Ace Education Mathematics O'level (Ace Education Book Series) 9982913905, 9789982913904

Citation preview

ACE EDUCATION MATHEMATICS O’LEVEL

Grade 10 − 12 GCSE, GCE

NSWANA CHING’AMBU

The Authorship and Career Network [email protected] +260976008283, +260972719373

i

The Authorship and Career Network ‘Impossibility Our Possibility’ The Authorship and Career Network is an organisation with a sole objective of promoting excellence in research, literacy, scholarship, education and skill development by supporting authors in Africa and beyond to publish their works.

Published for Africa by The Authorship and Career Network Indeco House, Cairo Road, Lusaka, Zambia.

Copyright © 2021 by Nswana Ching’ambu

All rights reserved. No part of this publication may be reproduced, stored or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning, or otherwise without written permission from the publisher. It is illegal to copy this book, post it to a website, or distribute it by any other means without permission.

Nswana Ching’ambu asserts the moral right as an author

First edition ISBN: 978-9982-913-90-4

ii

To the students Ace Education Book Series aims at giving detailed material in the simplest way to help students understand and recall information easily. The books also highlight the importance and application of each topic in real life so that students can understand why they are learning the material, how the material relates to or can be used in real life. Ace Education Mathematics O’level consists of 33 units. Each unit begins with the introduction and overview of the unit, and ends with the review questions and solutions, except few introductory units. To fully benefit, students are advised to cover everything in each unit considered. Students can also learn more from our social media platforms. On these platforms, students can find additional information, ask questions or participate in helping other students. With full confidence, this book will help a number of students, not just by boosting their scores, but also to understand Mathematics Ordinary Level.

Ace Education

About the author Nswana Ching’ambu has a bachelor’s degree in Human Biology from the University of Zambia and Teaching Methodology from Gideon Robert University. He has been lecturing Clinical Medicine courses at Gideon Robert University for many years. He lectures Anatomy and Physiology, Medical Biochemistry and Nutrition, and previously used to lecture Cellular Pathology and Basic Microbiology. He is the founder of Ace Education and the author of Ace Education Book Series for O’Levels: Biology, Chemistry, Physics and Mathematics. +260967744388 [email protected] [email protected] [email protected]

iii

TABLE OF CONTENTS

UNIT 1: NUMBERS………..……………………………………………………………………………...…..1 UNIT 2: SETS……………………………………..…………………………………………...………..…....12 UNIT 3: LAWS OF ARITHMETIC…………………...……………………………………………….…...21 UNIT 4: FRACTIONS………………………………………………………………………………….…….24 UNIT 5: PERCENTAGES…………………………………………………………...………………....…….29 UNIT 6: RATIO AND PROPORTION……………….………………………………………….....………34 UNIT 7: INDICES………………...………………………………………………………………………….40 UNIT 8: SEQUENCES AND SERIES………..………………………………………………….…………47 UNIT 9: ALGEBRA………………..……………………………………………………………….………..56 UNIT 10: LINEAR EQUATIONS………………...…………………………………………….…………..71 UNIT 11: COORDINATE GEOMETRY..……………………………………………………….………….82 UNIT 12: QUADRATIC EQUATIONS……………………………….……………………….……………92 UNIT 13: LINEAR INEQUALITIES……………………………….………………………..…………….108 UNIT 14: RELATIONS AND FUNCTIONS………………………...……………………………..……..123 UNIT 15: LINEAR FUNCTIONS…………….……………………………………………………………136 UNIT 16: GRAPHS OF LINEAR FUNCTIONS…………………………………………………………142 UNIT 17: GRAPHS OF CUBIC FUNCTIONS……………..……………………………………………156 UNIT 18: MOTION GRAPHS…………..………………………………………………………..………..166 UNIT 19: MATRICES…………..………………………………………………………………..…………174 UNIT 20: MENSURATION………………………..……………………………………………...………..190 UNIT 21: ANGLES…………………………………..……………………………………………..………204 UNIT 22: CIRCLE THEOREMS……………………………………………………………………..……211 UNIT 23: POLYGONS………………………….……………………………………………………...…...221 UNIT 24: TRIANGLES……………..…………………………………………………………………...….225 UNIT 25: SYMMETRY…………………………………………………………………………………….230 UNIT 26: CONSTRUCTION…………………………………………………………………….…………234 UNIT 27: EARTH GEOMETRY……………………...…………………………………………………...241 UNIT 28: PYTHAGORAS’ THEOREM……………………………………………………………...….. 257 UNIT 29: TRIGONOMETRY………..……………………………………………………….…………….260 UNIT 30: VECTORS…………………………………………………………………………………….….277 UNIT 31: TRANSFORMATIONS..………………………………………………………………………...297 UNIT 32: STATISTICS…………...………………………………………………………………….……..309 UNIT 33: PROBABILITY…..………………………………………………………………………………329

iv

1.0

NUMBERS

Introduction We always use mathematics in our everyday lives. For example, you go to a store to buy exercise books. How many will be purchased and for how much? Math. You wake up and want to plan your day. What time will you do that and for how long? Math. You are cooking. How much water, salt and seasoners to add? Math. You are building a house. How wide and high will it be? It is math again. In short, we knowingly or unknowingly apply mathematics in most of our everyday life activities. Undoubtedly, without mathematics life can be chaotic. We cannot solve even the simplest problem nor can we communicate with each other clearly. Mathematics is a language of science. It deals with numbers, quantities, shapes and how they relate to each other. Any language has the alphabet, words, sentences, punctuation marks and rules of grammar. Mathematics being a language also has these. The alphabet of mathematics are numbers. Words of mathematics are expressions, such as 1 cup of water, 3 teaspoons of sugar, etc. Sentences of mathematics are equations such as moving at 30km/hr we will cover 90km in 3 hours. The punctuation marks of mathematics are signs and symbols such as +, −, × and ÷. Mathematics also has sets of rules which govern how numbers, expressions, equations and signs have to be used. What are numbers? They are a way of expressing the quantity of something. This unit covers different sets of numbers. It also covers multiples, factors, squares and reciprocal of a number and many more.

Specific outcomes By the end of the unit, you will be able to: ❖ Describe the sets of numbers: integers, natural numbers, whole numbers, rational numbers, irrational numbers and real numbers ❖ Find the lowest common multiple (LCM) for the given numbers ❖ Find the highest common factor (HCF) for the given numbers ❖ Find the prime factors of a numbers ❖ Find a square number and cube number of a numb ❖ Find the square root and cube root ❖ Find the reciprocal of a number ❖ Work out significant figures ❖ Standard form: • Convert standard notation to standard form • Convert standard form to standard notation • Add and subtract standard form • Multiply and divide standard form

Numbers

1

NUMBERS 1.1

SETS OF NUMBERS 1. INTEGERS • Integers are numbers that are expressed without a fraction. • They are from negative infinity to positive infinity. • The symbol for integers: ℤ • Examples of integers: …, −3, −2, −1, 0, 1, 2, 3, … 2. POSITIVE INTEGERS • Positive integers begin from one to infinity. • The symbol for positive integers: ℤ+ • Examples of positive integers: 1, 2, 3, 4, 5, … 3. NEGATIVE INTEGERS • Negative integers begin from infinity to negative one. • The symbol for negative integers: ℤ− • Examples: −1, −2, −3, −4, −5, … 4. NATURAL NUMBERS • Natural numbers are positive integers from one to infinity. • They are used for counting. • The symbol for natural numbers: ℕ • Examples: 1, 2, 3, 4, 5, … 5. WHOLE NUMBERS • Whole numbers are positive integers including zero. • The symbol for whole numbers: 𝕎 • Examples of whole numbers: 0, 1, 2, 3, 4, 5, … 6. RATIONAL NUMBERS 𝑎 • Rational numbers are numbers that can be expressed as 𝑏 , where a and b are integers, and b ≠ 0 • The symbol for rational numbers: ℚ 1 5 7 • Examples of rational numbers: 2, 3, − 8, … 7. IRRATIONAL NUMBERS 𝑎 • Irrational numbers are numbers that cannot be expressed as 𝑏 , where a and b are integers, and b ≠ 0 • The symbol for irrational numbers: 𝕀 • Examples of irrational numbers: √2, √3, √5, π, e 8. REAL NUMBERS • Real numbers are a set of rational numbers, irrational numbers, and integers. • The symbol for real numbers: ℝ 5 • Examples of real numbers: −2, 0, 4, 13, √2, …

2

Numbers

❖

WORKED EXAMPLES 1. Identify the following numbers as rational, irrational or integer 2 A. 3 B. −10 C. √15 D. 2.5 E. 10000 F. √7 G. 0 H. −0.2

❖

SOLUTIONS 1. A. Rational number B. Integer C. Irrational number D. Rational number E. Integer F. Irrational number G. Integer H. Rational number

1.2

MULTIPLES • Multiples are numbers that are a product of a number by another integer. • Examples: i. Multiples of 2 are 2, 4, 6, 8, 10, …; 2 is a product of 2 and 1. 4 is a product of 2 and 2 and so on. ii. Multiples of 3 are 3, 6, 9, 12, 15, …; 6 is a product of 3 and 2. 12 is a product of 3 and 4. iii. Multiples of 4 are 4, 8, 12, 16, 20, …; 12 is a product 4 and 3, 20 is a product of 4 and 5. ❖ LOWEST COMMON MULTIPLE (LCM) • The lowest common multiple is the smallest number which is a multiple of both given numbers. • Examples: i. Common multiples of 2 and 3 are 6, 12, 18, 24, … The lowest common multiple of 2 and 3 is 6. ii. Common multiples of 4 and 5 are 20, 40, 60, … The lowest common multiple of 4 and 5 is 20. iii. Common multiples of 3 and 6 are 6, 12, 18, 24, … The lowest common multiple of 3 and 6 is 6.

1.3

FACTORS • A factor is a number that divides into another number without leaving a remainder. • Examples: i. Factors of 12 are 1, 2, 3, 4, 6 and 12. ii. Factors of 18 are 1, 2, 3, 6, 9 and 18. ❖ HIGHEST COMMON FACTOR • The highest common factor is the largest number which is a factor of both given numbers. • Examples: i. Common factors of 12 and 18 are 1, 2, 3 and 6. The highest common factor of 12 and 18 is 6. ii. Common factors of 20 and 30 are 1, 2, 3, 5 and 10. The highest common factor of 20 and 30 is 10. iii. Common factors of 9 and 18 are 1, 3 and 9. The highest common factor of 9 and 18 is 9. Numbers

3

1.4

PRIME FACTORISATION ❖ PRIME NUMBER • A prime number is a number that can be divided only by one and itself. • Examples of prime numbers: 2, 3, 5, 7, 13, 17, 19, 23, 31, … ❖ PRIME FACTOR • A prime factor is a factor that is also a prime number. • Prime factors of 15 are 3 and 5; 3 and 5 are factors of 15 and they are also prime numbers. • Prime factors of 30 are 2, 3 and 5. ❖ PRIME FACTORISATION • Prime factorisation is the process of expressing a number into products of prime numbers. • Examples: i. Prime factorisation of 30 = 2 × 3 × 5

ii. Prime factorisation of 48 = 2 × 2 × 2 × 2 × 3

iii. Prime factorisation = 280

4

Numbers

1.5

SQUARE AND CUBE NUMBERS 1. SQUARE NUMBERS • A square number is the product of a number multiplied by itself. • Examples of square numbers: i. 9 is a square number; a product of 3 × 3. ii. 36 is a square number; a product of 6 × 6. iii. 16 is a square number; a product of 4 × 4. iv. 144 is a square number; a product of 12 × 12. • Squares are written as 𝑎2 where a is a real number. i. 32 = 9 ii. 62 = 36 iii. 42 = 16 iv. 122 = 144 2. CUBE NUMBERS • A cube number is the product of a number multiplied by itself three times. • Examples of cube numbers: i. 27 is a cube number; a product of 3 × 3 × 3. ii. 8 is a cube number; a product of 2 × 2 × 2. iii. 64 is a cube number; a product of 4 × 4 × 4. iv. 125 is a cube number; a product of 5 × 5 × 5. • Cube numbers are written as 𝑎3 where a is a real number. i. 23 = 8 ii. 43 = 64 iii. 53 = 125

1.6

SQUARE AND CUBE ROOTS 1. SQUARE ROOTS • A square root is a number which when multiplied by itself twice gives a number. • Examples of square roots: i. 4 is a square root of 16; 4 × 4 gives 16. ii. 3 is a square root of 9; 3 × 3 gives 9. iii. 5 is a square root of 25; 5 × 5 gives 25. iv. 7 is a square root of 49; 7 × 7 gives 49. • Square roots are written as √𝑎 i. √16 = 4 ii. √9 = 3 iii. √25 = 5 iv. √49 = 7 2. CUBE ROOTS • A cube root is a number which when multiplied by itself three times gives a number. • Examples of cube roots: i. 2 is a cube root of 8; 2× 2 × 2 gives 8. ii. 3 is a cube root of 27; 3 × 3 × 3 gives 27. iii. 4 is a cube root of 64; 4 × 4 × 4 gives 64. iv. 5 is a cube root of 125; 5 × 5 × 5 gives 125. 3 • Cube roots are written as √𝑎 3 i. √8 = 8 3 ii. √27 = 3 3 iii. √64 = 4 3 iv. √125 = 5 Numbers

5

1.7

RECIPROCAL 1 • A reciprocal is a number which when multiplied by a fraction gives one, that is, a × 𝑎 = a. • •

The reciprocal of a number or a fraction is found by flipping over a number or a fraction. 𝑎 𝑏 1 The reciprocal of 𝑏 = 𝑎, reciprocal of b = 𝑏.

❖ WORKED EXAMPLES 1. Find the reciprocal of the following 2 A. 3 B.

50 7

C. 10 D. √2 ❖ SOLUTIONS 3 1. A. 2 B. C. D. 1.8

7 50 1 10 1 √2

SIGNIFICANT FIGURES 1. RULES OF SIGNIFICANT FIGURES 1. All non-zero digits are significant. For example, 157 has three significant figures. 2. Zeros between non-zero digits are significant. For example, 2001 has four significant figures. 3. Trailing zeros are not significant, but they are if there is a decimal point, for example: i. 200 has one significant figure while 200. has three significant figures. ii. 1500 has two significant figures while 5.100 has four significant figures. 4. Terminal zeros to the right of the decimal point are significant, for example: i. 0.100 has three significant figures. ii. 2.100 has four significant figures. iii. 1.00 has three significant figures. 5. Leading zeros are not significant, for example: i. 05 has one significant figure. ii. 0.0013 has two significant figures. ❖ WORKED EXAMPLES 1. How many significant figures are in each of the following? A. 1.0030 B. 28.20 C. 500720 D. 0.00021 E. 720000 ❖ SOLUTIONS 1. A. 1.0030 has five significant figures. B. 28.20 has four significant figures. C. 500720 has five significant figures. D. 0.00021 has two significant figures. E. 720000 has two significant figures.

6

Numbers

2.

HOW TO ROUND OFF TO SIGNIFICANT FIGURES 1. Count to the number of significant figures required. 2. If the first nonsignificant is less than 5, drop all nonsignificant digits, for example: i. 5726 to two significant figures = 5700 (2 is the first nonsignificant figure. It is less than 5). ii. 94530 to one significant figure = 90000 (4 is the first nonsignificant figure. It is less than 5). 3. If the first nonsignificant figure is five or greater, add one to the last significant digit and drop all the nonsignificant digits, for example: i. 47653 to 2 significant figures = 48000 (6 is the first nonsignificant figure. It is greater than 5). ii. 47.653 to 3 significant figures = 47.7 (5 is the first nonsignificant figure). iii. 47.653 to 2 significant figures = 48 (6 is the first nonsignificant figure. It is greater than 5).

1.9

STANDARD FORM • A standard form is a way of writing very large or small numbers. • Standard forms are written as the following general formula. A × 𝟏𝟎𝒏

where A = 1 to 9 n = integer

❖ HOW TO CONVERT FROM STANDARD NOTATION TO STANDARD FORM • To write in standard form, move the decimal point until A equals a number between 1 to 9. • Count the number of moves. • If the decimal point moves to the left n should have a positive sign. • If the decimal point moves to the right n should have a negative sign. ❖ HOW TO CONVERT FROM STANDARD FORM TO STANDARD NOTATION • Move the decimal point the number of places based on n. • If n is positive, move the decimal point to the right. • If n is negative, move the decimal point to the left. ❖ WORKED EXAMPLES 1. Convert the following standard notation to standard form. A. 5000 B. 4399 C. 542000000 D. 0.000542 E. 0.00000952 2. Convert the following standard form to standard notation. A. 1 × 102 B. 7.43 × 104 C. 7.43 × 10−4 D. 3 × 10−2

Numbers

7

❖

SOLUTIONS 1. A. B. C. D. E. 2. A. 1 × 102 = 100 B. 7.43 × 104 = 74300 C. 7.43 × 10−4 = 0.000743 D. 3 × 10−2 = 0.03

❖

ADDITION AND SUBTRACTION OF STANDARD FORMS • Standard forms can be added or subtracted when they have the same power. • For A × 10𝑛 and B × 10𝑛 . A × 𝟏𝟎𝒏 + B × 𝟏𝟎𝒏 = (A + B) × 𝟏𝟎𝒏 A × 𝟏𝟎𝒏 − B × 𝟏𝟎𝒏 = (A − B) × 𝟏𝟎𝒏

❖ WORKED EXAMPLES 1. Calculate the following standard forms. A. 4 × 109 + 2.5 × 109 B. 6.9 × 1020 + 1.5 × 1020 C. 4 × 109 − 2.5 × 109 D. 6.9 × 1020 − 1.5 × 1020 ❖ SOLUTIONS 1. A. 4 × 109 + 2.5 × 109 = (4 + 2.5) × 109 = 6.5 × 109 B. 6.9 × 1020 + 1.5 × 1020 = (6.9 + 1.5) × 1020 = 8.4 × 1020 C. 4 × 109 − 2.5 × 109 = (4 − 2.5) × 109 = 1.5 × 109 D. 6.9 × 1020 − 1.5 × 1020 = (6.9 − 1.5) × 1020 = 5.4 × 1020

8

Numbers

❖

MULTIPLICATION AND DIVISION OF STANDARD FORM 1. MULTIPLICATION • For A × 10𝑛 and B × 10𝑚 multiply A and B and add powers. (A × 𝟏𝟎𝒏 ) × (B × 𝟏𝟎𝒎 ) = (A × B) × 𝟏𝟎𝒏 + 𝒎

2. DIVISION • For A × 10𝑛 and B × 10𝑛 divide A and B and subtract powers. (A × 𝟏𝟎𝒏 ) ÷ (B × 𝟏𝟎𝒎 ) = (A ÷ B) × 𝟏𝟎𝒏 − 𝒎 ❖ WORKED EXAMPLES 1. Multiply the following standard forms. A. (4 × 109 ) × (2 × 103 ) B. (1.2 × 106 ) × (2.3 × 108 ) C. (4.5 × 104 ) × (8 × 106 ) 2. Find the value of the following standard forms. A. (4 × 109 ) ÷ (2 × 103 ) B. (2.3 × 108 ) ÷ (1.2 × 106 ) C. (4.5 × 104 ) ÷ (8 × 106 ) ❖ SOLUTIONS 1. A. (4 × 109 ) × (2 × 103 ) = (4 × 2) × 109 + 3 = 8 × 1012 B. (1.2 × 106 ) × (2.3 × 108 ) = (1.2 × 2.3) × 106 + 8 = 2.76 × 1014 C. (4.5 × 104 ) × (8 × 106 ) = (4.5 × 8) × 104 + 6 = 36 × 1010 = 3.6 × 1011

If A is more than 10, move the decimal point to the left to make it between 1 − 10. The power changes based on the number of moves

2. A. (4 × 109 ) ÷ (2 × 103 ) = (4 ÷ 2) × 109 − 3 = 2 × 106 B. (2.3 × 108 ) ÷ (1.2 × 106 ) = (2.3 ÷ 1.2) × 108 − 6 = 1.92 × 102 C. (4.5 × 104 ) ÷ (8 × 106 ) = (4.5 ÷ 8) × 104 − 6 = 0.56 × 10−2 = 5.6 × 10−3

If A is less than 1, move the decimal point to the right to make it between 1 − 10. The power changes based on the number of moves

Numbers

9

REVIEW QUESTIONS 1. Identify the following numbers as integers, rational or irrational. A. √17 B. −0.002 C. 5000 1 D. 100 2. Find the lowest common multiple of the following. A. 6 and 9 B. 7 and 8 C. 10 and 20 3. Find the highest common factor of the following. A. 12 and 16 B. 15 and 20 C. 3 and 6 4. What is the prime factorisation of the following? A. 20 B. 150 C. 200 5. Evaluate the following. A. 52 B. 53 C. √225 3 D. √512 6. Find the reciprocal of the following. 1 A. 3

7.

8.

9.

10.

11.

12.

10

B. 0.5 How many significant figures are in each of the following? A. 5.20 B. 520 C. 0.052 D. 0.00052 Round of the following to one significant figure. A. 76 B. 74 Convert the following to standard form. A. 0.00012 B. 12000 Convert the following to standard notation. A. 2.465 × 102 B. 2.465 × 10−2 Find the value of the following standard forms. A. 2.5 × 1015 + 1 × 1015 B. 2.5 × 1015 − 1 × 1015 Find the value of the following standard forms. A. 4 × 1012 × 5 × 1020 B. 4 × 1012 ÷ 5 × 1020

Numbers

SOLUTIONS 1. A. Irrational B. Integer C. Integer D. Rational 2. A. 18 B. 56 C. 20 3. A. 4 B. 5 C. 3 4. A. 20 = 2 × 2 × 5 B. 150 = 2 × 3 × 5 × 5 C. 200 = 2 × 2 × 2 × 5 × 5 5. A. 52 = 5 × 5 = 25 B. 53 = 5 × 5 × 5 = 125 C. √225 = 15 3 D. √512 = 8 6. A. 3 B. 2

0.5 =

1 2

7. A. 3 significant figures B. 2 significant figures C. 2 significant figures D. 2 significant figures 8. A. 80 B. 70 9. A. B.

1.2 × 10−4 1.2 × 104

10. A. 246.5 B. 0.02465 11. A. 3.1 × 1015 B. 1.5 × 1015 12. A. 2 × 1033 B. 8 × 10−9

Numbers

11

2.0

SETS

Introduction When you observe the kitchen, you will likely find that related objects are kept together. Plates are kept together, as well as pots and spoons. As they found together, we can collectively call them; a set of plates, a set of pots and a set of spoons. These sets are then put in the same or different compartments. We use sets in our everyday lives. Whenever we group similar things together, we are applying sets in real life. So, what is a set? A set is a collection of well defined objects. In this unit, you will learn about the types of sets, how to describe a set and many more.

Specific outcomes By the end of this unit, you will be able to: ❖ Define a set ❖ Represent a set in the following ways: • Descriptive form • Tabular form • Set builder notation ❖ Describe the following types of sets: • Empty set • Singleton set • Finite set • Infinite set ❖ Represent the following sets in a Venn diagram: • Subset • Complement of a set • Union of sets • Intersection of sets

12

Sets

SETS 2.1

SETS • A set is a collection of well-defined objects. • Objects in a set are called elements or members of a set. • Elements can be anything, for example, numbers, letters, words or people. • Sets are denoted by capital letters. • Members or elements can be represented by numbers, lowercase letters or words. • Elements are separated by commas and enclosed in curly brackets. • For example: A = {1, 2, 3, 4}; a set of natural numbers C = {red, yellow, blue, white}; a set of colours V = {𝑎, 𝑒, 𝑖, 𝑜, 𝑢}; a set of vowels

2.2

REPRESENTATION OF SETS • A set can be represented in various forms. • The commonest ways of representing sets are descriptive form, tabular form and set builder notation. 1. DESCRIPTIVE FORM • In descriptive form, a set is defined by describing the contents of a set. • For example: A = a set of first five natural numbers C = a set colours of the rainbow V = a set of vowels 2. TABULAR FORM • In tabular form (also called roster form), elements are listed separated by commas and enclosed in curly brackets. • For example: A = {1, 2, 3, 4, 5} C = {red, orange, yellow, green, blue, indigo, violet} V = {𝑎, 𝑒, 𝑖, 𝑜, 𝑢} 3. SET BUILDER NOTATION • In set builder notation, the common property of the elements is specified. • The notation is written in the form of: A = {𝑥 | common property that defines 𝑥} read as “set A is a set of all x’s such that x…” • For example: A = {𝑥 | 𝑥 is a natural number less or equal to 5} read as “set A is a set of all x’s such that x is a natural number less than or equal to 5” this can also be written as A = {𝑥 ∈ ℕ| 𝑥 ≤ 5} C = {𝑥 | 𝑥 is a colour of a rainbow} V = {𝑥 | 𝑥 is a vowel}

Sets

13

2.3

ELEMENTS OF A SET • The symbol ∈ means “is an element of” or “is a member of”. • For example, given that set N = {1, 2, 3, 4, 5}. 2 ∈ N, read as “2 is an element of set N” 5 ∈ N, read as “5 is an element of set N” • The symbol ∉ means “is not an element of” or “is not a member of”. • Given that set N = {1, 2, 3, 4, 5}. 0 ∉ N, read as “0 is not an element of set N”. 8 ∉ N, read as “8 is not an element of set N”. • n(A) indicates the number of elements in set A. For example, given that: N = {1, 2, 3, 4, 5}, then n(N) = 5, meaning that set N has 5 elements P = {0, 1}, then n(P) = 2, meaning that set P has 2 elements

2.4

TYPES OF SETS 1. EMPTY SET • An empty set is a set that does not contain any element. • It is also called a null set. • It is denoted by ∅ or { }. A={} 2. SINGLETON SET • A singleton set is a set that contains only one element. • For example: B = {4} P = { 2} R = {20} • Each of the sets above is a singleton set. Each set contains only one element. 3. FINITE SET • A finite set is a set that contains a countable number of elements. • For example: N = {1, 2, 3, 4, 5}; set N has a countable number of elements, that is, five elements. D = {days of the week}; set D has a countable number of elements. A week has seven days. 4. INFINITE SET • An infinite set is a set that contains an uncountable number of elements. • For example: A = {… , 2, 1, 0, 1, 2, … }; set A is the set of all integers. Integers are uncountable. The three dots (…) means “continue on”. P = {𝑥 | 𝑥 is a prime number greater than 2}; set P is a set of prime numbers. Prime numbers above 2 are uncountable.

14

Sets

2.5

UNIVERSAL SET AND SUBSETS ❖ UNIVERSAL SET • A universal set is a set that contains all elements under consideration. • It is denoted by U, and sometimes by 𝜉. • If A and B are two sets, such that set A = {1, 3, 5, 7, 9} and set B = {2, 4, 6, 8, 10}, then the universal set of set A and set B is U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}.

❖ SUBSETS • A subset is a set in which all its elements are members of another set. • It is denoted by ⊆, read as “is a subset of”. • Set A is a subset of set B if all the elements of set A are members of set B. A⊆B • For example, if set Q = {1, 2, 3, 4, 5} and set P = {2, 4} then set P is a subset of set Q, as all the elements of set P are also elements of set Q. This is written as shown: P ⊆ Q read as “set P is a subset of set Q”. • Set A can be equal to set B. However, if set A is not equal to set B then set A is a proper subset of set B. • A proper subset is a set in which all its elements are also members of another set but the other set contains elements it does not contain. • In the example above, set P is a proper set of set Q. All elements of set P are also members of set Q, but set Q contain members such as 1, 3 and 5, that are not members of set P. Proper subset is denoted by ⊂. • The set itself and the empty set are always subsets. • If a set contains at least one element that is not in the other set then the set is not a subset. • For example, if set C = {2, 4, 6} and D = {1, 2, 3, 4, 5}, set C is not a subset of set D. Set C has an element, element 6 that is not a member of set D. • The symbol ⊈ is used to indicate “not subset of”. Then C ⊈ D, and it is read “set C is not a subset of set D”.

❖ EXAMPLES 1. Write down the universal set of the following sets. A. J = {1, 2, 3} and K = {2, 4} B. L = {2, 5, 7, 9, 11, 13} and M = {2, 5, 7, 11} C. N = {2, 4, 6, 8} and O = {2, 4, 6, 8} D. P = { 1, 4, 3, 0, 1, 2, 5,6} and Q = {0, 3, 1, 7} 2. Indicate the relationship of each of the pair of set has a subset of, proper subset of or not subset of. A. J = {1, 2, 3} and K = {2, 4} B. L = {2, 5, 7, 9, 11, 13} and M = {2, 5, 7, 11} C. N = {2, 4, 6, 8} and O = {2, 4, 6, 8} D. P = {6, 1, 4, 3, 0, 1, 2, 5} and Q = {0, 3, 1, 7}

Sets

15

❖

SOLUTIONS 1. A. U = {1, 2, 3, 4} B. U = {2, 5. 7, 9, 11, 13} C. U = {2, 4, 6, 8} D. U = { 1, 0, 1, 2, 3, 4, 5, 6, 7} 2. A. B. C. D.

2.6

K ⊈ J (K is a proper subset of J) M ⊂ L (M is a proper subset of L) O ⊆ N (O is a subset of N) Q ⊈ P (Q is not a subset of P)

VENN DIAGRAM • A Venn diagram shows the relationship of sets in the form of a diagram. • It consists of a rectangle and circles within it. • The triangle represents the universal set and is usually indicated by the symbol 𝜉. • The circles within the triangle indicate subsets.

•

In the Venn diagram above, the universal set 𝜉 = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. Set A = {1, 2} and set B = {3, 4, 5}. Elements that are not in set A and set B but are members of the universal set are placed outside the circles but inside the triangle.

❖ SUBSETS • To represent a subset in a Venn diagram, a subset is placed inside another circle. The shaded region shows B ⊆ A

•

Example, given that 𝜉 = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. Set A = {1, 2, 3, 4, 5, 6} and set B = {1, 2, 3}. Show the sets in a Venn diagram. 7 to 10 elements are placed outside both circles but inside the rectangle. They are not members of set A and B, but are members of 𝜉.

16

Sets

❖

COMPLEMENT OF A SET • The complement of set A is a set of all elements in the universal set that are not in set A • It is denoted by A'. The shaded region shows A'

•

Example, given that 𝜉 = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} and set A = {1, 2, 3, 4}. Show the sets in a Venn diagram and find A'. 𝜉 = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} A = {1, 2, 3, 4}

A' = {5, 6, 7, 8, 9, 10}

❖

UNION OF SETS • A union of sets is a set formed by combining all elements of the given sets. • It is denoted by ∪. • Union of two sets A and B is obtaining by combining all elements of set A and set B, and it is denoted by A ∪ B. The shaded region shows A ∪ B

•

Example, given that 𝜉 = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} and set A = {2, 3, 4, 5, 6} and B = {5, 6, 7, 8}. Show the sets in a Venn diagram and find A ∪ B.

A ∪ B = {2, 3, 4, 5, 6, 7, 8} Sets

17

❖

INTERSECTION OF SETS • The intersection of sets is a set of common elements which belong to the given sets. • It is denoted by ∩. The shaded region shows A ∩ B

•

Example, given that 𝜉 = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} and set A = {2, 3, 4, 5, 6} and B = {5, 6, 7, 8}. Show the sets in a Venn diagram and find A ∩ B.

A ∩ B = {5, 6}

❖

18

EXAMPLES 1. Name the shaded region for the Venn diagram.

Sets

A.

B.

C.

D.

2.

3.

Name the region defined by the following shading. A.

B.

C.

D.

E.

F.

Given the following Venn diagrams

Diagram 1

Diagram 3

Diagram 2

Diagram 4 Sets

19

Diagram 5 Match the following description with the diagrams above A. P ∩ Q B. (P ∪ Q) ∩ R' C. P ∩ Q ∩ R D. Q ∩ (P ∪ R)' E. Q ∩ R ∩ P' F. Q ∪ R

❖

20

SOLUTIONS 1. A. A B. A' C. A ∪ B D. A ∩ B 2. A. B. C. D. E. F.

(A ∪ B)' or A' ∩ B' (A ∩ B)' or A' ∪ B' A ∩ B' A ∪ B' or (A' ∩ B)' (A' ∩ B') ∪ (A ∩ B) (A ∩ B') ∪ (A' ∩ B)

3. A. B. C. D. E. F.

Diagram 5 Diagram 4 Diagram 1 Diagram 6 Diagram 2 Diagram 3

Sets

Diagram 6

3.0

LAWS OF ARITHMETIC

Introduction In mathematics, addition (+), subtraction (−), multiplication (×) and division (÷) are called operations. These are used to add, subtract, multiply and divide numbers respectively. Arithmetic is the study of numbers using operations. We use arithmetic in our everyday lives. Whenever you add, subtract, multiply or divide something you are performing arithmetic. How do you work out arithmetic? What if there is more than one operation in the problem which should be performed first and should follow? This unit covers these questions.

Specific outcomes By the end of this unit, you will be able to: ❖ State: • cumulative law of arithmetic • associative law of arithmetic • distributive law of arithmetic ❖ Calculate arithmetic problems using BODMAS or PEDMAS

Laws of Arithmetic

21

LAWS OF ARITHMETIC 3.1

COMMUTATIVE LAW • Commutative law states that the order in which two numbers are added or multiplied does not affect the result. • When numbers are swapped around when added or multiplied the answer is still the same. • Addition and multiplication are commutative. • If two numbers are a and b, the: a+b=b+a a×b=b×a

3.2

•

Examples: 4 + 5 = 9 and 5 + 4 = 9, therefore 4 + 5 = 5 + 4 4 × 5 = 20 and 5 × 4 = 20, therefore 4 × 5 = 5 × 4

•

Subtraction and division are not commutative. 4 − 5 = −1 while 5 − 4 = 1, therefore 4 − 5 ≠ 5 − 4 4 ÷ 5 = 0.8 while 5 ÷ 4 = 1.25, therefore 4 ÷ 5 ≠ 5 ÷ 4

ASSOCIATIVE LAW • Associative law states that when three numbers are added or multiplied the grouping of the numbers does not affect the result. • Addition and multiplication are associative. • If three numbers are a, b and c, then: (a + b) + c = a + (b + c) (a × b) × c = a × (b × c)

3.3

•

Examples: (2 + 3) + 4 = 9 and 2 + (3 + 4) = 9, therefore (2 + 3) + 4 = 2 + (3 + 4) (2 × 3) × 4 = 24 and 2 × (3 × 4) = 24, therefore (2 × 3) × 4 = 2 × (3 × 4)

• •

Subtraction and division are not associative. Examples: (2 − 3) − 4 = −5 while 2 − (3 − 4) = 3, therefore (2 − 3) − 4 ≠ 2 − (3 − 4) 1 8 (2 ÷ 3) ÷ 4 = 6 while 2 ÷ (3 ÷ 4) = 3, therefore (2 ÷ 3) ÷ 4 ≠ 2 ÷ (3 ÷ 4)

DISTRIBUTIVE LAW • Distributive law states that multiplying a number by a group of numbers added together is the same as doing multiplication separately. • If three numbers are a, b and c, then: a × (b + c) = (a × b) + (a × c) = ab + ac a × (b − c) = (a × b) − (a × c) = ab − ac •

22

Examples: 2 × (3 + 4) = 2 × 7 = 14 and 2 × 3 + 2 × 4 = 6 + 8 = 8, therefore 2 × (3 + 4) = 2 × 3 + 2 × 4 2 × (4 − 3) = 2 × 1 = 2 and 2 × 4 − 2 × 3 = 8 − 6 = 2, therefore 2 × (4 − 3) = 2 × 4 − 2 × 3

Laws of Arithmetic

3.4

ORDER OF OPERATIONS • Addition, subtraction, multiplication and division are operations. • If a problem has more than one operation, the order of operations determine which operation should be performed first and which one should follow. • To easily remember the order of operations, use BODMAS rule or PEDMAS rule. ❖ BODMAS RULE • According the BODMAS rule, the order of operation should be Brackets ( ), Order or 𝑥 power (𝑎 𝑥 , √𝑎), Division (÷) or Multiplication (×), Addition (+) or Subtraction (−). ❖ PEDMAS RULE • According to PEDMAS rule, the order of operation should be Parenthesis ( ), Exponent 𝑥 (𝑎 𝑥 √𝑎), Multiplication (×) or Division (÷), Addition (+) or Subtraction (−). • •

Both BODMAS and PEDMAS are fundamentally the same, so whatever you choose is accurate. For multiplication or division and addition or subtraction, the operation that comes first is what should be performed.

