 on the 10-sided die? 4 A blue die and a green die are rolled. What is the probability of getting a 5 on the blue die and not a 5 on the green die? 4 5 Dean is an archer. The experimental probability that Dean will hit the target is 5 .

int-0085

a b c d

What is the probability that Dean will hit the target on two successive attempts? What is the probability that Dean will hit the target on three successive attempts? What is the probability that Dean will not hit the target on two successive attempts? What is the probability that Dean will hit the target on the first attempt but miss on the second attempt? Chapter 12 probability

415

stAtistiCs AND probAbility • ChANCe 6 MC A bag contains 20 apples, of which 5 are bruised.

Peter picks an apple and realises that it is bruised. He puts the apple back in the bag and picks another one. a The probability that Peter picks 2 bruised apples is: A

1 4

B

1 2

C

1 16

D

3 4

E

15 16

b The probability that Peter picks a bruised apple first but a

good one on his second attempt is: A

1 4

B

1 2

C

3 4

D

3 16

1 16 is 1 and 7

E

7 The probability that John will be late for a meeting

the probability that Phil will be late 3 for a meeting is 11 . What is the probability that: a John and Phil are both late b neither of them is late c John is late but Phil is not late d Phil is late but John is not late? 8 On the roulette wheel at the casino there are 37 numbers, 0 to 36 inclusive. Bidesi puts his chip on number 8 in game 20 and on number 13 in game 21. a What is the probability that he will win in game 20? b What is the probability that he will win in both games? c What is the probability that he wins at least one of the games? 9 Based on her progress through the year, Karen was given a probability of 0.8 of passing the Physics exam. If the probability of passing both Maths and Physics is 0.72, what is her probability of passing the Maths exam? 10 Suresh found that, on average, he is delayed 2 times out of 7 at Melbourne airport. Rakesh made similar observations at Brisbane airport, but found he was delayed 1 out of every 4 times. Find the probability that both Suresh and Rakesh will be delayed if they are flying out of their respective airports.

11 Bronwyn has 3 pairs of Reebok and 2 pairs of Adidas running shoes. She has 2 pairs of

Reebok, 3 pairs of Rio and a pair of Red Robin socks. Preparing for an early morning run, she grabs at random for a pair of socks and a pair of shoes. What is the probability that she chooses: a Reebok shoes and Reebok socks b Rio socks and Adidas shoes c Reebok shoes and Red Robin socks d Adidas shoes and socks that are not Red Robin? 12 We16 Two cards are drawn successively and without replacement from a pack of playing cards. Determine the probability of drawing: a 2 hearts b 2 kings c 2 red cards. 416

Maths Quest 10 for the Australian Curriculum

statistics AND probability • Chance 13 In a class of 30 students there are 17 girls. Two students are picked randomly to represent the

class in the Student Representative Council. Determine the probability that: a both students are boys b both students are girls c one of the students is a boy. Reasoning 14 Greg has tossed a tail on each of 9 successive coin tosses. He believes that his chances of

tossing a Head on his next toss must be very high. Is Greg correct? Justify your answer. 15 The multiplication law of probability relates to independent events. Tree diagrams can illustrate the reflection    sample space of successive dependent events and How are dependent events, the probability of any one combination of events can independent events and the be calculated by multiplying the stated probabilities multiplication law of probability reflected on a tree diagram? along the branches. Is this a contradiction to the multiplication law of probability? Explain.

12E

Conditional probability ■■ ■■ ■■

Conditional probability is when the probability of an event is conditional (depends) on another event occurring first. The effect of conditional probability is to reduce the event space and thus increase the probability of the desired outcome. For two events, A and B, the conditional probability of event B, given that event A occurs, is denoted by P(B|  A) and can be calculated using the formula: P ( A ∩ B) P(B|  A) = , P( A) ≠ 0 P( A)

Worked Example 17

A group of students was asked to nominate their favourite food, spaghetti (S) or lasagne (L). The results are illustrated in the Venn diagram at right. Use the Venn diagram to calculate the following probabilities relating to a student’s favourite food. a What is the probability that a randomly selected student prefers spaghetti? b What is the probability that a randomly selected student likes lasagne given that they also like spaghetti? Think a

1

From 40 students surveyed, shown in blue, 20 nominated their favourite food as ‘spaghetti’ or ‘spaghetti and lasagne’ as shown in red.

x

S 11

L 9

15 5

Write/draw a

x

S

L

11

9

15 5

2

The probability that a randomly selected student prefers spaghetti is found by substituting these values into the probability formula.

number of favourable outcomes total number of possible outcomes 20 P(spaghetti) = 40 P(event) =

=

1 2

Chapter 12 Probability

417

statistics AND probability • Chance b

The condition imposed ‘given they also like spaghetti’ alters the sample space to the 20 students described in part a, as shaded in blue. Of these 20 students, 9 stated their favourite foods as lasagne and spaghetti, as shown in red.

1

The probability that a randomly selected student likes lasagne, given that they like spaghetti, is found by substituting these values into the probability formula for conditional probability.

2

b

x

S 11

L 9

15 5

P(B|  A) = P(L|  S) = =

P ( A ∩ B) P( A) 9 40 1 2 9 20

Worked Example 18

If P(A) = 0.3, P(B) = 0.5 and P(A ß B) = 0.6, calculate: a  P(A ¶ B)              b  P(B | A) Think a

b

Write

1

State the addition law for probability to determine P(A ß B).

a P(A ß B) = P(A) + P(B) - P(A ¶ B)

2

Substitute the values given in the question into this formula and simplify.

0.6 = 0.3 + 0.5 - P(A ¶ B) P(A ¶ B) = 0.3 + 0.5 - 0.6 = 0.2

1

State the formula for conditional probability.

2

Substitute the values given in the question into this formula and simplify.

■■

b P(B|  A) =

P(B|  A) = =

P ( A ∩ B) , P(A) ò 0 P( A) 0.2 0.3 2 3

It is possible to transpose the formula for conditional probability to calculate P(A ¶ B): P ( A ∩ B) P(B |  A) = , P(A) ò 0 P( A) P(A ¶ B) = P(A) ì P(B| A) This is called the multiplication rule for probability. remember

1. Conditional probability is when the probability of an event is conditional (depends) on another event occurring first. 2. For two events, A and B, the conditional probability of event B, given that event A occurs, is denoted by P(B|  A) and can be calculated using the formula: P ( A ∩ B) P(B|  A) = , P( A) ≠ 0 P( A) 3. The multiplication rule for probability gives P(A ¶ B) = P(A) ì P(B| A) 418

Maths Quest 10 for the Australian Curriculum

stAtistiCs AND probAbility • ChANCe

exerCise

12e iNDiviDUAl pAthWAys eBook plus

Activity 12-E-1

Conditional probability flUeNCy 1 We17 A group of students was asked to nominate their favourite form of dance, hip hop (H)

or jazz (J ). The results are illustrated in the Venn diagram below. Use the Venn diagram given to calculate the following probabilities relating to a student’s favourite form of dance.

Introducing conditional probability doc-5122

x

H 35

Activity 12-E-2

J 12

29

Practice with conditional probability doc-5123 Activity 12-E-3

Tricky conditional probability problems doc-5124

14 a What is the probability that a randomly selected student prefers jazz? b What is the probability that a randomly selected student prefers hip hop, given that they

prefer jazz? 2 A group of students was asked which seats they found most comfortable, the seats in the

computer lab or the science lab. The results are illustrated in the Venn diagram below. Use the Venn diagram given to calculate the following probabilities relating to the most comfortable seats. x

C 15

S 8

5 2

a What is the probability that a randomly selected student prefers the science lab? b What is the probability that a randomly selected student prefers the science lab, given

that they might prefer the computer lab or the science lab? 3 We18 If P(A) = 0.7, P(B) = 0.5 and P(A ß B) = 0.9, calculate: a P(A ¶ B) b P(B | A). 4 If P(A) = 0.65, P(B) = 0.75 and P(A ¶ B) = 0.45, calculate: a P(B | A) b P(A | B). UNDerstANDiNg 5 A medical degree requires applicants to participate in two tests, an aptitude test and an

emotional maturity test. 52% passed the aptitude test, while 30% passed both tests. Use the conditional probability formula to calculate the probability that a student who passed the aptitude test also passed the emotional maturity test. 6 At a school classified as a ‘Music school for excellence’ the probability that a student elects to study Music and Physics is 0.2. The probability that a student takes Music is 0.92. What is the probability that a student takes Physics, given that the student is taking Music? 7 The probability that a student is well and misses a work shift the night before an exam is 0.045, while the probability that a student misses a work shift is 0.05. What is the probability that a student is well, given they miss a work shift the night before an exam? 8 Two marbles are chosen, without replacement, from a jar containing only red and green marbles. The probability of selecting a green marble and then a red marble is 0.67. The probability of selecting a green marble on the first draw is 0.8. What is the probability of selecting a red marble on the second draw, given the first marble drawn was green? Chapter 12 probability

419

statistics AND probability • Chance 9 Consider rolling a red and a black die and the probabilities of the following events:

Event A Event B Event C

the red die lands on 5 the black die lands on 2 the sum of the dice is 10. a   MC  The initial probability of each event described is: A P(A) =

1 6

B P(A) =

5 6

C P(A) =

5 6

P(B) =

1 6

P(B) =

2 6

P(B) =

2 6

P(C) =

1 6

P(C) =

7 36

P(C) = 18

D P(A) =

1 6

E P(A) =

1 6

P(B) =

1 6

P(B) =

2 6

1

P(C) = 12

P(C) = 12

b Calculate the following probabilities.    i P(A | B) ii P(B | A)

5

1



iii P(C | A)

iv  P(C | B)

10   MC  A group of 80 schoolgirls consists of 54 dancers and 35 singers. Each member of the

group is either a dancer or a singer, or both. The probability that a randomly selected student is a singer given that she is a dancer is: A 0.17 B 0.44 C 0.68 D 0.11 E 0.78 Reasoning 11 Explain how imposing a condition alters probability calculations. 12 At your neighbouring school, 65% of the students are male and 35% are female. Of the male

students, 10% report that dancing is their favourite activity; of the female students, 25% report that dancing is their favourite activity. Find the probability that: a a student selected at random prefers dancing and is female b a student selected at random prefers dancing and is male. 13 Using the information presented in Question 12 above, construct a tree diagram. From your diagram, calculate: reflection    a the probability that a student is male and does not prefer dancing How does imposing a condition b the overall percentage of students who prefer alter the probability of an event? dancing.

12F

Subjective probability ■■

■■

420

Consider the following claims: ‘I feel the Australian cricket team will win this year’s Test cricket series because, in my opinion, they have a stronger side than the opposition.’ Claims like this are often made by people who may not have all the facts, and may also be biased.   ‘I think this summer will be a cold one.’ A statement like this will have merit if it comes from an individual with relevant knowledge, such as a meteorologist or a scientist. However, often people make these remarks with limited observation. Subjective probability is usually based on limited mathematical evidence and may involve one or more of the following: judgements, opinions, assessments, estimations and conjectures by individuals. It can also involve beliefs, sentiments and emotions that may result in a certain amount of bias.

Maths Quest 10 for the Australian Curriculum

stAtistiCs AND probAbility • ChANCe

WorkeD exAMple 19

On Anzac Day Peter plays two-up, which involves tossing two coins. Heads win if both coins land Heads, while Tails win if both coins land Tails. If the coins land with one Head and one Tail they are called ‘odd’, and the coins are tossed again until either Heads or Tails wins. After observing for a while, Peter notices that the last five tosses had either Tails winning or were odd. This leads Peter to believe that Heads will win the next game, so he places $50 on Heads and loses. Peter questions the fairness of the game and states that the game is biased and favours Tails. Discuss the accuracy of Peter’s statement. thiNk

Write

Discuss the statement made and comment on the probability of obtaining Heads or Tails in this particular game.

Each game is independent and so five Tails or odd outcomes in the previous games have no effect on the outcome of the current game. The game is not biased. Peter took a risk and paid for it. He is wrong in suggesting that the game is not fair.

reMeMber

Subjective probability is based on judgements and opinions. It can also involve beliefs, emotions and bias. exerCise

12f iNDiviDUAl pAthWAys eBook plus

Activity 12-F-1

Subjective probability doc-5125

subjective probability UNDerstANDiNg 1 We19 Discuss the accuracy of these statements. a The team batting last can never win a cricket match at the MCG. b The Australian cricket team is so good that not even bad weather can stop it from winning. c Two children in John’s family are girls so the third one will be a girl, too. d The Wallabies defeated the All Blacks three times last year so they will win the first game

this year.

Activity 12-F-2

Harder subjective probability doc-5126 Activity 12-F-3

In-depth subjective probability doc-5127

e It rained heavily on the last three consecutive Fridays so do not organise sport on a

Friday. f According to the weather report only three in every twenty houses were damaged by the

cyclone, so my house will not be damaged. g New Zealand lost its cricket match against Australia because their team uniform looked

boring. h This coin is biased because we obtained six Heads in a row. i The USA topped the medal tally in the last Olympics so they will do the same again in

the next Olympics. Australian Rules football is the best sport in the world. 2 Comment on the accuracy of these statements. a I have bought only one ticket for the raffle, therefore I cannot win. b This particular horse has odds of 1–2. It is certain to win. c If you keep on betting on Heads, you cannot lose. d If you want to win at all times, bet on the favourites. e It is no use betting on the favourites as you cannot win a great deal of money, therefore you should bet on the outsiders. j

Chapter 12 probability

421

stAtistiCs AND probAbility • ChANCe 3 Assign a probability to each of the following, based on your experience or judgement. a The probability that you will be late for a class this week b The probability that your favourite sporting team will win its next match c The probability that two traffic lights in a row will be red when you approach successive

intersections d The probability that you will see a dog some time today reAsoNiNg

eBook plus

Digital doc

WorkSHEET 12.3 doc-5297

4 Comment on the contradictions involved in the following statements. a That job was hers but she did not do well in the interview. b The team had won the match but they became a little complacent towards the end. c ‘Makybe Diva’ was certain to win. I cannot believe she lost the race. 5 Compare and contrast experimental probability,

theoretical probability and subjective probability. 6 Describe a situation where subjective probability may

endanger people.

422

Maths Quest 10 for the Australian Curriculum

refleCtioN 

 

Is subjective probability reliable?

statistics AND probability • Chance

Summary Review of probability ■■ ■■

■■

■■

■■ ■■ ■■

Probabilities can be expressed as a percentage, fraction or decimal in the range 0 to 1, inclusive. number of times an event has occurred Experimental probability = total number of trials frequency of the score f or Relative frequency of a score = total sum of frequencies S f n(E ) Theoretical probability that an event, E, will occur is P(E ) = where n(E ) = number of n(x  ) times or ways an event, E, can occur and n(x  ) = the total number of ways all outcomes can occur. P(x  ) = 1 Venn diagrams provide a diagrammatic representation of sample spaces. b If the odds for an event are given as a–b, then P(the event occurs) = and a+b a . P(the event does not occur) = a+b Complementary and mutually exclusive events

■■ ■■ ■■ ■■

■■

■■ ■■

Complementary events have no common elements and together make up the universal set. If A and AÅ are complementary events then P(A) + P(AÅ) = 1. This may be rearranged to: P(AÅ) = 1 - P(A) or P(A) = 1 - P(AÅ). Mutually exclusive events have no common elements and cannot occur simultaneously. If events A and B are not mutually exclusive then: P(A or B) = P(A) + P(B) - P(A and B) or P(A ß B) = P(A) + P(B) - P(A ¶ B) where P(A ¶ B) is the probability of the intersection of sets A and B or the common elements in sets A and B. If events A and B are mutually exclusive then: P(A or B) = P(A) + P(B) or P(A ß B) = P(A) + P(B) since P(A ¶ B) = 0. Mutually exclusive events may or may not be complementary events. Complementary events are always mutually exclusive. Two-way tables and tree diagrams

■■ ■■

■■

Two-way tables give a clear diagrammatic representation of the sample space; however, they are limited to displaying two events. Tree diagrams are useful in working out the sample space and calculating probabilities of various events, especially if there is more than one event. On each branch of a tree diagram, the probability associated with the branch is listed. The products of the probabilities given on the branches are taken to calculate the probability for an outcome. The probabilities of all outcomes add to 1. Independent and dependent events

■■ ■■

■■

Events are independent if the occurrence of one event does not affect the occurrence of the other. If events A and B are independent, then P(A ¶ B) = P(A) ì P(B). This is the Multiplication Law of probability. Conversely, if P(A ¶ B) = P(A) ì P(B) then events A and B are independent. Dependent events affect the probability of occurrence of one another. Chapter 12 Probability

423

stAtistiCs AND probAbility • ChANCe Conditional probability ■

■

■

Conditional probability is when the probability of an event is conditional (depends) on another event occurring first. For two events, A and B, the conditional probability of event B, given that event A occurs, is denoted by P(B| A) and can be calculated using the formula: P ( A ∩ B) P(B| A) = , P( A) ≠ 0 P( A) The multiplication rule for probability gives P(A ¶ B) = P(A) ì P(B| A) Subjective probability

■

Subjective probability is based on judgements and opinions. It can also involve beliefs, emotions and bias.

MAPPING YOUR UNDERSTANDING

Homework Book

424

Using terms from the summary, and other terms if you wish, construct a concept map that illustrates your understanding of the key concepts covered in this chapter. Compare your concept map with the one that you created in What do you know? on page 379. Have you completed the two Homework sheets, the Rich task and two Code puzzles in your Maths Quest 10 Homework Book?

Maths Quest 10 for the Australian Curriculum

statistics AND probability • Chance

Chapter review b AÅ ¶ BÅ

Fluency

B x

A

1 Which of the following is always true for an event,

M, and its complementary event, MÅ? A P(M) + P(MÅ) = 1 B P(M) - P(MÅ) = 1 C P(M) + P(MÅ) = 0 D P(M) - P(MÅ) = 0 E P(M) ì P(MÅ) = 1 2 A number is chosen from the set {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. Which of the following pairs of events is mutually exclusive? A {2, 4, 6} and {1, 2, 3} B {1, 2, 3, 5} and {4, 6, 7, 8} C {0, 1, 2, 3} and {3, 4, 5, 6} D {multiples of 2} and {factors of 8} E {even numbers} and {multiples of 3} 3 Which of the following states the Multiplication Law of probability correctly? A P(A ¶ B) = P(A) + P(B) B P(A ¶ B) = P(A) ì P(B) C P(A ß B) = P(A) ì P(B) D P(A ß B) = P(A) + P(B) E P(A) = P(A ß B) ì P(B) 4 The odds 3-2 expressed as a probability are: A c e

1 5 1 2 1 3

B d

3 5 2 5

The following information relates to questions 5 and 6. x  = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} and A = {2, 3, 4} and B = {3, 4, 5, 8} 5 A ¶ B equals: A {2, 3, 3, 4, 4, 5, 8} B {3, 4} C {2, 3, 4} D {2, 3, 4, 5, 8} E {2, 5, 8} 6 A ¶ BÅ equals: A {3, 4} B {2} C {2, 3, 4, 5, 8} D {2, 3, 4} E {1, 2, 6, 7, 9, 10} 7 Shade the region stated for each of the following Venn diagrams. a AÅ ß B A B x

c AÅ ¶ BÅ ¶ C

A

B

x

C 8 Convert the following odds to probabilities. a 3–7 b 5–2 c 12–5 9 Convert the following probabilities to odds. a

7 11

b

6 7

25 33

c

problem solving 1 From past experience, it is concluded that there

is a 99% probability that July will be a wet month in Launceston (it has an average rainfall of approximately 80 mm). The probability that July will not be a wet month next year in Launceston is: A 99%

B 0.99

D 1

E 0

C

1 100

2 A card is drawn from a well-shuffled deck of

52 cards. What is the theoretical probability of not selecting a red card? A C

3 4 1 13

B

1 4

D

1 2

E 0 3 Which of the following events is not equally likely? A Obtaining a 5 or obtaining a 1 when a die is

rolled B Obtaining a club or obtaining a diamond when

a card is drawn from a pack of cards C Obtaining 2 Heads or obtaining 2 Tails when a

coin is tossed D Obtaining 2 Heads or obtaining 1 Head when a

coin is tossed twice E Obtaining a 3 or obtaining a 6 when a die is

rolled Chapter 12 Probability

425

statistics AND probability • Chance 4 The Australian cricket team has won 12 of the

last 15 Test matches. What is the experimental probability of Australia losing its next Test match? A C

4 5 1 4

B D

13 The Venn diagram below shows the results of a

survey completed by a Chinese restaurateur to find out the food preferences of his regular customers.