❖ WORKED EXAMPLES 1. 2 + 6 × 3 + 4 = 2 + 18 + 4 = 24 2. 2 + 6 ÷ 3 + 4 =2+2+4 =8 3. (2 + 6) × (3 + 4) =8×7 = 56 4. (6 − 2) ÷ (3 + 1) =4÷4 =1 5. 22 × 3 =4×3 = 12 6. 6 ÷ 2 × 3 =3×3 =9

If multiplication and division appear consecutively, perform the first operation to appear.

7. 6 × 2 ÷ 3 = 12 ÷ 3 =4 8. 3 + 5 − 2 =8−2 =6

If addition and subtraction appear consecutively, perform the first operation to appear.

9. 3 − 5 + 2 = −2 + 2 =0

Laws of Arithmetic

23

4.0

FRACTIONS

Introduction Half, three-quarters, one-third. These expressions are often used in our everyday life communication. All these expressions indicate a portion of the whole thing; for example, half of the bread, three quarters of pie, one third of the journey. Whenever are use these expressions, we are applying fractions in real life. So then, what is a fraction? A fraction is a number that represents a portion of the whole thing. How are fractions written? How do you add, subtract, multiply or divide fractions? This unit covers these questions.

Specific outcomes By the end of this unit, you will be able to: ❖ Define a fraction ❖ Know the types of fractions: • Proper fractions • Improper fractions • Mixed fractions ❖ Convert mixed fractions to improper fractions ❖ Add and subtract fractions ❖ Multiply and divide fractions ❖ Convert fractions to decimals

24

Fractions

FRACTIONS 4.1

FRACTION • A fraction is a number that represents part of a whole. • Fractions are written as follow:

𝒂

where b ≠ 0

𝒃 • •

a is called the numerator and b is called the denominator. For example, Handy has three pens. He lost one of them at school. What is the fraction of the pens he is left with? The total number of pens he had was three. The number of pens left is two. 2

Therefore, the fraction of pens left is . 3

4.2

EQUIVALENT FRACTIONS • Equivalent fractions are fractions with the same value. •

1

The following fractions are all equivalent to . 1

•

4.3

2

=

2 4

=

3 6

=

4

2

8

An equivalent fraction can be found by multiplying or dividing the numerator and denominator by the same number.

TYPES OF FRACTIONS 1. PROPER FRACTIONS • A proper fraction has a numerator that is smaller than the denominator. • Examples of proper fractions. 1

,

7

,

3

6 15 100

2. IMPROPER FRACTIONS • An improper fraction has a numerator that is greater than the denominator. • Examples of improper fractions. 5 19 125

,

4 15

,

99

3. MIXED FRACTIONS • A mixed fraction has a combination of a whole number and a proper fraction. • Examples of mixed fractions.

212, 579, 945 Fractions

25

❖

HOW TO CONVERT MIXED FRACTIONS TO IMPROPER FRACTIONS 1. Multiply the denominator and the whole number. 2. Add the product to the numerator. 3. Then, divide everything by the denominator.

❖ WORKED EXAMPLES 1. Convert the following mixed fractions into improper fractions. 1

A.

22

B.

35

C.

84

2 3

❖ SOLUTIONS 1. A.

4.4

1

22 2

B.

35

C.

84

3

2×2+1

=

2 5×3+2

=

5 4×8+3

=

4

= =

5 2 17 5 35 4

ADDITION AND SUBTRACTION OF FRACTIONS • Fractions can be added or subtracted. • To add or subtract fractions, follow the following steps. 1. Convert mixed fractions to improper fractions if necessary. 2. Find the common denominator by finding their lowest common multiple (LCM). 3. Divide the original denominator by the LCM and multiply the answer with the numerator on each fraction. 4. Add or subtract the numerator. In short, if 𝒂 𝒃

𝒂 𝒃

+

𝒄 𝒅

−

=

𝒄 𝒅

=

𝑎 𝑏

and

𝒂 𝒃

𝒂 𝒃 Fractions

+

−

𝒄 𝒃

𝒄 𝒃

=

=

𝑐 𝑑

𝒂𝒅 + 𝒃𝒄 𝒃𝒅

𝒂𝒅 − 𝒃𝒄 𝒃𝒅

5. If b = d, then:

26

=

𝒂+𝒄 𝒃

𝒂−𝒄 𝒃

are fractions, then:

❖

WORKED EXAMPLES 1. Solve the following fractions. A. B.

C.

D. E. F. G.

4.5

1 2 5 2 1 2 5 2

+ +

−

−

2 3 3 7 2 3 3 7

1

= =

=

= 2

22 + 3 1 3 1 3

+ −

5 3 5 3

= =

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

=

5

+

2

1+5 3 1−5 3

2 3

=

= 6 3

= −

= =

=

=

3+4 6

=

35 + 6 14 3−4 6

6

=

41 14

= −

35 − 6 14

(5 × 3) + (2 × 2) 2×3

7

= =

1 6

29 14 15 + 4 6

=

19 6

=2 4 3

MULTIPLICATION AND DIVISION OF FRACTIONS ❖ MULTIPLICATION OF FRACTIONS 1. Convert mixed fractions to improper fractions if necessary. 2. Multiply numerators and denominators of the fractions given. If

𝑎 𝑏 𝒂 𝒃

and ×

𝒄 𝒅

𝑐 𝑑

=

are fractions, then: 𝒂×𝒄 𝒃×𝒅

=

𝒂𝒄 𝒃𝒅

❖ DIVISION OF FRACTIONS 1. Convert mixed fractions to improper fractions if necessary. 2. Multiply the first fraction by the reciprocal of the second fraction. If

𝑎 𝑏 𝒂 𝒃

and ÷

𝒄 𝒅

𝑐 𝑑

=

are fractions, then: 𝒂 𝒃

×

𝒅 𝒄

=

𝒂𝒅 𝒃𝒄

Fractions

27

❖

WORKED EXAMPLES 1. Find the value of the following fractions. A. B.

2 3

×

1

=

5

2

2

33 × 7

2×1

=

3×5

=

11 3

×

2 7

2 15

=

11 × 2

=

3×7

22 21

2. Find the value of the following fractions. A. B.

4.6

2 3 2 3

÷ ÷

1

=

5 2

=

3

2 3 2 3

× ×

5 1 3 2

= =

2×5 3 ×1 2×3 3 ×2

= =

10 3 6 6

=1

FRACTIONS TO DECIMALS 1. Divide the numerator with the denominator. 2. If a and b are real numbers and b ≠ 0. 𝒂 𝒃

=a÷b

3. Use simple division or calculator to change a ÷ b into decimals. 4. If the decimal is a recurring decimal, draw a bar over the part that repeats. 𝒂 𝒃

𝒂 𝒃

= x.yyyyyy…, the recurring answer is written as x.𝑦̅

= x.yzyzyz…, the recurring answer is written as x.𝑦𝑧 ̅̅̅

❖ WORKED EXAMPLES 1. Change the following fractions into decimals. A. B. C. D. E.

28

Fractions

10

= 10 ÷ 5 = 2

5 1

= 1 ÷ 2 = 0.5

2

1

32 1 9

=

7 2

= 7 ÷ 2 = 3.5

= 0.11111… = 0. 1̅

1 11

̅̅̅ = 0.090909… = 0. ̅09

5.0

PERCENTAGES

Introductions Are you with a cellphone or computer? If so, what is the battery life? Likely, you will find that it isexpressed in percentage. Percentages play a significant role in our everyday lives. First, they help us make decisions. For example, if your cellphone battery life is about 10% you will likely decide to charge it. Second, percentages also help us to evaluate our performance. For example, if you score 80% on the math test, you know that you performed well. So then, what is a percentage? A percentage is a fraction with the denominator 100. The whole is 100. The percentage represents a portion of the whole; 100. If your cellphone battery life is 25%, in fraction form it can be written as 75

25 , 100

meaning it has

lost 100 or 75%. How do you convert a percentage to a fraction and vice versa? If you are given K 50 and told to spend 80%, how much should you use? This unit covers these questions and many more.

Specific outcomes By the end of this unit, you will be able to: ❖ Convert percentages to fractions ❖ Convert fractions to percentages ❖ Convert decimals to percentages ❖ Calculate the percentage of the amount ❖ Calculate reverse percentage

Percentages

29

PERCENTAGES 5.1

PERCENTAGE • A percentage is a fraction with the denominator 100. • If x is a percentage, that is x %, then: x% =

5.2

𝒙 𝟏𝟎𝟎

PERCENTAGE TO FRACTION • A percentage can be converted to a fraction by dividing it by 100%. •

If x is a percentage and x% =

𝒙% 𝟏𝟎𝟎 %

𝑎

=

𝑏

a fraction, x % can be changed into a fraction by the following.

𝒂 𝒃

❖ WORKED EXAMPLES 1. Convert the following percentages into fractions. A. 5 % =

5% 100 %

B. 25 % =

25 % 100 %

C. 100 % = D. 120 % =

5.3

1

=

100 % 100 % 120 % 100 %

20

=

1 4

= 1 = 1.2

FRACTION TO PERCENTAGE • A fraction can be converted to a percentage by multiplying by 100%. •

If

𝑎 𝑏

is a fraction and x % a percentage, then: 𝒂 𝒃

× 100 % = x %

❖ WORKED EXAMPLES 1. Convert the following fractions into percentages. 1 1 A. = × 100 % = 5 % 20

B.

1 4

20

=

1 4

× 100 % = 25 %

C. 1 = 1 × 100 % = 100 % D.

30

Percentages

112

=

112 × 100%

=

3 2

× 100 % = 150 %

5.4

PERCENTAGES TO DECIMALS • A percentage can be converted to a decimal number by dividing it by 100%. • First, convert a given percentage to a fraction then to a decimal. ❖ WORKED EXAMPLES 1. Convert the following percentages to decimals. A. 5 % =

5% 100 %

B. 25 % =

25 % 100 %

C. 250 % =

5.5

=

1

=

250 % 100 %

= 0.05

20 1 4

=

= 0.25

5 2

= 2.5

DECIMALS TO PERCENTAGES • A decimal number can be converted to percentages by multiplying it by 100%. ❖ WORKED EXAMPLES 1. Convert the following decimals to percentages. A. 0.05 = 0.05 × 100 % = 5 % B. 0.25 = 0.25 × 100 % = 25 % C. 0.002 = 0.002 × 100 % = 0.2% D. 1.5 = 1.5 × 100% = 150 %

5.6

PERCENTAGE OF AN AMOUNT 1. Convert the given percentage to a decimal. 2. Multiply the decimal by the total amount. ❖ WORKED EXAMPLES 1. What is 80 % of A. K 100 B. K 50 2. What is 60 % of A. K 100 B. K 50 3. Saviour brought a new item at the wholesale price of K 50 and later sold it at the retail price. After selling it he got a 15 % profit. A. How much was the profit? B. At what price did he sell the item? 4. Winnie has £ 200. She spends 25 % of it. How much money is left?

Percentages

31

❖

SOLUTIONS 1. A. 80 % of K 100 =

80 % 100 %

× K 100

= 0.8 × K 100 = K 80 B. 80 % of K 50 =

80 % 100 %

× K 50

= 0.8 × K 50 = K 40

2.

A. 60 % of K 100 =

60 % 100 %

× K 100

= 0.6 × K 100 = K 60 B. 60 % of K 50 =

60 % 100 %

× K 50

= 0.6 × K 50 = K 30

3.

A. Profit made =

15 % 100 %

× K 50

= 0.15 × K 50 = K 7.5 B. New price = original price + profit made = K 50 + K 7.5 = K 57.5

4.

Amount spent =

25 % 100 %

× £ 200

= 0.25 × £ 200 = £ 50 Amount left = total amount − amount spent = £ 200 − £ 50 = £ 150

32

Percentages

5.7

REVERSE PERCENTAGE • Reverse percentage is used to find the original amount after a percentage change. • The easiest way to find the original amount is by doing the ratio. 1. Find the percentage increase (if the price increased) or percentage decrease (if the price decreased). 2. If the amount increased, add the given percentage to 100%. 3. If the amount decreased, subtract the given percentage from 100%. 4. Then, do the ratio and cross multiple.

❖ WORKED EXAMPLES 1. Following a 10 % increment, petrol now costs K 15 per litre. How much did it cost before the increase? 2. Abigail pays $ 25 for a bag after getting a 20 % discount. How much did it initially cost?

❖ SOLUTIONS 1.

100 % +10 %

=

100 % 110 % K 15 100 %

=

K 15 𝑥

𝑥

x = K 13.6 Old price = K 13.6

2.

100 % − 20 %

=

100 % 80 % $ 20 100 %

=

$ 20 𝑥

𝑥

x = $ 25 Original price = $ 25

Percentages

33

6.0

RATIO AND PROPORTION

Introduction What is the number of boys and girls in your class? What is the size of this book to your cellphone or computer? Whenever you compare the quantities of two things you are applying ratios in real life. Ratio is simply the comparison of two numbers. For example, if the length of this book is 15cm and the length of the 15

laptop is 30cm then the ratio of the length of the book to the laptop is 30. Proportion helps us to know if two ratios are equal to each other. Imagine you are baking a cake. The recipes say; add 1 tablespoon of baking powder, 1 tablespoon of salt, 2 cups of milk, 2 cups of flour and 2 eggs. If you want to make a larger batch, what do you do? You use 3 tablespoons of baking powder, 3 tablespoons of salt, 6 cups of milk, 6 cups of flour and 6 eggs. The first ingredients are proportional to the second ingredients. Proportions can be direct or inverse. With direct proportion, when one thing increases the other also increases. For example, when you are alone you can just buy one biscuit for yourself. How about if you are two you and you have enough money? Likely, you can buy two biscuits, when three, three biscuits and so on. As seen, the more you are is directly proportional to the number of biscuits bought. When the number of your friends increases the number of biscuits also bought increases. With inverse proportion, when one thing increases the other thing decreases. For example, you can slash a portion of glass in one hour. How long can it take when you are two? How about three? Certainly, the more you are the less time it can take to slash the same portion. Therefore, the more you are is inversely proportional to the time spent slashing.

Specific outcomes This unit covers ratio and proportion. By the end of this unit, you will be able to: ❖ Work out ratio problems ❖ Work out direct proportion problems ❖ Work out inverse proportion problems

34

Ratio and proportion

RATIO AND PROPORTION 6.1

RATIO • A ratio is a comparison of two numbers. 𝑎

•

It is written as a:b or . It is read as a to b.

•

For example, a class has 20 boys and 25 girls. The ratio of boys to girls is 20:25.

𝑏

❖ EQUIVALENT RATIOS • Equivalent ratios are ratios with the same value. • A class with boys to girls ratio of 20:25 is equivalent to another class with boys to girls ratio of 4:5.

❖ WORKED EXAMPLES 1. What is the ratio of 5 books to 20 books? 2. What is the ratio of 30 minutes to 40 minutes?

❖ SOLUTIONS 1. 5 books to 20 books = 5:20 = =

5 20 1 4

= 1:4 2. 30 min to 45 min = 30:40 = =

30 40 3 4

= 3:4

6.2

PROPORTION • Proportion is an equation that indicates that two ratios are equal. 𝑎

𝑐

•

Proportion is expressed as

•

For example,

•

How to check if two ratios are proportional. 1. Cross multiply the ratios. 2. If the product gives the same numbers on both sides, the ratios are proportional. 3. If the product gives different numbers on sides, the ratios are not proportional.

1 2

=

5 10

𝑏

=

𝑑

or a:b = c:d.

indicate that these two ratios, that is, 1:2 and 5:10 are equal.

Ratio and proportion

35

❖

WORKED EXAMPLES 1. Are the following pairs of ratios proportional? A. B. C.

1 2 1

6

and

12 3

and

2 20 25

9 4

and

5

2. One box of boom detergent costs K 5. How much can ten boxes cost? 3. Solve the following proportion. 6 𝑥

❖

12

=

4

SOLUTIONS 1. A.

1 2 1 2

and =

6 12

6

Cross multiply, i.e. 1 × 12 and 2 × 6

12

12 = 12 The ratios are proportional B.

1 2 1 2

and =

3 9

3

Cross multiply, i.e. 1 × 9 and 2 × 3

9

9=6 The ratios are not proportional C.

20 25 20 25

and =

4 5

4 5

Cross multiply, i.e. 20 × 5 and 25 × 4

100 = 100 The ratios are proportional

2. 1:K 5 = 10:x 1 5

=

10

Cross multiply, i.e. x × 1 and 5 × 10

𝑥

x = 50 ∴ 10 boxes cost K 50

3.

6 𝑥

=

12 4

12x = 24 x=2

36

Ratio and proportion

Cross multiply, i.e. x × 12 and 6 × 4 Divide by 12 on both sides to get x

6.3

DIRECT AND INVERSE PROPORTIONAL ❖ DIRECT PROPORTIONAL • In direct proportional, as one variable increases the other variable also increases, as one variable decreases the other variable also decreases. • If y is directly proportional to x, then: y∝x or y = kx

where k is a constant

❖ WORKED EXAMPLES 1. Two variables x and y are directly proportional. When x = 4, y = 10. A. Find the formula which connects y and x. B. Using your formula, work out the value of y when x = 6. C. Using your formula, work out the value of x when y = 10. 2. y is proportional to the square of x. If y = 18 and x = 3, what is A. y, when x = 5 B. x, when y = 98

❖ SOLUTIONS 1. A. y = kx 10 = 4k 10

k= k=

4 5 2 5

y= x 2

B.

First find k by substituting y = 10 and x = 4 into the proportional equation.

To form an equation, insert the value of k into the proportional equation.

5

y= x =

2 5 2

×6

=15

C.

5

y= x 2 5

10 = x x=

2 20 5

=4

Ratio and proportion

37

2.

A.

y = k𝑥 2 18 = k(3)2 9k = 18 k=2

First find the equation connecting x and y

y = 2𝑥 2

The equation connecting x and y

y = 2𝑥 2 y = 2(5)2 = 2 × 25 = 50 B.

6.4

y = 2𝑥 2 98 = 2𝑥 2 𝑥 2 = 49 √𝑥 2 = √49 x=7

INVERSE PROPORTIONAL • With inverse proportional, as one variable increases the other variable decreases. As one variable decreases the other variable increases. • If y is inversely proportional to x, then: y∝ or y=

𝟏 𝒙 𝒌

where k is a constant

𝒙

❖ WORKED EXAMPLES 1. y is inversely proportional to x. When x = 4, y = 10. A. Find the equation connecting x and y. B. Find the value of y when x = 8. C. Find the value of x when y = 20. 2. If it takes ten days for four men to build a block wall, how many days will it take twenty men? ❖ SOLUTIONS 1. A.

y=

𝑘 𝑥 𝑘

10 =

First find k by substituting y = 10 and x = 4 into the proportional equation.

4

k = 40 y=

38

Ratio and proportion

40 𝑥

To form an equation, insert the value of k into the proportional equation.

B.

y= =

40 𝑥 40 8

=5 C. 20 = x=

40 𝑥 40 20

x=2 2. Take the number of men as x and the number of days as y Find the relationship. More men mean less time. Therefore, x and y are inversely proportional y= 10 =

𝑘 𝑥 𝑘 4

k = 40 y=

40 𝑥

Then, find the number of days for 20 men y= y=

40 𝑥 40 20

y=2

therefore, it can take two days for 20 men to build the block wall

Ratio and proportion

39

7.0

INDICES

Introduction Imagine you are told to multiply five by itself hundred times. Even with a calculator, it can take longer and the chances of making errors are high. In everyday life, you do not often do that. However, in maths, physics, chemistry and other fields multiplying a number by itself is very common! To reduce the amount of time spent and errors indices are used. Indices or exponents are simply a shorthand notation for multiplying the same number by itself several times. Multiplying 5 by itself a hundred times can be written as 5100. Working out 5100 with a calculator is fast and error free than typing 5 × 5 × 5 × 5… hundred times. How can you multiply or divide indices? This unit covers these and many more questions.

Specific outcomes By the end of this unit, you will be able to: ❖ Solve problems using: • multiplication of indices • division of indices • negative indices • power raise to a power • power of a fraction • zero indices

40

Indices

INDICES 7.1

INDICES • Index (or exponential) notation is a short way of writing the same number multiplied by itself many times. • For example, 5 × 5 × 5 × 5 can be written as 54 , 7 × 7 × 7 can be written as 73 . They are read “5 to the power 4” and “7 to the power 3” respectively. • Indices are denoted as follow: 𝒂𝒏

•

7.2

𝑎𝑛 is read as “a to the power n”.

where a is any real number n is an integer

a is called the base, and n is called index (or power, order, exponent).

LAWS OF INDICES 1. MULTIPLICATION OF INDICES • If a is a base, and m and n are powers, then: 𝒂𝒏 × 𝒂𝒎 = 𝒂𝒏 + 𝒎 •

Examples: 1. 22 × 23 = 22 + 3 = 25 = 32 2. 34 × 3 = 34 × 31 = 34 + 1 = 243 3. 55 × 54 = 55 + 4 = 59 4. 73 × 72 = 73 + 2 = 75

Use a calculator to find values for 59 and 75

2. DIVISION OF INDICES • If a is a base, and m and n are powers, then: 𝒂𝒏 ÷ 𝒂𝒎 = 𝒂𝒏

𝒎

• Examples: 1. 54 ÷ 52 = 54 2 = 52 = 25 2. 49 ÷ 43 = 46 3. 33 ÷ 3 = 33 ÷ 31 = 33 1 = 32 = 9

Use a calculator to find the value for 46

3. NEGATIVE INDICES • If a is a base and m is a negative power, then: 𝒂

𝒎

=

𝟏 𝒂𝒎

• Examples: 1. 3

4

2. 6

2

3. 8

1

= = =

1 34 1 62 1 81

= =

4. 43 ÷ 45 = 43

1 36 1 8 5

= 4

2

=

1 42

=

1 16 Indices

41

4.

POWER RAISED TO A POWER • If a is a base, and m and n are powers, with m raised to n, then: (𝒂𝒎 )𝒏 •

Examples: 1. (22 )3 = 22 × 3 = 26 = 64 2. (4

3. (33 )

5.

=(

2 )2 2

1 2 42

)

= 33 × (

1

=

=

42 × 2

2)

= 3

6

=

1 44 1 36

= =

1 256 1 729

𝐏𝐎𝐖𝐄𝐑 𝐎𝐅 𝐀 𝐏𝐑𝐎𝐃𝐔𝐂𝐓 • If a and b are bases, and the product of a and b is raised to power n, then: (𝒂𝒃)𝒏 = 𝒂𝒏 × 𝒃𝒏 •

6.

Examples: 1. (2 × 3)2 = 22 × 32 = 4 × 9 = 36 2. (3𝑥)3 = 33 × 𝑥 3 = 27 × 𝑥 3 = 27𝑥 3 3. (4𝑎𝑏)2 = 42 × 𝑎2 × 𝑏 2 = 16 × 𝑎2 × 𝑏 2 = 16𝑎2 𝑏 2

POWER OF A FRACTION • If a and b are bases, and the fraction of a and b is raised to power n, then: 𝒂 𝒏

(𝒃) = •

𝒃𝒏

Examples: 1. 2. 3.

7.

𝒂𝒏

2 3

(3)

𝑥 2

=

(3 )

=

𝑎

2

(4𝑏)

23 33 𝑥2 32

=

= =

8 27 𝑥2 9

𝑎2 42 × 𝑏2

=

𝑎2 16𝑏2

ZERO INDICES • If a is a base and n = 0, then: 𝒂𝟎 = 1 • •

42

Therefore, any number raised to the power zero is equal to one. Examples: 1. 90 = 1 2. 20 = 1 3. 1000 = 1 4. 83 ÷ 83 = 83 3 = 80 = 1 5. 4 5 × 45 = 4 5 + 5 = 40 = 1

Indices

8.

FRACTIONAL INDICES • If a is a base, and n and m are powers, then: 𝟏

𝒂𝒏 = 𝒏√𝒂 𝒎

𝒎

𝒏

𝒂 𝒏 = ( √𝒂) •

Examples: 1

1. 252 = √25 = 5 1

3

2. 643 = √64 = 4 3

3

4

3. 814 = ( √81) = 33 = 27 2

2

3

4. 1253 = (√125) = 52 = 25 2

5. 125−3

=

1 3

( √125)

2

=

1 52

=

1 25

Indices

43

REVIEW QUESTIONS 1. Evaluate the following 𝟑

A. 16𝟐 𝟐

B. 27𝟑 𝟏

C. 64𝟑 4

D. (√81)

4

2. Find the value of the following 𝟓

A. 4−𝟐 𝟒

B. 32−𝟓 3 −𝟐 4 𝟐 −

C. ( )

D. 243

𝟓

3. Evaluate the following 3 −2

A. (4) +

2 9

𝟏

81 −𝟒

B. (16) 𝟐

81 0

+ (16) 𝟏

C. 64𝟑 − 4𝟐 D. 32 + 23 × 20 4. Solve the following A. 22𝑥 + 1 = 1 B. 32𝑥 + 7 = 81 C. 82𝑥 = 42𝑥 − 1 D. 4𝑥 = 22 × 44

44

Indices

SOLUTIONS 𝟑

3

1. A. 16𝟐 = (√16) = 43 = 64 B.

𝟐

2

3

27𝟑 = (√27) = 32 = 9 𝟏

3

C. 64𝟑 = √64 = 4 4

4

𝟓

1

D. (√81) = 34 = 81

2. A. 4−𝟐 =

5 42

𝟒

B. 32−𝟓 =

3 −2

C. (4)

=

3. A.

3 −2

(4)

+

(√4)

=

4 325

1 2 2435

9

1 5

=

3 2

5

( √243)

3 2 ( ) 4

+

2 9

1 32 1

24

=

1 16

4 2

= 1×(

) 3

1

=

1

=

4

(4)

=1÷

3 2 ( ) 4

=

25

( √32)

1

2

1

=

5

1

𝟐

D. 243−𝟓 =

1

=

2

=

=

1 32

=

4 2

(3)

+

42 32

= 1×

16 9

=

16 9

1 9

2 9

𝟏

81 −𝟒 81 0 B. ( ) + ( ) = 16 16

= 1×

=

1

1

16 9

+

2 9

=

16 + 2 9

=

18 9

= 2

4

16 4 2 2+3 5 √16 +1 = +1 = = 1 +1 = ( ) +1 = 4 81 3 3 3 √81 81 ( )4 16

𝟐

𝟏

3

2

C. 64𝟑 − 4𝟐 = (√64) − √4 = 42 − 2 = 8 − 2 = 6 D. 32 + 23 × 20 = 9 + 8 × 1 = 9 + 8 = 17

Indices

45

4.

A. 22𝑥 + 1 = 1 22𝑥 + 1 = 20 2x + 1 = 0 2x = −1 1 x=−2 B. 32𝑥 + 7 = 81 32𝑥 + 7 = 34 2x + 7 = 4 2x = 4 − 7 2x = −3 3 x=−2 C. 82𝑥 = 42𝑥 − 1 (23 )2𝑥 = (22 )2𝑥− 1 26𝑥 = 24𝑥 − 2 6x = 4x − 2 6x − 4x = − 2 2x = − 2 x=−1

D. 4𝑥 = 22 × 44 4𝑥 = 4 × 16 4𝑥 = 64 4 𝑥 = 43 x=3

46

Indices

To solve equations using indices evaluate the equation so that both sides of the equation have the same base.

8.0

SEQUENCES AND SERIES

Introduction Most objects around us have patterns. Observe tiles, carpet or any fabric sheet. The chances are that you will see geometric patterns printed on them. When you observe the arrangement of chairs in a bus, you find that they are arranged in a pattern. The space between adjacent chairs is constant. In nature, sunflowers and many other flowers have their seeds arranged in a particular pattern. Designs that have patterns tend to follow a particular sequence. A sequence is a set of numbers that follows a particular rule. Examples of sequences include 2, 4, 6, 8, 10, … and 1, 2, 4, 8, 16, …. There are two types of sequences: arithmetic progression and geometric progression. What is arithmetic progression and geometric progression? A series is simply a sum of the sequence. Given the sequence: 2, 4, 6, 8. The series of this sequence is 2 + 4 + 6 + 8 which is equal to 20. How do you find the series of a very long sequence? This unit covers this and other questions.

Specific outcomes By the end of this unit, you will be able to: ❖ Define a sequence ❖ Arithmetic progression: • Define arithmetic progression • Find the common difference of a given arithmetic progression • Find the nth term of a given arithmetic progression ❖ Geometric progression: • Define geometric progression • Find the common ratio of a given geometric progression • Find the nth term of a given geometric progression ❖ Series: • Define series • Find the sum of arithmetic progression • Find the sum of geometric progression

Sequences and Series

47

SEQUENCES AND SERIES 8.1

8.2

SEQUENCES • A sequence is a set of numbers that follows a particular rule. • The word “term” is used to describe the number in the sequence. • If the set of numbers are T1, T2, T3, T4, T5, … The first (1st) term is T1, the fourth (4th) term is T4, 𝑛𝑡ℎ term is Tn. •

For example, given the sequence; 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, … What is A. T1 B. T3 C. T9 D. Tn

•

Solutions A. T1 = 3 B. T3 = 9 C. T9 = 27 D. Tn = 3n

ARITHMETIC PROGRESSION • Arithmetic progression is a sequence in which the difference of any two successive members is constant. • Example: 1. 1, 2, 3, 4, 5, 6, … the difference between two successive members is 1. 2. 5, 7, 9, 11, 13, … the difference between two successive members is 2. 3. 5, 10, 15, 20, … the difference between two successive members is 5. ❖ HOW TO FIND Tn (𝒏𝒕𝒉) TERM OF ARITHMETIC PROGRESSION ❖ Formula: Tn = a + (n − 1)d

where Tn = 𝑛𝑡ℎ term a = first term n = term number d = common difference

❖ EXAMPLES 1. Find the 9th term of the sequence 6, 9, 12, … 2. Find the 6th term of the sequence 2, 7, 12, … 3. What is the rule for the 𝑛𝑡ℎ term of the sequence 1,3, 5, … ❖ SOLUTIONS 1. Tn = a + (n − 1)d T9 = 6 + (9 − 1)3 = 6 + (8)3 = 6 + 24 = 30

48

Sequences and Series

To find the common difference (d) choose any term and subtract it from the next term, e.g. 9 − 6 = 3 or 12 − 9 = 3.

2.

Tn = a + (n − 1)d T6 = 2 + (6 − 1)5 = 2 + (5)5 = 2 + 25 = 27

3.

Tn = a + (n − 1)d Tn = 1 + (n − 1)2 = 1 + 2n − 2 = 2n − 1

8.3

GEOMETRIC PROGRESSION • Geometric progression is a sequence in which any two successive members differ from each other by a common ratio. • Examples: 1. 4, 8, 16, 32, 64, … differ by ×2 2. 10, 30, 90, 270, … differ by ×3 ❖ HOW TO FIND Tn (𝒏𝒕𝒉) TERM OF GEOMETRIC PROGRESSION • Formula: Tn = 𝒂𝒓𝒏 − 𝟏

where Tn = 𝑛𝑡ℎ term a = first term r = common ratio n = term number

❖ EXAMPLES 1. Find the 5th term in the sequence 4, 8, 16, … 2. Find the 6th term in the following geometric progression 256, 64, 16, … 3. What is the rule for the 𝑛𝑡ℎ term of the geometric progression 3, 15, 75, …? ❖ SOLUTIONS 1. Tn = 𝑎𝑟 𝑛 − 1 T5 = 4 × 25 − 1 = 4 × 24 = 4 × 16 = 64

To find the common ratio (r) choose any term and divide it by the previous term, e.g. 8 ÷ 4 = 2 or 16 ÷ 8 = 2.

2. Tn = 𝑎𝑟 𝑛 − 1 1 6−1

T6 = 256 × (4)

1 5 4 1 1024

= 256 × ( ) = 256 × 1

=4 3. Tn = 𝑎𝑟 𝑛 − 1 = 3 × 5𝑛 − 1 = 3∙ 5𝑛 − 1

Note: 3 × 5𝑛 − 1 ≠ 15𝑛 − 1 Sequences and Series

49

8.4

SERIES • Series is a sum of sequence to a certain number of terms. • For example: 1. The sum of the first four terms of the sequence 2, 4, 8, 10, 12, … is 2 + 4 + 6 + 8 = 20 2. The sum of the first three terms of the sequence 3, 8, 13, 18, 24, 29, … is 3 + 8 + 13 = 24 ❖ HOW TO FIND THE SUM OF ARITHMETIC PROGRESSION • Formula: 𝒏

1.

Sn = 𝟐 (𝒂𝟏 + 𝒂𝒏 )

2.

Sn = 𝟐 [𝟐𝒂𝟏 + (𝒏

𝒏

𝟏)𝐝]

where Sn = sum of first 𝑛𝑡ℎ term n = term number 𝑎1 = first term 𝑎𝑛 = 𝑛𝑡ℎ term d = common difference use the first formula if the last term is provided

❖ EXAMPLES 1. Calculate the sum of all even numbers from 2 to 100 2. Determine the sum of arithmetic series 3 + 8 + 11 + … + 73 3. Find the sum of the first six terms of the arithmetic sequence 2, 7, 12, … ❖ SOLUTIONS 𝑛 1. Sn = 2 (𝑎1 + 𝑎𝑛 ) =

50 2

Even numbers from 2 to 100 are 50, ∴ n = 50

(2 + 100)

= 25 × 102 = 2550 2. Tn = a + (n − 1)d 73 = 3 + (n − 1)5 73 = 3 + 5n − 5 5n = 73 − 3 + 5 5n = 75 n = 15

First find the number of terms (n) in a series

𝑛 2 15 S15 = 2 (3 + 73) 15 = 2 × 76

Sn = (𝑎1 + 𝑎𝑛 )

Use the number of terms (n) found to calculate the sum of series

= 570 𝑛

3. Sn = 2 [2𝑎1 + (𝑛

1)d]

S6 = [2(2) + (6

1)5]

6 2

= 3(4 + (5)5) = 3(4 + 25) = 3(29) = 87 50

Sequences and Series

❖

HOW TO FIND THE SUM OF GEOMETRIC PROGRESSION ❖ Formula: 1. 𝐒𝒏 = 2. 𝐒𝒏 = 3.

𝒂(𝒓𝒏 − 𝟏) 𝒓−𝟏 𝒂(𝟏− 𝒓𝒏 ) 𝟏−𝒓

𝐒∞ =

𝒂 𝟏−𝒓

where S𝑛 = sum of first 𝑛𝑡ℎ term S∞ = sum of infinite geometric series a = first term 𝑟 𝑛 = common ratio n = term number use • formula 1 if r > 1 • formula 2 if r < 1 • formula 3 for infinite geometric series. Infinite geometric series have common ratio (r) between −1 and 1

❖ EXAMPLES 1. For the series 2 + 6 + 18 + 54 + … find A. the 10th term B. sum of the first 8 terms 2. For the sequence 3, -6, 12, -24, … find A. the 6th term B. sum of the first 6 terms 3. Find the sum of the infinite geometric sequence 27, 18, 12, 8, … ❖ SOLUTIONS 1. A. Tn = 𝑎𝑟 𝑛 − 1 T10 = 2 × 310 − 1 = 2 × 39 = 39366 B. S𝑛 = S8 = =

𝑎(𝑟 𝑛 − 1) 𝑟−1 2(38 − 1) 3−1 2(6561 − 1)

= 6560

2

2. A. Tn = 𝑎𝑟 𝑛 − 1 T6 = 3 × ( 2)6 − 1 = 3 × ( 2)5 = 3 × ( 32) = − 96

Sequences and Series

51

2.

B. S𝑛 =

𝑎(1− 𝑟 𝑛 )

S6 = =

1−𝑟 3(1− (−2)6 ) 1 −(−2) 3(1− 64) 1+2

= − 63 3.

S∞ = =

𝑎 1−𝑟 27 1−

𝟐 𝟑

= 81

52

Sequences and Series

REVIEW QUESTIONS 1. The first three terms in an arithmetic progression are 2, 4, and 6. Find the A. common difference B. tenth term C. sum of the first 12 terms 2. For the sequence 3, 5, 7, 9, … write down A. the 12th term B. the expression for the nth term 3. The first three terms in geometric progression are 2, 6, and 18. Find the A. common ratio B. 5th term C. an expression for the nth term D. sum of the first 10 terms 4. For the sequence 2, -8, 32, -128, … A. Written a rule for the nth term B. Write the next number in the sequence C. Find the sum of the 5th term 5. For the sequence 3, 6, 12, 24, … find A. the sixth term B. an expression for the nth term 6. For the sequence 21, 18, 15, 12, … find A. the formula for the nth term B. sum of the first 16 terms 7. For the sequence -10, -7, -4, -1, … find A. 15th term B. Sum of the first 20 terms

Sequences and Series

53

SOLUTIONS 1. A. d = 𝑎2 − 𝑎1 =4−2 =2

To find the common difference (d) choose any term and subtract it from the subsequent term

B.

Tn = a + (n − 1)d T10 = 2 + (10 − 1)2 =2+9×2 = 20

C.

Sn = 2 [2𝑎1 + (𝑛 1)d]

𝑛

S12 =

12 2

[2(2) + (12 1)2]

= 6(4 + 11 × 2) = 6 × 26 = 156 2. A.

B.

3. A.

Tn = a + (n − 1)d T12 = 3 + (12 − 1)2 = 3 + 11 × 2 = 25 Tn = a + (n − 1)d = 3 + (n − 1)2 = 3 + 2n − 2 = 2n + 1 r = 𝑎2 ÷ 𝑎1 =6÷2 =3

To find the common ratio (r) choose any term and divide it by the previous term

B.

Tn = 𝑎𝑟 𝑛 − 1 T5 = 2 × 35 − 1 = 2 × 34 = 162

C.

Tn = 𝑎𝑟 𝑛 − 1 = 2 × 3𝑛 − 1 = 2⋅ 3𝑛 − 1

D. S𝑛 = S10 = =

𝑎(𝑟 𝑛 − 1) 𝑟−1 2(310 − 1) 3−1 2(59049 − 1) 2

= 59048

54

Sequences and Series

4. A.

Tn = 𝑎𝑟 𝑛 − 1 = 2 × ( 4)𝑛 − 1 = 2( 4)𝑛 − 1

B.

Tn = 𝑎𝑟 𝑛 − 1 T5 = 𝑎𝑟 5 − 1 = 2 × ( 4)5 − 1 = 2( 4)4 = 512

C.

S𝑛 = S5 = = =

𝑎(1− 𝑟 𝑛 ) 1−𝑟 2(1− (−4)5 ) 1 −(−4) 2(1−(−1024)) 1+4 2(1025)

= 410

5

5. A.

Tn = 𝑎𝑟 𝑛 − 1 T6 = 3 × 26 − 1 = 3 × 25 = 96

B.

Tn = 𝑎𝑟 𝑛 − 1 Tn = 3 × 2𝑛 − 1 = 3⋅ 2𝑛 − 1

6. A.

B.

Tn = a + (n − 1)d = 21 + (n − 1)(−3) = 21 − 3n + 3 = 24 − 3n 𝑛

Sn = 2 [2𝑎1 + (𝑛 1)d] S16 =

16 2

[2(21) + (16 1)( 3)]

= 8(42 + 15( 3)) = 8 × (−3) = − 24 7. A.

B.

Tn = a + (n − 1)d T15 = −10 + (15 − 1)3 = −10 + 14 × 3 = 32 𝑛 2 20 S20 = 2

Sn = [2𝑎1 + (𝑛 1)d] [2( 10) + (20

1)3]

= 10(−20 + 11 × 3) = 10 × 13 Sequences and Series

55

9.0

ALGEBRA

Introduction Algebra comes from an Arabic word meaning reunion of broken parts. Algebra is a branch of mathematics used to solve the quantity of unknown items. For example, a basket contains ten oranges. The unknown number of oranges were added. After recounting, you find sixteen oranges. To find the number of oranges added algebra could be used. Algebra uses letters to represent unknown values. Any letter can be used, but x and y are often used. From the orange scenario above, it can be said that; 10 oranges + unknown number of oranges = 16 oranges. This equation is correct but too long. Algebra can simplify it to 10 + x = 16; where x is the number of unknown oranges. It is also easier to solve 10 + x = 16 than the latter. This unit introduces you to algebra.

Specific outcomes By the end of this unit, you will be able to: ❖ Understand the components of algebraic expression ❖ Identify like and unlike terms ❖ Add and subtract terms ❖ Multiply and divide terms ❖ Expand and simplify algebraic expressions ❖ Expand binomial expressions ❖ Expand trinomial expressions ❖ Factorise algebraic expressions ❖ Factorise quadratic expressions ❖ Simplify algebraic expressions in fraction form

56

Algebra

ALGEBRA 9.1

ALGEBRAIC EXPRESSION • An algebraic expression is an expression consisting of variables, constants and algebraic operations. • Components of algebraic expression.

1. Variable: A symbol that represents a value that can change. It is usually written as a letter, for example; x, y. 2. Constant: A value that does not change. It is usually written as a number, for example, 2, 8, −5. 3. Coefficient: A number that multiplies a variable, for instance; 5x (x variable is multiplied 5 times), 3y (y variable is multiplied 3 times). 4. Power: Indicate a number of times a variable or constant multiply itself, for example; 𝑥 2 means x × x, 42 means 4 × 4. 5. Operation: Indicates an operation (+, −, ×, ÷) to be performed. 6. Term: variables, constants or both on which operations perform. Example of terms: x, 2y, 3. Terms are separated by positive (+) or negative (−) signs.

9.2

LIKE AND UNLIKE TERMS ❖ LIKE TERMS • Like terms are terms that have the same variables and power, for example; x and 2x are like terms, 3𝑥 2 and 5𝑥 2 are like terms, −3y and 7y are like terms. • Like terms can be added, subtracted, multiplied or divided. ❖ UNLIKE TERMS • Unlike terms are terms that do not have the same variables or power, for example; x and 𝑥 2 are unlike terms, x and y are unlike terms, 𝑦 3 and 𝑦 2 are unlike terms. • Unlike terms can be multiplied or divided but cannot be added or subtracted.

9.3

ADDITION AND SUBTRACTION OF TERMS • Like terms of algebraic expression can be added or subtracted. • The process of adding or subtracting like terms of an algebraic expression is known as simplifying algebraic expression. • To simplify an algebraic expression first group like terms together then perform the operation between like terms, that is, add or subtract. ❖ EXAMPLES 1. Simplify the following expressions. A. 7x + 2y − 5x B. 2𝑥 2 − 3x + 9xy + 𝑥 2 − 4xy C. z − 1 + 4 2. Simplify the expression 9ab + 2a − 5b − 5ba

Algebra

57

❖

SOLUTIONS 1. A. 7x + 2y − 5x = 7x − 5x + 2y = 2x + 2y

To simplify algebraic expression first group like terms together, then add or subtract.

B. 2𝑥 2 − 3x + 9xy + 𝑥 2 − 4xy = 2𝑥 2 + 𝑥 2 + 9xy − 4xy − 3x = 3𝑥 2 + 5xy − 3x C. z − 1 + 4 =z+3 2. 9ab + 2a − 5b − 5ba = 9ab − 5ba + 2a − 5b = 4ab + 2a − 5b 9.4

ab and ba are like terms, same as xy and yx etc.

MULTIPLICATION AND DIVISION OF TERMS • Both like and unlike terms can be multiplied or divided. • Dividing of algebraic expression involves cancelling common terms. • Multiplication and division of algebraic expression is also known as simplifying algebraic expression. ❖ EXAMPLES 1. Simplify the following expressions. A. x × y B. x × x C. 5x × 3y 2. Simplify the following expressions. A. 6xy𝑧 2 ÷ 2yz B. 2ab ÷ 4𝑏 2 3. Simplify the following expressions. A. B.

7 8𝑟 7 8𝑟

× ÷

2 5𝑟 2 2 5𝑟 2

❖ SOLUTIONS 1. A. x × y = xy B. x × x = 𝑥 2 C. 5x × 3y = 15xy 6𝑥𝑦𝑧 2

2. A. 6xy𝑧 2 ÷ 2yz = B. 2ab ÷ 4𝑏 2 = 3. A.

B. 58

Algebra

7 8𝑟 7 8𝑟

×

÷

2 5𝑟 2 2 5𝑟 2

=

=

2𝑦𝑧

2𝑎𝑏 4𝑏2

=

𝑎 2𝑏

7×2 8𝑟 7 8𝑟

× 5𝑟 2

×

5𝑟 2 2

Dividing of algebraic expression involves cancelling of common terms.

= 3xz

=

=

14 40𝑟 3

=

7 × 5𝑟 2 8𝑟 × 2

7 20𝑟 3

=

35𝑟 2 16𝑟

=

35𝑟 16

9.5

EXPANSION OF ALGEBRAIC EXPRESSION • Brackets are used to group terms together. • To expand brackets, multiply the term outside by each term inside.

❖ EXAMPLES 1. Expand the following expressions. A. 4(𝑥 + 𝑦) B. −2(3𝑥 4) C. 2(10𝑎 6𝑏 + 𝑐) D. 3𝑎2 (𝑎 + 1) 2. Expand and simplify the following expressions. A. 5(𝑎 + 7) + 2(𝑎 + 4) B. 𝑟 2 (3𝑟 + 4) + 6r(𝑟 2)

❖ SOLUTIONS 1. A. 4(𝑥 + 𝑦) =4×x+4×y = 4x + 4y B. −2(3𝑥 4) = −2 × 3x − 2 × (−4) = −6x + 8 = 8 − 6x C. 2(10𝑎 6𝑏 + 𝑐) = 2 × 10a − 2 × 6b + 2 × c = 20a − 12b + 2c

To expand, multiply the term outside by each term inside the bracket.

Multiplication sign rule +×+=+ −×−=+ +×−=− −×+=−

D. 3𝑎2 (𝑎 + 1) = 3𝑎2 × a + 3𝑎2 × 1 = 3𝑎3 + 3𝑎2 2. A. 5(𝑎 + 7) + 2(𝑎 + 4) =5×a+5×7+2×a+2×4 = 5a + 35 + 2a + 8 = 5a + 2a + 35 + 8 = 7a + 43 B. 𝑟 2 (3𝑟 + 4) + 6r(𝑟 2) = 𝑟 2 × 3𝑟 + 𝑟 2 × 4 + 6r × r + 6r × (−2) = 3𝑟 3 + 4𝑟 2 + 6𝑟 2 − 12r = 3𝑟 3 + 10𝑟 2 − 12r

Algebra

59

9.6

BINOMIAL EXPRESSION • A binomial expression is an expression with two unlike terms. • Examples of binomial expressions include; 2x + 1, x + y, 5a + 7b. • If one binomial expression is a + b and another is c + d, the product of these two binomial expressions is:

❖ EXAMPLES 1. Find the product of the following binomial expressions. A. (𝑥 + 1)(𝑥 + 2) B. (3𝑥 5)(5𝑥 + 8) C. (3𝑎2 + 2𝑏)(2𝑎 𝑏) 2. Expand and simplify the following binomial expressions. A. (𝑥 + 1)2 B. (2𝑎 3)2 C. (3𝑥 2 2𝑦)2 ❖ SOLUTIONS 1. A. (𝑥 + 1)(𝑥 + 2) =x×x+x×2+1×x+1×2 = 𝑥 2 + 2x + x + 2 = 𝑥 2 + 3x + 2 B. (3𝑥 2)(5𝑥 + 8) = 3x × 5x + 3x × 8 − 2 × 5x − 2 × 8 = 15𝑥 2 + 24x − 10x − 16 = 15𝑥 2 + 14x − 16 C. (3𝑎2 + 2𝑏)(2𝑎 𝑏) = 3𝑎2 × 2a − 3𝑎2 × b + 2b × 2a − 2b × b = 6𝑎3 − 3𝑎2 b + 4ab − 2𝑏 2 2. A. (𝑥 + 1)2 = (𝑥 + 1)(𝑥 + 1) =x×x+x×1+1×x+1×1 = 𝑥2 + x + x + 1 = 𝑥 2 + 2x + 1 B. (2𝑎 3)2 = (2𝑎 3)(2𝑎 3) = 2a × 2a − 2a × 3 − 3 × 2a − 3 × (−3) = 4𝑎2 − 6a − 6a + 9 = 4𝑎2 − 12a + 9 C. (3𝑥 2 2𝑦)2 = (3𝑥 2 2𝑦)(3𝑥 2 2𝑦) = 3𝑥 2 × 3𝑥 2 − 3𝑥 2 × 2y − 2y × 3𝑥 2 − 2y × (−2y) = 9𝑥 4 − 6𝑥 2 y − 6𝑥 2 y + 4𝑦 2 = 9𝑥 4 − 12𝑥 2 y + 4𝑦 2

60

Algebra

9.7

TRINOMIAL EXPRESSION • Trinomial expression is an expression with three unlike terms. • Examples of trinomial expressions include; x + y + z, 2a − b + 3c, 𝑎2 + b + c. • If one expression is a + b and another expression is c + d + e, the product of these two expressions, binomial and trinomial, is:

❖ EXAMPLES 1. Find the product of the expressions (𝑥 + 1)(𝑥 2 2𝑥 + 1). 2. Expand and simplify the following expressions (3 11𝑦)( 10𝑦 2 7𝑦 9). ❖ SOLUTIONS 1. (𝑥 1)(𝑥 2 2𝑥 + 1) = x × 𝑥 2 − x × 2x + x × 1 − 1 × 𝑥 2 − 1 × (− 2x) − 1 × 1 = 𝑥 3 − 2𝑥 2 + x − 𝑥 2 + 2x − 1 = 𝑥 3 − 2𝑥 2 − 𝑥 2 + x + 2x − 1 = 𝑥 3 − 3𝑥 2 + 3x − 1 2. (3 11𝑦)( 10𝑦 2 7𝑦 9) = 3 × (−10𝑦 2 ) + 3 × ( 7𝑦) + 3 × ( 9) − 11y × ( 10𝑦 2 ) − 11y × ( 7𝑦) − 11y × ( 9) = − 30𝑦 2 − 21y − 27 + 110𝑦 3 + 77𝑦 2 + 99y = 110𝑦 3 − 30𝑦 2 + 77𝑦 2 − 21y + 99y − 27 = 110𝑦 3 + 47𝑦 2 + 78y − 27

9.8

FACTORISATION • Factorisation is the opposite of expanding brackets. • A factorised expression is written as a product of its factors. • For example, 2(x + 1) is expanded to 2x + 2, factorisation rewrite 2x + 2 to 2(x + 1). • To factorise, find the highest common factor (HCF) of the given terms and place it in front of the brackets. Then divide each term by the HCF to obtain the expression inside the brackets. ❖ WORKED EXAMPLES 1. 2x + 6 = 2(x + 3)

HCF of 2x and 6 is 2, dividing 2 into 2x gives x and 2 into 6 gives 3, ∴ the expression inside brackets is (x + 3).

2. 𝑥 2 − x = x(x − 1) 3. 12𝑥 2 y − 8x𝑦 2 = 4xy(3x − 2y)

Algebra

61

9.9

FACTORING BY GROUPING METHOD • This method works on expressions with four terms where two pairs of terms have their common factors. • To factorise, create groups by grouping terms with common factors together. • Factor out the common factor from each group. ❖ EXAMPLES 1. Factor the following expressions. A. 𝑥 3 − 2𝑥 2 + 5x − 10 B. 6𝑦 3 + 3𝑦 2 + 8y + 4 C. 𝑘 3 + 3𝑘 2 + 2k + 6 2. Factor the following expressions. A. 3ab + 21a − 2b − 14 B. 3𝑝2 + pq − 12p − 4q ❖ SOLUTIONS 1. A. 𝑥 3 − 2𝑥 2 + 5x − 10 = (𝑥 3 − 2𝑥 2 ) + (5x − 10) = 𝑥 2 (x − 2) + 5(x − 2) = (𝑥 2 + 5)( x − 2) B. 6𝑦 3 + 3𝑦 2 + 8y + 4 = (6𝑦 3 + 3𝑦 2 ) + (8y + 4) = 3𝑦 2 (2y + 1) + 4(2y + 1) = (3𝑦 2 + 4)(2y + 1) C. 𝑘 3 + 3𝑘 2 + 2k + 6 = (𝑘 3 + 3𝑘 2) + (2k + 6) = 𝑘 2(k + 3) + 2(k + 3) = (𝑘 2 + 2)(k + 3)

2. A. 3ab + 21a − 2b − 14 = (3ab + 21a) + (− 2b − 14) = 3a(b + 7) − 2(b + 7) = (3a − 2)(b + 7) B. 3𝑝2 + pq − 12p − 4q = (3𝑝2 + pq) + (− 12p − 4q) = p(3p + q) − 4(3p + q) = (p − 4)(3p + q)

62

Algebra

9.10 FACTORISING A QUADRATIC EXPRESSION • A quadratic expression is an expression with the highest power raised to two. For example; 𝑥 2 + 1. • A quadratic expression is binomial if it has two terms such as 𝑥 2 + 1, 2𝑥 2 + 1. • It is trinomial if it has three terms such as 𝑥 2 + 2𝑥 + 1, 5𝑥 2 + 10x + 2. ❖ FACTORISING QUADRATICS WITH DIFFERENCE OF SQUARES • Quadratic expressions with a difference of squares are expressions in the form of 𝑎2 𝑏 2 . • To factorise quadratic expression with a difference of squares the following formula is used. 𝒂𝟐 𝒃𝟐 = (a + b)(a − b)

❖ EXAMPLES 1. Factorise the following A. 𝑥 2 − 4 B. 𝑥 2 − 16 C. 𝑥 2 − 49 2. Factorise the following A. 9𝑥 2 − 16 B. 𝑥 2 − 4𝑦 2 C. 4𝑥 2 − 81𝑦 2 ❖ SOLUTIONS 1. A. 𝑥 2 − 4 = 𝑥 2 − 22 = (x + 2)(x − 2) B. 𝑥 2 − 16 = 𝑥 2 − 42 = (x + 4)(x − 4) C. 𝑥 2 − 49 = 𝑥 2 − 72 = (x + 7)(x − 7)

2. A. 9𝑥 2 − 16 = (3𝑥 2 )2 − 42 = (3𝑥 2 + 4)(3𝑥 2 − 4) B. 𝑥 2 − 4𝑦 2 = 𝑥 2 − (2𝑦)2 = (x + 2y)(x − 2y) C. 4𝑥 2 − 81𝑦 2 = (2𝑥)2 − (9𝑦)2 = (2x + 9y)(2x − 9y)

Algebra

63

❖

FACTORISING A QUADRATIC TRINOMIAL • A quadratic trinomial is an expression in the form 𝑎𝑥 2 + 𝑏𝑥 + 𝑐. • Examples of quadratic expressions include; 𝑥 2 + 2𝑥 + 1, 2𝑥 2 6𝑥 + 9. • Factorising a quadratic trinomial is a trial and error process. ❖ Useful information to factorise quadratic trinomial when a = 1. 1. Convert quadratic expression into double brackets; (x )(x ). 2. Find two values that when multiplied gives c and when added gives b. 3. Insert values in brackets. ❖ EXAMPLES 1. Factorise the following. A. 𝑥 2 + 2𝑥 + 1 B. 𝑥 2 + 5𝑥 + 6 C. 𝑥 2 + 5𝑥 6 D. 𝑥 2 6𝑥 + 9 ❖ SOLUTIONS 1. A. 𝑥 2 + 2𝑥 + 1 = (x )(x ) = (x + 1)(x + 1)

64

Algebra

a = 1, b = 2, c = 1 Two numbers that when multiplied gives c and when added gives b are 1 and 1 1×1=1=c 1+1=2=b

B.

𝑥 2 + 5𝑥 + 6 = (x )(x ) = (x + 2)(x + 3)

a = 1, b = 5, c = 6 Two numbers that when multiplied gives c and when added gives b are 2 and 3 2×3=6=c 2+3=5=b

C.

𝑥 2 + 5𝑥 6 = (x )(x ) = (x + 6)(x − 1)

a = 1, b = 5, c = −6 Two numbers that when multiplied gives c and when added gives b are 6 and −1 6 × (−1) = −6 = c 6−1=5=b

D.

𝑥 2 6𝑥 + 9 = (x )(x ) = (x − 3)(x − 3)

a = 1, b = −6, c = 9 Two numbers that when multiplied gives c and when added gives b are −3 and −3 (−3) × (−3) = 9 = c −3−3=−6=b

❖

Useful information to factorise quadratic trinomial when a ≠ 1. 1. Multiply a and c. 2. Find two values that when multiplied gives the product of ac and when added gives b. 3. Rewrite the quadratic trinomial splitting the middle term using the values found in step 2. 4. Factorise the first two terms and last two terms. 5. Add up the factors found in step 4.

❖ EXAMPLES 1. Factorise the following. A. 2𝑥 2 + 6𝑥 + 4 B. 2𝑥 2 𝑥 − 15 C. 3𝑥 2 + 𝑥 − 4 D. − 5𝑥 2 + 𝑥 + 4

❖ SOLUTIONS 1. A. 2𝑥 2 + 6x + 4 = 2𝑥 2 + 2x + 4x + 4 = (2𝑥 2 + 2x) + (4x + 4) = 2x(x + 1) + 4(x + 1) = (2x + 4)(x + 1)

• • • • • •

B. 2𝑥 2 − x − 15 = 2𝑥 2 − 6x + 5x − 15 = (2𝑥 2 − 6x) + (5x − 15) = 2x(x − 3) + 5(x − 3) = (2x + 5)(x − 3)

• • • • • •

C. 3𝑥 2 + 𝑥 − 4 = 3𝑥 2 − 3x + 4x − 4 = (3𝑥 2 − 3x) + (4x − 4) = 3x(x − 1) + 4(x − 1) = (3x + 4)(x − 1)

• • • • • •

a = 2, b = 6, c = 4. Multiply a and c to get 8. Two values that when multiplied gives 8 and when added gives 6 are; 2 and 4. Rewrite the original expression replacing the middle term (6x) with two values found (2x + 4x). Factorise first two; 2𝑥 2 + 2x into 2x(x + 1), and last two terms; 4x + 4 into 4(x + 1) Add up factors.

a = 2, b = −1, c = −15 Multiply a and c to get −30 Two values that when multiplied gives −30 and when added gives −1 are; −6 and 5 Rewrite the original expression replacing the middle term (−x) with two values found (−6x + 5x ) Factorise first two; 2𝑥 2 − 6x into 2x(x − 3), and last two terms; 5x − 15 into 5(x − 3) Add up factors

a = 3, b = 1, c = −4 Multiply a and c to get −12 Two values that when multiplied gives −12 and when added gives 1 are; 4 and −3 Rewrite the original expression replacing the middle term (x) with two values found (−3x + 4x) Factorise first two; 3𝑥 2 − 3x into 3x(x − 1), and last two terms; 4x − 4 into 4(x − 1) Add up factors Algebra

65

1.

D.

−5𝑥 2 + 𝑥 + 4 = −5𝑥 2 + 5x − 4x + 4 = (−5𝑥 2 + 5x) + (−4x + 4) = −5x(x − 1) − 4(x − 1) = (−5x − 4)(x − 1)

• • • • • •

a = −5, b = 1, c = 4. Multiply a and c to get −20. Two values that when multiplied gives −20 and when added gives 1 are; 5 and −4. Rewrite the original expression replacing the middle term (x) with two values found (5x − 4x). Factorise first two; −5𝑥 2 + 5x into −5x(x − 1), and last two terms; −4x + 4 into −4(x − 1). Add up factors.

9.11 SIMPLIFICATION OF FRACTIONS • Some algebraic expressions in fractions need to be factorised before simplifying. • Different methods of factorisation can be applied then simplify. ❖ EXAMPLES 1. Simplify the following. A. B. C.

𝑥 2 + 3𝑥 𝑥+3 3𝑦 − 3 𝑦 2 − 2𝑦 + 1 𝑥 2 + 2𝑥 − 8 𝑥 2 + 7𝑥 + 12

❖ SOLUTIONS 1. A.

𝑥 2 +3𝑥 𝑥+3

𝑥(𝑥 + 3)

=

Factorise 𝑥 2 + 3𝑥 Cancel out x + 3

𝑥+3

= x

B.

3𝑦 − 3 𝑦2

= =

C.

(𝑦 − 1)

𝑥 2 + 7𝑥 + 12 (𝑥 − 2)(𝑥 + 4)

= Algebra

3(𝑦 − 1) (𝑦 − 1)(𝑦 − 1) 3

𝑥 2 + 2𝑥 − 8

=

66

− 2𝑦 + 1

𝑦 2 − 2y + 1 is a quadratic trinomial. Factorise by trial and error. Find two numbers that when multiplied gives 1 and when added gives −2; −1 and −1.

(𝑥 + 3)(𝑥 + 4) 𝑥−2 𝑥+3

Both 𝑥 2 + 2𝑥 8 and 𝑥 2 + 7𝑥 + 12 are quadratic trinomial. Factorise them.

REVIEW QUESTIONS 1. Simplify the following A. 6 − 2(y − x) − 2 B. 5(a + 2b) − (2a − 4b) C. 3m − (2n − 4m) − 6n D. 4a − 5b − 3(2b − a) E. 2(x + 4) − x(x + 1) 2. Simplify the following A. B. C. D.

15𝑚3 4𝑛2

÷

3𝑥𝑦 3 20𝑎3 𝑏2 8𝑎3 𝑏2 30𝑎3 𝑏4 11𝑝 18𝑞 2

×

5𝑚4 12𝑛3 10𝑎3 𝑏

× ÷

9𝑥 3 𝑦 2 4𝑎2 𝑏2 6𝑎4 𝑏2 9

44𝑝2

3. Factorise completely the following A. 4 − 16𝑝2 B. 3𝑥 2 − 3 C. 32𝑎2 − 50 D. x𝑎2 b − 4x𝑏 3 E. 𝑥 2 b + 2𝑥 2 − 9b − 18 4. Factorise the following expressions A. 𝑥 2 + 10x + 24 B. 2𝑥 2 − 7x + 3 C. 5𝑥 2 − x − 4 5. Simplify the following A. B. C. D.

𝑥2 − 1 𝑥2 − 𝑥 2𝑎2 − 8 𝑎+2 𝑦−1 𝑦2 − 1 2𝑥 2 − 3𝑥 − 5 𝑥2 − 1

Algebra

67

SOLUTIONS 1. A.

6 − 2(y − x) − 2 = 6 − 2y + 2x − 2 = 2x − 2y + 6 − 2 = 2x − 2y + 4

B.

5(a + 2b) − (2a − 4b) = 5a + 10b − 2a + 4b = 5a − 2a + 10b + 4b = 3a + 14b

C.

3m − (2n − 4m) − 6n = 3m − 2n + 4m − 6n = 3m + 4m − 2n − 6n = 7m − 8n

D.

4a − 5b − 3(2b − a) = 4a − 5b − 6b + 3a = 4a + 3a − 5b − 6b = 7a − 11b

E.

2(x + 4) − x(x + 1) = 2x + 8 − 𝑥 2 − x = −𝑥 2 + 2x − x + 8 = −𝑥 2 + x + 8

2. A.

15𝑚3 4𝑛2

=

5𝑚4

÷

12𝑛3 15𝑚3 12𝑛3

×

4𝑛2 5𝑚4 3𝑛

= 3× =

B.

20𝑎3 𝑏2 𝑦

=

C.

10𝑎3 𝑏

×

9𝑥 3 𝑦 2 1

×

2𝑏 3𝑥 2 𝑦 6𝑏𝑥 2

8𝑎3 𝑏2

÷

4𝑎2 𝑏2

30𝑎3 𝑏4 6𝑎4 𝑏2 8𝑎3 𝑏2 6𝑎4 𝑏2

= = = =

68

𝑚

3𝑥𝑦 3

=

𝑚

9𝑛

Algebra

×

30𝑎3 𝑏4 4𝑎2 𝑏2 4 3𝑎2 15𝑏2 2 5𝑏2 2𝑎2 5𝑏2

×

×𝑎

2 2

2. D.

11𝑝 18𝑞 2

= = 3. A.

44𝑝2 1 1

2𝑞 2 1

×

4𝑝

8𝑝𝑞 2

4 − 16𝑝2 = 22 − (4𝑝)2 = (2 + 4p)(2 − 4p)

B.

3𝑥 2 − 3 = 3(𝑥 2 − 1) = 3(𝑥 2 − 12 ) = 3(x + 1)(x − 1)

C.

32𝑎2 − 50 = 2(16𝑎2 − 25) = 2((4𝑎)2 − 52 ) = 2(4a + 5)(4a − 5)

D.

4.

9

×

x𝑎2 b − 4x𝑏 3 = xb(𝑎2 − 4𝑏 2 ) = xb(𝑎2 − (2𝑏)2 ) = xb(a + 2b)(a − 2b)

A. 𝑥 2 + 10x + 24 = (x + 6)(x + 4) B. 2𝑥 2 − 7x + 3 = 2𝑥 2 − x − 6x − 3 = (2𝑥 2 − x) + (−6x − 3) = x(2x − 1) −3(2x − 1) = (x − 3)(2x − 1) C. 5𝑥 2 − x − 4 = 5𝑥 2 − 5x + 4x − 4 = (5𝑥 2 − 5x) + (4x − 4) = 5x(x − 1) + 4(x − 1) = (5x + 4)(x − 1)

5. A.

𝑥2 − 1 𝑥2 − 𝑥 𝑥 2 − 12

= = =

𝑥(𝑥 − 1) (𝑥 + 1)(𝑥 − 1) 𝑥(𝑥 − 1) 𝑥+1 𝑥

Algebra

69

5. B.

2𝑎2 − 8 𝑎+2 2(𝑎2 − 4)

= = =

𝑎+2 2(𝑎2 − 22 ) 𝑎+2 2(𝑎 + 2)(𝑎 − 2) 𝑎+2

= 2(a − 2)

C.

𝑦−1 𝑦2 − 1 𝑦−1

= = =

D.

(𝑦 + 1)(𝑦 − 1) 1 𝑦+1

2𝑥 2 − 3𝑥 − 5

= = = =

70

𝑦 2 −12 𝑦−1

Algebra

𝑥2 − 1 (2𝑥 2 + 2𝑥) + (− 5𝑥 − 5) 𝑥 2 − 12 2𝑥(𝑥 + 1) −5(𝑥 + 1) (𝑥 + 1)(𝑥 − 1) (2𝑥 − 5)(𝑥 + 1) (𝑥 + 1)(𝑥 − 1) 2𝑥 − 5 𝑥−1

10

LINEAR EQUATIONS

Introduction A linear equation is an equation in the form of ax + b = c, where a, b and c are real numbers and x is a variable. Linear equations can also contain more than one variable such as ax + by + c = 0, where x and y are variables. In the first equation, that is, ax + b = c, b is a constant. In the second equation; that is, ax + by + c = 0, c is a constant. A constant represents the value that does not change. We use linear equations in real life unknowingly. For example, let’s say you are fifteen years old and your brother is four years old than you. Your years are variables because they change. However, four years between you and your brother is constant because it does not change. Therefore, your brother will always be four years older than you. Your age can be written as a linear equation x − y + 4 = 0 or y = x + 4, where x represents your age and y represents the age of your brother. With this equation, we can find the age of your brother by knowing your age. For example, when you turn 22, your brother is y = 22 + 4 or 28 years old.

Specific outcomes This unit will introduce you to linear equations. It covers linear equations in one variable and two variables. By the end of this unit, you will be able to. ❖ Identify a linear equation ❖ Solve linear equations in one variable ❖ Solve linear equations in two variables using: • Elimination method • Substitution method • Graphical method

Linear Equations

71

LINEAR EQUATIONS 10.1 LINEAR EQUATION • A linear equation is an equation with a variable raised to power one. • It is in form of ax + b = 0, where a and b are real numbers and x is a variable. • Examples of linear equations: 1. x + 2 = 0 2. 3y + 3 = 0 3. 2a + 1 = 7 • Linear equations have an equal sign indicating that what is on the left side is equal to the right side. ❖ HOW TO SOLVE LINEAR EQUATIONS IN ONE VARIABLE 1. Simplify the equation if necessary. 2. Move all terms with variables to the left side. If a term jumps to the other side of the equal sign it acquires an opposite sign; positive become negative, and negative become positive. 3. Move all terms without a variable to the right side. Change signs if a term jumps the equal sign. 4. Combine like terms. 5. Divide both sides of the equation by the numerical coefficient of the variable. ❖ EXAMPLES 1. Solve the following equations A. 2x + 1 = 7 B. 3x + 1 = 11 + x 2. Solve the linear equation A. 4(2x − 9) − 4x = 4 − 6x B.

𝑥−2 3𝑥 + 1

=3

❖ SOLUTIONS 1. A. 2x + 1 = 7 2x = 7 − 1 2x = 6 x=3

1 jumps the equal sign and changes from positive to negative. To leave only x on the left side divide both sides by the coefficient of x. In this question, divide both sides by 2.

B. 3x + 1 = 11 + x 3x − x = 11 − 1 2x = 10 x=5 2. A. 4(2x − 9) − 4x = 4 − 6x 8x − 36 − 4x = 4 − 6x 8x − 4x + 6x = 4 + 36 10x = 40 x = 10 B.

2−𝑥 3𝑥 + 1

=3

2 − x = 3(3x + 1) 2 − x = 9x + 3 − x − 9x = 3 − 2 −10x = 1 1 x = − 10

72

Linear Equations

First expand brackets

Cross multiply

Divide both sides by −10

10.2 LINEAR EQUATION IN TWO VARIABLES • A linear equation in two variables is an equation with two unknown variables raised to power one. • Is it in form of ax + by + c = 0, where a, b and c are real numbers. • Examples of linear equations in two variables: 1. 2x + y + 1 = 0 2. 3x − 5y = 2 3. y = x + 4 • Two equations are required in order to solve the two unknown variables. • These two equations are known as simultaneous equations, and they are solved simultaneously. • The simultaneous equations can be solved by three methods: elimination method, substitution method and graphical method. 1. ELIMINATION METHOD i. Place both equations in the form of Ax + By = C. ii. Determine which variable to eliminate by making its coefficient opposite. iii. Add straight down. iv. Solve the variable left. v. Substitute this result into either of the original equation to solve the other variable. vi. Check the solutions in both original equations. ❖ EXAMPLES 1. Solve the following simultaneous equations using elimination method. A. 3x − y = 12 2x + y = 13 B. 4x − 2y − 4 = 0 5x + 3y − 16 = 0 ❖ SOLUTIONS 1. A. 3x − y = 12 2x + y = 13

Step i. and ii.