1 5 3 4

x

Fried rice

5 A card is drawn from a well-shuffled pack of

10

52 cards. What is the theoretical probability of drawing: a an ace b a spade c a queen or a king d not a heart? 6 The odds for a horse to win a race are 4–3. a What is the probability that this horse will win

the race? b What is the probability that this horse will not

win the race? c Charlie bets $12 that this horse will win. If the

horse wins, what is Charlie’s payout? 7 A die is rolled five times. a What is the probability of rolling five 6s? b What is the probability of not rolling five 6s? 8 Alan and Mary own 3 of the 8 dogs in a race. What

is the probability that: a one of Alan’s or Mary’s dogs will win? b none of Alan’s or Mary’s dogs will win? 9 A die is rolled. Event A is obtaining an even

number. Event B is obtaining a 3. a Are events A and B mutually exclusive? b Calculate P(A) and P(B). c Calculate P(A ß B). 10 A card is drawn from a shuffled pack of 52 playing

cards. Event A is drawing a club and event B is drawing an ace. a Are events A and B mutually exclusive? b Calculate P(A), P(B) and P(A ¶ B). c Calculate P(A ß B). 11 Discuss the accuracy of the following statements. a It did not rain on Monday, Tuesday or

Wednesday, so it will not rain on Thursday. b A cricket team lost because two of its batsmen scored ducks. c The Rams family had a boy, then a girl and then another boy. They must have a girl next.

426

Maths Quest 10 for the Australian Curriculum

3

12 5

Chicken wings

8 Dim sims

Determine the number of customers: surveyed showing a preference for fried rice only showing a preference for fried rice s howing a preference for chicken wings and dim sims. b A customer from this group won the draw for a lucky door prize. Determine the probability that this customer:   i likes fried rice  ii likes all three — fried rice, chicken wings and dim sims iii prefers chicken wings only. c A similar survey was conducted a month later with another group of 50 customers. This survey yielded the following results: 2 customers liked all three foods; 6 preferred fried rice and chicken wings; 7 preferred chicken wings and dim sims; 8 preferred fried rice and dim sims; 22 preferred fried rice; 23 preferred chicken wings; and 24 preferred dim sims.   i Display this information on a Venn diagram.  ii What is the probability of selecting a customer who prefers all three foods, if a random selection is made? 14 A pair of dice is rolled and the sum of the numbers shown is noted. a Show the sample space in a two-way table. b In how many different ways can the sum of 7 be obtained? c Are all outcomes equally likely? d Complete the given table. a

  i  ii iii iv

Sum

12 Comment on the contradictions involved in these

statements. a I was defeated by a loser. b The slowest motocross racer in the competition won the race. c The most popular person did not get elected.

5

7

E 1

2 3 4 5 6 7 8 9 11 12

Frequency e What are the relative frequencies of the

following sums?

i 2



 ii 7



iii 11

stAtistiCs AND probAbility • ChANCe

15

16

17

18

f What is the probability of obtaining the

19 Determine the probability of drawing 2 aces from a

following sums? i 2 ii 7 iii 11 g If a pair of dice is rolled 300 times, how many times do you expect to obtain the sum of 7? A tetrahedral die is numbered 0, 1, 2 and 3. Two of these dice are rolled and the sum of the numbers (the number on the face that the die sits on) is taken. a Show the possible outcomes in a two-way table. b Are all the outcomes equally likely? c Which total has the least chance of being rolled? d Which total has the best chance of being rolled? e Which sums have the same chance of being rolled? An eight-sided die is rolled three times to see whether 5 occurs. a Draw a tree diagram to show the sample space. b Calculate: i P(three 5s) ii P(no 5s) iii P(two 5s) iv P(at least two 5s). A tetrahedral die (four faces labelled 1, 2, 3 and 4) is rolled and a coin is tossed simultaneously. a Show all the outcomes on a two-way table. b Draw a tree diagram and list all outcomes and their respective probabilities. c Calculate the probability of getting a Head on the coin and an even number on the die. A bag contains 20 pears, of which 5 are bad. Cathy picks 2 pears (without replacement) from the bag. What is the probability that: a both pears are bad? b both pears are good? c one of the two pears is good?

pack of cards if: a the first card is replaced before the second one is drawn b the first card drawn is not replaced. 20 On grandparents day at a school a group of grandparents was asked where they most like to take their grandchildren — the beach (B) or shopping (S). The results are illustrated in the Venn diagram below. Use the Venn diagram given to calculate the following probabilities relating to the place grandparents most like to take their grandchildren. ξ

B 5

S 8

2 10

a What is the probability that a randomly

selected grandparent preferred to take their grandchildren to the beach or shopping? b What is the probability that a randomly selected grandparent preferred to take their grandchildren to the beach, given that they preferred to take their grandchildren shopping? 21 Two marbles are chosen, without replacement, from a jar containing only red and green marbles. The probability of selecting a green marble and then a red marble is 0.72. The probability of selecting a green marble on the first draw is 0.85. What is the probability of selecting a red marble on the second draw if the first marble drawn was green? eBook plus

Interactivities

Test yourself Chapter 12 int-2858 Word search Chapter 12 int-2856 Crossword Chapter 12 int-2857

Chapter 12 probability

427

eBook plus

ACtivities

Chapter opener Digital doc

• Hungry brain activity (doc-5285): Chapter 12 (page 379) Are you ready? Digital docs

(page 380)

• SkillSHEET 12.1 (doc-5286): Set notation • SkillSHEET 12.2 (doc-5287): Simplifying fractions • SkillSHEET 12.3 (doc-5288): Determining complementary events • SkillSHEET 12.4 (doc-5289): Addition and subtraction of fractions • SkillSHEET 12.5 (doc-5290): Multiplying fractions for calculating probabilities 12A Review of probability Interactivity

• Random number generator (int-0089) (page 381) Digital docs

• Activity 12-A-1 (doc-5110): Review of probability (page 392) • Activity 12-A-2 (doc-5111): General probability problems (page 392) • Activity 12-A-3 (doc-5112): Tricky probability problems (page 392) • SkillSHEET 12.1 (doc-5286): Set notation (page 384) • SkillSHEET 12.2 (doc-5287): Simplifying fractions (page 392) • SkillSHEET 12.6 (doc-5291): Working with Venn diagrams (page 392) • SkillSHEET 12.7 (doc-5292): Writing odds as probabilities (page 395) • SkillSHEET 12.8 (doc-5293): Writing probabilities as odds (page 395) 12B Complementary and mutually exclusive events Digital docs

• Activity 12-B-1 (doc-5113): Complementary and mutually exclusive events (page 401) • Activity 12-B-2 (doc-5114): Harder complementary and mutually exclusive events (page 401) • Activity 12-B-3 (doc-5115): Tricky complementary and mutually exclusive events (page 401) • SkillSHEET 12.3 (doc-5288): Determining complementary events (page 401) • SkillSHEET 12.4 (doc-5289): Addition and subtraction of fractions (page 402) • SkillSHEET 12.9 (doc-5294): Distinguishing between complementary and mutually exclusive events (page 403) • WorkSHEET 12.1 (doc-5295): Introducing probability (page 403) 12C Two-way tables and tree diagrams elesson

• Games at Wimbledon (eles-1032) (page 405) 428

Maths Quest 10 for the Australian Curriculum

Digital docs

• Activity 12-C-1 (doc-5116): Review of two-way tables and tree diagrams (page 410) • Activity 12-C-2 (doc-5117): Practice with two-way tables and tree diagrams (page 410) • Activity 12-C-3 (doc-5118): Tricky two-way tables and tree diagrams (page 410) • SkillSHEET 12.5 (doc-5290): Multiplying fractions for calculating probabilities (page 410) • WorkSHEET 12.2 (doc-5296): Tree diagrams (page 412) 12D Independent and dependent events Interactivities

• A pack of cards (int-2787) (page 413) • Random numbers (int-0085) (page 415) Digital docs

(page 415)

• Activity 12-D-1 (doc-5119): Simple independent and dependent events • Activity 12-D-2 (doc-5120): Independent and dependent events • Activity 12-D-3 (doc-5121): Tricky independent and dependent events 12E Conditional probability Digital docs

(page 419)

• Activity 12-E-1 (doc-5122): Introducing conditional probability • Activity 12-E-2 (doc-5123): Practice with conditional probability • Activity 12-E-3 (doc-5124): Tricky conditional probability problems 12F Subjective probability Digital docs

• Activity 12-F-1 (doc-5125): Subjective probability (page 421) • Activity 12-F-2 (doc-5126): Harder subjective probability (page 421) • Activity 12-F-3 (doc-5127): In-depth subjective probability (page 421) • WorkSHEET 12.3 (doc-5297): Subjective probability (page 422) Chapter review Interactivities

(page 427)

• Test yourself Chapter 12 (int-2858): Take the end-ofchapter test to test your progress • Word search Chapter 12 (int-2856): an interactive word search involving words associated with this chapter • Crossword Chapter 12 (int-2857): an interactive crossword using the definitions associated with the chapter To access eBookPLUS activities, log on to www.jacplus.com.au

statistics aND probability • Data represeNtatioN aND iNterpretatioN

13

13A 13B 13C 13D 13E 13F

Measures of central tendency Measures of spread Box-and-whisker plots The standard deviation [suitable for 10A] Comparing data sets Skewness

What Do yoU kNoW ?

Univariate data

1 List what you know about data. Create a concept map to show your list. 2 Share what you know with a partner and then with a small group. 3 As a class, create a large concept map that shows your class’s knowledge of data. eBook plus

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opeNiNG QUestioN

A new drug for the relief of cold symptoms has been developed. To test the drug, 40 people were exposed to a cold virus. Twenty patients were then given a dose of the drug while another 20 patients were given a placebo. (In medical tests a control group is often given a placebo drug. The subjects in this group believe that they have been given the real drug but in fact their dose contains no drug at all.) All participants were then asked to indicate the time when they first felt relief of symptoms. The number of hours from the time the dose was administered to the time when the patients first felt relief of symptoms are detailed below. Group A (drug) 25 29 32 45 18 21 37 42 62 13 42 38 44 42 35 47 62 17 34 32 Group B (placebo) 25 17 35 42 35 28 20 32 38 35 34 32 25 18 22 28 21 24 32 36 Does the drug work? How do drug companies analyse these results?

statistics aND probability • Data represeNtatioN aND iNterpretatioN

are you ready?

Try■the■questions■below.■If■you■have■diffi■culty■with■any■of■them,■extra■help■can■be■obtained■by■ completing■the■matching■SkillSHEET.■Either■click■on■the■SkillSHEET■icon■next■to■the■question■ on■the■Maths■Quest■eBookPLUS■or■ask■your■teacher■for■a■copy. eBook plus

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Finding the mean of a small data set 1 Find■the■mean■of■the■following■data■sets. a 4,■6,■2,■6,■4,■3,■7,■3,■2,■9 b 10.4,■10.5,■10.7,■10.4,■10.1,■10.2,■10.4,■10.0 c 164,■136,■171,■144,■128,■130,■165,■170,■120,■124,■124,■143 Finding the median of a small data set 2 Find■the■median■of■each■of■the■following■data■sets. a 2,■6,■8,■4,■5,■6,■7 b 13,■10,■15,■12,■18,■17,■12,■12 c 52,■45,■23,■83,■9,■45,■71,■84,■90,■183 d 9.2,■9.3,■9.4,■9.3,■9.4,■9.5,■9.9,■9.4,■9.7,■9.8,■9.0,■10.0,■9.4,■9.2,■9.9 Finding the mode of a small data set 3 Find■the■mode■of■each■of■the■following■data■sets. a 2,■4,■3,■1,■6,■7,■3,■4,■3,■2,■7 b 34,■82,■94,■81,■70,■45,■32,■46,■48 c 1,■2,■4,■3,■5,■2,■6,■7,■3,■3,■2 Finding the mean, median and mode from a stem-and-leaf plot 4 Find■the■mean,■median■and■mode■of■the■data■presented■in■

the■stem-and-leaf■plot■shown■at■right.

Presenting data in a frequency distribution table 5 Place■the■following■set■of■scores■into■a■frequency■distribution■table■including■a■column■

for■f■■ì■x■and■cumulative■frequency. 8■ 7■ 4■ 9■ 6■ 7■ 9■ 6■ 5■ 4 7■ 9■ 8■ 6■ 5■ 8■ 9■ 4■ 5■ 8 Drawing statistical graphs 6 Use■the■data■from■question■5■to■draw■a■combined■frequency■histogram■and■polygon.

SkillSHEET 13.6 doc-5304

430

Key:■ 3■■|■■4■=■34 Stem Leaf 2 1■ 4■ 5■ 3 0■ 6 4 2■ 3■ 3■ 7 5 9

Maths Quest 10 for the australian curriculum

statistics AND probability • data representation and interpretation

13A

Measures of central tendency ■■

Measures of central tendency are summary statistics that measure the middle (or centre) of the data. These are known as the mean, median and mode. •• The mean is the average of all observations in a set of data. •• The median is the middle observation in an ordered set of data. •• The mode is the most frequent observation in a data set.

Ungrouped data Mean To obtain the mean of a set of ungrouped data, all numbers (scores) in the set are added together and then the total is divided by the number of scores in that set.

■■

Mean = ■■

Symbolically this is written x =

sum of all scores number of scores

∑x . n

Median ■■ The median is the middle value of any set of data arranged in numerical order. In the set of n numbers, the median is located at the n + 1 th score. The median is: 2 •• the middle score for an odd number of scores arranged in numerical order •• the average of the two middle scores for an even number of scores arranged in numerical order. Mode ■■ The mode is the score that occurs most often in a set of data. Sets of data may contain: 1. no mode; that is, each score occurs once only 2. one mode 3. more than one mode. Worked Example 1

For the data set 6, 2, 4, 3, 4, 5, 4, 5, find the: a  mean b  median c  mode. Think a

1

Write

Calculate the sum of the scores; that is, S x.

a S x = 6 + 2 + 4 + 3 + 4 + 5 + 4 + 5

= 33

2

Count the number of scores; that is, n.

n=8

3

Write the rule for the mean.

x=

4

Substitute the known values into the rule.

=

5

Evaluate.

= 4.125

6

Answer the question.

∑x n 33 8

The mean is 4.125. Chapter 13 Univariate data

431

statistics AND probability • data representation and interpretation b

b 23444556

1

Check that scores are arranged in numerical order.

2

Locate the position of the median using n +1 the rule , where n = 8. This places 2 the median as the 4.5th score; that is, between the 4th and 5th score.

Median =

Obtain the average of the two middle scores.

Median =

3

=

n+1 th score 2 8+1 2

th score

= 4.5th score 23444556

=

4+4 2 8 2

=4

c

4

Answer the question.

1

Systematically work through the set and make note of any repeated values (scores).

2

Answer the question.

The median is 4. éé c 23444556 å å å The mode is 4.

Calculating mean, median and mode from a frequency distribution table ■■ If data are presented in a frequency distribution table, the formula used to calculate the mean ∑( f × x ) . is x = n ■■ Here, each value (score) in the table is multiplied by its corresponding frequency; then all the f ì x products are added together and the total sum is divided by the number of observations in the set. ■■ To find the median we find the position of each score from the cumulative frequency column. ■■ The mode is the score with the highest frequency. Worked Example 2

For the table at right find the:    a  mean b  median c  mode.

Score (x) 4 5 6 7 8 Total

Think

432

1

Rule up a table with four columns titled Score (x), Frequency (  f  ), Frequency ì score (  f ì x) and Cumulative frequency (cf  ).

2

Enter the data and complete both the■ f ì x and cumulative frequency columns.

Maths Quest 10 for the Australian Curriculum

Frequency (  f )  1  2  5  4  3 15

Write

Frequency Cumulative Score Frequency ì score frequency ( x) (  f ) (  f ì x) (cf ) 4 1  4 1 5 2 10 1+2=3 6 5 30 3+5=8 7 4 28 8 + 4 = 12 8 3 24 12 + 3 = 15 n = 15 S(  f ì x) = 96

statistics AND probability • data representation and interpretation

a

b

c

x=

∑( f × x ) n

1

Write the rule for the mean.

2

Substitute the known values into the rule and evaluate.

x=

3

Answer the question.

The mean of the data set is 6.4.

1

Locate the position of the median n +1 using the rule , where n = 15. 2 This places the median as the 8th score.

2

Use the cumulative frequency column to find the 8th score and answer the question.

1

The mode is the score with the highest frequency.

2

Answer the question.

a

96 15 = 6.4

b The median is the

15 + 1 th or 8th score. 2

The median of the data set is 6.

c

The score with the highest frequency is 6. The mode of the data set is 6.

Grouped data Mean When the data are grouped into class intervals, the actual values (or data) are lost. In such cases we have to approximate the real values with the midpoints of the intervals into which these values fall. For example, when measuring heights of students in a class, if we found that 4 students had a height between 180 and 185 cm, we have to assume that each of those 4 students is 182.5 cm tall. The formula used for calculating the mean is the same as for data presented in a frequency table: ∑( f × x ) x= n Here x represents the midpoint (or class centre) of each class interval, f is the corresponding frequency and n is the total number of observations in a set.

■■

Median The median is found by drawing a cumulative frequency polygon (ogive) of the data and estimating the median from the 50th percentile.

■■

Modal class We do not find a mode because exact scores are lost. We can, however, find a modal class. This is the class interval that has the highest frequency.

■■

Worked Example 3

For the given data:          a  estimate the mean b  estimate the median c  find the modal class.

Class interval   60 – 4

–2 0

6 5 4 3 2 1

x

8

5 Note: The shaded region is the region required. a   i  y

y í 2x - 2

y-x>4

6

y

6 5 4 3 2 1

-1 -10 -2 -3 -4 -5 -6

-2

-1 -2 -3 -4 -5

4

2x - 3y í 18

y í 2x

7 6 5 4 3 2 1

-6

2

n

y

j

x+y>7

2000 r

Answers

811

10

8 a 100a + 75b í 450 b 50a + 75b í 300 b c

d Answers will vary.

6

2

0

1

2

3

4

5

6

a

Mass of chocolate chips in grams (c)

700

Note: The shaded region is the region required.

600 500

-4

-2

0

2

4

y

65

45

35

25

15

5

-5 -15 -25 -35

 

-10 -8 -6 -4 -2 0 -20 -40 -60 -80

11 a

6

  b

y

300

y = -5x + 15 (1, 10)

10

0 1 –2

100 p + c Ç 400

x 0 1

0 100 200 300 400 500 600700 Mass of peanuts in grams (p)

c

 d

y

1 0 -1

600 500

(3, -1)

y

x

3

x

4

y =–2–3 x + 1

700

0

400 200

5 y=

-3

7– x 5

-3

0.3p + 0.6c í 180

100 0 100 200 300 400 500 600700 Mass of peanuts in grams (p)

700

6

12 a x-intercept = 7 , y-intercept c = 6 b x-intercept =

40 3 21 16

1

(133), y-intercept c = -5

5 (116 ),

3 c x-intercept = y-intercept c = - 4 d x-intercept = -5.6, y-intercept c = 2.8

13 a

b

y

y

2x - 3y = 6

600 500

0 -2

400

3

x

3

-1 0

300

x y = -3x

200 100 0 100 200 300 400 500 600700 Mass of peanuts in grams (p)

Chapter review Fluency 1 A 2 D 3 B 4 A 5 C 6 E 7 C 8 D 9 A

Answers

10

y 15

(1, 1)

1

200

300

8

2 4 6 8 10

400

c

d

y

x

0

- –53

y x+y+3=0

5x + y = -3

f Answers will vary.

812

55

y = 3x - 2

c 0.3p + 0.6c í 180 d Mass of chocolate chips in grams (c)

-6

80 60 40 20

9 a p + c Ç 400 b

Mass of chocolate chips in grams (c)

-10 -8

Note: The shaded region is the region required.

4

e

x

-3

-3

-3

14 a

0

b

y

1– 2

(1, 1–2 ) 0

1 y = 1–2 x

x

y

x 0 1 y = -4x -4

x

x

c

y

x = -2

24 a (2, 7)

7 y=7 x

0

-2

15

y

d

d x

0

25 a (5, 2) d (1, 3) 26 a (0, 3) 27 a

y 7 (0, 7) - –27

 7 7  − ,  3 3

16 a y = 2x - 2 3

c (-3, -1) f (4, 2) c (2, 1) Note: The shaded region is the region required.

1

2y - 3x í 12

18

3

b y = - 32 x + 12

-6

20 Note: The shaded region is the region required. y a b y

c

-5 0

d

y

c

5x + y < 10

10

y í 2x + 10

x

-1 0

y 5

y = 5x

0 1

x y < 5x

x

x

-12 y

f

x=7

y 1

7

xí7

x

-2

h

y 9 2x + y í 9

21 a (3, 1) 22 a No 23 a (-2, 1) b (0, -2) c (5, 2)

0

y

x + 2y < 11

0

2

4

6

0

27

54

81

8

10

108 135

50 40 30 20 10

1 2 3 4 Hours worked (h)

d $91.13 2 a Number of rides Cost ($)

y > -12

x

b Pay = $13.50 ì (number of hours worked) c

0 • 12 x

–16• 4x - 3y í 48

0 -12

x

x

0

2

4

6

8

10

12.50 17.50 22.50 27.50 32.50 37.50

b Cost = $2.50 ì number of rides + $12.50 c

x

b (2, 3) b Yes

40 30 (12.50) 20 10 0

2

(10, 37.50)

4 6 8 10 Number of rides

Answers 4G ➜ 4G

i

0

y 5

x

–9 2

y Ç –21 x + 1

6

2 4 6 8 10 12 14 16

Pay ($)

Pay ($)

0

4

y

Problem solving 1 a Number of hours

Cost ($)

4

2

y + 3x > 0

10 9 8 7 6 5 4 3 2 1

0 -8 -6 -4 -2 -1 -2

y > 3x - 12 0

x

6

7 6 5 4 3 2 1

-2 -10 -2 -3 -4

-4

27 5

1 yÇx+1

4

y

d y = 5 x - 5

19 a y = -x + 8 +

b

b y = -3x + 4

c y = 2 x + 6

2

-2

-6

d y = 6

18 a y = 7x - 13

0

0

-2

-4

b y = -2x - 5

1 y = 2 x + 5

2 y = 5 x

-4

yÇx+4

f x = 5

17 a y = 3x - 4

g

5

4

d y = 4x

e y = - 4

e

f ( 2, -7)

y

6

-6

b y = -x - 4

c y = − 13 x + 2

c

e (-14, -53)

2

3(y - 5) = 6(x + 1)

c

c (-2, 2)

b (-2, 3) e (2, -2) b (-3, -3) yí3

x

0

b (-5, -3)

d $30 Answers

813

5 a

500

m

25è

•

A

100 0

0

2

4

6 8 10 12 Time

b Cost = 22.5 ì time + 160 c $435.63 4 a Numbers are 9 and 14. b Length = 11 metres, width = 6 metres c Chupa-chups cost 45 cents and Whizz fizzes cost 55 cents. 5 Milk $1.75, bread $2.35 6 13 kangaroos and 8 cockatoos 7 Rollercoaster ride $6, Ferris wheel ride $4, Gravitron ride $8 8 a d = b + 10 b 7000 = 70b + 40d c b = 60 and d = 70 d Number of seats in ‘Bleachers’ is 4200; the number of seats in the ‘Dress circle’ is 2800. e $644  000 9 a CG = 114 + 0.20k b CS = 90 + 0.32k c 200 km d 114 + 0.20k < 90 + 0.32k \ k > 200 e k < 200 10 a 5400 + 260d = CH b 61 days 11 a n < 16 800 km b Mick travelled less than 16  800 km for the year and his costs stayed below $16  000. CHAPTER 5

Trigonometry I Are you ready? 1 a 0.685 c 0.749 2 a i  15è33Å b i  63è16Å c i  27è10Å 3 a H O

A c

b 1.400 ii  15è32Å41ë ii  63è15Å32ë ii  27è10Å16ë b A q q

O H A

q

4 a x = 30 ì tan (15è) b x =

4.2 tan(28°)

c x = 5.3 ì tan (64è) 814

Answers

H O

S

B 120è

5k

Cost

N

300 •

20è

km

400

200

b

N

•

180

3 a

7.

5

70 km

N

km 60è

C

Exercise 5A — Pythagoras’ theorem 1 a 7.86 b 33.27 c 980.95 d 12.68 e 2.85 f 175.14 2 a 36.36 b 1.62 c 15.37 d 0.61 e 2133.19 f 453.90 3 23.04  cm 4 12.65  cm 5 a 14.14  cm b 24.04  cm c 4.53  cm 6 a 97.47  cm b 334.94  cm c 6822.90  cm2 7 a 6.06 b 4.24 c 4.74 8 18.03  cm 9 17.32  cm 10 19.23  cm 11 65.82  cm; 2501.16  cm2 12 39  m 13 Yes 14 4.34  km 15 38.2  m 16 20.61  m 17 130  mm 18 a 386.13  mm b 62.09  cm c 2.33  km d 16.15  cm e 541.70  cm f 2615.61  m g 478.97  mm h 369.87  km 19 54.67  mm 20 a 28  cm b 588  cm2 21 36.37  cm 22 552.86  cm2 23 21.46 diagonals, so would need to complete 22 24 1600  mm 25 5889.82  m 26 7.07  cm 27 $81.60 28 185  cm 29 Students own working. Exercise 5B — Pythagoras’ theorem in three dimensions 1 a 17.32 b 12.25 c 15.12 2 12.21, 12.85 3 4.84  m, 1.77  m 4 11.31, 5.66 5 31.62  cm 6 6  cm 7 12.65  cm 8 23  mm 9 No: maximum stick can be only 115  cm long. 10 3.41  cm 11 a i  283.02  m ii 240.21  m iii 150.33  m b 141.86  m

12 14.72  cm 13 13.38  cm 14 42.27  cm 15 1.3  m, 5.98  m2 16 Students’ own working Exercise 5C — Trigonometric ratios 1 a 0.5000 b 0.7071 c 0.4663 d 0.8387 e 8.1443 f 0.7193 2 a 0.6944 b 0.5885 c 0.5220 d -1.5013 e 0.9990 f 0.6709 g 0.8120 h 0.5253 i -0.8031



25 30

c tan (q ) = 45

b cos (q ) =

2.7 17 e sin (35è) = p t 7 20 g sin (15è) = h tan (q ) = 31 x 9 a a

d tan (q ) =

H

41è A

Exercise 5F — Angles of elevation and depression 1 8.74  m 2 687.7  m 3 a 176.42  m b 152.42  m 4 65è46Å 5 16.04  m 6 a h = x tan (47è12Å)  m; h = (x + 38) tan (32è50Å)  m b x = 76.69  m c 84.62  m 7 a h = x tan (43è35Å)  m; h = (x + 75) tan (32è18Å)  m b 148.40  m c 141.24  m 8 0.033  km or 33  m 9 21è 10 a 8.43  m b 56.54  m

11 44.88  m 12 a

.3 f sin (a ) = 14 17.5

O

b O = 34  mm, A = 39  mm, H = 51  mm c   i sin (41è) = 0.67 ii cos (41è) = 0.76 iii tan (41è) = 0.87 d a = 49è e   i sin (49è) = 0.76 ii cos (49è) = 0.67 iii tan (49è) = 1.15 f They are equal. g They are equal. h The sin of an angle is equal to the cos of its complement angle.

b 15.27  m 13 66  m 14 a 54è 15 a 2.16 m/s, 7.77 km/h

b 0.75  m b 54.5è

Exercise 5G — Bearings and compass directions 1 a 020èT b 340èT c 215èT d 152èT e 034èT f 222èT 2 a N49èE b S48èE c S87èW d N30èW e N86èE f S54èW 3 a 3  km 325èT b 2.5  km 112èT c 8  km 235èT d 4  km 090èT, then 2.5  km 035èT e 12  km 115èT, then 7  km 050èT f 300  m 310èT, then 500  m 220èT

4 a

b

N

135è

km

40è

N

m

0k

14

c

1.3

km

N

d

30è

50è

7k

32è

m 8k

N N 260è 120è 0.8 km N

240è

N 40è

m

Answers 5A ➜ 5G

100è 30 km

N

N

0

6 1.05  m 7 a x = 30.91 cm, y = 29.86 cm, z = 39.30 cm b 2941.54 cm2

15 m

23

Exercise 5D — Using trigonometry to calculate side lengths 1 a 8.660 b 7.250 c 8.412 2 a 0.79 b 4.72 c 101.38 3 a 33.45  m b 74.89  m c 44.82  m d 7.76  mm e 80.82  km f 9.04  cm 4 a x = 31.58  cm b y = 17.67  m c z = 14.87  m d p = 67.00  m e p = 21.38  km, q = 42.29  km f a = 0.70  km, b = 0.21  km 5 a 6.0  m b 6.7  m

1.76 m

3.1

i cos (a ) = 9.8

x

42è

m



5k

8 a sin (q ) = 12 15



km

e f i b i  sin (a) = g l c i  sin (b ) = k n d i  sin (g  ) = m b e i  sin (b ) = c v f i  sin (g  ) = u

7 a i  sin (q) =

Exercise 5E — Using trigonometry to calculate angle size 1 a 67è b 47è c 69è 2 a 54è47Å b 33è45Å c 33è33Å 3 a 75è31Å21ë b 36è52Å12ë c 37è38Å51ë 4 a 41è b 30è c 49è d 65è e 48è f 37è 5 a a = 25è47Å, b = 64è13Å b d = 25è23Å, e = 64è37Å c x = 66è12Å, y = 23è48Å 6 a r = 57.58, l = 34.87, h = 28.56 b 714 cm2 c  29.7è 7 a   i  29.0è ii  41.4è iii 51.3è b   i  124.42 km/h ii 136.57 km/h iii  146.27 km/h

km

-0.9613 0.1320 53è 41è 0è52Å 26è45Å 64è46Å59ë 88è41Å27ë 20.361 1.192 4909.913 14.814 e iii tan (q) = d i iii tan (a) = h l iii tan (b ) = j n iii tan (g  ) = o b iii tan (b ) = a v iii tan (g  ) = t

l o c f c f c f c f i l

2.1

0.9880 -0.5736 24è 86è 6è19Å 44è48Å 64è1Å25ë 36è52Å12ë 71.014 226.735 32.259 0.904 d ii cos (q) = f h ii cos (a) = g j ii cos (b ) = k o ii cos (g  ) = m a ii cos (b ) = c t ii cos (g  ) = u

k n b e b e b e b e h k

40

j 0.4063 m 1.7321 3 a 50è d 71è 4 a 54è29Å d 72è47Å 5 a 26è33Å54ë d 48è5Å22ë 6 a 2.824 d 2.828 g 7.232 j 0.063

S

Answers

815

N m 0k 22

70è180 km

30è

N

km

20è

320

S

5 a i  13.38  km ii 14.86  km b i  N B

N

130è 80

20 km

42è A

iii 222èT ii 51.42  km iii 61.28  km iv 310èT

ii 38.97  km iii 22.5  km iv 030èT

N B 130è

N

A

20

km

42è

80

km

N

210è

45

km

C

D

6 215èT 7 1.732  km 8 a 9.135  km b 2.305  km c 104è10Å T 9 684.86  km 10 a 60è43Å T b 69è27Å T c 204è27Å T 11 a q = 60è, a = 40è b 2.5 km c 2.198 km d 1.22 km Exercise 5H — Applications 1 a 36è52Å b 53è8Å c 2.4  m 2 a 14è29Å b 31  cm 3 6.09  m 4 19è28Å 5 62è33Å 6 a 11è32Å b 4è25Å 7 a   i  35.36  cm ii 51.48  cm iii 51.48  cm iv  57.23  cm v 29è3Å vi 25è54Å b   i  25.74  cm ii 12.5  cm iii 25è54Å iv  28.61  cm 8 a 77è b 71è56Å c 27.35  cm 9 a 7.05  cm b 60è15Å c 8.12  cm 10 a 28.74  cm b 40.64  cm c 66è37Å 11 a 26.88  cm b 11.07  cm d tan θ 12 a 90 m ì tan q2 b h = tan θ1 + tan θ 2 c 250 m Chapter review Fluency 1 E 2 D 3 E 7 E 8 B 9 B 11 a x = 113.06 cm 12 9.48  cm 14 17.6  m

816

Answers

Chapter 6

Surface area and volume

km

C

c i 

16 67.98  km 17 4.16  km 18 40è32Å Problem solving 1 a h = tan (47è48Å)x  m h = tan (36è24Å) (x + 64)  m b 129.10  m c 144.32  m 2 a 11.04  cm b 15.6  cm c 59è2Å 3 a 27.42 km b N43èW or 227èT 4 a 1280.6 m b 12:02:16.3 pm 5 33.29 m, 21.27 m

4 E 5 D 6 B 10 A b x = 83.46 mm 13 8.25  mm 15 26.86  m

Are you ready? 1 a 3.6 ì 106  mm2 b 2 ì 10-6  km2 4 2 c 5.2 ì 10 m 2 a 24  m2 b 30  cm2 c 4.9  cm2 2 2 3 a 150  cm b 232  cm c 1.22  m2 4 a 3.4 ì 106  cm3 b 2.5 ì 10-4  m3 c 6.5 ì 103  mm3 5 a 125  cm3 b 160  cm3 c 0.03  m3 Exercise 6A — Area 1 a 16  cm2 b 48  cm2 c 75  cm2 2 2 d 120  cm e 706.86  cm f 73.5  mm2 g 254.47  cm2 h 21  m2 i 75  cm2 2 Part e = 225p  cm2; part g = 81p  cm2 3 a 20.7  cm2 b 7.64  cm2 4 a 113.1  mm2 b 188.5  mm2 5 a i 12p  cm2 ii 37.70  cm2 b i 69π   mm2 ii 108.38  mm2 2 c i 261p  cm2 ii 819.96  cm2 6 E 7 D 8 a 123.29  cm2 b 1427.88  m c 52  cm2 d 30.4  m2 e 78  cm2 f 2015.5  cm2 9 a 125.66  cm2 b 102.87  cm2 c 13.73  m2 d 153.59  m2 e 13.86  m2 f 37.5  m2 10 11  707.93  cm2 11 21  m2 12 60 13 $840 14 a 260.87  m2 b 195.71  m2 c 130.43  m2 d 97.85  m2 e 37.5% f 18.75  m 15 a 50 = x + y b y = 50 – x c Area = 50x − x2 d x

0

Area (m2)

0 225 400 525 600 625 600 525 400 225 0

5

10

15

20

25

30

e No, impossible to  make a rectangle. f 600 Area

e

500 400 300 200 100 0 10 20 30 40 50 60 70 x

35

40

45 50

g x = 25 h y = 25 i Square j 625  m2 k r = 15.915  m l 795.77  m2 2 m 170.77  m 16 a Students’ work b 2020.83  m; horizontal 17 a Circular area, 1790.49 m2; rectangular area, 1406.25 m2  1 2 n m2; rectangular (square) b Circular area,  4π   1 2 4 area,  n  m2. Circular area is always or 1.27  16  π 1  1  4π ÷ 16  times larger. Exercise 6B — Total surface area 1 a 600  cm2 b 384  cm2 c 2 d 27  m 2 a 113.1  m2 b 6729.3  cm2 c 2 d 452.4  cm 3 a 1495.4  cm2 b 502.7  cm2 4 a 506.0  cm2 b 9.4  m2 c d 224.1  cm2 5 a 13.5  m2 b 90  m2 c d 9852.0  mm2 e 125.6  cm2 f 6 a 880  cm2 b 3072.8  cm2 c d 70.4  cm2 e 193.5  cm2 f 7 B 8 63 9 11  216  cm2 10 a 70.0  m2 b $455 11 a 3063.1  cm2 b $168.47

1440  cm2 8.2  m2 340.4  cm2 11  309.7  cm2 1531.4  cm2 75  cm2 1547.2  cm2

12 a q = 120 è

b x = 1; y = 3 c 3 3  cm2

d 6 3  cm2

e 32

13 a

8p  m2

7   m b 2

c 2 7π   m2 d 7 × 7 × 1 14 Calculation is correct. 15 a 6.6 m2 c Cheapest: 30 cm by 30 cm, $269.50; 20 cm by 20 cm (individually) $270; 20 cm by 20 cm (boxed) $276.50 16 r =

3 3a 2

Chapter review Fluency 1 D 4 A 5 a 84 cm2 d 56.52 cm2 6 a 300 cm2 7 a 499.86 cm2 8 a 18  692.48 cm2 b 1495.40 cm2 d 642 cm2 9 a 343 cm3 d 1.45 m3 g 297 cm3

2 C

3 E

b 100 cm2 e 60 cm2 b 224.52 cm2 b 44.59 cm2



c 6.5 cm2 f 244.35 cm2 c 160 cm2 c 128.76 cm2

c 804.25 cm2 e 873.36 mm2 b 672 cm3 e 1800 cm3 h 8400 cm3



f 760 cm2 c 153  938.04 cm3 f 1256.64 cm3 i 7238.23 mm3

Problem solving 1 a 62  m2 b $7290 3 2 2 V = π r h, the volume will be 1.5 times as large as the 2 original volume. 3 V = 3lwh, the volume will be 3 times as large as (or triple) the original volume. 4 a 3605.55  cm2 b $180.33 c 18062.1  cm3 d 9155.65  cm3 5 a 1.33  m b 910.81  m2 c $655.85 d 303.48  m3 e 11 trucks f 12 minutes

Chapter 7

Quadratic expressions Are you ready? 1 a 12x + 20 b 10x2 - 15x c -12x + 8x2 2 2 2 a x - 4 b 4x - 12x + 9 c 6x2 - 11x - 10 3 a 4x(x + 2) b -3x(5x + 3) c x(6x - 1) 4 a (x + 2)(3x + 4) b (x - 1)(4x - 1) c -(x + 3)(2x + 1)

Answers

Answers 5H ➜ 6C

Exercise 6C — Volume 1 a 27  cm3 b 74.088  m3 c 3600  cm3 d 94.5  cm3 2 a 450  mm3 b 360  cm2 3 a 6333.5  cm3 b 19.1  m3 c 280  cm3 d 288  mm3 4 a 7.2  m3 b 14 137.2  cm3 c 1436.8  mm3 d 523 598.8  cm3 5 a 377.0  cm3 b 2303.8  mm3 6 a 400  cm3 b 10  080  cm3 c 576  cm3 7 a 1400  cm3 b 10  379.20  cm3 c 41.31  cm3 d 48.17  cm3 e 218.08  cm3 f 3691.37  cm3 8 a Vnew = 27l3, the volume will be 27 times as large as the original volume. b Vnew = 18l2, the volume will be 18 of the original volume.

c Vnew = 2p r2h, the volume will be twice as large as the original volume. d Vnew = p r2h, the volume will remain the same. e Vnew = 3lwh, the volume will be 3 times as large as the original value. 9 E 10 7438.34  cm3 11 4417.9 L 12 10 215.05  cm3 13 a H = 6  m b 112  m3 c 19 bins d 112  000 L e 1.95  m from floor 14 a i 4.57  cm ii 262.5  cm2 b i 14.15  cm ii 323.27  cm2 c i 33.3  cm ii 434.28  cm2 d Sphere. Costs less for a smaller surface area. 15 Required volume = 1570.79 cm3; tin volume =1500 cm3; muffin tray volume = 2814.72 cm3. Marion should use the tin with approximately 70 cm3 mixture left over. 16 Increase radius of hemispherical section to 1.92 m. 17 Cut squares of side length, s = 0.3 m or 0.368 m from the corners. 18 Volume of water needed; 30.9 m3.