3x − y = 12 2x + y = 13 + 5x + 0 = 25 5x = 25 x=5

Step iii. Step iv.

2x + y = 13 2(5) + y = 13 10 + y = 13 y = 13 − 10 y=3

Step v.

3(5) − 3 = 12 15 − 3 = 12 12 = 12

Step vi.

2(5) + 3 = 13 10 + 3 = 13 13 = 13

Step vi.

Linear Equations

73

1.

B.

4x − 2y − 4 = 0 5x + 3y − 16 = 0 4x − 2y = 4 5x + 3y = 16 4x − 2y = 4 5x + 3y = 16

Step i. ×3 ×2

12x − 6y = 12 10x + 6y = 32 + 22x + 0 = 44

Step iii.

22x = 44 x=2

Step iv.

4(2) − 2y = 4 8 − 2y = 4 − 2y = 4 − 8 − 2y = − 4 y=2

Step v

4(2) − 2(2) = 4 8−4=4 4=4 5(2) + 3(2) = 16 10 + 6 = 16 16 = 16

2.

Step ii.

Step vi.

Step vi.

SUBSTITUTION METHOD i. Isolate a variable one of the equations. ii. Substitute the isolated variable in the other equation. iii. Solve the equation to find the value of the variable. iv. Substitute the result into the other equation to solve for the other variable. v. Check the solutions in both original equations. ❖ EXAMPLES 1. Solve the following simultaneous equations using the substitution method. A. 3x − y = 12 2x + y = 13 B. 4x − 2y − 4 = 0 5x + 3y − 16 = 0

74

Linear Equations

❖

SOLUTIONS 1. A. 3x − y = 12 2x + y = 13 2x + y = 13 y = 13 − 2x

Step i.

3x − (13 − 2x) = 12 3x − 13 + 2x = 12 3x + 2x = 12 + 13 5x = 25 x=5

Step ii. Step iii.

2x + y = 13 2(5) + y = 13 10 + y = 13 y = 13 − 10 y=3

Step iv.

3(5) − 3 = 12 15 − 3 = 12 12 = 12

Step v.

2(5) + 3 = 13 10 + 3 = 13 13 = 13

Step v.

B. 4x − 2y − 4 = 0 5x + 3y − 16 = 0 4x − 2y − 4 = 0 4x = 2y + 4 x=

Step i.

2𝑦 + 4 4

2𝑦 + 4 ) + 3y − 16 = 0 4 2𝑦 + 4 5( 4 ) = 16 − 3y

5(

5(2𝑦 + 4) = 4(16 − 3y) 10y + 20 = 64 − 12y 10y + 12y = 64 − 20 22y = 44 y=2

4x − 2y − 4 = 0 4x − 2(2) − 4 = 0 4x − 4 − 4 = 0 4x − 8 = 0 4x = 8 x=2

Step ii. Step iii.

Step iv.

Linear Equations

75

3.

GRAPHICAL METHOD • Simultaneous equations can also be solved graphically. • If equations are plotted on a graph, the point of intersection represents the values of variables, the solutions to the simultaneous equations. • This method should be used if only you are asked to and you have a graph paper. • Steps to follow to solve simultaneous equations using the graphical method. i. Create a table of values for each equation. Just two or three values for x and y are enough. ii. Draw the graph of both equations on the same plane. iii. From the graph, find the point of intersection. iv. The coordinates at the point of intersection are the solutions. •

EXAMPLE 1. Solve the simultaneous equation graphically 3x − y = 12 2x + y = 13

•

SOLUTION Table of values for 3x − y = 12 x y

3 −3

4 0

5 3

To create the table of values, choose any three values for x and find the value of y for each x value using the equation. For example, using the equation 3x − y = 12, when x = 3, y gives −3, when x = 4, y gives 0. Solve to prove.

Table of values for 2x + y = 13 x y

4 5

5 3

5 1

Point of intersection is 5, 3. Therefore x=5 y=3

76

Linear Equations

REVIEW QUESTIONS 1. Solve the following equations A. 4k − 7 = 5 B. 5a + 2 = 3a + 7 C. 8x − 1 = 6x − 5 D. 5b + 4 = 3b + 6 2. Solve the following equations A. 5(m − 2) = 3(m + 4) B. 8(x − 4) = 5(x − 4) C. −(−15 − a) = 5a − 1 D. 6 = 6(y + 7) + 5y 3. Solve the linear equations A. B. C. D.

2−𝑥 3𝑥 + 1 60

=2

=3

2−𝑦 5𝑎−2 3 2 3

=

4𝑎+1 2

x − 1 = 6x + 2

4. Solve the following simultaneous equations A. x + y = 13 x−y=5 B. x + 2y = 8 3x − 5y = −9 C. 3x + 4y − 23 = 0 2x − y − 8 = 0 D. 2x + 5y = 37 y = 11 − 2x

Linear Equations

77

SOLUTIONS 1. A. 4k − 7 = 5 4k = 5 + 7 4k = 12 k=3 B. 5a + 2 = 3a + 7 5a − 3a = 7 − 2 2a = 5 a=

5 2

C. 8x − 1 = 6x − 5 8x − 6x = −5 + 1 2x = −4 x = −2 D. 5b + 4 = 3b + 6 5b − 3b = 6 − 4 2b = 2 b=1

2. A. 5(m − 2) = 3(m + 4) 5m − 10 = 3m + 12 5m − 3m = 12 + 10 2m = 22 m = 11 B. 8(x − 4) = 5(x − 4) 8x − 32 = 5x − 20 8x − 5x = −20 + 32 3x = 12 x=4 C. −(−15 − a) = 5a − 1 15 + a = 5a − 1 a − 5a = −1 − 15 −4a = −16 a=4 D. 6 = 6(y + 7) + 5y 6 = 6y + 42 + 5y 6y + 5y = 6 − 42 11y = −36 y=−

78

Linear Equations

36 11

3. A.

2−𝑥 3𝑥 + 1

=2

2 − x = 2(3x + 1) 2 − x = 6x + 2 −x − 6x = 2 − 2 −7x = 0 x=0

B.

60 2−𝑦

=3

60 = 3(2 − y) 60 = 6 − 3y 3y = 6 − 60 3y = −54 y = −18

C.

5𝑎−2 3

=

4𝑎+1 2

2(5a − 2) = 3(4a + 1) 10a − 4 = 12a + 3 10a − 12a = 3 + 4 −2a = 7 a=−

D.

2 3

7 2

x − 1 = 6x + 2 2

x − 6x = 2 + 1

3 2𝑥−18𝑥 3

=3

2x − 18x = 9 −16x = 9 x=−

9 16

Linear Equations

79

4

A. x + y = 13 x−y=5 x=5+y (5 + y) + y = 13 5 + 2y = 13 2y = 13 − 5 2y = 8 y=4 x−y=5 x−4=5 x=5+4 x=9 B. x + 2y = 8 3x − 5y = −9 x + 2y = 8 x = 8 − 2y 3(8 − 2y) − 5y = −9 24 − 6y − 5y = −9 −6y − 5y = −9 − 24 −11y = −33 y=3 x + 2y = 8 x + 2(3) = 8 x+6=8 x=8−6 x=2 C. 3x + 4y − 23 = 0 2x − y − 8 = 0 3x + 4y = 23 2x − y = 8 y = 2x − 8 3x + 4(2x − 8) = 23 3x + 8x − 32 = 23 3x + 8x = 23 + 32 11x = 55 x=5 2x − y = 8 2(5) − y = 8 10 − y = 8 −y = 8 − 10 −y = −2 y=2

80

Linear Equations

4. D.

2x + 5y = 37 y = 11 − 2x 2x + 5(11 − 2x) = 37 2x + 55 − 10x = 37 2x − 10x = 37 − 55 −8x = −18 x=

9 4

y = 11 − 2x 9 4

y = 11 − 2( ) y = 11 − y=

13

9 2

2

Linear Equations

81

11

COORDINATE GEOMETRY

Introduction Our world is huge. What can help us locate the position of a certain city, find the distance between two places, determine which place is on a higher attitude? It is coordinate geometry. Printed and electronic maps, communication and satellite systems are some of the devices that use coordinate geometry.

Specific outcomes This unit introduces coordinate geometry. It covers how to calculate distance between two points, gradient of a line and many more. By the end of this unit, you will be able to: ❖ Define coordinate geometry ❖ Calculate distance between two points ❖ Calculate the midpoint of a line ❖ Calculate the gradient of a line ❖ Find the equation of a straight line

82

Coordinate Geometry

COORDINATE GEOMETRY 11.1 COORDINATE GEOMETRY • Geometry is the branch of mathematics that deals with the sizes, shapes, positions and dimensions of figures and their relationship. • Coordinate geometry is the study of geometry using coordinates. • Coordinates are a set of values that shows the exact position of a point. • A cartesian plane is a plane formed by the intersection of two perpendicular lines known as the x-axis and y-axis. • The x-axis is the horizontal line. • The y-axis is the vertical line. Point A has coordinates (2,4) Point B has coordinates (−3,1)

11.2 DISTANCE BETWEEN TWO POINTS • The distance between two points, A(x1,y1) and B(x1,y2), is calculated by the following formula. • The formula for calculating the distance between two points: Distance AB = √(𝒙𝟐

𝒙𝟏 )𝟐 + (𝒚𝟐 𝒚𝟏 )𝟐

❖ EXAMPLES 1. Find the distance between two points A. Point X(−2,3) and point Y(3,−9) B. Point P(2,−3) and point Q(5,1) ❖ SOLUTIONS 1. A. Point X(−2,3) and point Y(3,−9) Distance XY = √(3

2

( 2)) + ( 9 3)2

= √52 + ( 12)2 = √25 + 144 = √169 = 13

Coordinate Geometry

83

1.

B.

Point P(2,−3) and point Q(5,1) 2

Distance XY = √(5

2)2 + (1 ( 3)) = √32 + 42 = √9 + 16 = √25 =5

11.3 MIDPOINT OF A LINE • A midpoint is a point on a line that divides it into two equal parts. • For any two points, A(x1,y1) and B(x1,y2), the midpoint M(x,y) is given by the following formula. • Formula: 𝒙𝟏 + 𝒙𝟐 𝒚 𝟏 + 𝒚 𝟐 , ) 𝟐 𝟐

M(x,y) = (

❖ EXAMPLES 1. Find the midpoint of the following points A. C(−2,3) and D(4,−9) B. P(2,−3) and Q(8,1) ❖ SOLUTIONS 1. A. C(−2,3) and D(4,−9) −2 + 4 3 + (−9)

,

M(x,y) = (

2 2 −6

,

=(

2

)

2

)

2

= (1,−3) B. P(2,−3) and Q(8,1) 2 + 8 −3 + 1

,

M(x,y) = (

2 10 −2

=(

2

,

2

2

)

)

= (5,−1)

11.4 GRADIENT OF A LINE • Gradient is simply how steep the line is. • The steeper the line joining two points the higher the gradient. • The gradient of a line can be positive, zero or negative. 1. Positive gradient: the line between two points slopes upward. 2. Zero gradient: the line between two points is horizontal. 3. Negative gradient: the line between two points slopes downward. • The gradient (m) of a line joining two points, A(x1,y1) and B(x1,y2), is calculated by the following formula. • The formula for calculation gradient of a line: m= 84

Coordinate Geometry

𝒚𝟐 − 𝒚𝟏 𝒙𝟐 − 𝒙𝟏

or

𝒚𝟏 − 𝒚𝟐 𝒙𝟏 − 𝒙𝟐

❖

EXAMPLES 1. Find the gradient of the line joining the following points A. C(−2,3) and D(4,−9) B. P(2,−3) and Q(4,5)

❖

SOLUTIONS 1. A. C(−2,3) and D(3,−9) m= =

𝑦2 − 𝑦1 𝑥2 − 𝑥1 −9 − 3 3 − (−2) 12

=−

5

B. P(2,−3) and Q(4,5) m= =

−3 − 5 2 − 4 −8 −2

=4

11.5 EQUATION OF A STRAIGHT LINE • The equation of a straight line on a cartesian plane is in form of: y = mx + c

•

• • •

where y = y-coordinate x = x-coordinate m = gradient c = y-intercept, the point where the line crosses y-axis

When the line is horizontal, gradient (m) = 0, then, y = 0x + c y=c When the line is vertical, the gradient is undefined. This is because the line does not have a slope. When two lines are parallel, they have the same gradient. Therefore m is equal. 𝑚1 = 𝑚2 When two lines are perpendicular, that is, they are at 90o to each other, and given that 𝑚1 is the gradient of one line and 𝑚2 the gradient of another line, then, 𝑚1 = −

1

𝑚2

❖ HOW TO FIND THE EQUATION OF A STRAIGHT LINE 1. Calculate the gradient of the slope. When points are given, use the gradient formula. When parallel or perpendicular equation deduce the gradient from the equation. 2. Put the gradient of the slope and a set of the given points into the formula y − y1 = m(x − x1). 3. Simplify to find the equation of a straight line.

Coordinate Geometry

85

❖

EXAMPLES 1. Find the equation of the line passing through points (0,2) and (−2,0). 2. The line contains (1,1) and (−2,7). Find the equation of the line. 3. Find the equation of the line parallel to the x-axis and passing through (3,4). 4. Find the equation of the line parallel to the line 4x − 2y − 2 = 0 and passing through the point (3,2). 5. Find the equation of the line passing through the point (0,−2) and is perpendicular to y = 5x + 3.

❖

SOLUTIONS 1. m = = =

𝑦2 − 𝑦1 𝑥2 − 𝑥1 0 − 2

Find the gradient of the line

−2 − 0 −2 −2

=1 y − y1 = m(x − x1) y − 2 = 1(x − 0) y−2=x y=x+2 2. m = = =

Put the m value and a set of coordinates into the formula. Simplify

𝑦2 − 𝑦1 𝑥2 − 𝑥1 7 − 1

−2 − 1 −6 3

= −2 y − y1 = m(x − x1) y − 1 = −2(x − 1) y − 1 = −2x + 2 y = −2x + 3 y = 3 − 2x 3. 𝑚1 = 𝑚2 = 0 y − y1 = 𝑚2 (x − x1) y − 4 = 0(x − 3) y−4=0 y=4 4. 4x − 2y − 2 = 0 2y = 4x − 2 y = 2x − 1 𝑚1 = 𝑚2 = 2 y − y1 = 𝑚2 (x − x1) y − 2 = 2(x − 3) y − 2 = 2x − 6 y = 2x − 6 + 2 y = 2x − 4 86

Coordinate Geometry

Lines parallel to x-axis are horizontal, ∴ has m = 0

In short, when m = 0, y = c

Lines parallel to each other have the same m value

5.

𝑚2 = − =−

1

Lines perpendicular to each other. 𝑚2 ⊥ 𝑚1 of y = 5x + 3

𝑚1 1 5

y − y1 = 𝑚2 (x − x1) y − (−2) = − y+2=− y=−

1

5 1 5 1 5

(x − 0) x x − 2

Coordinate Geometry

87

REVIEW QUESTIONS 1. Given the following coordinates A(3,4) and C (9,4) A. Find the distance between point A and B. B. Find midpoint coordinates for point A and B. C. Find the gradient of the line connecting point A and B. D. Find the equation of the line. 2. Find the distance between point P(4,3) and Q(4,5) A. Find the distance between point A and B. B. Find midpoint coordinates for point A and B. 3. 4. 5. 6. 7.

The gradient of the line joining the points (−2,a) and (a,−14) is 2. Calculate the value of a. Find the gradient of the line 3y + 2x = 6. Find the gradient of the line 2x + 2y − 4 = 0. Find the equation of the line joining point A(0,7) to point B(7,0). The diagram below shows a cartesian plane with points A(0,6), B(6,0), C(5,5) and D(0,−3).

Find A. Equation of the line AB B. Distance CD

88

Coordinate Geometry

SOLUTIONS 1. A. Distance AC = √(𝑥2 𝑥1 )2 + (𝑦2 𝑦1 )2 = √(9 3)2 + (4 = √62 + 02 = √36 =6 𝑥1 + 𝑥2 𝑦1 + 𝑦2

B. M(AC) = (

,

2 2 3 + 9 4 + 4

=(

2

,

2

4)2

)

)

12 8

=(

2

, ) 2

= (6,4) C. m = = =

𝑦2 − 𝑦1 𝑥2 − 𝑥1 4 − 4 9 − 3 0 6

=0

D. y − y1 = 𝑚2 (x − x1) y − 4 = 0(x − 3) y−4=0 y=4

2. A. Distance PQ = √(𝑥2 = √(4

𝑥1 )2 + (𝑦2 𝑦1 )2 4)2 + (3

5)2

= √02 + ( 2)2 = √4 =2 𝑥1 + 𝑥2 𝑦1 + 𝑦2

B. M(PQ) = (

,

2 2 4 + 4 3 + 5

=(

2 8 8

=(

,

2

)

)

, )

2 2

= (4,4)

Coordinate Geometry

89

3.

m= 2= 2=

𝑦2 − 𝑦1 𝑥2 − 𝑥1 −14 − 𝑎 𝑎 − (−2) −14 − 𝑎 𝑎+2

2a + 4 = −14 − a 2a + a = −14 − 4 3a = −18 a = −6

4.

3y + 2x = 6 3y = −2x + 6 y=−

2 3

x+2

gradient (m) = −

5.

2 3

2x + 2y − 4 = 0 2y = −2x + 4 y = −x + 2 gradient (m) = −1

6.

m= = =

𝑦2 − 𝑦1 𝑥2 − 𝑥1 0 − 7 7 − 0 −7 7

= −1

y − y1 = 𝑚2 (x − x1) y − 7 = −1(x − 0) y − 7 = −x y = −x + 7

90

Coordinate Geometry

Divide everything by 2

7.

A. m = = =

𝑦2 − 𝑦1 𝑥2 − 𝑥1 6 − 0 0 − 6 6 −6

= −1

y − y1 = 𝑚2 (x − x1) y − 6 = −1(x − 0) y − 6 = −x y = −x + 6

B.

Distance CD = √(𝑥2

𝑥1 )2 + (𝑦2 𝑦1 )2 2

= √(5 0)2 + (5 ( 3)) = √52 + 82 = √25 + 64 = √89

Coordinate Geometry

91

12

QUADRATIC EQUATIONS

Introduction A quadratic equation is an equation in form of ax2 + bx + c = 0. These equations have several applications in real life. We can use the quadratic equation to find how far an object has reached when dropped. For example, a ball is dropped off from a 30m tall building. How far does it reach in 2 seconds? Let’s take d to be distance moved from the top of the building, u to be the ball's initial speed upon release, t to be time taken and g to be 1 1 gravity. Using a quadratic equation 2at2 + ut − d = 0 or rearranged equation d = ut + 2at2, we can calculate 1 2

the distance moved in 2 seconds. As the ball is dropped u = 0m/s. Therefore, d = 0(2) + (10) × 22 which gives 20m. so, in 2 seconds the ball travelled 20m.

Specific outcomes This unit introduces you to quadratic equations. You will learn how to solve quadratic equations using different methods. By the end of this unit, you will be able to. ❖ Identify a quadratic equation ❖ Solve quadratic equations using: • Factorisation methods • Completing the square methods • Quadratic formula method

92

Quadratic equations

QUADRATIC EQUATIONS 12.1 QUADRATIC EQUATION • A quadratic equation is an equation in the form a𝑥 2 + bx + c = 0. • Examples of quadratic equations includes: 1. 𝑥 2 + 2x + 1 = 0 2. 2𝑥 2 − x − 3 = 0 3. 𝑥 2 + 7x + 12 = 0 • Quadratic equations are solved using various methods such as the factorisation method, completing the square or quadratic formula.

12.2 FACTORISATION METHOD • To solve a quadratic equation using the factorisation method first factorise the quadratic expression. • To recapitulate how to factorise quadratic expressions refer to section 9.10 of this book. • If a quadratic equation a𝑥 2 + bx + c = 0 is factorised to (x + d)(x + e) = 0, then: (x + d)(x + e) = 0 x + d = 0 or x + e = 0 x = −d x = −e ❖ EXAMPLES 1. Solve the following equations A. 𝑥 2 + 3x − 10 = 0 B. 𝑥 2 − 6x + 8 = 0 C. 9𝑥 2 − 64 = 0 D. 3𝑥 2 − 75 = 0 E. 2𝑥 2 − 7x + 3 = 0 F. 5𝑥 2 − x − 4 = 0 ❖ SOLUTIONS 1. A. 𝑥 2 + 3x − 10 = 0 𝑥 2 + 5x −2x − 10 = 0 (𝑥 2 + 5x) + (−2x − 10) = 0 x(x + 5) −2(x + 5) = 0 (x + 5)(x − 2) = 0 x + 5 = 0 or x − 2 = 0 x = −5 x=2 B. 𝑥 2 − 6x + 8 = 0 𝑥 2 − 4x −2x + 8 = 0 (𝑥 2 − 4x) + (−2x + 8) = 0 x(x − 4) −2(x − 4) = 0 (x − 4)(x − 2) = 0 x − 4 = 0 or x − 2 = 0 x=4 x=2

Quadratic equations

93

1.

C.

9x2 − 64 = 0 (3𝑥)2 82 = 0 (3x + 4)(3x − 4) = 0 3x + 4 = 0 or 3x − 4 = 0 3x = −4 3x = 4 x=−

4 3

x=

D. 3𝑥 2 − 75 = 0 3(𝑥 2 − 25) = 0 3(𝑥 2 − 52 ) = 0 3(x + 5)(x − 5) = 0 (x + 5)(x − 5) = 0 x + 5 = 0 or x − 5 = 0 x = −5 x=5

E.

2𝑥 2 − 7x + 3 = 0 (2x − 1)(x − 3) = 0 2x − 1 = 0 or x − 3 = 0 2x = 1 x=3 x=

F.

1 2

5𝑥 2 − x − 4 = 0 (5x + 4)(x − 1) = 0 5x + 4 = 0 or x − 1 = 0 5x = − 4 x=1 x=−

94

4 5

Quadratic equations

4 3

12.3 COMPLETING THE SQUARE METHOD • This method changes the equation so that the left side of the equation is a perfect square trinomial. • A perfect square trinomial is an algebraic expression created by multiplying a binomial by itself. For example, 𝑥 2 + 6x + 9 is a perfect square created by multiplying x + 3 by itself or (x + 3)2. • Steps involved in factorising a quadratic equation which is in the form of a𝑥 2 + bx + c = 0. 1. Write the quadratic equation in the form a𝑥 2 + bx = −c 2. Multiply the equation by 1 𝑎

(a𝑥 2 + bx = −c) 𝑥2 +

𝑏 2

3. Add (

1 𝑎

2

)

𝑏𝑥

𝑐

=−

𝑎

𝑎

on both sides of the equation 𝑏𝑥

𝑥2 +

𝑎

𝑏 2

+(

) 2

=−

𝑏 2

𝑐

+( ) 𝑎 2

4. Factor the left side of the equation into a perfect square 𝑏 2

(𝑥 + 2)

= −

𝑏 2

𝑐

+( ) 𝑎 2

5. Square root both sides of the equation and solve for x 2

√(𝑥 + 𝑏) 2 𝑏

𝑥+

2

=

√−

± √−

=

𝑐 𝑎 𝑐 𝑎

𝑏 2

+ ( ) 2

𝑏 2

+ ( ) 2

❖

EXAMPLES 1. Solve the following quadratic equations using the completing the square method A. 𝑥 2 + 3x − 10 = 0 B. 𝑥 2 + 12x + 32 = 0 C. 3𝑦 2 − 5y + 2 = 0 D. 4𝑥 2 − 8x − 32 = 0 E. 𝑥 2 − 3x − 18 = 0

❖

SOLUTIONS 1. A. 𝑥 2 + 3x − 10 = 0 𝑥 2 + 3x = 10 3 2

𝑥 2 + 3x + (

) 2

𝑥 2 + 3x +

9 4

2

3

(𝑥 + 2) =

3 2

= 10 + (

)

2

= 10 +

9 4

49 4

2

√(𝑥 + 3) = √49 2 4 7

3

𝑥+ 2 = ± 2 𝑥=

7 2

x=2

−

3 2

or 𝑥 =

7 2

−

x = −5

3 2

Quadratic equations

95

1.

B. 𝑥 2 + 12x + 32 = 0 𝑥 2 + 12x = −32 12 2

𝑥 2 + 12x + (

12 2

) 2

= −32 + (

𝑥 2 + 12x + 62 = −32 + 62 𝑥 2 + 12x + 36 = −32 + 36

2

)

12 2

(𝑥 + 2 ) = 4 √(𝑥 + 6)2 = √4 𝑥+ 6 =±2 𝑥 = 2 − 6 or 𝑥 = −2 − 6 x = −4 x = −8

C.

3𝑦 2 − 5y + 2 = 0 3𝑦 2 − 5y = −2 1 3

(3𝑦 2 − 5y = −2)

𝑦2 − 𝑦2 − 𝑦2 − 𝑦2 −

5𝑦 3 5𝑦

3

+ (−

3 5𝑦

+ (−

3 5𝑦 3

(𝑦 −

2

=−

+

25 36

2

5

𝑦−

6 1 6

y=1

96

= −

5 2

2

) 6

= −

× ) 3 2

√(𝑦 − 5) 6

𝑦=

× ) 3 2

1 2

5

Quadratic equations

+

6

=

= −

2

+

3

3 25

2 3

+ (−

+ (−

36

1 36

1 = √

=± 5

1 2

5

36

1 6

or

𝑦 =− y=

2 3

1 6

+

5 6

5 2 6

)

5

1 2

× ) 3 2

1.

D.

4𝑥 2 − 8x − 32 = 0 4𝑥 2 − 8x = 32 1 4

(4𝑥 2 − 8x = 32)

𝑥 2 − 2x = 8 −2 2

𝑥 − 2x + ( 2

2

)

−2 2

= 8+(

2

)

𝑥 2 − 2x + (−1)2 = 8 + (−1)2 𝑥 2 − 2x + 1 = 8 + 1 2 2 ) = 9 2

(𝑥

√(𝑥 1)2 = √9 x−1 =±3 𝑥 = 3 + 1 or x=4

𝑥 = −3 + 1 x = −2

E. 𝑥 2 − 3x − 18 = 0 𝑥 2 − 3x = 18 −3 2

𝑥 2 − 3x + ( 𝑥 2 − 3x + 2

(𝑥

9

) 2

−3 2

= 18 + (

= 18 +

4 81

2

)

9 4

3 ) = 2 4

√(𝑥 𝑥 𝑥=

3 2 81 ) = √ 2 4 9 3 =± 2 2 9 3 2

x=6

+

2

or

𝑥 =−

9 2

+

3 2

x = −3

Quadratic equations

97

12.4 QUADRATIC FORMULA METHOD • A quadratic formula is a formula that provides solutions to quadratic equations. • Before using the quadratic equation make sure the equation is in the form of a𝑥 2 + bx + c = 0. • The formula for calculating quadratic equations.

𝒙=

where a, b, and c are coefficients in quadratic equation a𝑥 2 + bx + c = 0

−𝒃±√𝒃𝟐 − 𝟒𝒂𝒄 𝟐𝒂

❖ EXAMPLES 1. Solve the following equations using the quadratic formula A. 𝑥 2 + 3x − 10 = 0 B. 2𝑥 2 − 14x + 24 = 0 2. Solve the following equations, giving your answers in 2 decimal places A. 13 − 9x − 5𝑥 2 = 0 B. 2𝑦 2 = 6y + 3 C. 3𝑧 2 = 7z − 1 ❖ SOLUTIONS 1. A. 𝑥 2 + 3x − 10 = 0

x=

a=1 b=3 c = −10

−𝑏±√𝑏2 − 4𝑎𝑐

𝑥= 𝑥= 𝑥= 𝑥= 𝑥= 𝑥=

2𝑎 −3±√32 − 4(1)(−10) 2(1) −3±√9 + 40 2 −3±√49 2 −3 ± 7 2 −3 + 7 4

𝑥=

or

2

𝑥=

2

x=2

−3 − 7 2 −10 2

x = −5

B. 2𝑥 2 − 14x + 24 = 0

𝑥=

𝑥= 𝑥= 𝑥= 𝑥= 𝑥= x=4 98

Quadratic equations

a=2 b = −14 c = 24

−(−14)±√(−14)2 − 4(2)(24) 2(2) 14±√169 − 192 4 14±√4 4 14±2 4 14 + 2 16 4

4

or

𝑥= 𝑥=

14 − 2 12

x =3

4

2

2.

A.

13 − 9x − 5𝑥 2 = 0 −5𝑥 2 − 9x + 13 = 0

𝑥= 𝑥= 𝑥= 𝑥= 𝑥= 𝑥= 𝑥=

Write the equation in form of a𝑥 2 + bx + c = 0 a = −5 b = −9 c = 13

−𝑏±√𝑏 2 − 4𝑎𝑐 2𝑎 −(−9)±√(−9)2 − 4(−5)(13) 2(−5) 9±√81 + 260 −10 9±√341 −10 9 ± 18.466 −10 9 + 18.466

or

−10 27.466

𝑥= 𝑥=

−10

x = − 2.75

9 − 18.466 −10 −9.466 −10

x = 0.95

B. 2𝑦 2 = 6y + 13 2𝑦 2 − 6y − 13 = 0

𝑦= 𝑦= 𝑦= 𝑦= 𝑦= 𝑦= 𝑦=

Write the equation in form of a𝑥 2 + bx + c = 0 a = −5 b = −9 c = 13

−𝑏±√𝑏2 −4𝑎𝑐 2𝑎 −(−6)±√(−6)2 − 4(2)(−13) 2(2) 6±√36 + 104 4 6±√140 4 6 ± 11.832 4 6 + 11.832 4 17.832 4

y = 4.46

or

𝑦= 𝑦=

6 − 11.832 4 −5.832 4

y = −1.46

Quadratic equations

99

2.

C.

3𝑧 2 = 7z − 1 3𝑧 2 − 7z + 1 = 0

𝑧= 𝑧= 𝑧= 𝑧= 𝑧= 𝑧= 𝑧=

−𝑏±√𝑏 2 −4𝑎𝑐

a = −5 b = −9 c = 13

2𝑎 −(−7)±√(−7)2 − 4(3)(1) 2(3) 7±√49 − 12 6 7±√37 6 7 ± 6.083 6 7 + 6.083 6 13.083

z = 2.18

100

Write the equation in form of a𝑥 2 + bx + c = 0

Quadratic equations

6

or

𝑧= 𝑧=

7 − 6.083 6 0.917 6

𝑧 = 0.15

REVIEW QUESTIONS 1. Solve the following quadratic equations using factorisation A. 𝑥 2 + 6x = 0 B. 4𝑥 2 = 81 C. 𝑥 2 + 10x + 4 = −5 D. 2𝑥 2 − 3x − 4 = 5 2. Solve the following equations using completing the squares A. 𝑥 2 + 10x + 4 = −5 B. 2𝑥 2 − 3x − 4 = 5 C. 2𝑥 2 = 12x + 54 D. 𝑥 2 + 8x − 22 = −2 3. Solve the following equations using the quadratic formula A. 𝑥 2 + 10x + 4 = −5 B. 2𝑥 2 − 3x − 4 = 5 4. Solve the following equations using the quadratic formula, giving your answers in two decimal places A. 𝑛2 − n − 5 = 0 B. 2𝑝2 + 6p = 1 C. 1 − 2m − 5𝑚2 = 0 D. (2y − 1)(3y − 2) = 3

Quadratic equations

101

SOLUTIONS 1. A. 𝑥 2 + 6x = 0 x(x + 6) = 0 x = 0 or

x+6=0 x = −6

B. 4𝑥 2 = 81 4𝑥 2 − 81 = 0 (2𝑥)2 − 92 = 0 (2x + 9)(2x − 9) = 0 2x + 9 = 0 or 2x − 9 = 0 2x = −9 2x = 9 x=−

9

x=

2

9 2

C. 𝑥 2 + 10x + 4 = −5 𝑥 2 + 10x + 4 + 5 = 0 𝑥 2 + 10x + 9 = 0 (𝑥 2 + x) + (9x + 9) = 0 x(x + 1) + 9(x + 1) = 0 (x + 9)(x + 1) = 0 x + 9 = 0 or x + 1 = 0 x = −9 x = −1 D. 2𝑥 2 − 3x − 4 = 5 2𝑥 2 − 3x − 4 − 5 = 0 2𝑥 2 − 3x − 9 = 0 (2𝑥 2 − 6x) + (3x − 9) = 0 2x(x − 3) + 3(x − 3) = 0 (2x + 3)(x − 3) = 0 2x + 3 = 0 or x − 3 = 0 2x = −3 x=3 x=−

3 2

2. A. 𝑥 2 + 10x + 4 = −5 𝑥 2 + 10x = −5 − 4 𝑥 2 + 10x = −9 10 2

𝑥 2 + 10x + (

) 2

10 2

= −9 + (

2

𝑥 2 + 10x + 52 = −9 + 52 𝑥 2 + 10x + 25 = −9 + 25 (𝑥 +

10 2 ) = 16 2

√(𝑥 + 5)2 = √16 𝑥+ 5 =±4 𝑥 = 4 − 5 or 𝑥 = −4 − 5 x = −1 x = −9 102

Quadratic equations

)

2.

B. 2𝑥 2 − 3x − 4 = 5 2𝑥 2 − 3x = 5 + 4 2𝑥 2 − 3x = 9 1 3

(2𝑥 2 − 3x = 9) 3𝑥

𝑥2 −

=

2 3𝑥

𝑥2 −

𝑥2 −

2

+ (−

2

𝑥2 −

9

3𝑥

+

2

9 16

) 4

9

=

2

× ) 2 2 2

√(𝑥 − 3) 4 3

𝑥− 𝑥=

3 2

1 2

3

(𝑥 −

× ) 2 2

+ (−

2 3𝑥

1 2

3

4 9 4

+

= ± 3 4

=

= +

9 2 9

9 2

+ (−

+ (−

1 2

3

× ) 2 2

3 2

)

4

16

81 16

81 = √ 16

9 4

or

𝑥 =−

x=3

C.

=

x=−

9 4 3

+

3 4

2

2𝑥 2 = 12x + 54 2𝑥 2 − 12x = 54 1 3

(2𝑥 2 − 12x = 54)

𝑥 2 − 6x = 27 𝑥 2 − 6x + (−

6 2

) 2

= 27 + (−

6 2

)

2

𝑥 2 − 6x + (−3)2 = 27 + (−3)2 𝑥 2 − 6x + 9 = 27 + 9 6 2

(𝑥 − 2)

= 36

√(𝑥 3)2 = √36 x−3=±6 𝑥 = 6 + 3 or x=9

𝑥 = −6 + 3 x = −3

Quadratic equations

103

2.

D. 𝑥 2 + 8x − 22 = −2 𝑥 2 + 8x = − 2 + 22 𝑥 2 + 8x = 20 8 2

𝑥 2 + 8x + (

) 2

8 2

= 20 + (

)

2

𝑥 2 + 8x + 42 = 20 + 42 𝑥 2 + 8x + 16 = 20 + 16 8 2

(𝑥 + 2) = 36 √(𝑥 + 4)2 = √36 𝑥+4 =±6 𝑥 = 6 − 4 or x=2

3.