817

1 ( x − 3)( x + 7)

5 a x + 3

b

6 a 2 6

b 6 3

c

x+2 2( x + 3)

c 36 3

Exercise 7A — Expanding algebraic expressions 1 a 2x + 6 b 4x - 20 c 21 - 3x d -x - 3 e x2 + 2x f 2x2 - 8x g 15x2 - 6x h 10x - 15x2 i 8x2 + 2x j 4x3 - 6x2 k 6x3 - 3x2 l 15x3 + 20x2 2 2 2 a x - x - 12 b x - 2x - 3 c x2 - 5x - 14 d x2 - 6x + 5 e -x2 - x + 6 f x2 - 6x + 8 g 2x2 - 17x + 21 h 3x2 - x - 2 2 i 6x - 17x + 5 j 21 - 17x + 2x2 2 k 15 + 14x - 8x l 110 + 47x - 21x2 2 3 a 2x - 4x - 6 b 8x2 - 28x - 16 c -2x2 + 12x + 14 d 2x3 - 2x e 3x3 - 75x f 6x3 - 54x 3 2 g 2x - 12x + 18x h 5x3 - 30x2 + 40x i -6x3 - 6x2 + 120x 4 a x3 + 2x2 - x - 2 b x3 - 2x2 - 5x + 6 c x3 - 5x2 - x + 5 d x3 - 6x2 + 11x - 6 3 2 e 2x - 7x - 5x + 4 f 6x3 - 7x2 + 1 2 5 a x - x - 2 b -2x2 + 4x + 10 c 5x2 - 6x - 5 d 19x - 23 e -5x - 1 f -2x + 6

g x2 - 2x - 3 + 3x h 6 + 2 2 x − 3 3 x − 6 x 2 − 5 x 6 a A b C 7 B 8 a x2 - 2x + 1 b x2 + 4x + 4 c x2 + 10x + 25 d 16 + 8x + x2 e 49 - 14x + x2 f 144 - 24x + x2 2 g 9x - 6x + 1 h 144x2 - 72x + 9 2 i 25x + 20x + 4 j 4 - 12x + 9x2 k 25 - 40x + 16x2 l 1 - 10x + 25x2 9 a 2x2 - 12x + 18 b 4x2 - 56x + 196 2 c 3x + 6x + 3 d -4x2 - 12x - 9 e -49x2 + 14x - 1 f 8x2 - 24x + 18 g -12 + 108x - 243x2 h -45 + 330x - 605x2 2 i -16x - 16x - 4 10 a x2 - 49 b x2 - 81 c x2 - 25 2 2 d x - 1 e 4x - 9 f 9x2 - 1 g 49 - x2 h 64 - x2 i 9 - 4x2 11 a (x + 1)(x - 3) b x2 - 2x - 3 c 6  cm, 2  cm, 12  cm2 b 12 a (x + 1) m xm

(x + 2) m

c (x + 1)(x + 2) d x2 + 3x + 2 e 4  m2, 12  m2 13 a (x + 2)2 b 5(x + 2)2 c 5x2 + 20x + 20 3 2 d 500  cm e 100  cm , 100 tiles x 14 a x2 + x b 5x2 + 21x + 20 c i ii 1.50 m 2 818

Answers

15 a = 4, b = 4, c = -24, d = 0, e = 3 16 a (x + 2)(x – 1)(x – 3) b 56  cm3 c 0 d No; you can’t have a negative volume. e x > 3 f 18  cm3 g x = 6 h x3 - 2x2 – 5x + 6 Exercise 7B — Factorising expressions with three terms 1 a (x + 2)(x + 1) b (x + 3)(x + 1) c (x + 8)(x + 2) d (x + 4)2 e (x - 3)(x + 1) f (x - 4)(x + 1) g (x - 12)(x + 1) h (x - 6)(x + 2) i (x + 4)(x - 1) j (x + 5)(x - 1) k (x + 7)(x - 1) l (x + 5)(x - 2) m (x - 3)(x - 1) n (x - 4)(x - 5) o (x + 14)(x - 5) 2 a -2(x + 9)(x + 1) b -3(x + 2)(x + 1) c -(x + 2)(x + 1) d -(x + 10)(x + 1) e -(x + 2)(x + 5) f -(x + 12)(x + 1) g -(x + 3)(x + 4) h -(x + 2)(x + 6) i 2(x + 2)(x + 5) j 3(x + 1)(x + 10) k 5(x + 20)(x + 1) l 5(x + 4)(x + 5) 3 a (a - 7)(a + 1) b (t - 4)(t - 2) c (b + 4)(b + 1) d (m + 5)(m - 3) e (p - 16)(p + 3) f (c + 16)(c - 3) g (k + 19)(k + 3) h (s - 19)(s + 3) i (g + 8)(g - 9) j (v - 25)(v - 3) k (x + 16)(x - 2) l (x - 15)(x - 4) 4 a C b B

5 C 6   i d ii b iii b iv a v c vi d 7 a (2x + 1)(x + 2) b (2x - 1)(x - 1) c (4x + 3)(x - 5) d (2x - 1)(2x + 3) e (x - 7)(2x + 5) f (3x + 1)(x + 3) g (3x - 7)(2x - 1) h (4x - 7)(3x + 2) i (5x + 3)(2x - 3) j (4x - 1)(5x + 2) k (3x + 2)(4x - 1) l (3x - 1)(5x + 2) 8 a 2(x - 1)(2x + 3) b 3(3x + 1)(x - 7) c 12(2x + 1)(3x - 1) d -3(3x + 1)(2x - 1) e -30(2x + 1)(x - 3) f 3a(4x - 7)(2x + 5) g -2(4x - 3)(x - 2) h -(2x - 7)(5x + 2) i -(8x - 1)(3x - 4) j -2(3x - y)(2x + y) k -5(2x - 7y)(3x + 2y) l -12(5x + 3y)(10x + 7y) 9 a w2 + 5w - 6 b (w + 6)(w - 1) c (x + 5)(x - 2) 10 a x(x + 5) b x(x + 5) c (x - 1)2 d (x + 9)(x + 5) e (x - 15)(x - 6) f (x - 10)(x - 3) 11 (x - 0.5)(x + 1.5) 12 a (x - 5)(x + 1) b (x - 5) cm c x = 15 cm d 160  cm2 e 3000(x - 5)(x + 1)  cm2 or (3000x2 - 12 000x - 15 000)  cm2 13 a (2x + 3)(3x + 1) b P = 10x + 8 c x = 8 metres

14 a c e 15 a c

SA = 3x2 + 16x (3x + 4)(x + 4) 275  m2 Yellow = 3 cm ì 3 cm Black = 3 cm ì 6 cm White = 6 cm ì 6 cm

b Total area = 3x2 + 6x + 16 d l = 21  m; w = 7  m; d = 2  m f 294  m3 b Yellow = 0.36 m2 Black = 0.72 m2 White = 0.36 m2

Exercise 7C — Factorising expressions with two or four terms 1 a x(x + 3) b x(x - 4) c 3x(x - 2) d 4x(x + 4) e 3x(3x - 1) f 8x(1 - x) g 3x(4 - x) h 4x(2 - 3x) i x(8x - 11) 2 a (x - 2)(3x + 2) b (x + 3)(5 - 2x) c (x - 1)(x + 5) d (x + 1)(x - 1) e (x + 4)(x - 2) f (x - 3)(4 - x) 3 a (x + 1)(x - 1) b (x + 3)(x - 3) c (x + 5)(x - 5) d (x + 10)(x - 10) e ( y + k)( y - k) f (2x + 3y)(2x - 3y) g (4a + 7)(4a - 7) h (5p + 6q)(5p - 6q) i (1 + 10d)(1 - 10d) 4 a 4(x + 1)(x - 1) b 5(x + 4)(x - 4) c a(x + 3)(x - 3) d 2(b + 2d )(b - 2d ) e 100(x + 4)(x - 4) f 3a(x + 7)(x - 7) g 4p(x + 8)(x - 8) h 4(3x + 2)(3x - 2) i 3(6 + x)(6 - x) 5 a C b B c B

e (5p - 4t + 3t)(5p - 4t - 3t) f (6t − 1 + 5v )(6t − 1 − 5v ) 13 a E b A c D 14 B 15 a (x - 5)(x + 5) b (x - 5)  cm, (x + 5)  cm c 2  cm, 12  cm d 24  cm2 e 120  cm2 or 6 times bigger 16 a r metres b (r + 1) m c A1 = pr2  m2 d A2 = p(r + 1)2  m2 e A = p(r + 1)2 - pr2 = p(2r + 1) m2 f 34.56  m2 17 a Annie = (x + 3)(x + 2) m2 Bronwyn = 5(x + 2) m2 b (x + 3)(x + 2) – 5(x + 2) c (x + 2)(x - 2) = x2 – 4 d Width = 5  m e Annie has 30  cm2 and Bronwyn has 25  cm2.

6 C 7 a (x + 11)(x - 11)

b (x + 7)(x - 7)

Exercise 7D — Factorising by completing the square 1 a (x + 5)2 - 25 b (x + 3)2 - 9 c (x - 2)2 - 4 d (x + 8)2 - 64 e (x - 10)2 - 100 f (x + 4)2 - 16 g (x - 7)2 - 49 h (x + 25)2 - 625 i (x - 1)2 - 1

c (x + 15)(x - 15)

d (2x + 13)(2x - 13)

2 a (x - 2 + 11)(x - 2 - 11)

e (3x + 19)(3x - 19)

f 3(x + 22 )(x - 22)

g 5(x + 3)(x - 3)

h 2(x + 2)(x - 2) b (x - 4)(x + 6) d (x - 1)(x + 7) f (10 - x)(x + 2) h (7 - x)(5x + 1) b (x + y)(2 + a) d (x + y)(4 + z) f (n - 7)(m + 1) h 7(m - 3)(n + 5) j a(3 - b)(a + c) l m(m + n)(2 - n) b (m + 2)(n - 3) d (s + 3)(s - 4t) f (1 + 5z)(xy - z) b ( p - q)( p + q - 3) d (x + y)(7 + x - y) f (7g + 6h)(7g - 6h - 4)

c (x - 5 + 13)(x - 5 - 13) d (x + 3 + 19)(x + 3 - 19) e (x + 8 + 65)(x + 8 - 65) f (x - 7 + 6)(x - 7 - 6) g (x + 4 + 7)(x + 4 - 7) h (x - 2 + 17 )(x - 2 - 17) i (x - 6 + 11)(x - 6 - 11) 3 a (x - 1 +

5 )(x 2

b (x - 3 +

21 )(x 2

-3-

21 ) 2

c (x + 1 +

21 )(x 2

+1-

21 ) 2

d (x + 3 +

13 )(x 2

+3-

13 ) 2

e (x + 5 +

17 )(x 2

+5-

17 ) 2

f (x + 5 +

33 )(x 2

+5-

33 ) 2

g (x - 7 +

53 )(x 2

-7-

53 ) 2

h (x - 9 +

29 )(x 2

-9-

29 ) 2

i (x - 1 +

13 )(x 2

-1-

13 ) 2

2

2

2

2

2 2

2 2

2

-12

2

2

2

2

2

2

2

2

5 ) 2

Answers 7A ➜ 7D

i 12(x + 3)(x - 3) 8 a (x - 3)(x + 1) c (x - 5)(x + 1) e (6 - x)(x + 8) g 8(x - 3) i (x - 22)(9x + 2) 9 a (x - 2y)(1 + a) c (x - y)(a + b) e (f - 2)(e + 3) g 3(2r - s)(t + u) i 2(8 - j)(4 + k) k x(5 + y)(x + 2) 10 a (y + 7)(x - 2) c (q + 5)( p - 3) e (b + d)(a2 - c) 11 a (a - b)(a + b + 4) c (m + n)(m - n + l) e (1 - 2q)(5p + 1 + 2q) 12 a (x + 7 + y)(x + 7 - y) b (x + 10 + y)(x + 10 - y) c (a - 11 + b)(a - 11 - b) d (3a + 2 + b)(3a + 2 - b)

b (x + 1 + 3)(x + 1 - 3)

Answers

819

4 a 2(x + 1 + 3)(x + 1 - 3) b 4(x - 1 + 6)(x - 1 - 6) c 5(x + 3 + 2 2)(x + 3 - 2 2) d 3(x - 2 + 17)(x - 2 - 17 ) e 5(x - 3 + 7)(x - 3 - 7) f 6(x + 2 + 5)(x + 2 - 5) g 3(x + 5 + 2 3)(x + 5 - 2 3) h 2(x - 2 + 11)(x - 2 - 11) i 6(x + 3 + 14)(x + 3 - 14) 5    i  d ii b   iv  a v c vii  d viii e 6 a B b E

iii c vi d

7 E 8 a = 0.55; b = 5.45 Exercise 7E — Mixed factorisation 1 3(x + 3) 2 (x + 2 + 3y)(x + 2 - 3y) 3 (x + 6)(x - 6) 4 (x + 7)(x - 7) 5 (5x + 1)(x - 2) 6 5(3x - 4y) 7 (c + e)(5 + d) 8 5(x + 4)(x - 4) 9 -(x + 5)(x + 1) 10 (x + 4)(x - 3)

11 (m + 1)(n + 1) 13 4x(4x - 1) 15 3(3 - y)(x + 2) 17 4(x2 + 2) 19 (x + 5)(x - 5) 21 (x + 5)(x + 1) 23 (x + 2)(x - 2) 25 (y + 1)( x - 1) 27 7(x + 2)(x - 2) 29 (2 + r)( p - s)

12 (x + 7)(x - 7) 14 5(x + 10)(x + 2) 16 (x - 4 + y)(x - 4 - y) 18 (g + h)(f + 2) 20 5(n + 1)(2m - 1) 22 (x + 1)(x - 11) 24 (a + b)(c - 5) 26 (3x + 2)(x + 1) 28 -4(x + 6)(x + 1) 30 3(x + 3)(x - 3)

31 (u + v)(t - 3) 32 (x + 11 )(x - 11) 33 (4x - 1)(3x - 1) 34 (x + 1)(x - 3) 35 (x + 6)(x - 2) 36 4(x - 1)(x + 4) 37 3(x + 2)(x + 8) 38 (3 + x)(7 - x) 39 4(3 - x + 2y)(3 - x - 2y) 40 3(y + x) (y - x) 41 4(x + 2) 42 (3x - 4y)(x - 2y) 43 (x + 7)(x + 4) 44 (x + 2)(x - 5) 45 2(2x + 3)(x + 3) ( x + 5)( x − 2) ( x + 2)( x + 2) × 46 a ( x + 2)( x − 2) ( x − 4)( x + 2) b c 47 a d g j 820

( x + 5) ( x − 2) ( x + 2) ( x − 2)

×

( x + 2) ( x + 2) ( x − 4 ) ( x + 2)

x+5 x−4 x −1 x +1 b x−6 2x + 3 2x − 1 x+2 e x+4 x+5 4(b + 2) p( p + 7) h 5 ( p + 3)( p − 2) 5(d − 3 + 5e) 4(4 d + 3)

Answers

18 x ( x − 5) x−6 f x+3 5(m + 2 + n) i 2(2m − 5)

c

Chapter review Multiple choice 1 E 2 D 3 E 4 C 5 C 6 A 7 E 8 C 9 a 3x2 - 12x b -21x2 - 7x c x2 - 6x - 7 d 2x2 - 11x + 15 e 12x2 - 23x + 5 f 6x2 - 3x - 84 g 2x3 + 15x2 - 8x - 105 h 3x2 - 5x + 65 i 5x2 + 12x - 3 10 a x2 - 14x + 49 b 4 - 4x + x2 c 9x2 + 6x + 1 d -18x2 + 24x - 8 e -28x2 - 140x - 175 f -160x2 + 400x - 250 2 g x - 81 h 9x2 - 1 2 i 25 - 4x 11 a 2x(x - 4) b -4x(x - 3) c ax(3 - 2x) d (x + 1)(x + 2) e 2(2x - 5)(4 - x) f (x - 4)(x + 1) 12 a (x + 4)(x - 4) b (x + 5)(x - 5) c 2(x + 6)(x - 6) d 3(x + 3y)(x - 3y) e 4a(x + 2y)(x - 2y) f (x - 1)(x - 7) 13 a (x - y)(a + b) b (x + y)(7 + a) c (x + 2)( y + 5) d (1 + 2q)(mn - q) e (5r + 1)( pq - r) f (v - 1)(u + 9) g (a - b)(a + b + 5) h (d - 2c)(d + 2c - 3) i (1 + m)(3 - m) 14 a (2x + 3 + y)(2x + 3 - y) b (7a - 2 + 2b)(7a - 2 - 2b) c (8s - 1 + 3t )(8s - 1 - 3t ) 15 a (x + 9)(x + 1) b (x - 9)(x - 2) c (x - 7)(x + 3) d (x + 7)(x - 4) e -(x - 3)2 f 3(x + 13)(x - 2) g -2(x - 5)(x + 1) h -3(x - 6)(x - 2) i (4x - 1)(2x + 1) j (3x - 1)(2x + 1) k 4(2x + 3)(x - 1) l 5(7x - 3)(3x + 1) m -2(3x - 5)(2x - 7) n -3(3x - 1)(5x + 2) o -30(2x + 3)(x + 3)

16 a (x + 3 + 2 2)(x + 3 - 2 2) b (x - 5 + 2 7)(x - 5 - 2 7) c (x + 2 + 6)(x + 2 - 6) d (x - 25 +

17 )(x 2

- 25 -

17 ) 2

e (x + 72 +

53 )(x 2

+ 72 -

53 ) 2

f 2(x + 92 + 85 )(x + 92 - 85 ) 2 2 17 a 3x(x - 4) b (x + 3 + 7 )(x + 3 - 7 ) c (2x + 5)(2x - 5) d (2x + 5)(x + 2) e (a + 2)(2x + 3) f -3(x - 2)(x + 3) ( x − 2)( x − 1) 18 a 2( x + 4) b 7 c 8 x ( x − 4) 5( x + 1) Problem solving 1 a (x + 2)2 c 32x2 + 128x + 128 2 a 4r c 4pr    2 e 4p  (2r + 1)

b 32(x + 2)2 d 32  768  cm3 b 2r + 2 d (4pr    2 + 8r + 4)p f 28p m2

3 a (x – 7)(x + 2) c 35

b x – 7 cm d 1036 cm2

g - 32 +

4 Division by zero in Step 5

i

CHAPTER 8

Quadratic equations Are you ready? 1 a x(x - 3) b 4x(x + 3) c 12x(3x - 1) 2 a -2 and 3 b -2 and -3 c 2 and -3 3 a 2 6 b 10 2 c 36 2 4 a 0 b -16 c -38 5 a x = -2 b x = 3 c x = 1.5 Exercise 8A — Solving quadratic equations 1 a -7, 9 b -2, 3 c 2, 3 d 0, 3 e 0, 1 f -5, 0 g 0, 3 h -2, 0 i - 12 , 12

j -1.2, -0.5 2 a d g

l - 2 , 3

k 0.1, 0.75

1 , 1 2 - 67 , 1 12 0, 12 , 3

b -2, e h

- 23

3 2 , 5 3 0, 12 , - 25

c

1 , 4

f

- 85, 23

i 0, -3, 25

d - 23 , 0

e 0, 1 12

f 0, 13

h -  33 , 0

i 0, 1 14

4 a -2, 2

b -5, 5

c -2, 2

d -7, 7

e -1 13 , 1 13

f -2 12 , 2 12

g - 23 , 23

h - 12 , 12

i - 15 , 15

j -4, 4

k − 5, 5

l - 

5 a -2, 3 d 3, 5 g 5 j -3, 7 6 B 7 C 8 a - 12 , 3

b e h k

b 23, -1

c -2, 15

d 13, 1 12

e - 143 , 1

f 14 , 13

g -1 13 , 2 12

h -1 43 , -113

i - 25 ,

j 1 12, 2 23

k - 25 , 16

l 3, 4

-4, -2 1 -2, 5 -5, 6

c f i l

11 , 3

11 3

-1, 7 -1, 4 2, 6 3, 4

1 2

d 4 + 2 3 , 4 − 2 3

e 5 + 2 6 , 5 − 2 6

f 1 + 3 , 1 − 3

g −1 + 6 , −1 − 6

h −2 + 10 , − 2 − 10

i −2 + 15 , − 2 − 15 5 3 , 2 2

c

7 2

+

33 7 , 2 2

e 11 + 2

-

5 2

33 2

-

117 11 , 2 2

b - 25 +



-



117 2

d

1 2

+

f - 12 +

29 , 2 21 1 , 2 2 5 , 2

- 25 -

29 2

21 2

- 12 -

+

37 5 ,2 2

-

37 2

65 2

11 a -3, 1 b -4.24, 0.24 c -1, 3 d -0.73, 2.73 e 0.38, 2.62 f -0.30, 3.30 g -1.19, 4.19 h -2.30, 1.30 i -2.22, 0.22 12 No real solutions — when we complete the square we get the sum of two squares, not the difference of two squares and we cannot factorise the expression. 13 8 and 9 or -8 and -9 14 6 and 8, -6 and -8 15 9 or -10 16 2 or -2 23 17 8 or -10 12

18 6 seconds 19 a l = 2x b m 45

c

x cm

c + (2x)2 = 452, 5x2 = 2025 d Length 40 cm, width 20 cm 20 8 m, 6 m 21 a - 73 b x = 0 22 a

c x = ê  11 3

5 2

b (2 + x) m, (4 + x) m c (2 + x)(4 + x) = 24 d x = 2, 4 m wide, 6 m long 23 a (l - 4) cm b l - 8, l - 4 c (l - 8)(l - 4) = 620 d 31 cm e 836 cm2 24 a CAnnabel(28) = $364 800, CBetty(28) = $422 400 b 10 knots c Speed can only be a positive quantity, so the negative solution is not valid. Exercise 8B — The quadratic formula 1 a a = 3, b = -4, c = 1 b a = 7, b = -12, c = 2 c a = 8, b = -1, c = -3 d a = 1, b = -5, c = 7 e a = 5, b = -5, c = -1 f a = 4, b = -9, c = -3 g a = 12, b = -29, c = 103 h a = 43, b = -81, c = -24 i a = 6, b = -15, c = 1 −3 ± 13 5 ± 17 −5 ± 21 2 a b c 2 2 2 7 ± 45 d 2 ± 13 e −1 ± 2 3 f 2 9 ± 73 g h 3 ± 2 3 i −4 ± 31 2 1 ± 21 5 ± 33 j k l −1 ± 2 2 2 2 3 a -0.54, 1.87 b -1.20, 1.45 c -4.11, 0.61 d -0.61, 0.47 e 0.14, 1.46 f 0.16, 6.34 g -1.23, 1.90 h -1.00, 1.14 i -0.83, 0.91 j -0.64, 1.31 k -0.35, 0.26 l -1.45, 1.20 m 0.08, 5.92 n -0.68, 0.88 o -0.33, 2.00 4 C Answers

Answers 7E ➜ 8B

c −3 + 10 , − 3 − 10

+

5 2

4m

b −1 + 3 , −1 − 3

3 2

-

h

2m

9 a 2 + 2 , 2 − 2

10 a

65 9 ,2 2

37 2

2x cm

c 0, 7



3

- 2 -

x2

b -5, 0

7 2

+

7

3 a 0, 2 g 0,

9 2

37 , 2

821

6 B 5 C 7 C 8 a 0.5, 3 b 0, 5 c -1, 3 d 0.382, 2.618 e 0.298, 6.702 f 2, 4 g No real solution h -1, 8 i -4.162, 2.162 j -2, 1 k -7, 1.5 l No real solution m 2, 7 n - 12 , 13 o No real solution 9 a 2p r2 + 14p r - 231 = 0 b 3.5 cm c 154 cm2 10 a x(x + 30) b x(x + 30) = 1500 c 265 mm 11 a Pool A: 3 23 m by 6 23 m; Pool B: 3 13 m by 7 13 m b The area of each is 24 49 m2. 12 25 m, 60 m Exercise 8C — Solving quadratic equations by inspecting graphs 1 a x = -2, x = 3 b x = 1, x = 10 c x = -5, x = 5 d x = 2 e x = -1, x = 4 f x ö -1.4, x ö 4.4 g x = -25, x = 10 h x = 0 i x ö -2.3, x ö 1.3 j x ö -1.5, x = 1 2 a–j Confirm by substitution of above values into quadratic equations. 3 150 m 4 7 m 5 b x = -0.72, 1.39 c The answer for part b are the x-coordinates of the intersection of the quadratic in part a. Exercise 8D — Finding solutions to quadratic equations by interpolation and using the discriminant 1 a -4.5, 1.5 b -0.87, 1.5 c -4.6, 1.1 2 a -11 b 0 c 169 d 0 e 37 f 0 g 52 h -7 i -4 j 109 k 129 l 1 3 a No real solutions b 1 rational solution c 2 rational solutions d 1 rational solution e 2 irrational solutions f 1 rational solution g 2 irrational solutions h No real solutions i No real solutions j 2 irrational solutions k 2 irrational solutions l 2 rational solutions

4 a No real solutions

b 2 12

c -11, 2

d - 23

−3 ± 37 ö -4.541, 1.541 e 2 1 ± 13 1 ö -0.869, 1.535 f 5 g 3 h No real solutions i No real solutions

j k 822

−5 ± 109 ö -2.573, 0.907 6 −7 ± 129 ö -4.589, 1.089      l   5, 6 4

Answers

5 a b c 6 a

a = 3, b = 2, c = 7 -80 No real solutions a = -6, b = 1, c = 3

b 73 1 ± 73 d 12 8 C 10 k = -1

c 2 real solutions

7 A 9 C 11 m = 1, 8 12 n > - 49 13 p2 can only give a positive number which, when added to 24, is always a positive solution. 14 a 0.4 m b 0.28 m c 2.20 m d 2.5 m e i  Yes   ii  No f 1.25 m Exercise 8E — Solving a quadratic equation and a linear equation simultaneously 1 (-4, 1) and (1, 6) 2 a (-4, 12) and (-3, 10) b (-2, -5) and (6, 35) c (3, -2) and (5, 0) 3 (2, 4) 4 D = -8 5 a (-2, 4) and (5, 18) b (-2, -9) and (-1, -8) c (4, 10) d (-7, 18) and (-1, 6) e (1, 1) and (3, 9) f (1, 4) and (10, 22) 6 (-3, 1) and (-2, 1) 7 a (1, -5) b No, but the straight line is vertical and intersects at one point only. 8 (-2, 0) and (2, 0) y 9 a (3, 32)

(2, 21)

5 -3 -2

-1

x

1 — 11

b

y (2, 8)

-6

(1, 0) -6

c

-8

y

14 -7

-2 (-3, -4)

5 x

x

d

y

j

y

10

(-2, 28)

6

2

3– 2

5

x

10 6

k

6 — 11

2

e

y

x

5

21

y

(-2, 9) 3

-1 -6

-3

x

6

l

f

(5, 16)

11

y

y

7 x 54 — 13

x

6 8 (3, -15)

(3, 70) -48

-54

(-2, -80)

40

m

28

y 16

(-4, 0) -7

(-1, 7)

x

-4

g

-2 — - 16 9

y

(5, 14)

-9

4

6

(-4, -20)

n

x

-36

y (1, 20)

(-8, 20)

12

-61

h

y

-17–4

x

-4 -3

o

-2

-17

x

y

8 x -16

9

(1, -21) 9– 4

y

-5 (9, 39)

p

1

Answers 8C ➜ 8E

i

x

y

(6, 16) -4 (-3, -9)

- 3–4

3 6

x 4

-24

2 4 (2, 0)

x

Answers

823

10 1.322 km and 2.553 km x2 11 y = , (2, 2) and (-2, 2) 2

Exercise 9A — Plotting parabolas

1

4

2 B 4 D

g 2, 1

b e h b

-6, -1 5, -2 5, 6 -2, -1

e

1 - 2 , 4 5 5 , 3 2

h

2

c f i c

-8, -3 4, -7 7, -5 1 , -3 2

f

2 - 3,

2

i -7,

1 2 1 4

8 a -4 ê 17 b -1 ê 6 c -1, 9 4 10 a -0.651, 1.151 b -0.760, 0.188 c 0.441, -0.566 11 a -0.571, 0.682 b -0.216, 3.836 c -0.632, 0.632 12 -3, 7 13 -3, 1 14 a 2 irrational solutions b 2 rational solutions c No real solutions 15 a (-8, 22) and (2, 2) b (5, 10) c No solution Problem solving 1 -8 and 7 2 Length = 6 m, width = 3 m 3 a 2p r(r + 10) = 245 b 3.0 cm c 188 cm2 4 - 25 8 5 k > 9 and k < 1 6 a 6 m b 6 m 7 24 8 a y = 2x2 - 5x - 2 b No parabola is possible. The points are on the same straight line. 9 12( 5 + 2) cm

Chapter 9

2 a

824

2

Answers

2 3

y

1

y = –4 x2

1

-3 -2-1 0 1 2 3 x





-3 -2-1 0 1 2 3 x

x = 0, (0, 0) x = 0, (0, 0) 3 Placing a number greater than 1 in front of x2 makes the graph thinner. Placing a number greater than 0 but less than 1 in front of x2 makes the graph wider.

4 a

b

y 10 8 6

4 2 (0, 3)

x

-3-2-10 1 2 3

x = 0, (0, 1), 1 y

y = x2 - 3



-3-2-10 1 2 3

y y = x2 - 1 8

6

6

4

4

(0, -1)

x

-3-2-10 1 2 3 -2

x

x = 0, (0, 3), 3

d

2



y = x2 + 3

6

2

c

y

8

4



12 10

y = x2 + 1

2

-3-2-1 1 2 3 x -2

x = 0, (0, -3), -3 x = 0, (0, -1), -1 5 Adding a number raises the graph of y = x2 vertically that number of units. Subtracting a number lowers the graph of y = x2 vertically that number of units. b

y 16

+ 1)2

12

c -38 c x = 32 or x = 1.5

3 7  (x + + 1 b  x −  + c 2(x - 1)2 + 4  2 4 2± 2 −1 ± 5 b 2 2 1 ± 7 −1 ∓ 7 = −3 3 x = -2 or x = -3 b x = 1 or x = -2 x = 2 or x = -2 1 1 b x = 2 or x = - 3 x = - 2 or x = -2

b

2

(-5, 16) y = (x

b -16 b x = 3

x

y = 3x2

20

1)2

c x = 32 or x =

y

6 a

Are you ready? 1 a 0 2 a x = -2

5 a c 6 a

(0, 0)

-4 -3–2-1 1 2 3 -2

30 25 20 15 10 5

Functions

c

x = 0, (0, 0)

6

d 2, -7

4 a

y = x2

8

Chapter review Fluency 1 B 3 A 5 (3x + 4) m 6 a -5, -3 d 2, -6 g 3, 1 7 a -2, -6

3 a

y 10

8 4

c

(1, 4)

-6-5-4-3-2-1 01 2 x



x = -1, (-1, 0), 1 y 2 10 y = (x - 2) 8 6

0 12345 x

x = 2, (2, 0), 4

-6 -4 -2 0

2 x

x = -2, (-2, 0), 4

d

4 2



y y = (x + 2)2 16 12 8 4



y 2 10 y = (x - 1) 8 6 4 2 0 12345 x

x = 1, (1, 0), 1

7 Adding a number moves the graph of y = x2 horizontally to the left by that number of units. Subtracting a number moves the graph of y = x2 horizontally to the right by that number of units. 8 a

b

y 1



-3-2-1 01 2 3 4 x -2 -3 -4 -5 -6 -7 -8 y = -x2 + 1



x = 0, (0, 1), 1

c

d

y

y

-8

y = -(x + 2)2



x = -2, (-2, 0), -4

y y = (x - 5)2 + 1

y = -3(x - 1)2 + 2

x = -2, (-2, -9), min, -5

y y = x2 + 4x - 5 10 5 x

-6 -4 -2 0 -5 -10

f

x = -1, (-1, 16), max, 15

y 20 15 10

-10

5

-12 y = -x2 - 3

-6 -4 -2 0 -5

x = 0, (0, -3), -3

g

4 x y = -x2 - 2x + 15

2

-10



x = -1, (-1, 27), max, 24

y = -3x2 - 6x + 24 y

25

20 15 10 5 -6 -4 -2 -5 0 2 x -10 -15 -20 -25

h

26

x

4

-8

9 The negative sign inverts the graph of y = x2. The graphs with the same turning points are: y = x2 + 1 and y = -x2 + 1; y = (x - 1)2 and y = -(x - 1)2; y = (x + 2) and y = -(x + 2)2; y = x2 - 3 and y = -x2 - 3. They differ in that the first graph is upright while the second graph is inverted. 10 a

-25

2

-6

-6



e

x = 1, (1, 0), -1

-4

-4

-10 -15 -20

y = -(x - 1)2

0 12345 x -2

-6 -4 -2 0 1 x -2

-2 0 -5

y 0 -2-1 1 2 3 4 5 x -2 -3 -4 -5 -6 -7 -8 -9

x = 1, (1, 2), max, -1

y 5

d

x = 2, (2, 1) min, 5

y

20 y = (x - 2)2 + 1 16 12

b

1 0 12 3 4 5 6

x = 5, (5, 1), min, 26 y = 2(x + 2)2 - 3

y



x

x = -2, (-2, -3), min, 5 y 4 3 2 1 0 123456 -2 -3 -4 -5

4

x

14 C

15 A h 18 16 14 12 10 8 6 4 2

x

x = 3, (3, 4), max, -5

2

11 a I f the x2 term is positive, the parabola has a minimum turning point. If the x2 term is negative, the parabola has a maximum turning point. b If the equation is of the form y = a(x - b)2 + c, the turning point has coordinates (b, c). c The equation of the axis of symmetry can be found from the x-coordinate of the turning point. That is, x = b. 12 C 13 B 16 a

y = -(x - 3)2 + 4

-2 0

b

h = -(t - 4)2 + 16

0 1234567 8

i  16  m

t

Answers 9A ➜ 9A

c

-8 -6 -4 -2 0 -4

4



16 12 8 4



8

x

 ii   8  s Answers

825

17 a



c It is possible to have 0, 1 or 2 points of intersection. y

h 18 16 14 12 10 8 6 4 2

0

t

1 2 3

x

0

b   i  18  m

 ii  Yes, by 3  m

iii  1.5  s iv  3  s 18 a



y

y

x

0 0

x



y

y

x

0 0



x

y

0

x

Exercise 9B — Sketching parabolas using the basic graph of y = x 2 1 a Narrower, TP (0, 0) b Wider, (0, 0) c Narrower, TP (0, 0) d Narrower, TP (0, 0) e Wider, TP (0, 0) f Wider, TP (0, 0) g Narrower, TP (0, 0) h Narrower, TP (0, 0) 2 a Vertical 3 up, TP (0, 3) b Vertical 1 down, TP (0, -1) c Vertical 7 down, TP (0, -7) 1

1

d Vertical 4 up, TP (0, 4 ) 1



e Vertical 2 down, TP (0, - 12 ) f Vertical 0.14 down, TP (0, -0.14) g Vertical 2.37 up, TP (0, 2.37)

y

0

x

h Vertical 3 up, TP (0, 3) 3 a Horizontal 1 right, (1, 0) b Horizontal 2 right, (2, 0) c Horizontal 10 left, (-10, 0) d Horizontal 4 left, (-4, 0) 1

1

1

1

e Horizontal 2 right, ( 2 , 0) b An infinite number of points of intersection occur when the two equations represent the same parabola, with the effect that the two parabolas superimpose. For example y = x2 + 4x + 3 and 2y = 2x2 + 8x + 6. 826

Answers

f Horizontal 5 left, (- 5 , 0) g Horizontal 0.25 left, (-0.25, 0) h Horizontal 3 left, (− 3, 0) 4 a (0, 1), max b (0, -3), min c (-2, 0), max d (0, 0), min

e (0, 4), max f (0, 0), max g (5, 0), min h (0, 1) min 5 a Narrower, min b Narrower, max c Wider, min d Wider, max e Narrower, max f Wider, min g Narrower, min h Wider, max 6 a   i  Horizontal translation 1 left   ii  (-1, 0) iii  y y = (x + 1)2

f   i  Horizontal translation 4 right   ii  (4, 0) iii  y 2 y = (x - 4)

y = x2

0



x

(4, 0)

g   i  Reflected, wider (dilation)   ii  (0, 0) iii y 2

y = x2

y=x



x

(-1, 0) 0

b   i  Reflected, narrower (dilation)   ii  (0, 0) iii y

x y = - 2– x2

(0, 0)

5

h   i  Narrower (dilation)   ii  (0, 0) iii y y = 5x2

2

y=x

y = x2

x

0 y = -3x2

x

(0, 0)





c   i  Vertical translation 1 up   ii  (0, 1) iii  y y = x2 + 1

i   i  Reflected, vertical translation 2 up   ii  (0, 2) iii y y = x2 (0, 2)

y = x2

x

0

(0, 1) x

0



d   i  Wider (dilation)   ii  (0, 0) iii  y y = x2

y = -x + 2 j   i  Reflected, horizontal translation 6 right   ii  (6, 0) iii y y = x2 2

(6, 0) 0

(0, 0)



x

e   i  Vertical translation 3 down   ii  (0, -3) iii  y y = x2

y = -(x - 6) k   i  Reflected, vertical translation 4 down   ii  (0, -4) iii y y = x2 2

0

y = x2 - 3

x y = -x2 - 4

x

0



x

Answers 9B ➜ 9B

y = 1–3 x2

(0, -3)

Answers

827

l   i  Reflected, horizontal translation 1 left   ii  (-1, 0) iii y y = x2

p   i Narrower (dilation), reflected, horizontal translation 1 right, vertical translation 3 down 2   ii  (1, - 32 ) iii

(-1, 0)

y = x2

y

x

0

y = -(x + 1)2



0 (1, - 3 ) 2

m   i Narrower (dilation), horizontal translation 1 left, vertical translation 4 down   ii  (-1, -4) iii y

x y = - 74 (x - 1)2 - 32

7 a 10  cm b 5  cm c 5  cm d y = (x - 5)2

y = x2

x

0

Exercise 9C — Sketching turning point form 1 a (1, 2), min c (-1, 1), min e (5, 3), max

2

y = 2(x + 1) - 4



(-1, -4)

1

n   i Wider (dilation), horizontal translation 3 right, vertical translation 2 up   ii  (3, 2) iii y

(3, 2) 1

y = –2 (x - 3)2 + 2

y = x2

x

0

o   i Wider (dilation), reflected, horizontal translation 1 2 left, vertical translation 4 up   ii  (-2, 14 ) iii

y

y = x2

3

g (- 2 , - 4 ), min i (-0.3, -0.4), min 2 a i  (-3, -5) ii Min b i  (1, 1) ii Max c i  (-2, -4) ii Max d i  (3, 2) ii Min e i  (-1, 7) ii Max 1 1 f i  (- 5, - 2 ) ii Min 3   i b  y = -(x - 2)2 + 3 iii f  y = (x + 1)2 - 3 v c  y = x2 - 1 4 a A b C d C e B 5 a i  -3 b i  12 c i  -18 d i  -5 e i  4 f i  4 6 a   i  (4, 2) iv  18 vi  y

parabolas in



b (-2, -1), min d (2, 3), max f (-2, -6), min 1 2 h ( 3 , 3 ), min iii Narrower iii Same iii Narrower iii Wider iii Wider iii Wider

ii e  y = -x2 + 1 iv d  y = -(x + 2)2 + 3 vi a  y = (x - 1)2 - 3 c B

ii -3, 1 ii 2 ii No x-intercepts ii -1, 5 ii No x-intercepts ii -3 - 5, -3 + 5 (approx. -5.24, -0.76) ii Min iii Same width v No x-intercepts

y = (x - 4)2 + 2

18 (-2, 14 ) 0

x (4, 2)

  828

Answers

y = – 13 (x + 2)2 + 4

0 12 34

b   i  (3, -4) iv  5

ii Min v 1, 5

x

iii Same width

vi 

y

h   i  (1, 3)   ii  Min iii  Narrower iv  5   v  No x-intercepts

y = (x - 3)2 - 4

5 x

0 1 2 3 45 -4

(-1, 2)

i   i  (-2, 1) ii Max iii Narrower iv  -11 1 1   v  -2 (approx. -2.58, -1.42) , -2 + 3 3 vi  y x 0 1 -2 + — 3

1 -2 - — 3

x

iii Same width

  v  -5 - 3, -5 + 3 (approx. -6.73, -3.27) vi  y = (x + 5)2 - 3 y 22

 

-11 y = -3(x + 2)2 + 1

7 a 2(x - 43 )2 b x = 43 ê

73 8

=0

73 4

c ( 43 , - 73 ), minimum 8

-5 + 3

8 a y = -23(x + 4)2 + 6 b (-7, 0)

0x

(-5, -3)

ii Max

iii Same width

  v  1 - 2, 1 + 2 (approx. -0.41, 2.41) vi  y 1+ 2 x

1 2

y = -(x - 1) + 2   f   i  (-2, -3) ii Max iii Same width iv  -7 v No x-intercepts vi  y -2 (-2, -3)

9 a

b c d e

p ($) 1.9 1.4 1.0

0

(1, 2)

-1 0

x

0

(-2, 1)

0

1- 2 2 1

y = 2(x - 1)2 + 3

(1, 3)

 

3 2 1

-1   d   i  (-5, -3) ii Min iv  22

  -5 - 3 e   i  (1, 2) iv  1

y 5

(3, -4)

  c   i  (-1, 2) ii Min iii Same width iv  3 v No x-intercepts vi  y = (x + 1)2 + 2 y

0

x

-3

t (Hours

Exercise 9D — Sketching parabolas of the form y = ax 2 + bx + c 1 a y = (x + 2)2 - 6, (-2, -6) b y = (x + 6)2 - 40, (-6, -40) c y = (x - 4)2 - 10, (4, -10) d y = (x - 1)2 + 11, (1, 11) 5 5 3 3 e y = (x + 2 )2 - 4 , (- 2 , - 4 ) 1

9

x

7

g y = (x + 2 )2 -

41 , 4

7

41

(- 2 , -  4 )

h y = 2(x + 1)2 + 6, (-1, 6) i y = 3(x - 2)2 - 6, (2, -6) 2 a y = (x + 1)2 - 6, x-intercepts are -1 ê 6 (ö -3.4, 1.4)

y = x2 + 2x - 5 y

-1 - 6 -11

y = -(x + 3)2 - 2

-1 0

x -1 + 6

-5 (-1, -6) -6

Answers

Answers 9C ➜ 9D

2

(-3, -2)

5

after 12 pm.) 10 a 0.5 m b (15 + 4 15) m c Maximum height is 8 metres when horizontal distance is 15 metres.