A. 𝑥 2 + 10x + 4 = −5 𝑥 2 + 10x + 4 + 5 = 0 𝑥 2 + 10x + 9 = 0

𝑥=

−𝑏±√𝑏2 − 4𝑎𝑐

𝑥= 𝑥=

2𝑎 −10±√102 − 4(1)(9) 2(1) −10±√100 − 36 2

𝑥= 𝑥=

−10±√64 2 −10 ± 8

𝑥= 𝑥=

2 −10 + 8 2 −2 2

x = −1

104

𝑥 = −6 − 4 x = − 10

Quadratic equations

or

𝑥= 𝑥=

−10 − 8 2 −18 2

x = −9

3.

B. 2𝑥 2 − 3x − 4 = 5 2𝑥 2 − 3x − 9 = 0

𝑥= 𝑥=

−𝑏±√𝑏2 −4𝑎𝑐 2𝑎 −(−3)±√(−3)2 − 4(2)(−9) 2(2) 3±√9 + 72 4

𝑥= 𝑥=

𝑥=

3±√81 4 3±9

𝑥= 𝑥=

4 3+9 4 12

𝑥=

or

𝑥=

4

x=3

4.

3−9 4 −6

x=−

4 3 2

A. 𝑛2 − n − 5 = 0

𝑛=

−𝑏±√𝑏2 −4𝑎𝑐

𝑛= 𝑛= 𝑛= 𝑛= 𝑛= 𝑛=

2𝑎 −(−1)±√(−1)2 − 4(1)(−5) 2(1) 1±√1 + 20 2 1±√21 2 1 ± 4.583 2 1 + 4.583 2 5.583 2

n = 2.79

or

𝑛= 𝑛=

1 − 4.583 2 −3.583 2

n = −1.79

Quadratic equations

105

4.

B. 2𝑝2 + 6p = 1 2𝑝2 + 6p − 1 = 0

𝑝= 𝑝= 𝑝= 𝑝= 𝑝= 𝑝= 𝑝=

−𝑏±√𝑏2 − 4𝑎𝑐 2𝑎 −6±√62 − 4(2)(−1) 2(2) −6±√36 + 8 4 −6±√44 4 −6 ± 6.633 4 −6 + 6.633

𝑝=

or

4 0.633

𝑝=

4

p = 0.16

C.

4 −12.633 4

p = −3.16

1 − 2m − 5𝑚2 = 0 −5𝑚2 − 2m + 1 = 0

𝑚= 𝑚= 𝑚= 𝑚= 𝑚= 𝑚= 𝑚=

−𝑏±√𝑏2 − 4𝑎𝑐 2𝑎 −(−2)±√(−2)2 − 4(−5)(1) 2(−5) 2±√4 + 20 −10 2±√24 −10 2 ± 4.899 −10 2 + 4.899 −10 6.899 −10

m = 0.69

106

−6 − 6.633

Quadratic equations

or

𝑚= 𝑚=

2 − 4.899 −10 −2.633 −10

m = 0.29

4.

D.

(2y − 1)(3y − 2) = 3 6𝑦 2 − 4y − 3y + 2 = 3 6𝑦 2 − 7y + 2 − 3 = 0 6𝑦 2 − 7y − 1 = 0

𝑦= 𝑦=

−𝑏±√𝑏2 −4𝑎𝑐 2𝑎 −(−7)±√(−7)2 − 4(6)(−1)

𝑦= 𝑦= 𝑦=

2(6) 7±√49 + 24 12 7±√73 12 7 ± 8.544

𝑦= 𝑦=

12 7 + 8.544 12 15.544 12

y = 1.30

or

𝑦= 𝑦=

7 − 8.544 12 −1.544 12

𝑦 = −0.13

Quadratic equations

107

13

LINEAR INEQUALITIES

Introduction In unit 10, you covered linear inequalities. You learnt that linear equations can be in one variable such as x + 1 = 0 or two variables such as x + y = 1. Linear inequalities are more like linear equations. However, instead of having an equal sign (=), they have an inequality sign (>, = greater than x > 2, x is greater than 2

2. < = less than x < 2, x is less than 2

3. ≥ = greater than or equal to x ≥ 2, x is greater or equal to 2

4. ≤ = less than or equal to x ≤ 2, x is less or equal to 2

❖ HOW TO SOLVE LINEAR INEQUALITY IN ONE VARIABLE • Solving linear inequalities is similar to solving linear equations. • The only difference is when multiplying or dividing by a negative number the sign of the inequality reverses, for example: −2x > 4 =−

4 2𝑥 >− 2 2

=x 11 − 2x B. 3x − 7 ≥ 2(x − 6) C. 6 − x > 2

❖

SOLUTIONS 1. A. 6 − x > 11 − 2x 2x − x > 11 − 6 x>5

B. 3x − 7 ≥ 2(x − 6) 3x − 7 ≥ 2x − 12 3x − 2x ≥ − 12 + 7 x≥−5

C. 6 − x > 2 −x>2−6 −x>−4 x 2 on number appear as follow:

Overlap of solution can be written as −1 ≤ x < 2; read as x is greater than or equal to −1 but less than 2 •

110

Inequalities with the word “or” represent the union of solutions for each inequality, meaning one solution is true but the other solution is also true. For example, x ≤ −1 or x > 2

Linear Inequalities

❖

EXAMPLES 1. Solve the following inequalities and represent on a number line A. 5 ≤ x + 3 < 8 B. 2x − 5 < − 3 or 3x + 2 ≥ 17

❖

SOLUTIONS 1. A. 5 ≤ x + 3 < 8 5−3≤x or 5

Linear Inequalities

111

❖

SOLUTIONS 1. A. 4x − y ≤ 2 Table of values for 4x − y = 2 x y

0 −2

1 2

Graph as if it was an equation

2 6 To know which side to shade, choose any random point and test in inequality: E.g. test (3,2) 4(3) − 2 ≤ 2 12 − 2 ≤ 2 10 ≤ 2 not this side E.g. test (1,5) 4(1) − 5 ≤ 2 4−5≤2 −1 ≤ 2 shade this side

B. x + y > 5 Table of values for x + y = 5 x 1 2 3 y 4 3 2 To know which side to shade, choose any random point and test in inequality E.g. test (2,5) 2+5>5 7>5 shade this side E.g. test (2,2) 2+2>5 4>5 this side is not true for the inequality

112

Linear Inequalities

13.5 GRAPHING SYSTEMS OF INEQUALITIES • Systems of inequalities are a set of two or more inequalities with the same variables. • Systems of inequalities are mainly solved graphically. • The area overlapped by all the inequalities is the solution to the system. ❖ HOW TO GRAPH SYSTEMS OF INEQUALITIES i. Graph the boundary line for the first inequality. ii. Pick a point not on the line and test it in the inequality. iii. Shade the true side of the inequality. iv. Repeat steps i. to iii. for the remaining inequalities. v. The area where all the inequalities overlap is the solution to the system of inequalities. ❖ EXAMPLES 1. Solve each system of inequalities by graphing A. y > x − 1 y>3−x B. y ≥ 0 y x − 1 Table for y = x − 1 x y

0 −1

2 1

y>3−x Table for y = 3 − x x y

0 3

3 0 The overlapped region is the solution for the system of inequalities.

Linear Inequalities

113

1.

B. y ≥ 0

y 2x − 4 Table of values for y = 2x − 4 x y

0 −4

2 0

x+y −x + 8 −3x + x > 8 + 10 −2x > 18 x < −9

C. −3(2 − 5x) ≤ 9 −6 + 15x ≤ 9 15x ≤ 9 + 6 15x ≤ 15 x≤1 D. 4(2y − 1) ≥ y + 7 8y − 4 ≥ y + 7 8y − y ≥ 7 + 4 7y ≥ 11 y≥

11 7

2. A. −4 ≤ a + 3 < 9 −4 − 3 ≤ a < 9 − 3 −7 ≤ a < 6 B. −1 < 9 + x < 17 −1 − 9 < x < 17 − 9 −10 < x < 8 C. −4 < m − 5 ≤ −1 −4 + 5 < m ≤ −1 + 5 1 −3x + 5 Table of values x y

4.

−3 −2

0 5

A. y > 6 ………… i. y ≤ 2 ………… ii. Two points (−2,2) and (0,0) m=

𝑦2 − 𝑦1 𝑥2 − 𝑥1

=

2−0 −2 − 0

= −1

boundary line equation y − 𝑦1 = x(x − 𝑥1 ) y − 0 = −1(𝑥 0) y = −x inequality y < −x ………… iii. Two points (−2,0) and (0,4) m=

𝑦2 − 𝑦1 𝑥2 − 𝑥1

=

4−0 0 −(−2)

=

4 2

=2

boundary line equation y − 𝑦1 = x(x − 𝑥1 ) y − 0 = 2(𝑥 ( 2)) y = 2(x + 2) y = 2x + 4 inequality y ≥ 2x + 4 ………… iv.

Linear Inequalities

121

B.

Two points (0,0) and (5,5) m=

𝑦2 − 𝑦1 𝑥2 − 𝑥1

=

5−0 5−0

=1

boundary line equation y − 𝑦1 = x(x − 𝑥1 ) y − 0 = 1(𝑥 0) y=x inequality y ≥ x ………… i. Two points (0,1) and (7,0) m=

𝑦2 − 𝑦1 𝑥2 − 𝑥1

=

0−1 7−0

=−

1 7

boundary line equation y − 𝑦1 = x(x − 𝑥1 ) y−0=− y=−

1 7

1

(𝑥

7

7)

x+1 y=−

inequality;

1 7

x + 1 …………… (ii)

Two points (5,5) and (7,0) m=

𝑦2 − 𝑦1 𝑥2 − 𝑥1

=

0−5 7−5

=−

5 2

boundary line equation y − 𝑦1 = x(x − 𝑥1 ) y−0=− y=−

5 2

5 2

x+

inequality;

122

Linear Inequalities

(𝑥

7)

35 2

y≥−

5 2

x+

35 2

…………… (iii)

14

RELATIONS AND FUNCTIONS

Introduction The words relation and relationship are frequently used in our everyday communication. They are used to describe how two or more people or things are connected. The usage of relation in mathematics is more like in daily life usage. It describes the connection between two ordered pairs. For example, let’s take the number of 30-egg trays and the total number of eggs to be two ordered pairs. The number of trays is represented by x and the total number of eggs is represented by y. Therefore, the relation describing the number of trays and the total number of eggs can be written as y = 30x. If there is 1 tray, the total number of eggs should be 30: y = 30 (1) or y = 30. If there 2 trays, the total number of eggs should be 60: y = 30(2) or y = 60. If there are 5 trays, the total number of eggs should be 150: y = 30(5) or y = 150.

Specific outcomes What is a function? Are all relations functions? This unit covers these questions and many other. By the end of this unit, you will be able to ❖ Define relation, function, ordered pairs, domain and range ❖ Identify a: • One-to-one function • Many-to-one function ❖ Write a function notation ❖ Evaluate functions ❖ Find a composite function ❖ Find the inverse of a function ❖ Find the transpose of a function

Relations and Functions

123

RELATIONS AND FUNCTIONS 14.1 RELATION • Relation is the relationship between ordered pairs. • It indicates how ordered pairs are related. • For example, Gilbert is 4 years older than his brother Harry. Therefore, the relation between the age of Gilbert and Harry is G = H + 4. When Harry is 2 years old, Gilbert is 6 years old. When Harry is 4 years old, Gilbert is 8 years old. When Harry is 10 years old, Gilbert is 14 years old and so on. ❖ DOMAIN AND RANGE • In the above example, we put the age of Harry, and then relation produces the age Gilbert. 2 years for Harry gave 6 years for Gilbert, 4 years gave 8 years, 10 gave 14 years, etc. • The value that we put in the relation to produce a result is called the input or domain. • The value produced by the input of the relation is called the output or range. ❖ ORDERED PAIRS • An ordered pair is a set of input and output. • It is represented as (input, output). • From the above example, ordered pairs are (2,4), (4,8), (10,14). ❖ A relation can be represented as a table, a mapping diagram or a graph. 1. Table Harry’s age Gilbert’s age 2 6 4 8 10 14

2. Mapping diagram

3. Graph

124

Relations and Functions

14.2 FUNCTIONS • A function is a type relation that has one output for each input. • All functions are relations but not all relations are functions. A relation where an input as more than one output is not a function. •

The following are functions.

•

The above examples show that each domain or input value produces one range or output value. In example 2, −1 and 1 gave the same output but the relation is still a function because each input, even −1 or 1 produced one output.

•

The following relation is not a function.

•

The above relation is not a function because some inputs produced more than one output; 2 gave 6 and 7, 4 gave 7 and 9. As these inputs, 2 and 4, have more than one output the relation is not a function.

Relations and Functions

125

14.3 TYPES OF FUNCTIONS 1. ONE - TO - ONE FUNCTION • A function is one-to-one if no two inputs correspond to the same output. • Each input has exactly one output. • Below, function 1 is one-to-one while function 2 is not.

2. MANY - TO - 0NE FUNCTION • A function is many-to-one if two or more inputs have the same output. • Some inputs have the same output. • Below, function 2 is a many-to-one function while function 1 is not.

126

Relations and Functions

14.4 FUNCTION NOTATION • Function notation is a way of writing a function. • The most common function notation is f(x), which is read as “f of x”. f is a function name and x is the input value. • The equation y = x + 5 in function notation can be written as f(x) = x + 5, where x is the input value and x + 5 the output value. • Any letter can be used to name a function such as the following examples: f(x) = 𝑥 2 + 2x + 1 g(t) = 2t + 1 h(a) = 𝑎2

14.5 EVALUATING FUNCTIONS • To evaluate a function, substitute the input with the number or expression given. • Examples: 1. Given the function f(x) = x + 2, find f(5) 2. Given the function g(t) = 5𝑡 2 + 2, find g(2) 3. Find h(4k) when h(x) = 𝑥 2 + 4x + 2 4. Given f(x) = 2𝑥 2 + 5x + 8, find (2a − 4) in terms of a 5. Given f(x) = 2x + 5, if f(a) = 13 what is the value of a? 6. If f(x) = 𝑥 2 − 4, calculate h if f(h) = 45 •

Solutions 1. f(x) = x + 2 f(5) = 5 + 2 f(5) = 7 2. g(t) = 5𝑡 2 + 2 g(2) = 5(2)2 + 2 g(2) = 5(4) + 2 g(2) = 20 + 2 g(2) = 22

Substitute x with 5 and simplify Substitute t with 2 and simplify

3. h(x) = 𝑥 2 + 4x + 2 h(4k) = (4𝑘)2 + 4(4k) + 2 h(4k) = 16𝑘 2 + 16k + 2

Substitute x with 4k and simplify

4. f(x) = 2𝑥 2 + 5x + 8 f(2a − 4) = 2(2𝑎 4)2 + 5(2a − 4) + 8 f(2a − 4) = 2(4𝑎2 − 16a + 16) + 10a − 20 + 8 f(2a − 4) = 8𝑎2 − 32a + 32 + 10a − 20 + 8 f(2a − 4) = 8𝑎2 − 32a + 10a + 32 − 20 + 8 f(2a − 4) = 8𝑎2 − 22a + 20 5. f(x) = 2x + 5 f(a) = 13 2(a) + 5 = 13 2a + 5 = 13 2a = 13 − 5 2a = 8 a=4

Substitute x with 2a − 4, expand brackets and simplify

The function, f(a), is equal to 13, therefore, 2(a) + 5 = 13

Relations and Functions

127

6.

f(x) = 𝑥 2 − 4 f(h) = 45 𝑥 2 − 4 = 45 𝑥 2 − 49 = 0 𝑥 2 − 72 = 0 (x + 7)(x − 7) = 0 x + 7 = 0 or x − 7 = 0 x = −7 x=7

14.6 COMPOSITE FUNCTIONS • A composite function is formed when one function is substituted into another function. • If one function is f(x) = 2x + 1 and another function is g(x) = x + 3. The input of f(x) = 2x + 1 is x and output is 2x + 1. The input of g(x) = x + 3 is x and output is x + 3. • When the output of f(x) = 2x + 1, which is 2x + 1, becomes the input of g(x) = x + 3 the function created is a composite function and it can be written as g(𝑓(𝑥)) = (2x + 1) + 3. • In short, a composite function is created when the output of one function becomes the input of another function. • The composite function g(𝑓(𝑥)) can also be written as (g ∘ f)(x) or gf(x). • Composite function, for example, g(𝑓(𝑥)) is read as “g of f of x. • Remember that g(𝑓(𝑥)) ≠ f(𝑔(𝑥)). ❖ HOW TO FIND A COMPOSITE FUNCTION 1. Write the outer function. 2. Substitute x of the outer function with the output of the other function. 3. Simplify the answer. ❖ EXAMPLES 1. If f(x) = 2x + 1 and g(x) = x + 3, find A. (g ∘ f)(x) B. (f ∘ g)(x) C. (g ∘ f)(2) D. (f ∘ g)(2) E. (g ∘ f)(x) = 10 F. (f ∘ g)(x) = 10 2. Given that g(x) = 6x and h(x) = 2x − 5, find A. gh(2) B. hg(−3) 3. Given that f(x) = 2x + 1 and g(x) = A. fg(x) B. fg(3)

3𝑥 − 5 2

, find

4. The functions f and g are defined as f(x) = 8x and g(x) = A. an expression for fg(x) B. the value of x for which fg(x) = 20

128

Relations and Functions

3𝑥 − 2 4

, find

❖

SOLUTIONS 1. A. (g ∘ f)(x) = g(𝑓(𝑥)) g(x) = x + 3 g(𝑓(𝑥)) = (2x + 1) + 3 = 2x + 1 + 3 = 2x + 4 B. (f ∘ g)(x) = f(𝑔(𝑥)) f(x) = 2x + 1 f(𝑔(𝑥)) = 2(x + 3) + 1 = 2x + 6 + 1 = 2x + 7 C. (g ∘ f)(x) = g(𝑓(𝑥)) g(𝑓(𝑥)) = 2x + 4 g(𝑓(2)) = 2(2) + 4 =4+4 =8 D. (f ∘ g)(x) = f(𝑔(𝑥)) f(𝑔(𝑥)) = 2x + 7 = 2(2) + 7 =4+7 = 11 E. (g ∘ f)(x) = g(𝑓(𝑥)) g(𝑓(𝑥)) = 10 2x + 4 = 10 2x = 10 − 4 2x = 6 x=3 F. (f ∘ g)(x) = f(𝑔(𝑥)) f(𝑔(𝑥)) = 10 2x + 7 = 10 2x = 10 − 7 2x = 3 x=

3 2

2. A. g(x) = 6x gh(x) = 6(2x − 5) gh(x) = 12x − 30 gh(3) = 12(3) − 30 = 36 − 30 =6 B. h(x) = 2x − 5 hg(x) = 2(6x) − 5 hg(x) = 12x − 5 hg(−3) = 12(−3) − 5 = −36 − 5 = −41 Relations and Functions

129

3.

A. f(x) = 2x + 1 3𝑥

5

fg(x) = 2( )+1 2 = 3x − 5 + 1 = 3x − 4 B. fg(x) = 3x − 4 fg(3) = 3(3) − 4 =9−4 =5 4.

A. f(x) = 8x 3𝑥

2

fg(x) = 8( ) 4 = 2(3x − 2) = 6x − 4 B. fg(x) = 6x − 4 fg(x) = 20 6x − 4 = 20 6x = 20 + 4 6x = 24 x=4

14.7 INVERSE FUNCTION • An inverse function is a function that reverses another function. • It is denoted by f −1 . • If f(x) turns input A into output B, then f −1 (x) must turn output B into input A.

•

For example, a function f(x) = 2x + 1, if the input is 2 then the output is 5. f(x) = 2x + 1 f(2) = 2(2) + 1 =5 With the inverse of f(x), that is f −1 , f −1 (5) gives back 2

❖ HOW TO FIND THE INVERSE OF A FUNCTION 1. Replace f(x) with y 2. Swap x and y 3. Solve for y 4. Replace y with f −1 (x) ❖ EXAMPLES 1. Find the inverse function for f(x) = 3x − 2 2. Find the inverse function of f(x) = − 3. Given that f(x) =

130

Relations and Functions

7𝑥 + 1 𝑥+2

, find f −1 (x)

𝑥

2

+2

❖

SOLUTIONS 1. f(x) = 3x − 2 y = 3x − 2 x = 3y − 2 3y = x + 2 y= f

−1

𝑥+2

2. f(x) = −

3

𝑥 2 𝑥

y=− x=− 2

2. Swap x and y 3. Solve for y

3 𝑥+2

(x) =

𝑦

1. Change function into an equation

2 𝑦 2

4. Replace y with f −1 (x)

+2 +2 +2

= −x + 2

y = 2(−x + 2) y = −2x + 4 f −1 (x) = −2x + 4

3. f(x) = y= x=

7𝑥 + 1 𝑥+2 7𝑥 + 1 𝑥+2 7𝑦 + 1 𝑦+2

x(y + 2) = 7y + 1 xy + 2x = 7y + 1 xy − 7y = −2x + 1 y(x − 7) = −2x + 1 y=

−2𝑥 + 1 𝑥−7 −2𝑥 + 1

f −1 (x) =

❖

𝑥−7

NOTE • f −1 (𝑓(𝑥)) = x, read as f inverse of f of x equals x • For example; given a function f(x) = x + 1, find f −1 (𝑓(𝑥)) f(x) = x + 5 y=x+5 x=y+5 y=x−5 f −1 (x) = x − 5 f −1 (x) = x − 5 f −1 (𝑓(𝑥)) = (x + 5) − 5 =x+5−5 =x

Relations and Functions

131

REVIEW QUESTIONS 1. Given that f(x) = 2 − 3x, find A. f(−2) B. f −1 (−2) 2. A function g is defined as g(x) = 2x − 5, find A. g(3) B. the value of x for which g(x) = 7 C. g −1 (x) 3. A function is defined by f(x) = 5x + 9, find A. f(1) B. k, if f(x) = k C. f −1 (x) 4. A function h is defined as h(x) =

3𝑥 + 2 4

, find

A. h(−2) B. the value of x for which h(x) = 4 C. h−1(x)

5. The function f and g are defined be f(x) = 2x + 1 and g(x) = 5x − 1, find A. g −1 (x) B. fg(x) C. fg(3) 6. Given that f(x) =

5𝑥 + 4

−1

5

and g(x) = x − 1, find

A. f (x) B. f −1 (−2) C. fg(x), in its simplest form 7. The function g and f are defined as g(x) = −1

A. g (x) B. x, if g(x) = f(x) C. g −1 f(x)

132

Relations and Functions

𝑥−1 2

, and f(x) = 3x − 5, find

SOLUTIONS 1. A. f(x) = 2 − 3x f(−2) = 2 − 3(−2) f(−2) = 2 + 6 f(x) = 8 B. f(x) = 2 − 3x y = 2 − 3x x = 2 − 3y 3x = 2 − x 2−𝑥

x= f

−1

3

(x) =

2−𝑥 3

2. A. g(x) = 2x − 5 g(3) = 2(3) − 5 g(3) = 6 − 5 g(3) = 1 B. g(x) = 7 2x − 5 = 7 2x = 7 + 5 2x = 12 x=6 C. g(x) = 2x − 5 y = 2x − 5 x = 2y − 5 2y = x + 5 y= g

−1

𝑥+5 2

(x) =

𝑥+5 2

3. A. f(x) = 5x + 9 f(1) = 5(1) + 9 f(1) = 5 + 9 f(1) = 14 B. f(k) = k 5k + 9 = k 5k − k = −9 4k = −9 k=−

9 4

C. f(x) = 5x + 9 y = 5x + 9 x = 5y + 9 5y = x − 9 y= f

−1

𝑥−9 5

(x) =

𝑥−9 5

Relations and Functions

133

4.

A. h(x) =

3𝑥 + 2

h(−2) = h(−2) =

4 3(−2) + 2 4 −6 + 2

h(−2) = − h(−2) = −

4 2 4 1 2

B. h(x) = 4 3𝑥 + 2 4

=4

3x + 2 = 8 3x = 8 − 2 3x = 6 x=2 C. h(x) = y= x=

3𝑥 + 2

4 3𝑥 + 2

4 3𝑦 + 2 4

4x = 3y + 2 3y = 4x − 2 y=

5.

4𝑥 − 2 3

A. g(x) = 5x − 1 y = 5x − 1 x = 5y − 1 5y = x + 1 y= g

−1

𝑥+1 5

(x) =

𝑥+1 5

B. fg(x) = 2(5x − 1) + 1 fg(x) = 10x − 2 + 1 fg(x) = 10x − 1 C. fg(x) = 10x − 1 fg(3) = 10(3) − 1 fg(3) = 30 − 1 fg(3) = 29

134

Relations and Functions

6.

A. f(x) = y= x=

5𝑥 + 4 5 5𝑥 + 4 5 5𝑦 + 4 5

5x = 5y + 4 5y = 5x − 4 y=

5𝑥 − 4 5

f −1 (x) = B. f −1 (x) =

f −1 (−2) = f

5 5(−2) − 4 5 −10 − 4

(−2) = −

C. fg(x) = fg(x) = fg(x) =

7.

5

5𝑥 − 4

f −1 (−2) =

−1

5𝑥 − 4

A. g(x) = y= x=

5 14 5

5(𝑥 − 1) + 4 5 5𝑥 − 5 + 4 5 5𝑥 − 1 5

𝑥−1 2 𝑥−1 2 𝑦−1 2

2x = y − 1 y = 2x + 1 g −1 (x) = 2x + 1 B. g(x) = f(x) 𝑥−1 2

= 3x − 5

x − 1 = 2(3x − 5) x − 1 = 6x − 10 x − 6x = −10 + 1 −5x = −9 x= C.

9 5

g −1 (x) = 2x + 1 g −1 f(x) = 2(3x − 5) + 1 g −1 f(x) = 6x − 10 + 1 g −1 f(x) = 6x − 9 Relations and Functions

135

15

LINEAR FUNCTIONS

Introduction Linear functions are functions in the form of f(x) = ax + b. when plotted on a graph, linear functions always produce a straight line. In real life, we can use a linear function to predict the outcome. For example, you consume 5 units of electricity per day. How many units will be consumed in 2 days, 3 days, 15 days and 30 days? Let’s take x to be the number of days and f(x) to be the number of units consumed. Therefore, the linear function for the number of units consumed can be written as f(x) = 5x. The number of units consumed in 2 days is 10; f(2) = 5(2). The number of units consumed in 3 days is 15; f(3) = 5(3). The number of units consumed in 15 days is 75; f(15) = 5(15). Finally, the number of units consumed in 30 days is 150; f(30) = 5(30). When you plot the number of units consumed against the number of days, you will find that the function produces a straight line.

Specific outcomes This unit covers the features of linear functions and how to sketch graphs of linear functions. By the end of this unit, you will be able to: ❖ Identify a linear function ❖ Find the x and y - intercept of the graph ❖ Sketch the graph of linear functions

136

Linear Functions

LINEAR FUNCTIONS 15.1 LINEAR FUNCTION • A linear function is a function in the form of f(x) = ax + b, where x is the input and ax + b is the output, and a and b are real numbers. • A linear function always produces a straight line on the graph. 15.2 GRAPH OF A LINEAR FUNCTION

❖ a in f(x) = ax + b • a is the gradient of the function. • The larger a is, the steeper the graph is. • If a > 0, the graph slopes upward. • If a < 0, the graph slopes downward. ❖ b in f(x) = ax + b • b is the y-intercept. • It indicates where the graph passes the y-axis.

15.3 HOW TO FIND INTERCEPTS OF A GRAPH ❖ y-intercept • y-intercept is found by letting x = 0 • Examples 1. Find the y-intercept of f(x) = 2x + 4 2. Find the y-intercept of f(x) = 3x − 1 • Solutions 1. f(x) = 2x + 4 y = 2x + 4 y = 2(0) + 4 y=4 2. f(x) = 3x − 1 y = 3x − 1 y = 3(0) − 1 y = −1

Rewrite the function into equation, let x = 0 and found the value of y.

Rewrite the function into equation, let x = 0 and found the value of y.

Linear Functions

137

❖

x-intercept • x-intercept is found by let y = 0. • Examples: 1. Find the x-intercept of f(x) = 2x + 4 2. Find the x-intercept of f(x) = 3x − 1 •

Solutions 1. f(x) = 2x + 4 y = 2x + 4 0 = 2x + 4 2x = −4 x = −2 2. f(x) = 3x − 1 y = 3x − 1 0 = 3x − 1 3x = 1 x=

Rewrite the function into equation, let y = 0 and found the value of x

Rewrite the function into equation, let y = 0 and found the value of x

1 3

15.4 HOW TO SKETCH A LINEAR FUNCTION 1. Rewrite the function f(x) = ax + b, into an equation y = ax + b. 2. Determine the sign of a. If a is positive, the graph slopes upward. If a is negative, the graph slopes downward. 3. Find y-intercept by letting x = 0. Note that y = b. 4. Find x-intercept by letting y = 0. 5. Draw a straight line connecting the two intercepts. ❖ EXAMPLES 1. Sketch the graph of the following functions A. f(x) = 2x + 2 B. f(x) = −2x + 2 C. f(x) = 2x − 2 D. f(x) = x ❖ SOLUTIONS 1. A. f(x) = 2x + 2 y = 2x + 2 a > 0, ∴ graph slopes upward y-intercept y = 2x + 2 y = 2(0) + 2 y=2 x-intercept y = 2x + 2 0 = 2x + 2 2x = −2 x = −1

138

Linear Functions

1.

B.

f(x) = −2x + 2 y = −2x + 2 a < 0, ∴ graph slopes downward y-intercept y = −2x + 2 y = −2(0) + 2 y=2 x-intercept y = −2x + 2 0 = −2x + 2 2x = 2 x=1

C.

f(x) = 2x − 2 y = 2x − 2 a < 0, ∴ graph slopes downward y-intercept y = 2x − 2 y = 2(0) − 2 y = −2 x-intercept y = 2x − 2 0 = 2x − 2 2x = 2 x=1

D.

f(x) = x y=x a > 0, ∴ graph slopes upward y-intercept y=x y=0 x-intercept y=x x=0 As x and y are both intercepting at 0, find another point by letting x equal to any number except 0, say x = 2, then y=x y=2 Linear Functions

139

REVIEW QUESTIONS 1. Sketch the graph of the following linear functions A. f(x) = 3x B. f(x) =

1 2

C. f(x) = −

x+3 5 4

x+1

D. f(x) = −x − 2

SOLUTIONS 1. A. f(x) = 3x y = 3x a 0, the graph opens down, the graph has minimum point

B. x = − =−

a=1 b = −4 c=5

𝑏 2𝑎 −4 2(1)

= − (−2) =2 y = f(−

𝑏 2𝑎

)

y = 𝑥 2 − 4x + 5 y = 22 − 4(2) + 5 =4−8+5 =1 Turning point (2,1)

C. f(x) = 𝑥 2 − 4x + 5 y = 𝑥 2 − 4x + 5 y = 02 − 4(0) + 5 =5 y-intercept = 5

Change quadratic function into an equation To find y-intercept, let x = 0 and solve for y.

D. f(x) = 𝑥 2 − 4x + 5 y = 𝑥 2 − 4x + 5 𝑥 2 − 4x + 5 = 0 𝑥 2 − 4x = −5 −4

𝑥 2 − 4x + (

2

−4

) = −5 + ( 2 )

𝑥 2 − 4x + (−2)2 = −5 + (−2)2 𝑥 2 − 4x + 4 = −5 + 1 (𝑥 2)2 = −1 x−2=√ 1 √ 1 has no value ∴ the function has no x-intercept

E.

146

Equation of axis of symmetry; y = 1

Graphs of Quadratic Functions

Change quadratic function into an equation To find x-intercept, let y = 0 and solve for x Solve the equation using completing the square To recapitulate how to solve quadratic equations using completing the square refer to section 12.3 of this book.

Any quadratic function with a square root of negative number has no x-intercept, meaning the graph does not cross x-axis.

Axis of symmetry passes through the turning point and is parallel to y-axis, ∴ it is the y-coordinate of the turning point.

3.

A. f(x) = −𝑥 2 + 4 f(x) = −𝑥 2 + 4 a < 0, the graph opens up, the graph has maximum point

B. x = − =−

a = −1 b=0 c=4

𝑏 2𝑎 0 2(1)

=0 y = f(−

𝑏 2𝑎

)

y = 02 + 4 =4 Turning point (0,5) C. f(x) = −𝑥 2 + 4 y = −𝑥 2 + 4 y = 02 + 4 =4 y-intercept = 4

D. f(x) = −𝑥 2 + 4 y = −𝑥 2 + 4 −𝑥 2 + 4 = 0 𝑥2 = 4 √𝑥 2 = √4 x=2 or x = −2 x-intercept is −2 and 2

16.4 HOW TO SKETCH A QUADRATIC FUNCTION 1. Rewrite the quadratic function, f(x) = a𝑥 2 + bx + c, into a quadratic equation, y = a𝑥 2 + bx + c. 2. Determine whether the graph opens up or down. 3. Find the turning point. 4. Find the y and x-intercept. 5. Draw to smooth curve connecting points. ❖ EXAMPLES 1. Sketch the graph of the function f(x) = 𝑥 2 + 4x + 3 2. Sketch the quadratic function f(x) = −𝑥 2 − 6x − 5 3. Given the quadratic function f(x) = (𝑥 1)2 + 2, sketch the graph 4. Sketch the graph of the quadratic function f(x) = 𝑥 2

Graphs of Quadratic Functions

147

❖

SOLUTIONS 1. f(x) = 𝑥 2 + 4x + 3 y = 𝑥 2 + 4x + 3 a > 0, the graph opens up, the graph has minimum point x=− =−

a=1 b=4 c=3

𝑏 2𝑎 4 2(1)

= −2 y = f(−

𝑏 2𝑎

)

y = 𝑥 2 + 4x + 3 = ( 2)2 + 4(−2) + 3 =4−8+3 = −1 Turning point (−2,−1) f(x) = 𝑥 2 + 4x + 3 y = 02 + 4(0) + 3 y=3 y-intercept = 3 f(x) = 𝑥 2 + 4x + 3 y = 𝑥 2 + 4x + 3 𝑥 2 + 4x + 3 = 0 (𝑥 2 + x) + (3x + 3) = 0 x(x + 1) + 3(x + 1) = 0 (x + 3)(x + 1) = 0 x + 3 = 0 or x + 1 = 0 x = −3 x = −1 x-intercept is −3 and −1 a > 0, graph opens up Turning point (−2,−1) y-intercept is 3 x-intercepts are −3 and −1

148

Graphs of Quadratic Functions

2.

f(x) = −𝑥 2 − 6x − 5 y = −𝑥 2 − 6x − 5 a < 0, the graph opens down, the graph has a maximum point x=− =− =−

a = −1 b = −6 c = −5

𝑏 2𝑎 −6 2(−1) −6 −2

=−3 y = f(−

𝑏 2𝑎

)

y = −( 3)2 − 6(−3) − 5 = −9 + 18 − 5 =4 Turning point (−3,4) f(x) = −𝑥 2 − 6x − 5 y = −𝑥 2 − 6x − 5 y = −(0)2 6(0) − 5 = −5 y-intercept = −5 f(x) = −𝑥 2 − 6x − 5 y = −𝑥 2 − 6x − 5 −𝑥 2 − 6x − 5 = 0 −𝑥 2 − x − 6x − 5 = 0 (−𝑥 2 − x) + (− 6x − 5) = 0 −x(x + 1) −5(x + 1) = 0 (−x − 5)(x + 1) = 0 −x − 5 = 0 or x + 1 = 0 x = −5 x = −1 x-intercept is −5 and −1 a < 0, graph opens down Turning point (−3,4) y-intercept is −5 x-intercepts are −5 and −1

Graphs of Quadratic Functions

149

3.

f(x) = (𝑥 1)2 + 2 y = (𝑥 1)2 + 2 a > 0, the graph opens down, the graph has maximum point Turning point x=1 y=2

The function is in vertex form, f(x) = a(𝑥 ℎ)2 + k

When the function is in vertex form, turning point is (h,k) Take note of the signs f(x) = (𝑥 1)2 + 2; h = 1, k = 2 f(x) = (𝑥 + 2)2 − 1; h = −2, k = −1

y-intercept y = (𝑥 1)2 + 2 y = (0 1)2 + 2 =1+2 =3 x-intercept y = (𝑥 1)2 + 2 (𝑥 1)2 + 2 = 0 (𝑥 1)2 = −2 x−1=√ 2 √ 2 has no value there is no x-intercept a > 0, graph opens up Turning point (1,2) y-intercept is 3 no x-intercept

150

Graphs of Quadratic Functions

4.

f(x) = 𝑥 2 y = 𝑥2 a > 0, the graph opens down, the graph has a maximum point x=− =−

a=1 b=0 c=0

𝑏 2𝑎 0 2(1)

=0 y = f(−

𝑏 2𝑎

)

y = 02 =0 Turning point (0,0) y-intercept y = 𝑥2 y = 02 =0 x-intercept y = 𝑥2 𝑥2 = 0 x=0

a > 0, graph opens up Turning point (0,0) y-intercept is 0 x-intercept is 0

Graphs of Quadratic Functions

151

REVIEW QUESTIONS 1. A function y = (1 + x)(x − 2) A. Sketch the graph of the function B. Find the coordinates for the turning point 2. The diagram below shows a sketch of the graph y = x(x − 2), passing through O and P

Find A. The coordinates of the points where the curve cuts the x-axis B. The minimum value of the function C. The equation of the axis of symmetry of the graph 3. The diagram below shows a sketch of the graph of y = −𝑥 2 − 2x + 3, passing through A, B and C Find

A. The coordinates of points A, B and C B. The coordinates of the turning point of the graph 4. The sketch below represents a graph that cuts the x-axis at −2 and 4 Find

A. The equation of the graph B. The minimum value of the graph C. The coordinates of point A

152

Graphs of Quadratic Functions

SOLUTIONS 1. A. x-intercept y = (1 + x)(x − 2) (1 + x)(x − 2) = 0 1+x=0 or x = −1

x−2=0 x=2

y-intercept y = (1 + x)(x − 2) y = x − 2 + 𝑥 2 − 2x y = 𝑥2 − x − 2 y = (0)2 − (0) − 2 y = −2 a > 0, graph opens up

𝑏

B. x = −

2𝑎 −1

=− =

2(1)

1 2

y = f(−

𝑏 2𝑎

1 2

) 1

y = (2) − (2) − 2 =

1 4

=−

−

1 2

−2

9 4 1

9

2

4

Turning point ( , − )

Graphs of Quadratic Functions

153

2.

A. y = x(x − 2) x(x − 2) = 0 x=0 or

x−2=0 x=2 coordinates where the curve curs the x-axis are (0,0) and (2,0)

B. y = x(x − 2) y = 𝑥 2 − 2x x=− =− =

𝑏 2𝑎 −2 2(1)

2 2

=1 y = 𝑥 2 − 2x = 12 − 2(1) =1−2 = −1 Minimum value = −1 C. Equation of symmetry; y = −1

3.

A. y = −𝑥 2 − 2x + 3 x-intercept −𝑥 2 − 2x + 3 = 0 −𝑥 2 + x − 3x + 3 = 0 (−𝑥 2 + x) + (− 3x + 3) = 0 −x(x − 1) − 3(x − 1) = 0 (−x − 3)( x − 1) = 0 −x − 3 = 0 or x = −3 A is (−3,0) and C is (1,0) y-intercept y = −𝑥 2 − 2x + 3 = −(0)2 − 2(0) + 3 =3 B is (0,3)

154

Graphs of Quadratic Functions

x−1=0 x=1

3.

B.

y = −𝑥 2 − 2x + 3 x=− =− =−

𝑏 2𝑎 −2 2(−1) 2 2

= −1 y = −𝑥 2 − 2x + 3 = −( 1)2 − 2(−1) + 3 = −1 + 2 + 3 =4 Turning point is (−1,4)

4.

A. x-intercept are at −2 and 4, therefore (x + 2)(x − 4) = 0 𝑥 2 − 4x + 2x − 8 = 0 𝑥 2 − 2x − 8 = 0 The equation of the graph is y = 𝑥 2 − 2x − 8

B. x = − =− =

𝑏 2𝑎 −2 2(1)

2 2

=1 y = 𝑥 2 − 2x − 8 = (1)2 − 2(1) − 8 =1−2−8 = −9 Minimum value = −9 C. y-intercept y = 𝑥 2 − 2x − 8 = (0)2 − 2(0) − 8 = −8 A is (0,−8)

Graphs of Quadratic Functions

155

17

GRAPHS OF CUBIC FUNCTIONS

Introduction Cubic functions are functions in form of ax3 + bx2 + cx + d = 0, where x is the input value and f(x) the output value. In this unit you learn how to sketch cubic functions and many other.

Specific outcomes

v By the end of this unit, you will be able to: ❖ Identify a cubic function ❖ Sketch the graph of a cubic function ❖ Find the gradient at a point ❖ Find the solutions to simultaneous equations using the graph ❖ Estimate the area under a curve

156

Graphs of Cubic Functions

GRAPHS OF CUBIC FUNCTIONS 17.1 CUBIC FUNCTIONS • A cubic function is a function in form of f(x) = a𝑥 3 + b𝑥 3 + cx + d, where a, b, c, and d are real numbers and a ≠ 0. • x- and y - intercepts: • y-intercept is a point where the graph crosses the y-axis. For cubic functions, the y-intercept is d. y = f(0) = d • x-intercepts are points where the graph crosses the x-intercepts. A cubic function can have one, two or three x-intercepts. ❖ EXAMPLES 1. Plot the graph of y = 𝑥 3 for −2 ≤ x ≤ 2 2. Sketch the graph of y = 𝑥 3 + 3𝑥 2 − x − 4 for −4 ≤ x ≤ 4. Use your graph to find A. The value of y when x = 2.6 B. The value of x when y = 10 3. Sketch y = 𝑥 3 − 9x + 5 for −4 ≤ x ≤ 4. Use your graph to find A. The value of y when x = −3.4 B. The value of x when y = 20 ❖ SOLUTIONS 1. y = 𝑥 3 Table of values x

−2

−1

0

1

2

y

−8

−1

0

1

8

Graphs of Cubic Functions

157

2.

y = 𝑥 3 + 3𝑥 2 − x − 4 Table of values x

−4

−3

−2

−1

0

1

2

3

y

−16

−1

2

−1

−4

−1

14

47

A. y = 30 B. x = 1.8

158

Graphs of Cubic Functions

3.

y = 𝑥 3 − 9x + 5 Table of values x y

−4 −3 −23 5

−2 15

−1 13

0 5

1 −3

2 −5

3 5

4 33

A. y = −4 B. x = 3.6

Graphs of Cubic Functions

159

17.2 GRADIENT AT A POINT • The gradient of a graph at a certain point of a curve is found by drawing a tangent at that point and calculate its gradient. • Choose any two points on the tangent line and calculate the gradient using the gradient formula. • Gradient formula: m=

𝒚𝟐 −𝒚𝟏 𝒙𝟐 −𝒙𝟏

16.6 GRAPHICAL SOLUTION TO EQUATIONS • Simultaneous equations can be solved by sketching the graphs of the equations. • The solution to the equations is the points of intersection of the two equations.

16.7 ESTIMATING AREA UNDER A CURVE • The area under a curve can be estimated by dividing the curve into triangles, rectangles, and/or trapezium shapes. • Calculate the area of these shapes using the formula below.

160

1

i.

Triangle area = 2 bh

ii.

Rectangle area = bh

iii.

Trapezium area = 2(a + b)h

Graphs of Cubic Functions

1

REVIEW QUESTIONS 1. The values of x and y are connected by the equation y = 𝑥 3 − 6x + 2 as shown in the table below x

−3

−2

−1

0

1

2

3

y

p

6

7

2

−3

−2

11

A. Calculate the value of p. B. Using the scale of 2cm to represent 1 unit on the x-axis and 1cm to represent 2 units on the y-axis for −3 ≤ x ≤ 3 and −8 ≤ y ≤ 12, draw the graph of y = 𝑥 3 − 6x + 2. C. Use your graph to solve the equation 𝑥 3 − 6x + 2 = 0. D. On the same axes, draw the graph of y = 2x − 2. E. Use your graph to solve 𝑥 3 − 6x + 2 = 2x − 2. 2. The diagram below shows the graph of y = 𝑥 3 + 3𝑥 2 − x − 3

A. Use the graph to find the solutions to the equations i. 𝑥 3 + 3𝑥 2 − x − 3 = 0 ii. 𝑥 3 + 3𝑥 2 − x = 5 B. Calculate the estimate of i. the gradient of the curve at the point (−3,0). ii. the area bounded by the curve, x = 0, y = 0 and y = 20.

Graphs of Cubic Functions

161

3.

The diagram below shows the graph of y = 𝑥 3 + 𝑥 2 − 12x

A. Use the graph to solve the equation i. 𝑥 3 + 𝑥 2 − 12x = 0 ii. 𝑥 3 + 𝑥 2 − 12x = x + 10 B. Calculate the estimate of the i. gradient of the curve at the point where x = −3. ii. area bounded by the curve, x = −3, x = −1 and y = −10. C. Find the equation of the tangent to the curve y = 𝑥 3 + 𝑥 2 − 12x at the point where x = −3.

162

Graphs of Cubic Functions

SOLUTIONS 1. A. y = 𝑥 3 − 6x + 2 p = 𝑥 3 − 6x + 2 p = ( 3)3 − 6(−3) + 2 p = −7 B. table of values x

−3

−2

−1

0

1

2

3

y

−7

6

7

2

−3

−2

11

C. The solutions to 𝑥 3 − 6x + 2 = 0 are x-coordinates where the graph crosses the x-axis; which are x = −2.55, x = 0.35 and x = 2.2 E. The solutions to 𝑥 3 − 6x + 2 = 2x − 2 are x-coordinates where the two graphs intersect; which are x = 0.55 and x = 2.5

Graphs of Cubic Functions

163

2.

A i. The solutions to 𝑥 3 + 3𝑥 2 − x − 3 = 0 are x-coordinates where the graph crosses the x-axis; which are x = −3, x = −1 and x = 1 ii. 𝑥 3 + 3𝑥 2 − x = 5 is same as 𝑥 3 + 3𝑥 2 − x − 3 = 2, therefore the solutions to 𝑥 3 + 3𝑥 2 − x = 5 are where the graph crosses y = 2; which are x = −2.7, x = −1.5 and x = 1.2 B. i. To find the gradient at (−3,0) create a tangent. Pick any two coordinates along the tangent such as (−3,0) and (−3.5,−5) and their gradient: m=

𝑦2 − 𝑦1 𝑥2 − 𝑥1

=

−5 − 0 −3.5 − (−3)

= 10

ii. Area bounded by the curve, x = 0, y = 0 and y = 20 is depicted above by the shaded region. It consists of a triangle and a rectangle, therefore A= =

1

2 1 2

bh + bh × 2 × 3 + 2 × 10

= 3 + 20 = 20

164

Graphs of Cubic Functions

Triangle b = −1 − (−3) = 2 h=3 Rectangle b = −1 − (−3) = 2 h = 0 − (−10) = 10

3.

A.

i. The solutions to 𝑥 3 + 𝑥 2 − 12x = 0 are x-coordinates where the graph crosses the x-axis; which are x = −4, x = 0 and x = 3 ii. To find 𝑥 3 + 𝑥 2 − 12x = x + 10, draw the graph of y = x + 10. The points of intersection are the solutions; which are x = −3.8, x = −0.8 and x = 3.7

B.

i. To find the gradient at x = −3 create a tangent. Pick any two coordinates along the tangent such as (−3,18) and (−4,10) and find their gradient m=

𝑦2 − 𝑦1 𝑥2 − 𝑥1

=

18 − 10 −3 − (−4)

=8

ii. Area bounded by the curve, x = −3, y = −1 and y = −10 is depicted above by the shaded region. It consists of a trapezium, therefore area is: 1

A = (a + b)h 2 1

= × (22 + 28) × 2 2

a = 12 − (−10) = 22 b = 18 − (−10) = 28 h = −1 − (−3) = 2

= 50 C.

The gradient of the tangent is m = 8 y − 𝑦1 = m(𝑥 𝑥1 ) y − 18 = 8(x − (−3)) y − 18 = 8(x + 3) y − 18 = 8x + 24 y = 8x + 24 + 18 y = 8x + 42

Graphs of Cubic Functions

165

18

MOTION GRAPHS

Introduction Motion is the change in position of an object with repeat to time. An object in motion moves from one point to another. Describing the motion of an object with words is a bit hard to grasp. Therefore, motion graphs are used to depict the motion of an object. The following are common terms used when describing the motion of an object: • Distance: The total length taken between two points. • Displacement: The shortest length taken by an object in a specific direction. • Speed: Distance covered by an object per unit time. Speed describes the rate at which an object is moving. • Velocity: The rate of change of displacement per unit time. Velocity describes the speed and direction of a moving object. • Acceleration: The rate of change of velocity per unit time. It describes the rate at which an object is changing speed or direction.

Specific outcomes There are two types of motion graphs: distance (or displacement) − time and speed (or velocity). This unit covers these types of motion graphs. By the end of this unit, you will be able to ❖ Distance − time graph: • Identify a distance − time graph • Describe features of a distance − time graph • Calculate speed from distance − time graph ❖ Speed − time graph: • Identify a speed − time graph • Describe features a speed − time graph • Calculate acceleration and distance from speed − time graph

166

Motion Graphs

MOTION GRAPHS 18.1 DISTANCE (DISPLACEMENT) – TIME GRAPH • Distance (or displacement) – time graph describes the distance covered by an object per unit time. • Features on the distance-time graph: 1. Gradient (slope line) = speed 2. Horizontal (flat) line = stop 3. Curved line = acceleration

❖ EXAMPLES 1. A car travels 60km for 2 hours. It stops at a lay-by for 30 minutes. Then it continues at a steady speed of 40km/hr for 1 hour. Draw a distance-time graph.

Distance covered after stopping d=S×t d = 40km × 1hr d = 40km

2. David walks at a uniform speed of 2m/s for 30 seconds. He stops for 2 seconds, then walks back from where he came from for 20 seconds. Draw a distance-time graph

Distance covered in first 30s d=S×t d = 2 × 30 d = 60m

Motion Graphs

167

3.

Gilbert walks at a uniform speed of 3m/s for 20 seconds. He stops for 20 seconds, then continues to walk for another 30 seconds at 2m/s. Draw a distance-time graph Distance covered in first 20s d=S×t d = 3 × 20 d = 60m Distance covered in last 20s d=S×t d = 2 × 30 d = 60m

18.2 SPEED (VELOCITY) – TIME GRAPH • Speed (or velocity) – time graph describes the speed of an object with the time taken. • Features of the speed − time graph: 1. Gradient (slope line) = constant acceleration • Slant upward = acceleration • Slant downward = deceleration 2. Horizontal (flat) line = constant speed, zero acceleration 3. Curved line = changing acceleration 4. Area of a graph = distance covered • To find the distance covered by an object on a speed-time graph calculate the area of the graph using the following formulae: 1

i.

Triangle area =

ii.

Rectangle area = bh

2

bh

1

iii. Trapezium area = (a + b)h 2

168

Motion Graphs

From the graph, total distance = total area Total area = area A (triangle) + area B (rectangle) + area C (trapezium) + area D (rectangle) + area E (triangle)

❖

EXAMPLES 1. A motorcyclist accelerates from rest to 5m/s2 for 8 seconds, then travels at a constant speed for 10 seconds. He then decelerates at 2m/s2 for 5 seconds and continues at a constant for 5 seconds A. Draw a speed-time graph. B. Find the total distance covered by the motorcyclist.

❖

SOLUTIONS 1.

A.

First phase t = 8s v = u + at v=0+5×8 v = 40m/s

B.

Second phase v = 40m/s t = 8 + 10 t = 18s

Third phase t = 18 + 4 t = 22s v = u + at v = 40 + (− 2)5 v = 40 − 10 v = 30m/s

Fourth phase v = 30m/s t = 23 + 5 t = 28s

Total distance = total area = area A + Area B + area C + area D Area A =

1 2

× 8 × 40 = 160m

Area B = 10 × 40 = 400m Area C =

1 2

(40 + 30) × 4 = 140m

Area D = 5 × 30 = 150m Total distance = 160 + 400 + 140 + 150 Total distance = 850m

Motion Graphs

169

REVIEW QUESTIONS 1.

The diagram below shows the speed-time graph of a 100m runner who accelerates uniformly for 3 seconds until he reaches a speed of 12m/s. he maintains the speed for 7 seconds and then uniformly retards for a further 4 seconds and comes to a stop.

Calculate the A. acceleration during the first 3 seconds B. retardation at the end of this race C. distance he covered in the first 10 seconds 2.

The diagram below shows the speed-time graph of a particle. The particle started off from rest and accelerated uniformly for 10 seconds. It then travelled at a constant speed for 20 seconds and then decelerated to rest

A. Find the speed v the particle reached if its acceleration was 2m/s2 in the first 10 seconds. B. Given that the total distance covered was 750m, find the value of t in the diagram. C. What distance was covered in the first 30 seconds?

170

Motion Graphs

3.

The diagram below shows a speed-time graph of an object. It starts from rest and accelerates uniformly for 2 seconds until it reaches a speed of 10m/s. It moves at this constant speed for 6 seconds then accelerates until it reaches a speed of v m/s after 5 seconds. Finally, it retards for the next 8 seconds until it comes to rest.

Calculate the A. acceleration during the first 2 seconds B. value of v if the retardation in the last 8 seconds is 3m/s2 C. average speed for the whole journey

Motion Graphs

171

SOLUTIONS 𝑣−𝑢

1. A. a =

𝑡 12 − 0

=

3

= 4 m/s2 𝑣−𝑢

B. a = = =

𝑡 0 − 12 14 − 10 −12 4

= −3 m/s2 C. Distance covered is equal to area of trapezium d= =

1 2 1 2

(a + b)h (7 + 10) × 12

= 103m

2. A. v = u + at = 0 + 2 × 10 = 20 m/s2 B. d =

1 2

(a + b)h

750 =

1 2

(20 + t) × 20

400 + 20t = 1500 20t = 1500 − 400 20t = 1100 t = 55s C. d = =

1 2 1 2

(a + b)h (20 + 30) × 12

= 300m

172

Motion Graphs

3. A. a = =

𝑣−𝑢 𝑡 10 − 0 2

= 5 m/s2 B. v = u + at 0 = v + (−3) × 8 v = 24 m/s C. Total distance is equal to the total area Total area = area of trapezium + area of trapezium + area of triangle A= =

1 2 1 2

(a + b)h +

1 2

(a + b)h +

(6 + 8) × 10 +

1 2

1 2

bh

(10 + 24) × 5 +

1 2

× 8 × 24

= 70 + 85 + 96 = 251m Average speed = =

total distance time taken 251 21

= 11.9524 = 12 m/s

Motion Graphs

173

19

MATRICES

Introduction A matrix is a rectangular arrangement of numbers in rows and columns. Numbers in horizontal lines of a matrix are called rows. Numbers in the vertical lines of a matrix are called columns. Matrices have several real life applications. For example, in physics, they are applied in the study of electrical circuits. They are also applied in computer science, robotics and other fields.

Specific outcomes This unit introduces matrices. You will learn about the types of matrices, addition and subtraction of matrices and many more. By the end of this unit, you will be able to: ❖ Know matrix order and matrix elements ❖ Know the types of matrices ❖ Work out addition and subtraction of matrices ❖ Work out multiplication of matrices ❖ Find the determinant of a matrix ❖ Find the inverse of a matrix ❖ Solve simultaneous equations using matrices

174

Matrices

MATRICES 19.1 MATRIX • A matrix is a rectangular arrangement of numbers into rows and columns. 7 2 A=( 5 1

2 ) 9

2 rows

3 columns ❖ MATRIX ORDER • Matrix order or dimension is the number of rows and columns that a matrix has. • Rows and listed first and columns second. • In general, a matrix is defined by the order of r × c, where r is the number of rows and c is the number of columns. • A matrix with 2 rows and columns is written as 2 × 3 and read as “two by three”. • Worked examples: 1. What is the order of the following matrices? 7 2 A. P = ( ) matrix order = 2 × 2 3 1 7 2 9 B. Q = ( ) matrix order = 2 × 3 3 1 5 7 2 C. R = (3 1) matrix order = 3 × 2 9 5 ❖ MATRIX ELEMENTS • Matrix elements are individual items in a matrix. • It can be numbers, symbols or expressions.

19.2 TYPES OF MATRICES 1. ROW MATRIX • A row matrix is a matrix that has one row. • The matrix order of a row matrix is 1 × c, where c is the number of columns. • Examples: A. T = (7 5 9) matrix order = 1 × 3 B. S = (2 4) matrix order = 1 × 2 2. COLUMN MATRIX • A column matrix is a matrix that has one column. • The matrix order of a column matrix is r × 1, where r is the number of rows. • Examples: 7 A. U = (5) matrix order = 3 × 1 9 2 B. V = ( ) matrix order = 2 × 1 4

Matrices

175

3.

SQUARE MATRIX • A square matrix is a matrix that has the same number of rows and columns. • Examples: 2 0 A. L = (9 5 0 3 4 B. M = ( 7

4.

1 1) matrix order = 3 × 3 8

2 ) 5

matrix order = 2 × 2

DIAGONAL MATRIX • A diagonal matrix is a matrix in which all elements are zeros except those in the main diagonal. • Examples: 4 0 A. X = (0 1 0 0

0 0) 1

1 0 B. Y = ( ) 0 2

5.

ZERO (OR NULL) MATRIX • A zero matrix is a matrix in which all elements are zeros • It is usually denoted by O 0 0 A. O = ( ) 0 0 B. O = (0 0

6.

0)

IDENTITY MATRIX • An identity matrix is a matrix in which all elements of the main diagonal are ones and all others are zeros • It is usually denoted by I 1 0 A. I = ( ) 0 1 1 0 B. I = (0 1 0 0 •

176

0 0) 1

Any matrix multiplied by an identity matrix results in itself; I × A = IA = A

Matrices

19.3 ADDITION AND SUBTRACTION OF MATRICES • Only matrices with the same matrix order, that is, the same number of rows and columns can be added or subtracted. • To add or subtract matrices, add or subtract the corresponding elements of the matrices. 𝑞 𝑟 𝑠 𝑎 𝑏 𝑐 • If matrix A = (𝑑 𝑒 𝑓 ) and matrix B = ( 𝑡 𝑢 𝑣 ), then: 𝑥 𝑦 𝑧 𝑔 ℎ 𝑖 𝒒 𝒓 𝒔 𝒂+𝒒 𝒃+𝒓 𝒄+𝒔 𝒂 𝒃 𝒄 A + B = (𝒅 𝒆 𝒇) + ( 𝒕 𝒖 𝒗) = ( 𝒅 + 𝒕 𝒆 + 𝒖 𝒇 + 𝒗) 𝒙 𝒚 𝒛 𝒈 𝒉 𝒊 𝒈+𝒙 𝒉+𝒚 𝒊+𝒛 𝒒 𝒓 𝒔 𝒂−𝒒 𝒃−𝒓 𝒄−𝒔 𝒂 𝒃 𝒄 A − B = (𝒅 𝒆 𝒇) − ( 𝒕 𝒖 𝒗) = ( 𝒅 − 𝒕 𝒆 − 𝒖 𝒇 − 𝒗) 𝒙 𝒚 𝒛 𝒈 𝒉 𝒊 𝒈−𝒙 𝒉−𝒚 𝒊−𝒛

❖ EXAMPLES 1. Add the following matrices 2 1 2 i. A = ( ) and B = ( 7 4 9 2 3 ii. P = ( 0 1) and Q = ( 2 1

1 ) 3 1 3 5

2 2) 2

2. Subtract the following matrices 2 1 2 i. A = ( ) and B = ( 7 4 9 2 3 ii. P = ( 0 1) and Q = ( 2 1

1 ) 3 1 3 5

2 2) 2

❖ SOLUTIONS 1. i. A + B = (

2+2 2 1 2 1 ) + ( ) = ( 7+2 7 4 9 3

2 ii. P + Q = ( 0 2 2. i.

A−B = (

2 7

2 ii. P − Q = ( 0 2

3 1 1) + ( 3 1 5 1 2 ) − ( 4 9

2 + ( 1) 2 2) = ( 0 + ( 3) 2+5 2

1 2 ) = ( 3 7

3 1 1) − ( 3 1 5

1+1 0 2 ) = ( ) 4+3 9 7

2 1 2 4

3+2 3 1+2 ) = ( 3 1 + ( 2) 7

5 3) 3

1 4 0 ) = ( ) 3 5 1

2 ( 1) 2 0 ( 3) 2) = ( 2 5 2

3 2 1 1 2 ) = ( 3 1 ( 2) 3

1 1) 1

Matrices

177

19.4 MULTIPLICATION OF MATRICES ❖ SCALAR MULTIPLICATION • In matrices, real numbers are called scalars. • Scalar multiplication is the product of a scalar and a matrix. • Each element of the matrix is multiplied by a scalar. 𝑎 𝑏 • If A = ( ) is a matrix and k is a scalar, the product of A and k is: 𝑐 𝑑 𝒂 𝒃 𝒌𝒂 𝒌𝒃 kA = k⋅( )=( ) 𝒄 𝒅 𝒌𝒄 𝒌𝒅

❖ EXAMPLES 1 2 Given a matrix A = ( ), find 3 4 A. 2A B. −A ❖ SOLUTIONS 1 2 2×1 2×2 2 4 A. 2A = 2⋅( ) = ( ) = ( ) 3 4 2×3 2×4 6 8 1 2 1×1 1×2 1 B. −A = −1⋅( ) = ( ) = ( 3 4 1×3 1×4 3

❖

MULTIPLICATION OF TWO MATRICES • To multiply two matrices, the number of columns of the left matrix must be equal to the number of rows of the right matrix. • The result matrix will have the same number of rows as the left matrix and the same number of columns as the right matrix. 𝑒 𝑓 𝑎 𝑏 • For matrices A = ( ) and B = ( ), the product of A and B is: 𝑔 ℎ 𝑐 𝑑 𝒂 𝒃 𝒆 𝒇 AB = ( )( ) 𝒄 𝒅 𝒈 𝒉 𝒂𝒆 + 𝒃𝒈 𝒂𝒇 + 𝒃𝒉 =( ) 𝒄𝒆 + 𝒅𝒈 𝒄𝒇 + 𝒅𝒉 • • •

178

Matrices

2 ) 4

The direction of multiplication is left matrix × right matrix ×

Multiplication is not commutative. Therefore, AB ≠ BA. Multiplication of matrices is associative. Therefore, (AB)C = A(BC). If the number of columns in the left matrix is not equal to the number of rows in the right matrix the product of the matrix is undefined.

❖

EXAMPLES 1. Simplify the following matrices. Write undefined if the product is not defined. 1 4 6 3 5 A. ( )( ) 2 7 5 7 1 2 B. (6 4 9) ( 5) 1 2 C. ( 5) (6 4 9) 1 5 0 3 4 3 D. ( )( 6 0) 2 4 5 6 1 5 0 3 4 3 E. ( 6 0) ( ) 2 4 5 6 1

❖

SOLUTIONS 1. A. Undefined. It has 3 columns but 2 rows 2 B. (6 4 9) ( 5) = (6 × ( 2) + 4 × 5 + 9 × 1) = (17) 1 2 C. ( 5) (6 1 D. (

2×6 ) 4 9 = ( 5×6 1×6

2×4 5×4 1×4

5 0 3 )( 6 0) 5 6 1 3 × 5 + ( 4) × 6 + ( 3) × ( 6) = ( 2 × 5 + 4 × 6 + 5 × ( 6) 15 24 + 18 0 + 0 + 3 = ( ) 10 + 24 30 0 + 0 + 5 21 0 = ( ) 16 5 3 2

2×9 12 5 × 9) = ( 30 1×9 6

8 20 4

18 45) 9

4 4

5 0 3 4 3 E. ( 6 0) ( ) 2 4 5 6 1 5 × ( 3) + 0 × ( 2) = ( 6 × ( 3) + 0 × ( 2) 6 × ( 3) + 1 + ( 2) 15 + 0 20 + 0 = ( 18 + 0 24 + 0 18 2 24 + 4 15 20 15 = ( 18 24 18) 16 28 23

3 × 0 + ( 4) × 0 + ( 3) × ( 1) ) 2 × 0 + 4 × 0 + 5 × ( 1)

5 × ( 4) + 0 × 4 6 × ( 4) + 0 × 4 6 × ( 4) + 1 × 4

5 × ( 3) + 0 × 5 6 × ( 3) + 0 × 5) 6 × ( 3) + 1 × 5

15 + 0 18 + 0) 18 + 5

Matrices

179

19.5 DETERMINANT OF A MATRIX • A determinant is a real number that can be calculated from a square matrix. • It is denoted by det(A) or |A|. • A determinant is calculated from a 2 × 2 matrix by subtracting the product of the minor diagonal from the major diagonal. 𝑎 𝑏 • For a matrix A = ( ), 𝑐 𝑑 |A| = ad − bc ❖ EXAMPLES Find the determinant of the following matrices 2 3 A. P = ( ) 4 5 5 2 B. Q = ( ) 4 3 ❖ EXAMPLES 2 3 A. P = ( ) 4 5 |P| = 2 × 5 − 3 × 4 = 10 − 12 =−2 5 2 B. Q = ( ) 4 3 |Q| = 5 × 3 − 2 × 4 = 15 − 8 =7 19.6 INVERSE OF A MATRIX • The inverse of a matrix A is a matrix which when multiplied by A results in an identity matrix • It is denoted by A−1 A × A−𝟏 = I

where A = matrix A A−1 = inverse of matrix A I = identity matrix

❖ HOW TO FIND THE INVERSE OF A 2 × 2 MATRIX 1. Swap the position of elements in the main diagonal. 2. Put negative signs in front of elements in the minor diagonal. 1

3. Multiply the matrix by determinant. 𝑎 𝑏 If matrix A = ( ), then: 𝑐 𝑑 A−𝟏 = = 180

Matrices

1 𝒅 ( |A| 𝒄 1 ( 𝒂𝒅 𝒃𝒄

𝒃 ) 𝒂

𝒅 𝒄

𝒃 ) 𝒂

❖

EXAMPLES Find the inverse of the following matrices 4 7 A. X = ( ) 2 6 1 4 B. Y = ( ) 2 3

❖

SOLUTIONS 4 7 A. X = ( ) 2 6 1 6 X−1 = |X| ( 2

7 ) 4 1 6 = 4×6 7×2 ( 2 1 6 7 = 10 ( ) 2 4 6

(

=

−

10

=

1 B. Y = ( 2

( −

5 1

=

(

−

12

=

10

−

5

7 − 10

7 )( 5 6 −1

2 5

5

10 4)

( 56

7

5

10 2)

− −

1 = ( 0

5

7

−

5 6

−

5

28

10 14 10

)

+ +

4 5 12) 5

0 ) 1

Y × Y−1 = I

4 ) 1 1 3 4 = 1×3 4×2 ( ) 2 1 1 3 4 = − 5( ) 2 1 1

4 = ( 2

7

4 ) 3

Y−1 = |Y| (

X × X−1 = I 3

−

10 2 3

7 ) 4

3 2

3

4

5 2

5 1)

5

−

5

3

=

(

3

4 5 ) 1 − 5 5 8 4 4

5 6

5 6

5 8

5

5

5

4 −5 )( 2 3

1 = ( 2

− + − +

1 = ( 0

− −

5 3) 5

0 ) 1

Matrices

181

19.7 TRANSPOSE OF A MATRIX • Transpose of a matrix is the flipping of a matrix over its diagonal. • It is denoted as AT . • If matrix A has m rows and n columns, a transpose will have n rows and m columns. 𝑎 𝑏 𝑐 • For a matrix A = (𝑑 𝑒 𝑓 ), the transpose of a matrix A is: 𝑔 ℎ 𝑖 𝒂 A𝐓 = (𝒃 𝒄

𝒅 𝒈 𝒆 𝒉) 𝒇 𝒊

❖ EXAMPLES 1. Find the transpose of the following matrices 1 2 A. M = (3 4) 5 6 1 2 3 B. N = ( ) 4 5 6 ❖ SOLUTIONS 1.

1 2 A. M = (3 4) 5 6 1 3 5 MT = ( ) 2 4 6 1 B. N = ( 4

2 3 ) 5 6 1 4 NT = (2 5) 3 6

19.8 SOLVE SIMULTANEOUS EQUATION USING MATRIX i. Write the equation in matrix form. ii. Find the inverse matrix of a 2 × 2 matrix. iii. Multiply both sides of the matrix equation with the inverse matrix. ❖ EXAMPLES Solve the following simultaneous equations using matrices A. x + y = 2 2x − y = 1

182

B.

3x + 2y = 11 2x + y = 8

C.

3x − y = 5 2x + y = 5

Matrices

D.

2x + 3y = 8 x − 2y = −3

❖

SOLUTIONS A. x + y = 2 2x − y = 1 1 𝑥 2 ) (𝑦) = ( ) 1 1

1 ( 2

1 1 1

1 ( 2 2

1 (32 3 1 (23 3

1 ) = 2

1 1 3 1) (2 3

+

2 3 2 3

1 ( 3

1 2

1 ) = (32 1 3

1

1 3 1) 3

1 𝑥 ) ( ) = (32 1 𝑦 3

1 3 2 + 3

1 𝑥 3 1) (𝑦) 3

=

2 (43 3

+

1 3 1) 3

2 ( ) 1

1 3 1) 3

1 0 𝑥 1 ( )( ) = ( ) 0 1 𝑦 1 x=1 y=1 B.

3x + 2y = 11 2x + y = 8 3 2 𝑥 11 ( )( ) = ( ) 2 1 𝑦 8 1 1 2 1 2 1 ( ) = 1( ) = ( 3 4 2 3 2 3 2 1 2 3 2 𝑥 1 2 11 ( )( )( ) = ( )( ) 2 3 2 1 𝑦 2 3 8 3+ 4 2+2 𝑥 11 + 16 ( ) (𝑦) = ( ) 6 6 4 3 22 24 1 0 𝑥 5 ( ) (𝑦) = ( ) 0 1 2 x=5 y = −2

2 ) 3

Matrices

183

C.

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

3 ( 2

1 1 ) = 2 3

1 ( ( 2)

3

(

1 5 2 5

(

3 5 6 5

1 3 5 3) (2 5

+ +

2 5 6 5

1 ( 5

1 2

1 ) = ( 3 1 5 2 5

1 𝑥 )( ) = ( 1 𝑦 1 5 2 5

1

+5

𝑥 3) (𝑦) = ( + 5

1 5 2 5

1 5 3) 5

1 5 5 3) (5) 5 5 5 +5 5 10 15) + 5 5

1 0 𝑥 2 ( )( ) = ( ) 0 1 𝑦 1 x=2 y=1

D.

2x + 3y = 8 x − 2y = −3 3 𝑥 8 )( ) = ( ) 2 𝑦 3

2 ( 1

2 1 4

2 ( 3 1

2 (71 7

3 ) = 2

3 2 7 2) (1 7

2 3 + 7 7 (2 2 −7 7

6 7 3 7

+

y=

184

Matrices

7 2 7

7

7

6 𝑥 7 4 ) (𝑦) 7

=

0 𝑥 ) ( ) = ( 72 ) 1 𝑦 25

3 ) = (71 2 2

7

x=

2 1

3 𝑥 ) ( ) = (71 2 𝑦

25

1 ( 0

1 ( 7

16 ( 87 7

9

+7

6) 7

3 7 2) ( 7

8 ) 3

3 7 2) 7

REVIEW QUESTIONS 2 1. Given that P = ( 1 2. 3. 4. 5. 6. 7. 8. 9.