9

  y = -(x + 2) - 3 g   i  (-3, -2) ii Max iii Same width iv  -11 v No x-intercepts vi  y -3 -2 -1 0 -2

3

$1.90 $1 3  pm $1.40

f y = (x + 12 )2 - 4 , (- 2 , - 4 )

-7

 

vi 

829

b y = (x - 2)2 + 3, no x-intercepts

h y = -5(x - 1)2 - 30, no x-intercepts y 0

y y = x2 - 4x + 7

7 x

0



(1, -30)

-30 -35

(2, 3)

y = -5x2 + 10x - 35



c y = (x + - 12, x-intercepts are −6 ± 48 = -3 ê 2 3 (ö -6.5, 0.5) 2 3)2

x

1

1

3

i y = -7(x + 2 )2 + 50 4 , x-intercepts are −1 ± 29 (ö -3.2, 2.2) 2

y

y (-1– , 50 3–) 49 2

4

-3 + 2√3 0

-3 - 2√3

x

–3

-1 - √29 2

(–3, –12)

1 5 d y = (x - 2 )2 - 5 4 , x-intercepts

y

5 ± 21 are (ö 0.2, 4.8) 2 1 0

y = x2 -5x + 1 5 - √21 2

(2 1–2, -5 1–4)

1

5

e y = -(x + 2 )2 + 7 4 , x-intercepts are −5 ± 29 (ö -5.2, 0.2) 2 -5 - √29 2

(-21–2 , 71–4 )

2

0

-4

(- 1–, -12 1–)



2

c

3

y

y

y = x2 - 12x + 32

32

x

0

-12

4



4

y = x2 - 8x - 9

-1 0

d

x

y (-3, 1) x

-4 -2 0

x

9

8 (6, -4)

(0, -9) -5 + √29 2 x

1

1 0

-8

(4, -25)



e

y = -x2 - 6x - 8

f

(-3, 36) y

y = -x2 -5x + 1 1

b

y = x2 + x - 12 y

5 + √21 2 x

71–4

x

0 y = -7x2 - 7x + 49

y

-2 –2



1

2 –2

3 a

-1 + √29 2

– 1–

y (1, 36) 35

27 3

f y = -(x - 2 )2 - 2 4 , no x-intercepts y 0

x

1– 2



( 1–2, -2 3–4)

-3

g

y = x2 + 4x - 5

3

1

g y = 3(x + 2 )2 - 12 4 , x-intercepts are −1 ± 17 (ö -2.6, 1.6) 2 y y = 3x2 + 3x - 12 - 1–

830

Answers

0



0 -5 y = -x2 + 2x + 35

1

x

4 a

(-2, -9) -1–

b

y

2

x

-9

0

y = 2x2 - 17x - 9 9 x

y 14

y = 3x2 - 23x + 14

02 – 3

-12 (- 1–, -12 3–) 2

4

7 x

-5

2

-3 - √17 2



-1 + √17 2

x

y 0

-5

y = -x2 + x - 3



0 3 -9 y = -x2 - 6x + 27



(4 1–4, -451–8 )



7 x 1) (3 5–6, -30— 12

c

y = 5x2 + 27x + 10

d

y

y = 6x2 + 7x - 3

y 14

10 a

y

12

10 x - 2–5 0

-5

e

2

9) 7 , -26— (-2 — 20 10

y

0 1 –

- 3–

f

(1 3– , 10 1–8 ) 4

10

x

3

-3

8

1) 7 , -5 — (-— 24

12

6

y

(2 3– , 36 1–8 ) 4

4 2

21 4 - 1– 0

y = -2x2 + 7x + 4

g

y

h

5 — 1 (— 12, 724)

- 3– 0 7 x 2 y = -2x2 + 11x + 21 y

31 25 — (1— 36, 4872)

6

0

- 2– 3

i

x

3 – 2

y = -6x2 + 5x + 6

y



x 0 2– 7– 2 -14 9 2 y = -18x + 67x - 14

y = 2x2 - 7x + 8

5x

0

-5

x

4

2

-2 y = x2 + 2x + 5 y=

x2 +

4x + 5

y = x2 - 2x + 5 y = x2 - 4x + 5

b If p < 0, the turning point is on the right side of the y-axis. If p > 0 the turning point is on the left side of the y-axis. As the magnitude of p increases the turning point moves away from the y-axis. All graphs have the same y-intercept (0, 5). 11 a b h = 0 h (25, 2500) c 2500  m 2500 d 25  s after launching e 50  s

8 50 t

0

(13–4 , 17–8 )



x

0

f1(x) = x2 + 6·x + 5

200

0

13 a

h 17



-1.72 0

h = -4.9t2 + 1.5t + 17

b 2  s c 0.15  s d 17.11  m

(10, 200)

20 x

t

2.02

14 a A = 2x(150 - x)  m2 b A 11 250

f2(x) = 2·x + 1

b 2x + y = 40  m d A = 2x(20 - x)  m2 f y

Answers 9C ➜ 9D

5 a B b C 6 a iv b vii c vi d iii e i f viii g ii h v 7 a A: y = 2(x - 3)2 + 4, y = 2x2 - 12x + 22; B: y = -3(x + 1)2 - 1, y = -3x2 - 6x - 4 b Translated 4 units to the left and 5 units down, reflected in the x-axis, and dilated by 32 in the y-direction. 8 (-0.32, 3.18) and (-4.68, -1.18) 9 a (-2, -3) y b

12 a A = xy  m2 c y = (40 - 2x)  m e (10, 200) g Maximum area is 200  m2, paddock is 10  m wide and 20  m long.

(75, 11 250)

1 0

1

(-2, -3)

x

0

150 x

c 11  250  m2, 75  m and 150  m −5 45 15 a a = 64 , b = 16 , c = −85 b (18, 20) 16 −5 c y = 64 (x - 18)2 + 20

Answers

831

y

6

Exercise 9E — Exponential functions and their graphs y 1 a

8

y = 10x

100

y = 3 ì 2x y ì 2x

10 6 4

y = 1–5 ì 2x x 1 2 3

2 -3 -2 -1 0

10

(0, 1)

2 x

-1

-4

-3

-2

-1

0

1

1 1 10 000 10 00

1 10 0

1 10

1

10

x y

0

-2

1

2

3

4



100 1000 10 000

b Providing a realistic scale is difficult. y y = 4x 2 a 100 80 60 40 (0, 1)

20

(1, 4)

-4 -3 -2 -1 0 1 2 3 4 x

y

b

x

2x

3 ì 2x

1 5

-3

0.125

0.375

0.025

-2

0.25

0.75

0.05

-1

0.5

1.5

0.1

 0

1

 3

0.2

 1

2

 6

0.4

 2

4

12

0.8

 3

8

24

1.6

7 The coefficient, k, affects the steepness of the graph: the larger the value of k, the steeper the graph. 8 y -x

y = 5x

100 80

y=2

60

10 8

40 (0, 1)

20

6

(1, 5)

4

-4 -3 -2 -1 0 1 2 3 4 x

c

y

ì 2x

(0, 1)

2 -3 -2 -1 0

y = 6x

x

1 2 3

100 80 60

9

40 (0, 1) 20

y-intercept at (0, 1) Equation of horizontal asymptote is y = 0.

(1, 6)

y

y = 3-x

y = 3x

10

-4 -3 -2 -1 0 1 2 3 4 x

8 6

3

y = 4-x

y = 3-x

y = 4x

y 100

4

y = 3x

(0, 1)

2

80

x

0

60

y = 2-x

40 20 -4 -3 -2 -1 0

y = 2x

(0, 1) 1

2

3

4 x

10 The negative index reflects the graph in the y-axis. 11 a y 1 x y = ( –2 )

10 8

4 Increasing the value of a increases the steepness of the graph where x is positive and flattens the graph where x is negative. y 5 a x 60

6 4 2 -3 -2 -1 0

y=2ì3

x

1 2 3

50 40 30 20 (0, 2)

10

832

Answers

-3

-2

-1

0

1

2

3

y

8

4

2

1

0.5

0.25

0.125

x

-4 -3 -2 -1 0 1 2 3 4 x

b 2

x

c y = 0

 1 b   = (2−1 ) x = 2− x  2

12

y y = (1.8)x

y 10

b

y = (1.5)x

y = 2x

8 y = (1.2)x

6

1

13 a

4

x

0

y

y = 2x - 1

(0, 1) y = 10 ì (1.3)x

-6

-4

2 (0, 0) 2

0

-2

4

6

8

10 x

-2 10

c

y

(0, 16)

15 x

0

y = 2x + 4

b 10 c y = 0 14 a, b, c 

10

y = 2x

5

(0, 1)

-10 -5 0 -5 y

(0, 1) 2

6

1 -6 -5 -4 -3 -2 -1

2

-12 -10 -8 -6 -4 -2 0 -2 y = 3x - 3 -4

2

4

x

6

b

y = 4x - 3

n V

4

b

2

-8 -6 -4 -2 0 -2

2

4

6

8

A

2

3

x

A = 1000 ì (1.1)n

1

4

5

6 n

x

40 000

V = 40 000 ì (0.85)n 1

2

3

4

5 n

c As n increases, the value of the car decreases. d $17 748 25 a 190 s (bacteria A); 110 s (bacteria B) b Bacteria A starts at 20 000; bacteria B starts at 260 000. c Bacteria A d 240 s 26 a 65 536 b 2.3 ì 1018 x c 2048 ì ( 2 ) d   i  18 days   ii  25 days   iii  38 days Answers

Answers 9E ➜ 9E

1 2 3 4 5x

3

V

0

9 8 7 6 5 4 3 2 1

2

0 1 2 3 4 5 40 000 34 000 28 900 24 565 20 880 17 748

15 000

21 Moves the graph horizontally 22 a y

-2 -3

1

c $1331 24 a

y = 4x

y = 2x 0 -9 -8 -7 -6 -5 -4 -3 -2 -1 -1

(0, 0.5)

0 1 2 3 4 5 6 1000 1100 1210 1331 1464.10 1610.51 1771.56

0

8

y = 2x + 4

-1

y = 2x - 1

1000

10

y = 4x + 1

y = 2x

23 a n A

15 a y = 0 b y = 2 c y = -3 16 Moves the graph vertically 17 a ç iv b ç i c ç ii d ç iii 18 B 19 B 20 a, b, c  y

6

y

8 4

y = 3x

x

10

d

10

y = 3x + 2

5

833

Exercise 9F — The hyperbola 1 See the table at the bottom of the page*. y

1 2 3

y = -10 —– x

1 2 3 2 3

-5 -4 -3 -2 -1 0

x

0

y 10

— y = 10

10 -3 -2 -1

5

x

-10

x

-10

6 2 a   i 

y (1, 6)

y

6 y=— x

5 y=—

5

0

x

0

(1, -6) — y = -6

x

1

x

x

7 The negative reflects the curve y =

ii 

8

y

x — y = 20

20 0

x

y

-2

-1

0

1

-0.25 -0.33 -0.5 -1 Undefined



x

1

-3

k in the x-axis. x 2 1

3

4

0.5 0.33

y 1

y = —— x-1 1

-1

iii 

0 -1

y —– y = 100

100 0

x

x=1 x

1

x

1 2

Equation of vertical asymptote is x = 1. 9 a y 1

y = —— x-2

b   i  x = 0, y = 0    ii  x = 0, y = 0 iii  x = 0, y = 0 y 3

1 0 1 -— 2

x

2 3

x=2 (1, 4) (1, 3) (1, 2) 0

b

4 y=— x 3 y=— x 2 y=— x x

y 1

y = —— x-3 1 1 -— 3

0

x

34

x=3

4 It increases the y-values by a factor of k and hence dilates the curve by a factor of k.

1 *

834

x

-5

-4

-3

-2

-1

0

1

2

3

4

5

y

-2

-2.5

-3.3

-5

-10

Undefined

10

5

3.3

2.5

2

Answers

y

c

y 5

e 1

y = —— x+1 -2 -1

1 -1

10 The a translate the graph left or right, and x = a becomes the vertical asymtote. 11 a y



y 8

d

e

x

4

0 1 -2

-1 -3

x

2

4

12 x

8

-13

f



y -2

(2, 2)

y -4 -5

1 -3 -6

2

4 x

3

7

y = —— x-1

-2 6 -3

-8

-9

c

y

y 3

6 x

-3

-10

b

b



y 7

-4

-4

x

Centre (0, 0), radius 103

5 (1, 2)

y = —— x+1

3

1 3

-3 –

Centre (0, 0), radius 5

2 a

1 3– x

1 3

-3 –

-5



-1 0 -4

1 3

5 x

x

1

y

f

3–

-5

0

x = -1

(-2, 4)



y 10

2 x

10

-5

15 x

5

-10

-5

3 a (x + 2)2 + (y + 4)2 = 22 b (x - 5)2 + (y - 1)2 = 42 y y c

y 1

2— 2 -2

0

-3 1

-4

5 y = —— x+2

12 Check with your teacher.

1 1    b  y = x−3 x + 10

b

Centre (0, 0), radius 7

c



y 6 -6

6 x -6



4 x

Centre (0, 0), radius 6

Centre (0, 0), radius 4

d

y 9 -9

9 x -9

Centre (0, 0), radius 9

-2

12

4

x

e x2 + (y - 9)2 = 102 f (x - 1)2 + (y + 2)2 = 32 y y 1 -2

9

-4

-7

-10

6

-10 -1

4 10

x

-2

1

4 x

Answers 9F ➜ 9G

-4

7 x

14

x

19

y 4

7



x

14

7

-3

-7

9

c (x - 7)2 + (y + 3)2 = 72 d (x + 4)2 + (y - 6)2 = 82 y y 4

Exercise 9G — The circle 1 a y

5

-6

x

(-3, -5)

Possible answers: a  y =

5 1

x

-4 -2 -2

-5

4 D 5 B 6 (x - 5)2 + (y - 3)2 = 16 Chapter review Fluency 1 D 2 A 3 D 4 A 5 B

Answers

835

6 a (4, -15) b (-2, 9) 7 x -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 y 16 7 0 -5 -8 -9 -8 -5 0 7 16 y 16 14 12 10 8 6 4 2



y

y = 10 ì 3x

450 400 350 300 250 200 150 (0, 10) 100 50 -3 -2 -1 0

12

y = 10-x

x

0 -8 -6-4 -2 -2 2 4 6 8 -4 -6 -8 (-4, -9) -10

1

(0, 1) 1

13 a

y

3 x

2

y 140 120 100 80 60 40 20

-3 -2 -1 0

TP (-4, -9); x-intercepts: -7 and -1 8 a TP (3, 1); no x-intercepts; y-intercept: (0, 10) 10

11

2

y 10

y = (x - 3)2 + 1

8 6

(3, 1)

y = (1.2)x

2

x

-3 -2 -1 0

5 5 , −1 + ; b TP (-1, -5); x-intercepts: −1 − 2 2 y-intercept: (0, -5) y

5– 2

-1 -



(-1, -5)

9 TP (-1, 16); x-intercepts: -5 and 3; y-intercept: (0, 15) y (-1, 16)

15

y = -x2 - 2x + 15

0

-5

y = 5 ì 3x y = 2 ì 3x y = 1–2 ì 3x

-3 -2 -1 0

1

x

-3

y

0.008

b

-2 0.04

-1 0.2

y

y = 5x

160 140 120 100 80 60 40 (0, 1) 20 -4 -3 -2 -1 0

Answers

0

1

2

3

1

5

25

125

1

2

3 4x

1

2

3

x

b Changing the sign of the index reflects the graph in the y-axis. 16 a y (1, 4)

0 (1, 5)

x

3

y = (2.5)

45 40 35 30 25 20 15 10 5

-3 -2 -1 0

10 a

2

b Increasing the value of k makes the graph steeper. 15 a y -x x y = (2.5)

x

3

3 x

2

36 32 28 24 20 16 12 8 4

5– 2

x

0 -3

1

b Increasing the value of a makes the graph steeper for positive x-values and flatter for negative x-values. 14 a y

y = 2(x + 1)2 - 5 -1 +

836

y = (1.5)x

4

(0, 1)

0



x

3

4 y=— x

x

y

b

2 a x

0

x (1, -2)

-3 y = —— x-2

0

0

b 4  m c 2 s d 4 s 3 a

y 3 — 2

(2, 4)

h = 4t - t2

— y = -2

17

h

4 t

h 21 h = -x + 4x + 21

(2, 25)

-3 0

7

2

x

2

x

(3, -3)

1 . 18 Check with your teacher. Possible answer is y = x+3 19 a y 2 2 4 x + y = 16

-4

4 a

4 x

0

b 25  m c 2  m d 7  m h

(2, 20) h = -5t2 + 20t

-4

b 

20 a

y

(x - 5)2 + (y + 3)2 = 64

0

8 (5, -3)

x

x 2 + 4x + y 2 - 2y = 4

(-2, 1) 3

b

x

y 8 0 (-4, -4)

x

x 2 + 8x + y 2 + 8y = 32

4 2 -2

0

2

4x

-2 -4 y = -x2 + 4x - 1 y = x2 - 4x + 7

5 a [0, 12] b 32 m c 11:41 am to 6:19 pm 6 a P < x2 - 5x b 6.25 m c i Check with your teacher. ii Dilation by a factor of 0.48 d 28.6% 7 a Check with your teacher. b When x = 0.3, b = 10.7. Therefore if p is greater than 10.7 cm the platform would hit the bridge. Chapter 10

Deductive geometry Are you ready? 1 a DABC 2 a CA c ±BCA

b ±ACB c AC b PQ d ±QPR AD AE DE 3 DADE ~ DABC, = = AB AC BC 4 a Parallelogram b Trapezium c Kite Exercise 10A — Congruence review 1 a I and III, SAS b I and II, AAS c II and III, RHS d I and II, SSS

Answers

Answers 10A ➜ 10A

21 x2 + y2 = 36 Problem solving 1 a y = -(x - 2)2 + 3 = -x2 + 4x - 1 b y 6

4 t

b 4 s c 2 s d The ball is never above a height of 20  m.

y

0

0

837

2 a x = 3 cm b x = 85è c x = 80è, y = 30è, z = 70è d x = 30è, y = 7 cm e x = 40è, y = 50è, z = 50è, m = 90è, n = 90è 3 a Use SAS b Use SAS. c Use ASA. d Use ASA. e Use SSS. 4 C, D 5 a x = 110è, y = 110è, z = 4 cm, w = 7 cm b x = 70è c x = 30è, y = 65è 6 The third sides are not necessarily the same. 7 Corresponding sides are not the same. 8 Use SSS. Exercise 10B — Similarity review 1 a i and iii, RHS b i and ii, SAS c i and iii, SSS d i and iii, AAA e i and ii, SSS 2 a Triangles PQR and ABC b Triangles ADB and ADC c Triangles PQR and TSR d Triangles ABC and DEC e Triangles ABC and DEC 3 Check with your teacher. AB BC AC 4 a = = AD DE AE b f = 9, g = 8 5 x = 4 1 6 x = 20è, y = 2 4 7 x = 3, y = 4 Exercise 10C — Congruence and proof 1 Use AAS. 2 Check with your teacher. 3 Use SAS; then corresponding sides in congruent triangles are equal. NO = OP. 4 Check with your teacher. 5 Use SAS; then alternate angles in congruent triangles are equal. Hence AB || CD. 6 Use AAS. 7 Use RHS. 8 Use AAS. 9 Use RHS or AAS; then corresponding sides and angles in congruent triangles are equal. 10 Use RHS. 11 a B

y x x A

y D

C

b ±ADB = 90è given ±ABC = 90è given ±BAD = xè given ±BAC = xè given À ±ABD = 90è - x and ±ACB = 90è - x. À DBAD ~ DCAB. 838

Answers

AD AB = AB AC \AB2 = AD ì AC d ±BDC = 90è given ±ABC = 90è given ±ACB = 90è - x ±DCB = 90è - xè À ±DBC = xè ±BAC = xè given \ DBCD ~ DACB CD BC e = BC AC \ BC2 = CD.AC f AB2 + BC2 = AD ì AC + CD ì AC = AC (AD + CD) = AC ì AC \ AB2 + BC2 = AC2. g Students to do.

c

Exercise 10D — Quadrilaterals: definitions and properties 1 a True b True c True d False e True f False g False h False 2 None are true. 3 a, c, f 4 a, b, c, f, g, h 5 a, c, d, e, f 6 a, b, c, d, e, f, g, h 7 Rhombus, square 8 Rectangle, square 9 Parallelogram, rhombus, rectangle, square 10 Square 11 a

b 6 sides c 7 sides d Table size

Number of sides hit

5 cm ì 3 cm

6

7 cm ì 2 cm

7

4 cm ì 3 cm

5

4 cm ì 2 cm

1

6 cm ì 3 cm

1

9 cm ì 3 cm

2

12 cm ì 4 cm

2

e If the ratio of the sides is written in simplest form then the pattern is m + n - 2.

f There are two routes for the ball when hit from B. Either 2 or 3 sides are hit. The ball does not end up in the same hole each time.   A suitable justification would be a diagram — student to draw. g Isosceles triangles and parallelograms. The triangles are congruent. h The shapes formed are parallelograms. There is only one possible path although the ball could be hit in either of two directions initially. i Given m : n is the ration length to width in simplest form. When m is even and n is odd the destination pocket will be the upper left. When m and n are both odd, the destination pocket will be the upper right. When m is odd and n is even the destination pocket will be the lower right. j Students to investigate.