3 5 9 ) and Q = ( ), find P 2Q. 6 7 1 8 1 7 6 Given that A = ( ) and Q = ( ), find 2A + 2Q. 2 3 2 1 1 3 5 If M = ( ) and N = ( ), calculate MN. 3 1 4 9 2 2 Given that C = ( ) and D = ( ), express CD as a single matrix. 2 3 3 4 3 Express (2 3 1) ( 5 2) as a single matrix. 3 1 2 Find the value of a given that (𝑎 4) ( ) = 6 1 2 𝑥 4 14 Given that ( )( ) = ( ), find the value of x. 5 1 3 17 𝑎 2 0 3 15 Given that ( ) ( 4) = ( ), find the value of a and b. 0 3 𝑏 2 1 2 3 1 4 0 1 Given that M = ( ) and N = (4 1 6), find 0 3 5 5 2 3 T A. M B. NT 4 1 ), find the determinant of T. 2 2 3 1 3 4 Given that B = ( ) and C = ( ), find the value of x if B and C have equal determinants. 2 0 4 𝑥 4 2 12 4 If matrix P = ( ) and Q = ( ), find the value of k if P and Q have equal determinants. 0 3 9 𝑘 3 2 Given that matrix B = ( ), find the 𝑥 4 A. The value of x, given that the determinant of B is 2 B. Inverse of Q 8 12 The determinant of matrix S = ( ) is 8, find 𝑥 4 𝑥 A. The value of x B. The inverse of Q

10. Given that matrix T = ( 11. 12. 13.

14.

Matrices

185

SOLUTIONS 2 1. P − 2Q = ( 1 2 =( 1 2 =( 1

3 5 9 ) − 2( ) 6 7 1 3 10 18 )−( ) 6 14 2 10 3 18 ) 14 6 2 8 21 =( ) 13 4 8 2 16 =( 4 16 =( 14 2 =( 18

2. 2A + 2Q = 2(

1 7 6 ) + 2( ) 3 2 1 2 14 12 )+( ) 6 4 2 14 2 + 12 ) 4 6+2 14 ) 8

3 5 1 3. MN = ( )( ) 1 4 3 3 + ( 15) =( ) 1 + 12 12 =( ) 13 9 2 2 )( ) 2 3 3 18 + ( 6) =( ) 4 + ( 9) 12 =( ) 13

4. CD = (

5. (2

6. (𝑎

7. (

186

4 3 1) (5 3

3 2) = (8 + ( 15) + 3 1 = ( 14 3)

2 4) ( ) = 6 1 2a − 4 = 6 2a = 6 + 4 a = 10 a=5 2 𝑥 4 14 )( ) = ( ) 5 1 3 17 8 + 3x = 14 3x = 14 − 8 3x = 6 x=2

Matrices

6 + 6 + 1)

8.

𝑎 3 15 ) ( 4) = ( ) 𝑏 2 1 2a + 0 + 3 = 15 0 + (−12) + b = 2

2 0 ( 0 3

2a + 3 = 15 2a = 15 − 3 2a = 12 a=6 −12 + b = 2 b = 2 + 12 b = 14 9.

4 0 A. M T = ( 0 3) 1 5 2 4 5 B. N T = (3 1 2) 1 6 3

10.

|B| = (4 × 2) − (2 × 1) =8−2 =6

11.

|B| = (3 × 0) − ((−2) × (−1)) =0−2 = −2 |C| = (3 × x) − (4 × (−4)) = 3x − (−16) = 3x + 16 |B| = |C| −2 = 3x + 16 3x = −2 − 16 3x = −18 x = −6

Matrices

187

12.

|P| = (4 × 3) − (2 × 0) = 12 − 0 = 12 |Q| = (12 × k) − (4 × (−19)) = 12k − (−36) = 12k + 36 |B| = |C| 12 = 12k + 36 12k = 12 − 36 12k = −24 k = −2

13.

A.

|B| = (3 × 4) − (x × (−2)) = 12 − (−2x) = 12 + 2x |B| = 2 12 + 2x = 2 2x = 2 − 12 2x = −10 x = −5

B.

3 B=( 𝑥

2 ) 4 3 2 =( ) 5 4

1 4 2 B-1 = |B| ( ) 5 3 1 4 2 = 2( ) 5 3 2 1 = (5 3) 2

14.

A.

2

|S| = (8 × x) − ((x − 4) × 12) = 8x − (12x − 48) = 8x − 12x + 48 = −4x + 48 |S| = 8 −4x + 48 = 8 −4x = 8 − 48 −4x = −40 x = 10

188

Matrices

14.

B.

8 12 S=( ) 𝑥 4 𝑥 8 12 =( ) 10 4 10 8 12 =( ) 6 10 1 8 12 S-1 = |S| ( ) 6 10 1 10 12 = ( ) 8 6 8

10 8 6

=(

−8 5

=(

4 3

−4

−

12 8 8) 8

3

−2

) 1

Matrices

189

20

MENSURATION

Introduction Mensuration is the study of geometric figures and their parameters. We apply mensuration to solve numerous real life problems. For example, we use it to demarcate plots, find the size of an object, measure the volume of liquids, find the amount of carpet needed for our room, find the distance from one place to another and so on.

Specific outcomes In this unit, you will learn how to find the perimeter, area and volume of common figures. By the end of this unit, you will be able to calculate the perimeter, area and volume of the following figures ❖ 2D objects • Square • Rectangle • Triangle • Parallelogram • Rhombus • Trapezium • Circle • Semicircle • Ring • Sector • Segment of a circle ❖ 3D objects • Cuboid • Cube • Cylinder • Cone • Sphere • Hemisphere • Pyramid • Triangular prism

190

Mensuration

MENSURATIONS 20.1 MENSURATION • Mensuration is the study of geometric figures and their parameters. • Some important parameters include perimeter, area and volume. 1. Perimeter is the distance around a figure. 2. Area is the amount of space surrounding a figure. Surface area is the amount of space surrounding 3-dimensional solid objects. The area is measured in squared units such as cm2, m2. 3. Volume is defined as the amount inside a solid figure. It is measured in cubic units such as m3, cm3.

20.2 MENSURATION FIGURES (2D OBJECTS) 1. SQUARE Perimeter = a + a + a + a = 4a Area = a × a = 𝑎2

2. RECTANGLE Perimeter = a + a + b + b = 2a + 2b = 2(a + b) Area = a × b = ab

3. TRIANGLE Perimeter = a + b + c 1

Area = 2 × b × h 1

= 2 bh

4. PARALLELOGRAM Perimeter = a + a + b + b = 2a + 2b = 2(a + b) Area = a × h = ah

Mensuration

191

5.

RHOMBUS Perimeter = a + a + a + a = 4a Area = a × h = ah

……. 1

1

Area = 2 × 𝑑1 × 𝑑2 1

= 2 𝑑1 𝑑2

6.

…… 2

TRAPEZIUM Perimeter = a + b + c + d 1 2

Area = (a + b)h

7.

CIRCLE Circumference = 2πr Area = π𝑟 2

8.

SEMICIRCLE Perimeter = πr + 2r = r(π + 2) 1

Area = 2 π𝑟 2

9.

RING (SHADED AREA) Area = π(𝑅 2

192

Mensuration

𝑟2)

where r = radius 22 π = pi = 7 = 3.142

10.

SECTOR Perimeter = s + r + r = s + 2r θ

Arc length (s) = 360 × 2πr

(θ in degrees)

where s = arc length θ = theta π = pi

= rθ (θ in radians) θ

Area = 360 × π𝑟 2 (θ in degrees) 1

= 2 𝑟 2 θ (θ in radians)

11.

SEGMENT OF A CIRCLE θ

1

𝑟 2 sin θ 360 2 𝑟2 π = (180 θ sin θ) (θ in degrees) 2

Area =

Area =

π𝑟 2

1 2 𝑟 (θ 2

sin θ)

(θ in radians)

20.3 SOLID FIGURES (3D OBJECTS) 1. CUBOID Surface area = 2(lb + lh + bh) Volume = l × b × h

where l = length b = breadth h = height

2. CUBE Surface area = 6𝑎2 Volume = 𝑎3

Mensuration

193

3.

CYLINDER Lateral surface area = 2πrh Total surface area = 2πrh + 2π𝑟 2 = 2πr(h + r) Volume = π𝑟 2 h

4.

CONE Lateral surface area = πrl l

Total surface area = πrl + π𝑟 2 = πr(l + r)

where r = radius h = height l = slant height

1

Volume = 3 π𝑟 2 h

5.

SPHERE Surface area = 4π𝑟 2 4

Volume = 3 π𝑟 3

6.

HEMISPHERE Surface area = 2π𝑟 2 + π𝑟 2 = 3π𝑟 2 2

Volume = 3 π𝑟 3

7.

PYRAMID 1

Lateral surface area = 2 ps 1 2

Total surface area = ps + lb 1

Volume = 3 lbh

194

Mensuration

where p = perimeter of the base = 2l + 2b s = slant height

8.

TRIANGULAR PRISM Surface area = bh + 2ls + lb 1 2

Volume = bhl

Mensuration

195

REVIEW QUESTIONS 1. The diagram below shows a pentagon ABCDE. AB = CB = CD = DE, AE = 12m, DE = 5m and ∠ABC = ∠BCD = ∠DEA = 90o .

A. Find AD B. Calculate the perimeter of the pentagon C. Calculate the area of the pentagon 2. The diagram below shows a cylinder of height 10cm and diameter 7cm. (Take π to be

22 ). 7

A. Calculate the lateral surface area of the cylinder. B. Calculate the volume of the cylinder. 3. The diagram below is a sector with a radius of 21cm and ∠AOB = 120o . (Take π to be

A. Calculate the arc length s B. Calculate the perimeter of the sector C. Calculate the area of the sector

196

Mensuration

22 ). 7

4. The diagram below is a sector with a radius of 21cm and ∠AOB = 120o . (Take π to be

22 ). 7

A. Calculate the arc length s B. Calculate the perimeter of the sector C. Calculate the area of the sector 5. The diagram below shows a sector, ∠ AOB = θ and radius is 4.2cm. Given that the area of the sector is 22 9.24cm2, find the value of θ. (Take π to be 7 ).

6. The diagram below shows a cone with a base radius of 5cm, height 12cm and slant height 13cm. 22 (Take π to be 7 ).

A. Calculate the lateral surface area of the cone B. Calculate the volume of the cone 7. The figure below shows a frustum of a cone that opens on both sides. The smaller diameter is 2cm, the Large diameter is 8cm, the longitudinal length is 20cm and slant length is 24cm. (π = 3.142)

A. Calculate the curved surface area B. Calculate the volume

Mensuration

197

8. The figure below shows a glass cup. The base diameter and top diameter are 4cm and 8cm respectively, while the height is 14cm. (π = 3.142).

A. Calculate the surface area of glass B. Calculate the volume of a glass 9. The diagram below shows a cross section of a circular metal rod of radius 15cm. The metal has a central hole which is a square of 5cm. (π = 3.142).

A. Calculate the area of the shaded part B. Given that the length of the metal rod is 20cm, calculate the volume of the metal rod 10. The figure below shows a circle enclosed in a sector. The radius of the sector is 60cm and the radius of the circle is 15cm. (π = 3.142).

A. Calculate the perimeter of the sector AOB B. Calculate the area of the shaded part

198

Mensuration

11. A cylindrical geyser of radius 25cm and length 80 cm is placed with its curved surface on an horizontal ground. It is filled partially with water and the segment. The segment ABX shows the cross section of water in the geyser, O is the centre of the circular end of the geyser, X is vertically below O and ∠AOB = 120o . (π = 3.142).

A. Calculate the total surface area of the geyser B. Calculate the volume of water in the geyser

Mensuration

199

SOLUTIONS 1. A.

AD = √AE2 + DE2 = √122 + 52 = √169 = 13m

B. P = 13 + 13 + 13 + 5 + 12 = 56m C. Area of pentagon = area of square + area of triangle 1 = 𝑙 2 + 2 bh 1

= 352 + 2 × 5 × 12

= 169 + 30 = 199m2 2. A. A = 2πrh 22 = 2 × × 7 × 10 7

= 440cm2 B. V = πr 2h 22 = 7 × 72 × 10 = 1540cm2

θ × 2πr 360 120 22 = ×2× 360 7

3. A. s =

× 21

= 44cm B. P = s + 2r = 44 + 2(21) = 86cm θ

C. A = 360 × π𝑟 2 120

= 360 ×

22 × 7 2

= 462cm

200

Mensuration

212

θ

4. A. s = 360 × 2πr 45

= 360 × 2 ×

22 7

× 14

= 11cm B. P = s + 2r = 11 + 2(14) = 86cm θ

C. A = 360 × π𝑟 2 45

= 360 ×

22 7 2

× 142

= 77cm θ

5. A = 360 × π𝑟 2 360A = π𝑟 2 θ θ= θ=

360A π𝑟 2 360 × 9.24 22 × 4.22 7

θ = 60o

6. A. A = πrl = 3.142 × 5 × 13 = 2.4.23cm2 1 3 1 = 3

B. V = π𝑟 2 h × 3.142 × 52 × 12

= 314.2cm3 7. A. Lateral surface area = πl(R + r) = 3.142 × 24 × (4 + 1) = 377.04cm2 1

B. V = 3 πh(R + r + Rr) 1

= 3 × 3.142 × 20 × (4 + 1 + 4 × 1) = 188.52cm3

Mensuration

201

8. A. Surface area = πl(R + r) + π𝑟 2 = 3.142 × 16 × (4 + 2) + 3.142 × 22 = 314.2cm2 1

B. V = 3 πh(R + r + Rr) 1

= 3 × 3.142 × 14 × (4 + 2 + 4 × 8) = 205.2773 = 205.28cm3 9. A. Area of metal rod = π𝑟 2 = 3.142 × 52 = 78.55cm2 Area of square hole = 𝑙 2 = 22 =4 Area of shaded part = area of metal rod − area of square hole = 78.55 − 4 = 74.55cm2 B. Volume of metal rod = area of shaded part × height = 74.55 × 20 = 1491cm3

θ

10. A. Arc length AB = 360 × 2πr 30

= 360 × 2 × 3.142 × 60 = 31.42cm

Perimeter = arc length AB + OA + OB = 31.42 + 60 + 60 = 151.42cm θ

B. Area of sector = 360 × π𝑟 2 120

= 360 × 3.142 × 602 = 942.6cm2 Area of circle = π𝑟 2 = 3.142 × 152 = 706.95cm2 Area of shaded part = Area of sector − Area of circle = 942.6 − 706.95 = 235.65cm3

202

Mensuration

11. A. A = 2πr(h + r) = 2 × 3.142 × 25 × (60 + 25) = 13353.5cm2 𝑟2 π ( θ sin θ) 2 180 252 3.142 = ( × 120 2 180

B. Area of segment =

sin 120)

= 384.9504 = 384.95cm2 Volume of water = area of segment × height = 384.95 × 60 = 23037cm3

Mensuration

203

21

ANGLES

Introduction Angles are used in our everyday lives. Engineers and architectures apply angles to design buildings, rods and devices. A lot of things at home apply angles such as doors, windows, laptops and clocks.

Specific outcomes This unit covers angles. By the end of this unit, you will be able to know: ❖ Types of angles • Acute angles • Right angles • Obtuse angles • Straight angles • Reflex angles • Complete run angles ❖ Special angles • Supplementary angles • Complementary angles • Vertically opposite angles • Adjacent angles ❖ Angle properties of parallel lines • Corresponding angles • Alternate interior angles • Alternate exterior angles • Consecutive interior angles

204

Angles

ANGLES 21.1 ANGLES • An angle is a figure formed when two straight lines meet a common point. • The point at which two lines meet is called vertex.

21.2 TYPES OF ANGLES 1. ACUTE ANGLE • An acute angle is an angle that is less than 90o .

2. RIGHT ANGLE • A right angle is an angle that is exactly 90o .

3. OBTUSE ANGLE • An obtuse angle is an angle that is greater than 90o but less than 180o.

4. STRAIGHT ANGLE • A straight angle is an angle that is exactly.

5. REFLEX ANGLE • A reflex angle is an angle that is greater than 180o but less than 360o.

6. COMPLETE TURN ANGLE • A complete turn angle is an angle that is exactly 360o.

Angles

205

21.3 SPECIAL ANGLE PAIRS 1. SUPPLEMENTARY ANGLES • Supplementary angles are two angles whose sum is 180o. a + b = 180o

2. COMPLEMENTARY ANGLES • Supplementary angles are two angles whose sum is 90o .

a + b = 90o

3. VERTICALLY OPPOSITE ANGLES • Vertically opposite angles formed by the intersection of two straight lines are equal. • The adjacent angles on a straight line are supplementary. a=c b=d a + b = 180o c + d = 180o

4. ADJACENT ANGLES • Adjacent angles are two angles sharing the common side and vertex.

a and b are adjacent angles

21.4 ANGLE PROPERTIES OF PARALLEL LINES • A Transversal is a line that intersects two or more lines.

206

Angles

1.

CORRESPONDING ANGLES (F ANGLES) • Corresponding angles are angles that lie on the same side of the transversal in corresponding positions. • Corresponding angles are equal.

a=b

2.

ALTERNATE INTERIOR ANGLES (Z ANGLES) • Alternate interior angles are interior angles that lie on the opposite side of the transversal. • These angles are equal.

c=d

3.

ALTERNATE EXTERIOR ANGLES • These are exterior angles that lie on opposite sides of the transversal. • Alternate exterior angles are equal.

e=f

4.

CONSECUTIVE INTERIOR ANGLES (C ANGLES) • These are interior angles that lie on the same side of the transversal. • Consecutive interior angles add up to 180o.

g + h = 180o

Angles

207

❖

EXAMPLES 1. Find the missing angles in the diagram

2. Find the missing angles

❖

SOLUTIONS 1. a + 50o = 180o a = 180o − 50o = 130o b = 50o

Vertically opposite angles

c=a = 130o

Vertically opposite angles

d=c = 130o

Corresponding angles

e=b = 50o

Alternate interior angles

f =a = 130o

Corresponding angles

g=e = 50o

208

Supplementary angles

Angles

Vertically opposite angles

2.

a = 55o

Alternate interior angles

b + 123o = 180o b = 180o − 123o = 57o c=b c = 57o

Corresponding angles

Alternate interior angles

d + b + 55o = 180o d = 180o − b − 55o = 180o − 57o − 55o = 68o

Supplementary angles

REVIEW QUESTIONS 1. Find the missing angles

2. Calculate angle x

Angles

209

3. Calculate the labelled angles

SOLUTIONS 1. a = 68o b = 52o a + b + c = 180o c = 180o − a − b = 180o − 68o − 52o = 60o d = 60o d + e = 180o e = 180o − d = 180o − 60o = 120o 2.

x + 147o + 164o = 360o x = 360o − 147o − 164o x = 49o

3.

90o + p = 128o p = 128o − 90o = 38o q + 128o = 180o q = 180o − 128o = 52o

210

Angles

22

CIRCLE THEOREMS

Introduction A circle is a round shaped figure formed when all the points are equidistant from the centre point. When we draw some lines inside a circle, patterns and angles are formed. Circle theorems describe the properties of these patterns and angles.

Specific outcomes This unit covers circle theorems. By the end of this unit, you will be able to: ❖ Know parts of a circle ❖ Know angle properties of a circle ❖ Use angle properties of a circle to solve problems

Circle Theorems

211

CIRCLE THEOREMS 22.1 PARTS OF A CIRCLE

1. 2. 3. 4. 5. 6. 7. 8.

Circumference. The distance around a circle. Centre. The point that is equivalent to all points on the circle. It is usually denoted by O. Radius. The distance from the centre to any point of the circle. It is denoted by r. Diameter. The distance from the side of the circle to the other side passing through the centre. It is denoted by Ø. Chord. A line joining any two points of the circle. Segment. A region of a circle that is cut off by the chord. Secant. A line intercepting a circle at two points. Tangent. A line intercepting a circle at one point.

22.2 ANGLE PROPERTIES OF A CIRCLE 1. The angle at the centre is twice the angle at the circumference.

2. The angle in a semicircle is 90o .

∠ABC = 90o

3. The angles in the same segment are equal.

∠a = ∠b

212

Circle Theorems

4.

Two radii make an isosceles triangle.

∠a = ∠b OA = OB

5.

Opposite angles in a cyclic quadrilateral add up to 180o. ∠a + ∠c = 180o ∠b + ∠d = 180o

6.

The angle between the chord and tangent is equal to the angle in the alternate segment.

∠a = ∠c ∠b = ∠d

7.

The exterior angle of a cyclic quadrilateral is equal to the opposite interior angle.

∠a = ∠c

8.

The angle between a tangent and a radius is 90o at the point of the tangent.

∠ABO = 90o

Circle Theorems

213

9.

If CA and CB are tangents to the circle at point A and B respectively, then: ∠a = ∠b ∠c = ∠d ∠e = ∠ f = 90o ∠AOB = ∠2a = ∠2b = ∠a + ∠b ∠ACB = ∠2c = ∠2d = ∠c + ∠d ∠a + ∠b + ∠c + ∠d + ∠e + ∠ f = 360o

214

Circle Theorems

REVIEW QUESTIONS 1. Find the labelled angles

2. A circle centre passes through the points A, B, C, and D as shown in the diagram below. Angle AOC = 110o and AO is parallel to BC.

Calculate A. ∠OCA B. ∠ACB C. ∠ABC 3. In the diagram below, points A, B, C and D are on a circle. BD is the diameter of the circle. ∠ACB = 42o , ∠CAD = 33o and the lines AC and BD intersect at X.

Calculate A. ∠CBD B. ∠ACD C. ∠AXB

Circle Theorems

215

4. In the diagram, TR is the diameter of the circle with centre O, and QR and PS are parallel. Angle TOS = 80o and angle TRP = 10o .

Find A. ∠RPS B. ∠PST C. ∠PSO 5. The diagram below shows a circle with a tangent RWS. The points V, W, X and Y are on the circle such that ∠XYW = 44o , ∠VWY = 54o and ∠SWV = 39o .

Find A. ∠RWX B. ∠XVW C. ∠YXW 6. In the diagram below, O is the centre of the circle ABCD. AC is a diameter, AD and BC are produced to meet at E, angle CED = 27o and angle ACB = 43o .

Calculate A. ∠ADB B. ∠CAD C. ∠DCE

216

Circle Theorems

7. In the diagram below, PR is a diameter with centre O. Q and S are points on the circumference. The tangent to the circle at the point P meets QS produced at T, ∠PQS = 26o and ∠QPR = 42o .

Calculate A. ∠PRS B. ∠PRQ C. ∠PTQ

Circle Theorems

217

SOLUTIONS 1. ∠OAC + ∠OCA + ∠AOC = 180o ∠OCA = ∠ACO r + r + ∠ACO = 180o 2r = 180o − ∠AOC 2r = 180o − 120o 2r = 60o r = 30o 1

∠ABC = 2 ∠AOC 1

s = 2 × 120o s = 60o ∠BAC = ∠BAO + ∠OAC = 23o + 30o = 53o ∠BCD = ∠BAC + ∠ABC t = 53o + 60o t = 113o

2. A. ∠OAC + ∠OCA + ∠AOC = 180o ∠OAC = ∠OCA 2∠OAC + ∠AOC = 180o 2∠OAC = 180o − ∠AOC 2∠OAC = 180o − 110o 2∠OAC = 70o ∠OAC = 35o B.

∠ACB = ∠OAC ∠OAC = ∠OCA ∠ACB = ∠OCA = 35o

C.

∠ADC = ∠AOC ∠ADC

1 2 1 = 2

× 110o

∠ADC = 55o ∠ABC + ∠ADC = 180o ∠ABC = 180o − ∠ADC = 180o − 55o = 125o

218

Circle Theorems

3. A. ∠CBD = ∠CAD = 33o B. ∠ACD + ∠ACB = 90o ∠ACD = 90o − ∠ACB = 90o − 42o = 48o C. ∠AXB = ∠ACB + ∠CBD = 42o + 33o = 75o

4. A. ∠ROS = 180o − ∠TOS = 180o − 80o = 100o 1

∠RPS = 2 ∠ROS 1

= 2 × 100o = 50o B.

∠PST = ∠PRS = 10o

C.

∠OST + ∠OTS + ∠TOS = 180o ∠OST = ∠OTS 2∠OST + ∠TOS = 180o 2∠OST = 180o − ∠TOS 2∠OST = 180o − 80o 2∠OST = 100o ∠OST = 50o ∠PSO + ∠PST = 50o ∠PSO = 50o − ∠PST = 50o − 10o = 40o

5. A.

∠RWX = ∠XYW = 44o

B.

∠XVW = ∠XYW = 44o

C.

∠YXW = ∠YXV + ∠VXW ∠VXW = ∠SWV ∠YXW = 54o + 39o = 93o

Circle Theorems

219

6. A.

∠AOB = ∠OCB + ∠OBC ∠OCB = ∠OBC ∠AOB = 2∠OCB = 2 × 43o = 86o

B.

∠CAD + ∠ACB + ∠ABC = 180o ∠CAD + 43o + 90o = 180o ∠CAD = 180o − 43o − 90o = 47o

C.

∠CDE + ∠CDA = 180o ∠CDE + 90o = 180o ∠CDE = 180o − 90o = 90o ∠DCE + ∠CDE + ∠CED = 180o ∠DCE + 90o + 27o = 180o ∠DCE = 180o − 90o − 27o = 63o

7.

A.

∠PRS = ∠PQS = 26o

B. ∠PRQ + ∠RPQ + ∠PQR = 180o ∠PRQ + 42o + 90o = 180o ∠PRQ = 180o − 42o − 90o = 48o C. ∠TPQ = ∠RPQ + ∠TPR = 42o + 90o = 132o ∠PTQ + ∠TQP + ∠TPQ = 180o ∠PTQ + 132o + 26o = 180o ∠PTQ = 180o − 132o − 26o = 22o

220

Circle Theorems

23

POLYGONS

Introduction A polygon is a closed figure made up of straight line segments joined together. To form a polygon at least three straight line segments are needed. Therefore, polygons with the least number of line segments are triangles that have three straight line segments. Most structures around us are polygons. Book pages, cell phones and tiles are polygons with four straight line segments or sides. The Pentagon building and black sections on soccer balls are polygons with five sides. Metal nuts, stop traffic signs and white sections on soccer balls are polygons with six sides.

Specific outcomes This unit introduces polygons. It covers how to name polygons, types of polygons and many more. By the end of this unit, you will be able to: ❖ Identify a polygon ❖ Name a polygon ❖ Know the following types of polygons: • Regular polygons • Irregular polygons • Convex polygon • Concave polygon ❖ Calculate the sum of interior angles of a polygon ❖ Calculate each interior angle of a polygon

Polygons

221

POLYGONS 23.1 POLYGONS • A polygon is a closed figure made up of straight line segments joined together. • Polygons contain sides and angles. • They are named according to the number of sides. Number of sides Name of the polygon 3

Triangle

4

Quadrilateral

5

Pentagon

6

Hexagon

7

Heptagon

8

Octagon

9

Nonagon

10

Decagon

23.2 TYPES OF POLYGONS 1. REGULAR POLYGON • A regular polygon is a polygon with equal sides and equal angles. • Examples of regular polygons: equilateral triangles, squares.

Equilateral triangle ∠a = ∠b = ∠c

Square ∠r = ∠s = ∠t = ∠u

2. IRREGULAR POLYGON • An irregular polygon is a polygon with unequal sides and angles. • Examples of irregular polygons: scalene triangle, pentagon with unequal sides.

Scalene triangle ∠a ≠ ∠b ≠ ∠c

222

Polygons

Pentagon with unequal sides and angles ∠r ≠ ∠s ≠ ∠t ≠ ∠u ≠ ∠v

3.

CONVEX POLYGON • A convex polygon is a polygon with no interior angle greater than 180o. ∠a, ∠b and ∠c less than 180o

4.

CONCAVE POLYGON • A concave polygon is a polygon with at least one interior angle greater than 180o.

∠a is greater than 180o

23.3 ANGLES OF THE POLYGON ❖ EXTERIOR ANGLES • The exterior angle of the polygon is formed by the continuation of the side of the polygon. • The exterior angle and the adjacent interior angle add up to 180o.

∠a + ∠b = 180o

❖ SUM OF INTERIOR ANGLES OF A REGULAR POLYGON • The following formula calculates the sum of interior angles of an n-sided regular polygon: Sum of interior angles = (𝑛 2) × 180o

where n = number of sides of a polygon

• The following formula calculates each angle of an n-sided regular polygon: Each interior angle =

(𝑛 − 2)×180o 𝑛

where n = number of sides of a polygon

Polygons

223

❖

EXAMPLES 1. Find the sum of interior angles of a regular triangle. What is the value of each interior angle? 2. Find the sum of interior angles of a pentagon. What is the value of each angle? 3. Find the sum of interior angles of a 12-sided polygon. What is the value of each angle?

❖

SOLUTIONS 1. Sum of interior angles = (𝑛 2) × 180o = (3 2) × 180o = 1 × 180o = 180o Each interior angle = = =

Number (n) of triangle = 3

(𝑛 − 2) × 180o 𝑛 (3 − 2) × 180o 180o

3

3

= 60o 2. Sum of interior angles = (𝑛 2) × 180o = (5 2) × 180o = 3 × 180o = 540o Each interior angle = = =

(𝑛 − 2) × 180o 𝑛 (5 − 2) × 180o 5 3 × 180o 5

= 108o

3. Sum of interior angles = (𝑛 2) × 180o = (12 2) × 180o = 10 × 180o = 1800o Each interior angle = = =

(𝑛 − 2) × 180o 𝑛 (12 − 2) × 180o 12 10 × 180o 12

= 150o

224

Polygons

Number (n) of pentagon = 5

24

TRIANGLES

Introduction A triangle is a polygon with three sides. Triangle structures are extensively used in architectures with high tension such as bridges, towers and tall buildings. This is because, unlike other polygons, triangles do not deform easily under stress.

Specific outcomes This unit introduces triangles. It covers the types of triangles and many more. By the end of this unit, you will be able to: ❖ Identify a triangle ❖ Know the types of triangles ❖ Calculate the missing angle based on the type of a triangle ❖ Know how to test for congruent triangles

Triangles

225

TRIANGLES 24.1 TRIANGLES • A triangle is a three-sided polygon. • The sum of interior angles of a triangle is 180o.

24.2 TYPES OF TRIANGLES ❖ BY SIDES 1. Equilateral triangle. It has three equal sides and angles. 2. Isosceles triangle. It has two equal sides. Therefore, the angles opposite the equal sides are equal. 3. Scalene triangle. This triangle has no equal sides.

Equilateral triangle All three equal sides

Isosceles triangle Two equal sides

Scalene triangle No equal sides

❖ BY ANGLES 1. Acute triangle. It has all three angles that are less than 90o . 2. Right angled triangle. It has one right angle, a 90o angle. 3. Obtuse triangle. It has one angle greater than 90o .

Acute triangle All angles less than 90o

Right angled triangle One angle at 90o

❖ EXAMPLES Find the missing angle in the following 1.

3.

226

Triangles

2.

Obtuse triangle One angle greater than 90o

❖

SOLUTIONS 1. 40o + a + b = 180o a + b = 2a = 2b 40o + 2a = 180o 2a = 180o − 40o 2a = 140o a = 70o a=b b = 70o

2. Let adjacent angle to x be a 70o + 30o + a = 180o 100o + a = 180o a = 180o − 100o a = 80o a + x = 180o 80o + x = 180o x = 180o − 80o x = 100o 3. x + 55o = 180o x = 180o − 55o x = 125o 65o + 55o + y = 180o 120o + y = 180o y = 180o − 120o y = 60o

24.3 CONGRUENT TRIANGLES • Congruent triangles are triangles with the same three sides and angles. • Congruent triangles are still congruent even if flipped, turned or rotated. ❖ TESTS FOR CONGRUENT TRIANGLES 1. SSS (SIDE - SIDE - SIDE) • If three sides of a triangle are equal to three sides of another triangle then two triangles are congruent. 2. SAS (SIDE - ANGLE - SIDE) • If two sides and one included angle of one triangle are equal to two sides and one included angle of another triangle then the triangles are congruent. 3. AAS (ANGLE - ANGLE - SIDE) • If two angles and one non included side of one triangle are equal to two angles and one non included side of another triangle then the triangles are congruent.

Triangles

227

4.

ASA (ANGLE - SIDE - ANGLE) • If two angles and one included side of one triangle are equal to two sides and one included side of another triangle then the triangles are congruent.

5.

RHS (RIGHT ANGLE - HYPOTENUSE - SIDE) • If the hypotenuse and one side of a right-angled triangle are equal to the hypotenuse and one side of another right-angled triangle then two triangles are congruent.

SSS

SAS

AAS

RHS

24.4 SIMILAR TRIANGLES • Two triangles are similar if 1. All three corresponding sides are proportional. 2. All three corresponding angles are equal.

228

Triangles

SAS

❖

EXAMPLE Which pair of the following triangles are congruent by ASA?

❖

SOLUTION The answer is B. Triangles in choice A are not congruent. Triangles in choice C are congruent by SSS. Triangles in choice D are congruent by SAS.

Triangles

229

25

SYMMETRY

Introduction Objects that can be divided into identical halves are said to be symmetrical. A lot of objects around us are symmetrical. Symmetry is pleasing to the eye that why designing and art extensively apply it. Take a look at the carpet, tiles or kitchen utensils. You will likely find that they have many symmetrical drawings. Apart from being beautiful, symmetrical objects are very stable. Therefore, architecture heavily relies on symmetry.

Specific outcomes This unit covers symmetry. By the end of this unit, you will be able to: ❖ Identify symmetry objects ❖ Know the types of symmetry: • Line symmetry • Rotational symmetry point symmetry

230

Symmetry

SYMMETRY 25.1 SYMMETRY • Symmetry is when an object looks identical to its original shape after being flipped or turned. • Objects that can be divided into identical halves are symmetrical objects. • Objects that cannot be divided into identical halves are asymmetrical objects. • Symmetrical objects have a symmetry plane, a plane that divides an object into two equal halves.

25.2 TYPES OF SYMMETRY 1. LINE SYMMETRY • Line symmetry is a type of symmetry where one half is a reflection of the other. • It is also called reflection symmetry or mirror symmetry. • A line of symmetry is an imaginary line where an object can be folded into identical halves. • Objects have different numbers of lines of symmetry.

Rectangle Two lines of symmetry

Square Four lines of symmetry

Parallelogram No line of symmetry

2. ROTATIONAL SYMMETRY • A rotational is a type of symmetry where an object can be rotated about its centre and fits exactly over its original position. • The order of rotational symmetry is the number of times an object matches its position in one full rotation. • Objects with rotational symmetry have different orders of symmetry.

Equilateral triangle Order of rotational symmetry: 3

Square Order of rotational symmetry: 4

Symmetry

231

3.

POINT SYMMETRY • Point symmetry is a type of symmetry where every part of an object matches another object with the same distance from the central point but in the opposite direction. • To find out if an object exhibit point symmetry, rotate the object 180o about the central point. If it retains its original shape then an object has point symmetry.