4 a Similar, scale factor = 1.5 b Not similar c Similar, scale factor = 2 5 a x = 48è, y = 4.5 cm b x = 86è, y = 50è, z = 12 cm c x = 60è, y = 15 cm, z = 12 cm 6 Use equiangular test. 7 Use equiangular test. 8 A

Exercise 10E — Quadrilaterals and proof 1 Use AAS to show DZWX @ DZYX. 2 Use AAS to show DAED @ DCEB and hence AE = EC and DE = EB. 3 a Use SAS. b ±AED = ±CED (corresponding angles in congruent triangles equal) and ±AED + ±CED = 180è (angle sum of straight lines is 180è) \ ±AED = ±CED = 90è c Corresponding angles in congruent triangles are equal. 4 Use SAS to show DDAE @ DBAE. Hence, DE = EB. (See previous question.) 5 Use co-interior angles and parallel lines. 6 Use SAS. AC = BD (corresponding sides in congruent triangles are equal). 7 AX || DY because ABCD is a parallelogram AX = DY (given) \ AXYD is a parallelogram since opposite sides are equal and parallel. 8 a Use SAS. b Use SAS. c Opposite sides are equal. 9 AC = DB (diameters of the same circle are equal) AO = OC and OD = OB (radii of the same circle are equal) \ ABCD is a rectangle. (Diagonals are equal and bisect each other.) 10 Check with your teacher. 11 PS = QR (corresponding sides in congruent triangles are equal) PS || QR (alternate angles are equal) \ PQRS is a parallelogram since one pair of opposite sides are parallel and equal. 12 MP = MQ (radii of same circle) PN = QN (radii of same circle) and circles have equal radii. \ All sides are equal. \ PNQM is a rhombus.



C

D

Bisect ±BAC AB = AC (given) ±BAD = ±DAC AD is common. \ DABD @ DACD (SAS) \ ±ABD = ±ACD (corresponding sides in congruent triangles are equal) 9 A, B, C, D 10 a False b True c True 11 a Use SAS. b Use SAS. c Use SAS. d They are all the same length. e B and C 12 Use SAS. PQ = PS (corresponding sides in congruent triangles are equal) 13 Rhombus, square 14 A quadrilateral is a rhombus if: a all sides are equal b the diagonals bisect each other at right angles c the diagonals bisect the angles they pass through. 15 WZ || XY (co-interior angles are supplementary) and WZ = XY (given) \ WXYZ is a parallelogram since one pair of sides is parallel and equal.

16

A

D

B

C

±ABD = ±ADB (angles opposite the equal sides in an isosceles triangle are equal) ±ABD = ±BDC (alternate angles equal as AB || DC) \ ±ADB = ±BDC \ Diagonals bisect the angles they pass through. 17 Corresponding sides are not the same. 18 ±FEO = ±OGH (alternate angles equal as EF || HG) ±EFO = ±OHG (alternate angles equal as EF || HG) ±EOF = ±HOG (vertically opposite angles equal) \ DEFO ~ DGHO (equiangular) 19 A rhombus is a parallelogram with two adjacent sides equal in length. 20 Rectangle, square

Answers

Answers 10B ➜ 10E

Chapter review Fluency 1 a I and III, ASA or SAS b I and II, RHS 2 a x = 8 cm b x = 70è c x = 30è, y = 60è, z = 90è 3 a Use SAS. b Use ASA.

B

839

CHAPTER 11

Problem solving I 1 23.83 cm 2 81x4 - 216x3y + 216x2y2 - 96xy3 + 16y4 y 3 3 + ø 9 - 8g g

4

3 - ø 9 - 8g 4



2

( 43 , g - 98 )

x-intercepts: x =

1 e (y-intercept - 45 ) and h (y-intercept = 14 ), gradient = - 20 ;

19 a A = 80p cm2 b If radius and height are both halved, the surface area is one-quarter its original value.

x

0

16 400 students 17 11.75 cm 18 a (y-intercept -12) and f (y-intercept = 2), gradient = –2; b (y-intercept 5) and d (y-intercept = 6), gradient = -1; c (y-intercept - 13 ) and g (y-intercept =  72 ), gradient = 3;

20 y =

3 ± 9 − 8g ; 4

− 4k  h  x − 2  + k h2  2

9 3 y-intercept: y = g; turning point:  , g −  4 8 4 1231.5 cm3 5 54.28è 6 8.3 cm by 1.7 cm f + 3be  f + 3be  ; y = a 7 x =  − 3b ae + d  ae + d 

8 a x = 8 b x = -4 c x = 12 9 No, Mary will need 64.5 cm of ribbon. 10 17.05è 11 -q Ç x Ç p 2 3  L − l 4π  L − l  12 π  l+ cm 3     2  3  2  13 a This is a quadratic equation, which means that there is a possibility of two different answers. Marlon has one of the two parts of the answer correct. b No. x(x - 3) = 10 x2 - 3x = 10 x2 - 3x - 10 = 0 (x - 5)(x - 2) = 0 (x - 5) = 0   or   (x - 2) = 0 x = 5     x = 2 y 14

2

e f 21   ≤ A ≤    4  4 7 18 22 y = -  x + 5 5 23 229.1 m P P 24 by 4 4 25 Any false statement that occurs during the solving of simultaneous equations indicates the lines are parallel, and have no points of intersection. 26 47 cm for the circle and 53 cm for the square 27 a y = x2 + 2x b y = -4x2 + 14x - 15 c y = x2 - 4x + 4 28 a 1072 cm3 b 9.4 cm 29 a 28.3 m b The image width doubles. 30 a Total length = 4l - 8, where l is the length of the lawn. b Cost = 23(4l - 8) + 100 c $1296 31 a a = 2 b b = 4 c c = -3 d The equation of the quadratic is y = 2x2 + 4x - 3. y

25 20 15

0

1

10

x

5 -3

-2

-1

-6

32 a L =

840

1





= 2 ì 2b ì 4 2b





= 4 2b2 cm2.

Answers

1

-5 y = -8

15 a 4 2b b 19.5è, 70.5è, 90è. 1 c Area = 2 base ì height

0

130 − x 2 2x

130 x − x 3 2 c x = 6.6 cm, L = 6.55 cm 33 Check with your teacher.

b V =

1

34 V = 3 p r2 s 2 − r 4

2

3x

35 a b2 + 8b + 28a + 4c - 4ac - 12 < 0 b b2 + 8b + 28a + 4c - 4ac - 12 = 0 c b2 + 8b + 28a + 4c - 4ac - 12 > 0 36 a 55 bottles b 20r cm c 17.32r cm d 3nr cm 37 132 passengers — 72 Virgin Green passengers, 60 Qintas passengers 38 a 75è58Å b 73 cm 39 a 3rd is (x + 2) cm; 4th is (x + 3) cm; 5th is (x + 4) cm b 5 cm 100 % larger than the c Circumference of 4th circle is x +3 3rd circle’s circumference. 19 40 a 24 b

5 24

c 32 3 cm 1 ± 10 41 x = 3 42 a 7 3 metres b 35 15 m2 7 3 metres 6 ii  18 43 a A = nx + 96.25n b $3500 44 a B (12.5, 0) and C (37.5, 0) 22 2 94 x - x + 90 = 0 b i  375 15 ii  (17.1, 18) c i  Translated 25 units to the right

c i 

ii  ( 75 , 125 ) 2 6 45 a x 2 = b

10

100 3

π c The cylinder has larger surface area by 57.25 cm2. 46 a 97.2 km/h b 140.4 km/h in 2 seconds 47 a r

b 43 r

163è20Å 16è40Å b km

c Check with your teacher. 48 a At t = 0, h = h0 = the starting height. b h 30

25 km N 121è40Å a km d km

19 km

10 1

2

3 t

c 22.25 m d 0.375 seconds e 1.55 seconds (2 x − 1)(3 x + 1) 1 49 a = 6(2 x − 1)(3 x + 1) 6

b 23.3 km east and 33.9 km south of its starting point c 145è30ÅT 62 a (x - 5)(x + 2) b (x - 5) is the shorter length. c x = 17 d 228 cm2 63 The rug is 80 cm wide and 400 cm long. 64 a 4r cm b (2r + 20) cm c 4p r2 cm2 Answers

Answers 11 ➜ 11

c km 58è20Å

20

0

b Solving 36x2 - 6x - 6 = 97 026 using any method gives x = -51.833 metres and x = 52. Ignore the negative solution because x > 0 for measurement units. Possible dimensions could be: 3(2x - 1) by 2(3x + 1) 2(2x - 1) by 3(3x + 1) 6(2x - 1) by 1(3x + 1) (2x - 1) by 6(3x + 1). Or any possible combination for numbers whose product is 6, such as 1.5 and 4 50 a i  Width: 2.95 m to 3.05 m, length: 4.45 m to 4.55 m ii  0.67% b 49.23 m2 c $760 51 a p = 72è, s = 108è b ABCD is trapezoidal with AD||BC. ±BAD = ±CDA = 72è ±ABC = ±BCD = 108è 52 The square numbers are 1, 4, 9, 16, 25, 36, ………… The difference between these numbers is 3, 5, 7, 9, 11... If this continues to 75, it is the 37th number, so 382 – 372 = 75. So, the two natural numbers are 37 and 38. 53 a Mr Barnes has (x + 2)(x + 5) m2; Mr Snowdon has 4(x + 2) m2. b (x + 2)(x + 5) - 4(x + 2) c (x + 2)(x + 1) d The carpet has a width of 3 m. e Mr Barnes bought 18 m2 and Mr Snowdon bought 12 m2. 54 The side of the square is 4 5 cm. 55 Check with your teacher. 56 Check with your teacher. 57 Use similar triangles. 58 240 students 59 a V = x(x + 10)(x + 7) b 44 400 cm3 c 40 cm 60 a k = 12 k = 12 makes the two equations represent the same line, giving an infinite number of solutions. All other values of k generate two parallel lines. b k ≠ 12 N 61 a

841

d (4p r2 + 80p r + 400p ) cm2 e 80p (r + 5) cm2 f 1.131 m2 g 30 cm 65 The factors of 24 are: 1 and 24; 2 and 12, 3 and 8 and 4 and 6. To make the first bracket equal 1, then x must be 7 and to make the second bracket equal 24, then y must be 28. This pattern continues until all possibilities are found. They are: Factors  1  2  3  4  6  8 12 24

x  7  8  9 10 12 14 18 30

24 12  8  6  4  3  2  1

y 28 16 12 10  8  7  6  5

69 a r = -4 b s < 70 Dan is 25 years old. 71 19.85 m xm 1m 72 a

c t =

c (2 73 + 30 ) m of pipe is required. 75 Approximately 98.3 cm 76 a b = -8, c = 12 b y = x2 - 8x + 12 c (4, -4) d y 15 10 5 0

2

4

6

10 x

8

-5

77 a 8 m b 57è c 7.6 m 78 a First ripple’s radius is 3 cm, second ripple’s radius is 15 cm. y b

66 5704 mm 67 360 adults and 190 children 68 a The cloth is x cm wide and 4x cm long. b P = 10x, A = 4x2 c Length = 120 cm, width = 30 cm. d Perimeter = (10x + 48) cm, area = (4x2 + 60x + 144) cm2 e The area has increased by 1944 cm2. 25 8

73 m long. 2 b The chain length is (2 3 − 1) m, so (8 3 − 4) m of chain is required.

74 a The struts each need to be

15 10 5

-15 -10

0

-5

9 - 32

5

10

-10 -15

c 2.4 cm/s d 1 minute 23 seconds after it is dropped 79 The centre of the racket travels 5.24 m. 80 a i  6 cm

xm

3m 10 cm

Temperature degrees celsius

b (x + 1)(x + 3) d 5.24 m by 3.24 m T 73 a

c 5 m by 3 m

8 cm

10 cm

12 cm



25

ii  10 cm

20

8 cm

15 10 cm

10

2

4 6 Hours

b 21èC c Decreasing d Increasing e 5 èC after 4 hours f 21èC Answers

8 cm 12 cm

5 0

842

15 x

-5

8h

iii 

6 cm

10 cm

8 cm

12 cm

10 cm

6 cm

b Area of rectangle = 12 ì 8 = 96 cm2. Area of parallelogram = 12 ì 8 = 96 cm2, 12 × 16 = 96 cm2, Area of triangle = 2 1 Area of trapezium = 2 (18 + 6) ì 8 = 96 cm2.

90 a y = -2x2 - 12x - 14 b a = -2, b = -12, c = -14 c y (-3, 4)

4 2 (-1.6, 0) (-4.4, 0) x 0 -5 -4 -3 -2 -1 -2 1 2 3

c Perimeter of rectangle = 40 cm. Perimeter of parallelogram = 44 cm, Perimeter of triangle = 48 cm, Perimeter of trapezium = 44 cm. The triangle has the largest perimeter, while the rectangle has the smallest. 81 a 0 b (6, 6) h c

-4 -6 -8 y = -2x2 - 12x - 14 -10 -12 (0, -14) -14 -16

Height (metres)

8 6 4

2 0

2

12 d

4 6 8 10 Distance (metres)

d 12 m e 6 m f 6 m 82 a 60è b 3.98 km c 71è d 1.34 km 83 True: the tip travels 30.2 m. 84 a 56 941 cm3 b 11 938 cm2 y 85 a 12

Height (metres)

10 8 6 4 2 0

1 2 3 4 5 6 Horizontal distance (metres)

x

b 8 m c 11 m above the water d 5.83 m 86 a 24.5 m3 b The dimensions of the smaller skip are half those of the larger one. 87 7.6è 88 x = 13.75, y = 11.4 89 a Centre is (1, 2). b Radius is 2. c x-intercept 1; y-intercepts (− 3 + 2) and ( 3 + 2) d y



4

(1, 2)



1 0 -1 -2

1

2

3

4

5x

1 12 1 12 1 12 1 12 1 12 1 12 1 12 1 12

1 1 = 13 + 156 1 = 141 + 84 1 = 15 +

= = = = =

1 60 1 1 + 48 16 1 1 + 36 18 1 1 + 30 20 1 1 + 28 21 1 1 + 24 24

104 The factors of a number are generally written in pairs, producing an even number of factors. With a perfect square, one of these factors will be paired with itself, producing an odd number of factors. This occurs for all perfect squares. 105 15.9 cm 106 The total number of tiles needed for r rows is r2. Answers

Answers 11 ➜ 11



2

-1



5

3

-2

91 180è 92 The area of material required is 1.04 m2. If Tina is careful in placing the pattern pieces, she may be able to cover the footstool. 93 a Prove the equation is y = 15 - 0.15x2. b The height at the edge of the road is 5.4 m. c The road needs to be 15.5 m wide. 94 - 49 95 a 50 minutes b 8.03 am or 8.07 am c 8.11 am d 8.05 am e 25 minutes f Between 8.03 am and 8.07 am 96 For example: take the two numbers 48 and 60. Their HCF is 12. Difference = 60 - 48 = 12 60 ó 12 = 5 and 48 ó 12 = 4 The two numbers are exactly divisible by 12. This theorem says, then, that the HCF of 48 and 60 is 12, which is the case. 97 Matt is travelling at 45 km/h and Steve is travelling at 60 km/h. 98 The perimeter of the octagon is 61 cm. 99 Check with your teacher. 100 17.4 circuits 101 Approximately 40 400 years from now 102 Check with your teacher. 103 The 8 different ways are:

843

 et the cost of an apple be a cents, and the cost of a 107 a L banana be b cents. 6a + 4b = 700 [1] 1a + 9b = 450 [2] Subtract [2] from [1]. 5a - 5b = 250 Divide through by 5. a - b = 50 b An apple costs 50c more than a banana. 108 a 1.25 mL b 1.47 mL c Answers may vary. The two answers are only slightly different. Because of an inability to measure to that degree of accuracy in the home, they both provide a good guide to a safe amount to administer. d Approx 1 and 10 years 109 y = 2x2 - 5x + 1 110 11

12

1 2

10 9

3

8

4 7

6

5

The total of the numbers here in each region is 26. 111 Call the people A, B, C and D who take times of 1, 2, 5 and 10 minutes respectively. A and B go over 2 min B returns 2 min C and D go over 10 min A returns 1 min A and B go over 2 min Total time = 2 + 2 + 10 + 1 + 2 = 17 min 112 35 tennis balls 113 58 railings 114 If the cut was vertical, a rectangle (or straight line if the cut was on the very edge) would result. A horizontal cut would result in a circle. A cut which goes through the sides at an angle would produce an ellipse or a parabola. An ellipse results if the cut is simply through the sides. A parabola results if the cut is through the side and the base (or top).

115 Answers may vary. The value of n must be greater than 1 because the second number would be 0 if n were 1. 116 The deck should be 5 m wide and 10 m long on each side. 117 The original number could have been 187, 781, 286, 682, 385, 583 or 484. 118 There is no solution to the equation. It is not possible for the square root of a number to be negative. Squaring both sides of the equation has produced an invalid solution. 844

Answers

119 a There are 5 different shapes.

b They all have a perimeter of 10 units, except for the last one which has a perimeter of 8 units. 120 112 km 121 Check with your teacher. 122 505 1 1 1 123 = + n n + 1 n( n + 1) 124 Fold the left vertical line forward, so that page 3 sits on top of page 4. Fold the entire bottom half backwards, so that page 5 sits behind page 4. Fold forward the left vertical fold, so that page 3 sits on top of page 2. Finally, fold forward the left vertical fold, so that page 6 sits on page 7. The pages are now in order from page 1 to page 8. 125 9 trains 126 The claim appears to be true. Further investigation would need to be conducted to determine if it worked in all cases — for example, if the lines were parallel, at right angles, vertical or horizontal. 127 6450 multi-packs 128 Take the total, subtract 16, then divide by 4 to get the first number in the square. The other numbers are 1, 7 and 8 larger. 129 11 times 130 a x2 but no y2 gives vertical parabola, y2 but no x2 gives a horizontal parabola. b The value of a represents the horizontal stretching factor. Positive a, open to right, negative a open to left.

c

x + 4 = (y + 1)2

5

y

8 6

2

9 a

1 -4

-3

-2

10 a i  20

ii 

1

ii 

-1

b i  16 11 A

-2

12 a

0

-1

(-4, -1)

1

x

2

(0, -3)

-3

d 13 a

-4

c

Probability b 4

1

1 4

2 3 1 12

c 13

5

c 8

1

2

5

1 4

or 0.25

3

6

3 10

or 0.3

4

3

3 20

or 0.15

5

4

1 5

ii

1

6 1 2

12 3 iv  20 = 5 4 a A ¶ B c AÅ ¶ BÅ

b   i 

5 a

1 6

=

8

10

ii 

4 20

=

1 5

v 

8 20

=

2 5

b

4 17

9 10

1

b 10

17 6

Tennis



iii

Volleyball

x

Walking 10

15

38

8 2

17 6

b 96

20

c i 

35 96

14

d i 

63 96

18

8

23

21

= 32

ii 96

16 a iii 

2 20

=

x = 30

Volleyball

1 10

c

Soccer 1

7 2

b XÅ ¶ Y d A ¶ C ¶ BÅ 1 8

1

ii 96 = 12

4

5

4

7 1

1

b i  2 ii 6

6 A 7 a

12

38

1

c i  2

Tennis 1

iii  30

2

iv  5

7

v  15

Answers 12A ➜ 12A

19

2 10 20

15

4 16

10

Tennis

x

x

Walking

2

3

1 9

6

8

1.00

B 5 11

17

15

c 10

A

38

Volleyball

or 0.2

“ f = 20

b 10

10 2



or 0.1

x

Walking

Tennis

1 10

17

7 20

8

7

2

7 13

1

Volleyball 15

1

3

f 2

1

d 0

c 2

133

b 312

3 a

1

iii  16

e 0 b

b Drawing a red card

b 156

1 5

iv  40

c 2

1 5 11 20

Exercise 12A — Review of probability 1 x Relative frequency f

2 a

7

iii  80

1 4

b

c 23

b 9

3 a Not drawing an ace c Obtaining a 4 or a 5 5 a

9

1 5 17 80

1 13 12 13

15 a i 

Are you ready? 1 a Set A: 3, Set B: 4, Set C: 4

4 a

c 5

14 a C b D c E

Chapter 12

2 a

b 10 3

(0, 1)

(-3, 0)

4

1

17 30

8

ii 15 Answers

845

17 a

7 C

x = 35

Calculator

1 10

b 10

9

c 50

10 a

9 50

b 50

41

c 25

11 a

15 16

b 9

8 a

Graph book

47

9 D 7

18

5 5

b i  25 d i  18 a

12 No, getting 1 Tail is possible too. 1 13 a 8

ii 23

18 35 1 5

ii 7

5

iii 

6 7

iii 

ii

3 10

12

1 7 1 7

iv  35

b 15 a

c 9–4

16 a

19 4–3 1

A

Overlaying AÅ and BÅ shows AÅ ¶ BÅ as the area surrounding A and B

b 47

1 a P(A) 2 a M

d 1

N

b P(M ) + P(N ) - P(M ¶ N ) c i  False ii  True 3 a

1 13

4 20% or

b 1 5

1

5 a  i  2 16

6 25 846

iii Cannot be determined

12 13

Answers

1

ii 2

b Yes

18 a

5 9

19 a

8 14

20 a

1 13

b

The union of A and B is shown in brown, leaving the surrounding area as (A ß B)Å

c 0.35

b 25

19

c 25

9

2

c 13

3

b 13

4

b 7 or

4 7

2

10

1

b 14 or 7

c 14 or

1

5 7

4

b 4

c 13

21 a Yes

B

Exercise 12B — Complementary and mutually exclusive events

9 25 7 26

17 C

20 a No. P(Azi rolls a 5) = 8 and P(Robyn rolls a 5) = 16 1 b Yes. P(Azi wins) = 2 and P(Robyn wins) = 12 21 Yes. Both have a probability of 12. 22 a The person with the 6-sided die has less chance of winning. 1 For the 8-, 12- & 16-sided dice: P(mult 4) = 4 ; for 1 6-sided die, P(mult 4) = 6 b Answers may vary, check with your teacher.   i Example — rolling an even number ii Example — rolling a 3 iii Example — rolling a number greater than 3 23 P(Alex wins) = 15 ; P(Rene wins) = 25 24 Yes, equivalent fractions; 166 = 83 25 B

7 8

2

14 3

b $50

A

d 25

7

1 2

22 Answers may vary, check with your teacher. a i No. There are many other foods one could have. ii Having Weet Bix and not having Weet Bix b i No. There are other means of transport; for example, catching a bus. ii Walking to a friend’s place and not walking to a friend’s place c i No. There are other possible leisure activities. ii Watching TV and not watching TV d i No. The number 5 can be rolled too. ii Rolling a number less than 5 and rolling a number 5 or greater e Yes. There are only two possible outcomes; passing or failing. 23 No. The number 2 is common to both events. 24 a i T

ii F

v F

vi F

b i

3 16

ii 1

c i

5 16

ii

iii T

iv F

5

iv 16

8

iii 8

3 16

iii

3

3 4

Exercise 12C — Two-way tables and tree diagrams 1 a 0.2 b 0.1 c 0.2 d 0.5 e 0.4 f 0.8 Card outcomes 2 i Coin outcomes

c i 

13

12

ii a  18

b  0.5 c  0.5

Club,

Spade,

Diamond,

Heart,

H

H,

H,

H,

H,

T

T,

T,

T,

T,

3 a 

Die 1 outcomes

1

3

4

5

2

(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)

3

(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)

1– 2

1– 2

B

1– 2

R

1– 2

B

1– 2

R

1– 2

B

5

(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)

a

1 8

b 8

6

(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)

d

1 8

e 8

7 a

1 12

Coin outcomes

2

3

4

5

6

7

R

9

10

1– 3

H (H, 1) (H, 2) (H, 3) (H, 4) (H, 5) (H, 6) (H, 7) (H, 8) (H, 9) (H, 10)

c

3

4

1 4

5

6

7

8

(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6) (2, 7) (2, 8)

3

(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6) (3, 7) (3, 8)

4

(4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6) (4, 7) (4, 8)

Yellow octahedron outcomes

2

5

(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6) (5, 7) (5, 8)

6

(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6) (6, 7) (6, 8)

7

(7, 1) (7, 2) (7, 3) (7, 4) (7, 5) (7, 6) (7, 7) (7, 8)

8

(8, 1) (8, 2) (8, 3) (8, 4) (8, 5) (8, 6) (8, 7) (8, 8)

Green octahedron outcomes 3

4

5

6

7

8

(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6) (1, 7) (1, 8)

2

(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6) (2, 7) (2, 8)

3

(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6) (3, 7) (3, 8)

4

(4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6) (4, 7) (4, 8)

5

(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6) (5, 7) (5, 8)

1– 2

1– 2

(7, 1) (7, 2) (7, 3) (7, 4) (7, 5) (7, 6) (7, 7) (7, 8)

8

(8, 1) (8, 2) (8, 3) (8, 4) (8, 5) (8, 6) (8, 7) (8, 8)

R B

RBR RBB

R B

BRR BRB

R B

BBR BBB

1– 8 1– 8 1– 8 1– 8 1– 8 1– 8

— 1

3

c 8

3

7

f

1– 6

1– 2

1– 6

1– 2 1– 3

1– 6

1 2

2 Outcomes Probability 1– RR 9

R

G

RG

B

RB

R

GR

G

GG

B

GB

R

BR

G

BG

B

BB

1 — 18 1– 6 1 — 18 1 — 36 1 — 12 1 — 6 1 — 12 1– 4

— 1

  b {(R, R), (R, G), (R, B)} 1 c 3 d 8 a

7 18

1

2

B

1– 2

1– 2

G

1– 2

B

1– 2

G

1– 2

B

1– 2

1– 2 1– 2

1– 2

1– 4 1– 2

1– 4

2 1– 4

1– 4

1– 2

1– 4

3 1– 4

  b No 1

1

1– 2

1– 2

1– 4

1

1– 4

1– 2

1– 2

G

1– 2

1– 2

1– 2

  b 3 3 c 8 d They are equally likely. 7 e 8 9 a 1

(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6) (6, 7) (6, 8)

7

1 8

1– 2

Outcomes Probability 1– RRR 8 1– RRB 8

3 B G

Outcomes Probability 1– BBB 8 1– BBG 8

B G

BGB BGG

B G

GBB GBG

B G

GGB GGG

1– 8 1– 8 1– 8 1– 8 1– 8 1– 8

— 1

2 Outcomes Probability 1– 11 4

1 2

12

3

13

1

21

2

22

3

23

1

31

2

32

3

33

3

1– 8 1– 8 1– 8 1 — 16 1 — 16 1– 8 1 — 16 1 — 16

— 1

Answers 12C ➜ 12C

1

6

1– 2

1– 2

(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6) (1, 7) (1, 8)

2

1– 2

B

1

1

G

1– 2

Green octahedron outcomes 2

1– 2

1– 3

1– 6

T (T, 1) (T, 2) (T, 3) (T, 4) (T, 5) (T, 6) (T, 7) (T, 8) (T, 9) (T, 10)

1 5

1– 2

1– 3

8

3 R B

1– 2

1

Die outcomes

b

Yellow octahedron outcomes

R

(4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)

1

c

2

4

5 a 

 

1

6

(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)

1

b

2

1

  b 4 a

6

Die 2 outcomes

9

c i  4     ii  2     iii  8     iv  16 Answers

847

10

1 1– 6

t

5– 6

Outcomes Probability 1 — tt 36

2 1– 6 5– 6

1– 6

t' 5– 6

t t'

tt'

5 — 36

t

t't

5 — 36

t'

14 a

1

b c d

B Outcomes Probability 1 — SS 16 S

2

1– 4 3– 4

S

SS'

3 — 16

S

S'S

3 — 16

S'

S'S'

S'

3– 4

1– 4

S' 3– 4

9 —

16 — 1

S = outcome of spade Sample space = {SS, SSÅ, SÅS, SÅSÅ}

b c d

1 16 9 16 3 8

12

1 1– 3

X

1– 4 1– 4

Y

1– 4

1– 3

1– 3

1– 3

1– 3

W 1– 4

Z

1– 3

1– 3

1– 3

  a 13 a

1 4

2 Outcomes Probability 1 — Y XY 12 1– 1 — 3W XW 12 1 — XZ Z 12 1 — X YX 12 1– 1 — 3W YW 12 1 — YZ Z 12 1 — X WX 12 1– 1 3Y — WY 12 1 — WZ Z 12 1 X — ZX 12 1– 1 3 — ZY Y 12 1 — ZW W 12 — 1

1

b 4

4 — 10

6 — 10

B

G

3– 9

B

BB

2 — 15

6– 9

G

BG

4 — 15

d 848

2 15 8 15

Answers

1– 3

1– 3

1– 3

1– 3

E

 

ADF

1– 3

H 1– 3

AEF AEG AEH BCF BCG BCH

F G

BDF BDG BDH BEF

H

1– 3 1– 3

1 — 18 1 — 18 1 — 18 1 — 18 1 — 18 1 — 18 1 — 18 1 — 18 1 — 18 1 — 18 1 — 18 1 — 18 1 — 18 1 — 18 1 — 18

ADG ADH

F G

H F –1 3 G

C

D

F G

1– 3

1 — 18 1 — 18 1 — 18

ACG ACH

H

1– 3

1– 3

F G H

BEG BEH

— 1

Sample space = {ACF, ACG, ..., BEG, BEH} 1 1 b 2 c 6 15 There is a 50% chance that a third room will be needed. 16 Susan would have 16% chance of passing the exam if the last three questions had the standard 4 choices. This chance is reduced to 48% with the inclusion of two questions offering 6 possible answers. 17 i a Outcomes Probability

R

1– 2

1– 2

G

1– 2

R

RR

1– 4

1– 2

G

RG

1– 4

1– 2

R

GR

1– 4

1– 2

G

GG

1–

4 — 1

  1

1

b 2 c 2 ii As the first counter is not replaced, the probability of drawing the second counter is altered. This is reflected in the probabilities along the branches of the tree 1 diagram; P(2 counters of the same colour) = 3; 2

P(2 counters of different colours) = 3 .

c 1 2– 4

2– 4

R

G

Outcomes Probability 1– RR 6

1– 3

R

2– 3

G

RG

1– 3

2– 3

R

GR

1– 3

1– 3

G

GG

1–

6 —

c 3

1   Exercise 12D — Independent and dependent events 1 a 0.28 b 0.12 c 0.42 d 0.18 1 1 1 2 a Yes b i  2 ii 6 c 12

e No

3 40

4– 9

B

GB

5– 9

G

GG

4 — 15 1–

3 — 1

  b

1– 3

2 Outcomes Probability

1

1– 3

1– 3

D

ACF

F G H

1– 3

E

1– 3

1– 3

1– 3

1– 3

1– 2

1– 4

 

1– 3

1– 2

1

1– 3

A

36 — 1

1 36 5 18 25 36 11 36

11 a

C

1– 3

t = outcome of 3

a

3 Outcomes Probability 1– 3

25 —

t't'

2

1

1

5

4 36

7 a

12 a

c 25

48

c 77

b 125 b D

3 77 1 37

b 77

1 5 1 17 26 145

b

1

8 a 9 0.9 1 10 14 11 a

64

16

5 a 25 6 a C

b 1369

b

1 5 1 221 136 435

1

8

4

d 25 d

18 77

73

c 1369

1

c 10

1

d 3

25

c 102 221

13 a b c 435 14 No. Coin tosses are independent events. No one toss affects the outcome of the next. The probability of a Head or Tail on a fair coin is always 0.5. Greg has a 50% chance of tossing a Head on the next coin toss as was the chance in each of the previous 9 tosses. 15 No. As events are illustrated on a tree diagram, the individual probability of each outcome is recorded. The probability of a dependent event is calculated (altered according to the previous event) and can be considered as if it was an independent event. As such, the multiplication law of probability can be applied along the branches to calculate the probability of successive events. Exercise 12E — Conditional probability 41

1 a P(J) = 90

12

b P(H  |  J) = 41 2 a P(S) = 13 30 b P(S  |  (C ß S)) = 13 28 3 a 0.3 b

3 7

4 a

9 13

b

3 5 15

5 0.58 or 26 6 0.22 or

5 23

6



ii  P(B  |  A) = 16

iii  P(C  |  A) =

1 6

iv  P(C  |  B) = 0 10 A 11 Conditional probability is when the probability of one event depends on the outcome of another event. 12 a 0.0875 b 0.065 13 a 0.585 b 0.1525 or 15.25%.

Chapter review Fluency 1 A 2 B 5 B 6 B 7 a A

3 B

4 D B

x

Answers

Answers 12D ➜ 12F

7 0.9 8 0.8375 9 a D b i  P(A  |  B) = 1

Exercise 12F — Subjective probability 1 a The outcome depends upon whether it is a Test match or a one-day game and how effective the bowlers and batsmen are; not forgetting the pitch usually favours spin bowling. b The outcome depends on which team is better on the day and which team can adjust to the conditions. c No. The third one has an equal chance of being a girl or a boy. d This is not necessarily true. Current position and form of both teams should be used as a gauge. e It does not mean it will rain again on Friday. f There is no certainty about that. It depends upon the condition and location of your house. g Cricket games are not won or lost by the attractiveness of the uniform. h It is possible to get 6 Heads in a row on a normal coin. i They will have a good chance but there is no certainty. The country with the best competitors on the day of each event will win. j This is dependent on the person’s own interests. 2 a You still have a chance. b No horse is certain to win. Lots of problems can occur on the track. c This is not true. Even though Heads and Tails have equal chances, it does not mean half the results will show Heads. d Favourites do not always win. e Sometimes outsiders pay well, if you back the right one! You can lose more money than you win. 3 Answers will vary. Class discussion required as there are many factors to consider. 4 a There is a contradiction. The job was never hers. She had to do well to win the position. b The team may have had a lead but a match is only won when finished. c No horse is certain to win. 5 Experimental probability is based on data collected from trials. The more trials undertaken, the closer the experimental probability will reflect theoretical probability.   Theoretical probability is based on mathematical models. A theoretical probability does not guarantee a particular outcome in real life situations.   Subjective probability is based on judgements and opinions and hence may be biased. Subjective probability may approach theoretical probability if the assigned probability is based on real experiences and judgements made from an objective and educated position. 6 Answers will vary. Class discussion may be required. Example only: medical — our town is so far away from any major airports that it is unlikely our residents will need immunisation from swine flu.

849

b

A

B

14 a

x

Die 2 outcomes 2

1

x

A

B

8 a 107 9 a 4–7 b 1–6 c 8–25

b 27

c 175

3 7

b 47

c $28

f i  g 50 15 a

6

(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)

1 36 1 36

ii ii

1 52

11 a W  hether it rains or not on Thursday is not determined by what happened on Monday, Tuesday or Wednesday. It can still rain on Thursday. b The team’s win or loss depends upon how other players bat and bowl or how the other team plays. c There is an equal chance of having a boy or a girl. 12 a If you were defeated, the opponent was the winner. b The slowest motocross rider could not win the race if he/she crossed the finish line first. c The person elected was the most popular choice for the position. 13 a i  50 ii 7 iii 25 iv 8 3 6 b i  12 ii 50 iii 25 c i 

n(x ) = 50 Fried rice

Chicken wings

4

10 6

2 11

0

Die 1 outcomes

and P(B) =

1 6

1 ii 25

b c d e 16 a

Dim sims

6

7

8

9

10 11 12

4

5

6

5

4

3



iii 181 iii 181

1 6 1 6



2

2

1

3

(0, 0) (0, 1) (0, 2) (0, 3)

1 (1, 0) (1, 1) (1, 2) (1, 3) 2

(2, 0) (2, 1) (2, 2) (2, 3)

3

(3, 0) (3, 1) (3, 2) (3, 3)

No 0 and 6 3 0 and 6, 1 and 5, 2 and 4 1

1– 8

7– 8

f

fÅ

2 1– 8

f

7– 8

fÅ

1– 8

f

7– 8

fÅ

3 f fÅ

1– 8 7– 8 7– 8 7– 8

1– 8 1– 8 1– 8

7– 8

ffÅf ffÅfÅ

f fÅ

fÅff fÅffÅ

f fÅ

fÅfÅf fÅfÅfÅ

ii 343 512

21 512

11 iv 256

iii  17 a

Outcomes Probability 1 —– fff 512 7 —– fffÅ 512

f fÅ

f = outcome of 5 1 b   i  512

Die outcomes

12 5

5

Die 2 outcomes

1 Coin outcomes

1 2

4 13

Answers

(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)

0

c 23 10 a No b P(A) = 14 , P(B) = 131 , P(A ¶ B) =

850

5

3

d 43

c

(4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)

2

c 132

b 85 9 a Yes b P(A) =

4

Frequency 1

b 14

8 a

(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)

4

5 a  131

7775 7776 3 8

3

3

e i 

b

(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)

2

4 B

7 a

2

Sum

3 D

1 7776

6

b 6 c No. Frequency of numbers is different. d

Problem solving 1 C 2 D

6 a

5

(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)

 

C

4

1 Die 1 outcomes

c

3

2

3

4

H

(H, 1) (H, 2) (H, 3) (H, 4)

T

(T, 1) (T, 2) (T, 3) (T, 4)

7 —– 512 49 —– 512 7 —– 512 49 —– 512 49 —– 512 343 —– 512

1

b 1– 4

H 1– 2

1– 4 1– 4

1– 4

1– 4

1– 2

T

1– 4 1– 4

1– 4

b

1 4 1 19 1 169 1 221

20 a

15 25

=

3 5

b

8 10

=

4 5

c 18 a 19 a

4 a Mean = 2.5, median = 2.5 b Mean = 4.09, median = 3 c Median 2 5 a 72 3 b 73 c 70 –