Object retain its original shape when rotated 180o

232

Symmetry

REVIEW QUESTIONS 1. The diagram below shows a regular hexagonal prism

A. How many planes of symmetry does it have? B. What is the order of rotational symmetry? 2. The diagram below shows a cuboid with square edges

A. How many planes of symmetry does it have? B. What is the order of rotational symmetry? 3. The diagram below shows an equilateral triangular prism

A. How many planes of symmetry does it have? B. What is the order of rotational symmetry?

SOLUTIONS 1. A. 7 B. 6 2. A. 5 B. 4 3. A. 4 B. 3

Symmetry

233

26

CONSTRUCTION

Introduction Construction involves drawing lines, angles and shapes using a pencil, a pair of compasses and a straight edge (ruler). Construction has numerous applications in real life. It is applied to design the blueprint of buildings, roads, bridges and computer designs. Art is another field that heavily applies construction.

Specific outcomes This unit introduces you to angle construction and bisection. By the end of this unit, you will be able to: ❖ Construct 60o ❖ Construct 30o ❖ Construct 15o ❖ Construct 90o ❖ Construct 45o ❖ Bisect a given angle ❖ Construct a given polygon

234

Construction

CONSTRUCTIONS 26.1 ANGLE CONSTRUCTION 1. HOW TO CONSTRUCT 𝟔𝟎o i. Draw the arm OA with a ruler. ii. Place the compass at O and draw an arc that passes through A. iii. Place the compass at A and draw an arc that passes through O and intersects the arc in step ii. and label the point of intersection as B. iv. Join points O and B. The angle AOB is 60o .

∠AOB = 60o

2. HOW TO CONSTRUCT 𝟑𝟎o i. First construct 60o , angle AOB. ii. Place the compass at point B and draw an arc passing through point A. iii. Place the compass at point A and draw an arc passing through point B and intersecting the arc in step ii. and label the point of intersection as C. iv. Join points O and C. The angle AOC is 30o .

∠AOC = 30o

Construction

235

3.

HOW TO CONSTRUCT 𝟏𝟓o i. First construct 60o , then 30o angle AOC. ii. Place the compass at point O and draw an arc passing through point C and intersecting line A at P. iii. Place the compass at point C and draw an arc passing through point P. iv. Place the compass at point P and draw an arc passing through point C and intersecting the arc in step iii. and label the point of intersection as D. v. Join points O and D. The angle AOD is 15o .

∠POD = 30o

4.

HOW TO CONSTRUCT 𝟗𝟎o i. Draw the arm OA with a ruler. ii. Place the compass at O and draw an arc. iii. Place the compass at A and draw an arc to intersect the arc in step ii. and label the point of intersection as B. iv. Place the compass at point B and draw an arc to intersect the arc in step ii. and label the point of intersection as C. v. With the compass at point B, extend the arc from point C. vi. Place the compass at point C and draw an arc to intersect the extended arc in step v. and label the point of intersection as D. vii. Join points O and D. The angle AOD is 90o .

∠AOD = 90o 236

Construction

5. HOW TO CONSTRUCT 𝟒𝟓o i. First construct 90o , angle AOD. ii. Place the compass at O and draw an arc from point D to intersect arm OA at point P. iii. Place the compass at point P and draw an arc. iv. Place the compass at point D at draw arc to intersect the arc in step iii. and label the point of intersection as E. v. Join point O and E. The angle POE is 45o .

∠POE = 45o

26.2 ANGLE BISECTOR ❖ HOW TO BISECT ANY ANGLE i. Draw the given angle, for example, ∠AOB. ii. Place the compass at point A and draw an arc. iii. Place the compass at point B and draw an arc that intersects with the arc in step ii, and label the point of intersection as C. v. Join points O and C to bisect ∠ AOB.

Construction

237

REVIEW QUESTIONS 1.

A. Construct a triangle JKL in which KL = 8cm, KJ = 6cm and JL = 10cm. B. Measure and write angle JKL. C. Within the triangle JKL, draw the locus of points which are i. 5cm from J ii. 3cm from JL iii. equidistant from JK and JL D. A point Q, within the triangle JKL, is such that it is greater than or equal to 5cm from J, less than or equal to 3cm from JL and nearer to JK than to JL. Indicate by shading the region in which Q must lie.

2.

A. B. C.

D. E.

3.

A. B. C.

D.

4.

A. B. C. E.

238

Construct triangle PQR in which PQ = 10cm, QR = 8cm and ∠PQR = 50o . Measure and write the length of PR. On your diagram, within triangle PQR, construct the locus of points which are i. equidistant from P and Q ii. equidistant from PR and PQ iii. 5cm from R A point T within triangle PQR is such that it is 5cm from R and equidistant from P and Q. Label the point T. Another the X is such that it is less than or equal to 5cm from R, nearer to Q than P and near PQ than PR. Indicate clearly, by shading, the region in which X must lie. Construct triangle PQR in which PQ is 9cm, angle PQR = 60o and QR = 10cm. Measure and write the length of PR. On your diagram, draw the locus of points within the triangle PQR which are i. 3cm from PQ ii. 7cm from R iii. equidistant from P and R A point M, within triangle PQR, is such that it is nearer to R than P, less than or equal to 7cm from R and less than or equal to 3cm from PQ. Shade the region in which M must lie. Construct a quadrilateral ABCD in which AB = 10cm, angle ABC = 120o, angle BAD = 60o , BC = 7cm and AD = 11cm. Measure and write the length of CD. A point P, within the quadrilateral ABCD, is such that it is 8cm from A and equidistant from BC and CD. Label point P. Another point Q, within the quadrilateral ABCD, is such that it is nearer to CD than BC and greater than or equal to 8cm from A. Indicate, by shading, the region in which Q must lie.

Construction

SOLUTIONS 1.

A. B. ∠JLK = 143o

2.

A. B. PR = 7.8cm

Construction

239

3. A. B.

PR = 9.5cm

4. A. B.

CD = 8.7cm

240

Construction

27

EARTH GEOMETRY

Introduction Earth geometry is the study of measurements of the earth. In this unit, you will learn how to calculate the distance between two points on earth. You will also learn how to calculate the time difference between two points on earth.

Specific outcomes By the end of this unit, you will be able to: ❖ Know great circles and small circles ❖ Know latitudes and longitudes ❖ Describe a location on earth using latitude and longitude ❖ Calculate great circle distance in kilometres and nautical miles ❖ Calculate speed in km/hr and knots ❖ Calculate the time difference between two places

Earth Geometry

241

EARTH GEOMETRY 27.1 EARTH GEOMETRY • Earth geometry is the study of measurements of the earth. • The Earth is a sphere; therefore, the distance travelled between any two points, for example, from Lusaka to London, is not a straight line distance but a curved distance. • The earth has two poles. The northmost part of the earth is called the North pole. The southmost part of the earth is called the South pole. • The halfway between the north pole and south pole lies an imaginary line called the Equator. • When the earth is sliced along the equator, it is divided into two equal halves called hemispheres. • The half of the earth that lies north of the equator is called the Northern hemisphere. • The other half of the earth that lies south of the equator is called the Southern hemisphere.

27.2 GREAT CIRCLES AND SMALL CIRCLES ❖ GREAT CIRCLES • Great circles are imaginary planes that pass through the centre of the earth. • The radius of the great circle is the same as the radius of the earth. • Examples of the great circles of the earth include the equator and all longitudes of meridians (lines running from north to south). ❖ SMALL CIRCLES • Small circles are imaginary lines that do not pass through the centre of the earth. • Examples of small circles include all other lines parallel to the equator.

242

Earth Geometry

27.3 LATITUDES AND LONGITUDES ❖ LATITUDES • The imaginary lines that run east to west of the earth are called the parallels of latitudes. • All parallels of latitudes are parallel to each other. • The main parallel of latitude is the equator. It is the only great circle among the parallel of latitudes. • The other parallel of latitudes are small circles and lie north or south of the equator • The parallel of latitudes ranges from 90oN to 90oS. The north pole being 90oN and the south pole 90oS. • The equator is labelled as 0o, and the other parallel of latitudes are measured with reference from the equator.

• The angle of latitude is the angle that a line from the centre of the earth to the parallel of latitude makes with the equator. • Each 1o of longitude represents 60 nautical miles (approximately 111 km). • The diagram below illustrates the 30 oN latitude.

Earth Geometry

243

❖

LONGITUDES • The imaginary lines that run north-south of the earth are called meridians of longitude. • All meridians of longitude are great semi-circles. • They all converge at the poles and farthest apart at the equator. • Unlike the latitudes, meridians of longitude are not parallel to each other. • Meridians of longitude range from 0o to 180o east and 0o to 180o west. • The main meridian of longitude is the Greenwich meridian (also called prime meridian). It is labelled as 0o, and divide the earth into the eastern and western hemisphere. • The other meridians of longitude lie to the east or west of the Greenwich meridian.

• • •

244

The angle of meridian is the angle a particular meridian of longitude makes with Greenwich meridian. Each 1o of longitude represents 60 minutes. The diagram illustrates a 60o E meridian of longitude.

Earth Geometry

27.4 POSITION COORDINATES • The intersection of parallels of latitudes and meridians of longitudes drawn on a map produces a geographic grid • The geographic grid is used to identify locations on the earth

• •

Locations are described using latitude (oN or oS) and longitude (oE or oW). For example, Lusaka, Zambia, has coordinates (15o S, 28o E). This means it is located 15o south of the equator and 28o east of the Greenwich meridian.

❖ EXAMPLES 1. The figure below shows some latitudes and longitudes of the earth.

A. B. C. D.

Find the coordinates of the points labelled A to H. What is the difference in latitude between point A and B? What is the difference in longitude between point A and E? What is the difference in longitude between point E and G?

Earth Geometry

245

❖

SOLUTIONS 1. A. A is (70o N, 0o ) B is (70o N, 40o W) C is (0o N, 0o E) D is (0o N, 40o W) E is (30o S, 0o E) F is (30o S, 40o W) G is (60o S, 0o E) H is (60o S, 40o W) B. Difference in latitude between A and B = 40o + 0o = 40o C. Difference in longitude between A and E = 70o + 30o = 100o

When locations lie on opposite side of the hemisphere add the coordinates.

D. Difference in longitude between E and G = 60o − 30o = 30o

When locations lie on the same side of the hemisphere subtract the coordinates.

27.5 NAUTICAL MILES AND KNOTS ❖ NAUTICAL MILES • A nautical mile is a unit of distance. • Symbol: M, NM or nm. • It is chiefly used for sea and air travel. • An angle of 1o on a great circle is equivalent to 60 nautical miles. • 1 nautical mile is equal to 1852 metres (or 1.852 km). ❖ EXAMPLES 1. Convert the following kilometres into nautical miles A. 1 km B. 10 km C. 500 km D. 1200 km 2. Convert the following nautical miles into kilometres A. 1 nm B. 10 nm C. 60 nm D. 1500 nm

246

Earth Geometry

❖

SOLUTIONS 1. A. x =

1 1.852

= 0.53996 = 0.54 nm B. x =

∴ 1 km = 0.54 nm

10 1.852

= 5.3996 = 5.4 nm C. x =

∴ 10 km = 5.4 nm

500 1.852

= 269.978 = 270 nm D. x =

∴ 500 km = 270 nm

1200 1.852

= 647.948 = 648 nm

2.

❖

∴ 1200 km = 648 nm

A. x = 1 × 1.852 = 1.852 km

∴ 1 nm = 1.852 km

B. x = 10 × 1.852 = 18.52 km

∴ 10 nm = 18.52 km

C. x = 60 × 1.852 = 111.12 = 111 km

∴ 60 nm = 111 km

D. x = 1500 × 1.852 = 2778 km

∴ 1 nm = 1.852 km

KNOTS • A knot is a unit of speed. • 1 knot = 1 nm/hr (nm/hr is nautical mile per hour). • 1 knot = 1.852 km/hr

27.6 SPEED • Speed is the distance travelled per unit time. • It can be calculated in kilometres per hour (km/hr) or in knots, where 1 knot is equal to 1 nautical mile per hour (nm/hr). • Note: 1 knot = 1.852 km/hr Speed =

distance time

When distance is in nautical miles, speed should be calculated in knots. When distance is in kilometres, speed should be calculated in km/hr.

Earth Geometry

247

❖ EXAMPLES 1. A cruise ship travelled 336 km from 09 am to 3pm. Calculate its average speed in km/h. 2. If an airplane travels north along the Greenwich meridian 1368 nautical miles in 4 hours, what is its average speed? A. in knots B. in km/hr 3.

A ship travels northeast at 22 knots for 3 hours. A. How far did it travel in nautical miles? B. How far did it travel in kilometres?

❖ SOLUTIONS distance 1. Speed = time 336 = 6 = 56 km/hr

2.

A. Speed = =

distance time 1368

4 = 342 knots B. 1 knot = 1.852 km/hr 342 knots = x x = 342 × 1.852 = 633.384 = 633 km/hr

3. A. distance = speed × time = 22 × 3 = 66 nm B. 1 nm = 1.852 km 66 nm = x x = 66 × 1.852 = 122.232 = 122 km

248

Earth Geometry

27.7 GREAT CIRCLE DISTANCE • Since the earth is a sphere, a straight line between two points on a map is not the shortest distance. • The shortest distance corresponds to an arc length between the points of the earth’s surface. This distance is called the great circle distance. • The angle made at the centre of the earth by the two points is called the angular distance. • The great circle distance is calculated by the formula below. d=

θ

where d = great circle distance θ = angular distance r = radius of the earth (approx. 6400 km)

o × 2πr

360

This formula is used to calculate distance in kilometres

Or

d = 60θ

where d = great circle distance θ = angular distance

This formula is used to calculate distance in nautical miles

❖ EXAMPLES 1. Given that the angular distance between two points is 50o and the radius of the earth is 6400 km, what is the great circle distance between two points? 2. What is the shortest distance from Cape Town, South Africa (34o S, 18o W) to Stockholm, Sweden (59o N, 18o W). The radius of the earth is 6400 km. 3. Find the distance in nautical miles from Chile (33o S, 71o W) and Boston, USA (42o S, 71o W). 4. Calculate the distance from Minsk, Belorussia (54o S, 28o W) to Pretoria, South Africa (25o S, 28o E).

❖ SOLUTIONS 1. d = =

θ 360o 50 360o

× 2πr × 2π(6400)

= 5585.0536 = 5585 km 2. θ = 34o + 59o = 93o d= =

θ 360o 93 360o

When locations lie on opposite side of the hemisphere add the coordinates.

× 2πr × 2π(6400)

= 10388.1997 = 10388 km Earth Geometry

249

3.

θ = 54o + 25o = 79o

When locations lie on opposite side of the hemisphere add the coordinates.

d = 60θ = 60 × 79o = 4740 nm

4.

θ = 33o + 42o = 75o

When locations lie on opposite side of the hemisphere add the coordinates.

d = 60θ = 60 × 75o = 4500 nm

27.8 LONGITUDE AND TIME DIFFERENCES • The earth makes a complete revolution (360o ) in 24 hours. • As the earth is spinning, time is the same on all parts of the world lying along the same longitude but different on other parts of the worlds lying in different longitudes. • There is a one hour time difference in local time for every 15o of longitude (or 4 minutes per 1o ). • For the longitudes toward the east, add 1 hour for each 15o of longitude (or 4 minutes for each 1o ). • For the longitudes toward the west, subtract 1 hour for each 15o of longitude (or 4 minutes for each 1o ).

250

Earth Geometry

❖

EXAMPLES 1. Lusaka, Zambia, has coordinates (15o S, 28o E) while Sydney, Australia has (34o S, 151o E). A. Calculate the difference in longitude between the two places. B. Calculate the time difference between the places. C. What is the time in Sydney when it is 14:35 in Lusaka? 2. New Delhi, India, has coordinates (28o N, 77o E) while Los Angeles, USA has (34o N, 119o W). A. Calculate the difference in longitude between the two places. B. Calculate the time difference between the places. C. What is the time in Los Angeles when it is 21:00 in New Delhi?

❖

SOLUTIONS 1. A. Difference in longitude = 151o − 28o = 123o B. 1o = 4 min o 123 = x x = 4 × 123 = 492 min = 8 hours 12 minutes ∴ time difference = 8 hours 12 minutes C. Time in Sydney = 14:35 + 8:12 = 22:47

2. A. Difference in longitude = 77o + 119o = 196o B. 1o = 4 min 196o = x x = 4 × 196 = 784 min = 13 hours 04 minutes ∴ time difference = 13 hours 04 minutes C. Time in Los Angeles = 22:00 − 13:04 = 07:56

Earth Geometry

251

REVIEW QUESTIONS 1. The diagram below shows a wire model of the earth. The circle of latitude in the north is 50o N and the circle of latitude in the south is 60o S. A and C are on the longitude 55o W while B and D are on longitude 50o E. (Take π = 3.142 and radius = 3437nm).

A. B. C. D.

Write the positions, using longitudes and latitudes, of points A and D. Calculate the difference in longitudes between A and B. Given that the time at D is 09:20, what would be the time at town C? Calculate the distance BD along the longitude 50o E in nautical miles.

2. A, B, C and D are points on the surface of the earth as shown in the diagram below. (Take π = 3.142 and radius = 3437nm).

A. B. C. D.

Using latitudes and longitudes, write the positions of the points A and B. Find the difference in longitude between points C and D. Calculate the distance CD in nautical miles. Given that the local time at D is 13:05, find the time at C.

3. The points K, L and M are on the surface of the earth as shown in the diagram below. (Take π = 3.142 and radius = 6370km).

252

Earth Geometry

A. Find the difference in longitude between points K and L. B. Find, in kilometres, the distance i. LM ii. KL 4. W, X, Y and Z are four points on the surface of the earth as shown in the diagram below. (Take π = 3.142 and radius = 3437nm).

A. Calculate the difference in latitude between W and Y. B. Calculate the distance in nautical miles between i. X and Z along the longitude 105o E. ii. Y and Z along the circle of latitude 30o S. 5. The diagram below shows point A(60o N,30o W), B(30o S,30o W) and C(30o S,75o E). (Take π = 3.142 and radius = 3437nm).

A. If the local time at B is 15:00, what is the local time at C? B. It takes a plane 6 hours to fly from A to B. What is its speed in knots?

Earth Geometry

253

SOLUTIONS 1. A. A(50o N,55o W) D(60o S,50o E) B. Difference in longitude = 55o + 50o = 105o C. Difference in latitude = 55o + 50o = 105o Time difference between C and D 1o

105o

=

4 min 𝑥

x = 4 × 105 = 420 min = 7 hours

Local time at C = 09:00 − 7 hours = 02:00 D. θ = 50o + 60o = 110o d = 60θ = 60 × 110o = 6600nm

2. A. A(60o N,51o W) B(80o S,51o E) B. Difference in longitude = 61o + 31o = 30o C. d = 60θ = 60 × 30o = 1800nm D. Time difference between C and D 1o

30o

=

4 min 𝑥

x = 4 × 30 = 120 min = 2 hours

Local time at C = 13:05 − 2 hours = 11:05

254

Earth Geometry

3.

A. Difference in longitude = 30o + 60o = 90o B. i. θ = 50o + 55o = 95o θ

d= =

360o 95 360o

× 2πr × 2(3.142) × 6370

= 10563.22944 = 10563.23 km C. ii. θ = 30o + 60o = 90o d= =

θ 360o 90 360o

× 2πr × 2(3.142) × 6370

= 10007.27km

4.

A. Difference in latitude = 80o + 30o = 110o B. i. θ = 80o + 30o = 110o d = 60θ = 60 × 110o = 6600nm ii. θ = 15o + 105o = 120o d = 60θ = 60 × 120o = 7200nm

Earth Geometry

255

5.

A. Difference in longitude = 30o + 75o = 105o Time difference between B and C 1o

105o

=

4 min 𝑥

x = 4 × 105 = 420 min = 7 hours

Local time at C = 15:00 + 7 hours = 22:00

B. θ = 30o + 75o = 105o d = 60θ = 60 × 105o = 6300nm Speed = =

distance time

6300 7

= 900 knots

256

Earth Geometry

28

PYTHAGORAS’ THEOREM

Introduction Pythagoras’ theorem is a statement that shows the relationship between the lengths of the sides of a rightangled triangle. It states that in a right-angled triangle the square of the hypotenuse side is equal to the sum of squares of the other sides. Pythagoras theorem has many real life applications. It is used in construction and architecture. It is also used in navigation to find the shortest distance between two points. This unit covers the Pythagoras’ theorem.

Specific outcomes By the end of this unit, you will be able to: ❖ Know parts of a right-angled triangle ❖ State Pythagoras’ theorem ❖ Use Pythagoras’ theorem to solve problems

Pythagoras’ Theorem

257

PYTHAGORAS’ THEOREM 28.1 PARTS OF A RIGHT - ANGLED TRIANGLE

• • •

Hypotenuse. The side of the triangle opposite to the right angle. It is the longest side. Opposite. The part of the triangle that forms the right angle and is opposite to the angle θ. Adjacent. The part of the triangle that forms the right angle and is adjacent to θ.

28.2 PYTHAGORAS’ THEOREM If ∠ABC = 𝟗𝟎o , then 𝒄𝟐 = 𝒂𝟐 + 𝒃𝟐 If 𝒄𝟐 = 𝒂𝟐 + 𝒃𝟐 , then ∠ABC = 𝟗𝟎o where c = hypotenuse a = opposite b = adjacent

❖

EXAMPLES Solve the missing side

1.

2.

3.

258

Pythagoras’ Theorem

❖

SOLUTIONS 1. 𝑐 2 = 𝑎2 + 𝑏 2 52 = 32 + 𝑥 2 𝑥 2 = 52 − 32 𝑥 2 = 25 − 9 √𝑥 2 = √16 x=4

2. 𝑐 2 = 𝑎2 + 𝑏 2 352 = 𝑥 2 + 282 𝑥 2 = 352 − 282 𝑥 2 = 1225 − 784 √𝑥 2 = √441 x = 21

3. 𝑐 2 = 𝑎2 + 𝑏 2 𝑥 2 = 302 + 402 𝑥 2 = 900 + 1600 √𝑥 2 = √2500 x = 50

Pythagoras’ Theorem

259

29

TRIGONOMETRY

Introduction In the previous unit, you learned that given the length of two sides of a right-angled triangle you could calculate the length of the other side. How about calculating the angles of a triangle when given sides? We can use trigonometry. Trigonometry is a branch of mathematics that describes the relationship between the angles and lengths of a triangle. Trigonometry has many real life applications. For example, as triangles do not easily deform under stress, much of architecture and engineering rely on triangular structures to strengthen the building. To form these structures, trigonometry is applied. Navigation also uses trigonometry to found the shortest distance and angle between two places.

Specific outcomes This unit covers trigonometry. By the end of this unit, you will be able to: ❖ Know parts of a right-angle triangle ❖ Use the following trigonometric ratios to solve problems: • Sine ratio • Cosine ratio • Tangent ratio ❖ Find special angles using trigonometric ratios ❖ Calculate angles and lengths of the right angled triangle using trigonometric ratios ❖ Know the angle of elevation and depression ❖ Sine rule: • State sine rule • Use sine rule to find angles and lengths of a triangle ❖ Cosine rule: • State cosine rule • Use cosine rule to find angles and lengths of a triangle ❖ Calculate the area of a triangle using trigonometry

260

Trigonometry

TRIGONOMETRY 29.1 TRIGONOMETRY • Trigonometry is derived from a Greek word that means triangle measure. • It involves the study of relationships between sides and angles of a triangle. ❖ PARTS OF A RIGHT - ANGLED TRIANGLE • Hypotenuse. The side of the triangle opposite to the right angle. It is the longest side. • Opposite. The part of the triangle that forms the right angle and is opposite to the angle θ. • Adjacent. The part of the triangle that forms the right angle and is adjacent to θ.

29.2 TRIGONOMETRIC RATIOS • Trigonometric ratios are ratios of sides of a right-angled triangle. • There are three basic trigonometric ratios: sine, cosine and tangent, abbreviated as sin, cos and tan respectively. • These ratios can be used to find the angle or side of a triangle. Sin θ =

opposite hypotenuse

Cos θ =

Tan θ =

adjacent hypotenuse opposite adjacent

29.3 SPECIAL ANGLES • For most angles, the value of the trigonometric ratio cannot be calculated easily without a calculator. • However, some angles give value that can easily be remembered. • Some of these angles are listed in the table below with their trigonometric values. Sin θ Cos θ Tan θ

0o 0 1 0

30o 1 2 √3 2 1

45o 1 √2 1 √2 1

60o √3 2 1 2

90o 1

180o 0

0

−1

√3

_

0

√3 261

❖

EXAMPLES Find the exact value of the following without a calculator. A. Sin 30o B. Cos 0o C. Tan 45o

❖

SOLUTIONS A. Sin 30o =

1 2

B. Cos 0o = 1 C. Tan 45o = 1

29.4 HOW TO FIND THE VALUE OF NON - SPECIAL ANGLES • The value of non-special angles, even of special angles, can be found using a calculator. • Use the sin, cos, and tan on the scientific calculator. • To find, for example, sin of angle 50o (sin 50o ) place sin, 50 then = sign. You have to get 0.77 (2 decimal places). ❖ EXAMPLES 1. Find the exact value of the following using a calculator. A. Sin 30o B. Cos 0o C. Tan 45o 2. Find the value of the following using a calculator, leaving your answer in two decimal places. A. Sin 25o B. Cos 134o C. Tan 74o ❖ SOLUTIONS 1 1. A. sin 30o = 2

B. Cos 0o = 1 C. Tan 45o = 1 2. A. sin 25o = 0.42 B. Cos 134o = −0.69 C. Tan 74o = 3.49

29.5 HOW TO FIND ANGLES • To find an angle when given the value use the inverse button of your calculator, sin −1 for sin angles, cos−1 for cos angles and tan−1 for tan angles. • For example, when given that sin θ = 0.42 to find the θ place sin −1 (0.42) on your calculator. You have to get 24.83o ≈ 25o . • For most scientific calculators to have sin −1, cos−1 or tan−1 place shift button then sin, cos or tan button respectively.

262

Trigonometry

❖

EXAMPLES 1. Find the angle of the following A. Sin θ =

1 2

B. Cos θ = 1 C. Tan θ = 1 2. Find the angle of the following leaving your answer to one decimal place A. Sin θ = −0.75 B. Cos θ = 0.75 C. Tan θ = 0.75 ❖

SOLUTIONS 1. A. Sin θ =

1 2 1

θ = sin −1 ( θ = 30

o

)

2

B. Cos θ = 0 θ = cos −1(1) θ = 90o C. Tan θ = 1 θ = tan−1 (1) θ = 45o 2. A. Sin θ = −0.75 θ = sin −1 (−0.75) θ = −48.6o B. Cos θ = 0.75 θ = cos −1(0.75) θ = 41.4o C. Tan θ = 0.75 θ = tan−1 (0.75) θ = 36.9o

29.6 SOLVING TRIGONOMETRIC PROBLEMS ❖ EXAMPLES Solve the following 1.

2.

Trigonometry

263

3.

❖

4.

SOLUTIONS 1. cos θ =

adj

Sides give adj = 6 hyp = 9 Then use cosine

hyp adj

cos (x) =

hyp 6

cos (x) =

9 2

cos (x) =

3 2

x = cos −1 (3) x = 48.2o

2. cos θ =

adj

hyp y o

cos 38 =

12

y = cos 38o × 12 y = 9.46 cm opp

3. tan θ =

adj 10 o

tan 51 = a=

𝑎 10

Sides give adj = y hyp = 12 Then use cos

For a sides give are opp = a adj = 10 Then use tan

tan 51o

a = 8.1 m

sin θ =

opp

hyp 10 o

sin 51 = b=

𝑏 10 sin 51o

b = 12.9 m

264

Trigonometry

For b sides give are opp = 10 hyp = b Then use tan

4.

sin θ = sin a =

opp

Sides give opp = 70 hyp = 100 Then use sin

hyp 70 100 7

a = sin−1 (10) a = 44.4o

29.7 ANGLES OF ELEVATION AND DEPRESSION ❖ ANGLE OF ELEVATION • The angle of elevation is the angle upward from the horizontal. • The observer’s line of sight would be above the horizontal.

❖ ANGLE OF DEPRESSION • The angle of depression is the angle downward from the horizontal. • The observer line of light would be below the horizontal.

29.8 SINE AND COSINE RULE • The sine and cosine rule are used to solve non right-angled triangles. ❖ SINE RULE • Sine rule is used when given i. Two angles and one side of a triangle. ii. Two sides and a non-included angle. • For a given a triangle 𝒂 sin A

=

𝒃 sin B

=

𝒄 sin C

or

sin A 𝒂

=

sin B 𝒃

=

sin C 𝒄

Trigonometry

265

❖

COSINE RULE • Cosine rule is used when given i. Three sides. ii. Two sides and an included angle. • For a given triangle 𝒂𝟐 = 𝒃𝟐 + 𝒄𝟐 − 2abcosA or

❖

Area = Area =

•

1 2 1

𝟐𝒃𝒄

2

× base × height ah

Since sin C =

ℎ 𝑏

or h = bsin C, therefore

A=

1 2

absin C

EXAMPLES 1. Solve the following

A.

B.

C.

D.

2. Find the area of a triangle

266

𝒃𝟐 + 𝒄𝟐 − 𝒂𝟐

AREA OF A TRIANGLE •

❖

cos A =

Trigonometry

❖

SOLUTIONS 𝑎

1. A.

sin A 3.5

=

sin 43

𝑏 sin B 2.3

=

sin x =

sin 𝑥 2.3 × sin 43 3.5

sin x = 0.448 x = sin −1 (0.448) x = 26.6o B.

𝑎 sin A 6

=

sin 34

𝑏 sin B 𝑦

=

y=

sin 15 6 × sin 15 sin 34

y = 2.78 cm C. 𝑎2 = 𝑏 2 + 𝑐 2 − 2abcosA 𝑎2 = 102 + 92 − 2(10)(9)cos 47 𝑎2 = 100 + 81 − 122.76 𝑎2 = 58.24 √𝑎2 = √58.24 a = 7.63 D. cos A = cos a = cos a =

𝑏2 + 𝑐 2 − 𝑎2 92

2𝑏𝑐 + 192 − 162 2(9)(19)

81 + 361 − 256 342

cos a = 0.544 a = cos −1(0.544) a = 57o

2. A = =

1 2 1 2

absin C × 12 × 14 × sin 65

= 76.1 cm2

Trigonometry

267

REVIEW QUESTIONS 1. Solve the following equations A. sin θ = 0.5 for 0o ≤ θ ≤ 360o B. sin x = 0.6792 for 0o ≤ x ≤ 180o C. cos a = 0.95 for 0o ≤ a ≤ 360o D. 13cos θ = 5 for 0o ≤ θ ≤ 360o E. tan y = 0.7 for 0o ≤ y ≤ 180o F. 6tan θ = 9 for 0o ≤ θ ≤ 360o 2. The diagram below shows triangle ABC in which AC = 18cm, ∠CAB = 80o and ∠ABC = 90o .

Calculate the length of BC. 3. The figure below shows triangle ABC in which AC = 5cm. Given that sin θ = 0.5

Calculate the length of BC. 4. In the diagram below, BCD is a straight line. It is given that AB = 12cm, AC = 13cm and ∠ABC = 90o

Find ∠ACD 4

5. The figure below shows a right angled triangle. QR = 4cm and tan (∠QPR) = 3

Find sin ∠PQR 268

Trigonometry

6. The diagram below shows triangle ABC in which AB = 16cm, BC = 8cm and angle ABC = 150o.

Calculate the area of triangle ABC. 7. The diagram below shows triangle XYZ in which XY = 12cm, angle XYZ = 150o and its area is 30cm2.

Calculate the length of YZ. 8. The diagram below shows an isosceles triangle ABC where AB = BC = (x − 3)cm and ∠ABC = 90o .

Find the value of x for which the area of triangle ABC is 18cm2. 9. The figure below shows a triangle ABC. AB = 15cm, angle ABC = 79o and angle ACB = 40o .

Calculate A. distance AC B. area of triangle ABC C. shortest distance from B to AC

Trigonometry

269

10. In triangle ABC below, AC = 275 km, angle BAC = 125o and angle ACB = 40o .

Calculate A. the distance BC B. the area of triangle ABC C. the shortest distance from A to BC 11. The figure below shows a triangle JKL in which JK = 5km, KL = 3km and angle JKL = 110o.

Calculate A. the distance JL B. the area of triangle JKL C. the shortest distance from K to JL 12. In the triangle below AB = AC = 43km and BC = 37km.

Calculate A. angle BAC B. angle ACB C. the area of triangle ABC D. the shortest distance from C to AB

270

Trigonometry

SOLUTIONS 1. A. sin θ = 0.5 θ = sin−1(0.5) = 30o all possible solutions for 0o ≤ θ ≤ 360o sin is positive in 1st and 2nd quadrant θ = 30o , 180o − 30o θ = 30o , 150o B. sin x = 0.6792 x = sin−1(0.6792) = 42.8o all possible solutions for 0o ≤ x ≤ 180o sin is positive in 1st and 2nd quadrant x = 42.8o, 180o − 42.8o x = 30o , 137.2o C. cos a = 0.95 a = cos−1(0.95) = 18.2o all possible solutions for 0o ≤ a ≤ 360o cos is positive in 1st and 4th quadrant a = 18.2o, 360o − 18.2o a = 18.2o, 341.8o

S T

A C

S T

A C

S T

A C

D. 13cos θ = 5 5

cos θ = 13

5 13

θ = cos −1 ( ) = 67.4o all possible solutions for 0o ≤ θ ≤ 360o cos is positive in 1st and 4th quadrant θ = 67.4o, 360o − 67.4o θ = 67.4o, 292.6o E. tan y = 0.7 y = tan−1 (0.7) = 34.9920o = 35o all possible solutions for 0o ≤ y ≤ 180o tan is positive in 1st and 3rd quadrant y = 35o , 180o + 35o (not needed) y = 35o

S T

A C

S T

A C

Trigonometry

271

F.

6tan θ = 9 tan θ =

9 5

9 5

θ = tan−1 ( ) = 56.3o all possible solutions for 0o ≤ θ ≤ 360o tan is positive in 1st and 3rd quadrant θ = 56.3o, 180o + 56.3o θ = 56.3o, 236.3o 2.

opp

sin θ = hyp sin (30) =

BC 18

BC = 18sin (30) = 9cm 3.

opp

sin θ = hyp 5

0.5 = BC 5

BC = 0.5

BC = 10cm 4.

sin θ =

opp hyp

sin (∠ACB) =

12 13 12

∠ACB = sin−1 (13) = 67.380o = 67.4o

∠ACD = 180o − ∠ACB = 180o − 67.4o = 112.6o 5.

opp

tan θ = adj

4

tan (∠QPR) = 3

∴ PR = 3cm

PQ = √PR2 + QR2 = √32 + 42 = √9 +16 = √25 = 5cm opp

sin θ = hyp 3

sin (∠PQR) = 5

272

Trigonometry

S T

A C

6.

1

A = 2 absin C 1

= 2 × 16 × 8 × sin (150) = 32 cm2 7.

1

A = 2 absin C 1

30 = 2 × 12 × YZ × sin (150) 60 = 12YZ sin (150) 60

YZ = 12sin (150) = 10cm 8.

1

A = 2 absin C 1

18 = 2 × (x − 3)(x − 3) × sin (90) 36 = (x − 3)(x − 3) × 1 (x − 3)(x − 3) = 36 𝑥 2 − 3x − 3x + 9 = 36 𝑥 2 − 6x + 9 − 36 = 0 𝑥 2 − 6x − 27 = 0 𝑥 2 + 3x − 9x − 27 = 0 (𝑥 2 + 3x) + (− 9x − 27) = 0 x(x + 3) − 9(x + 3) = 0 (x − 9)(x + 3) = 0 x−9=0 or x+3=0 x=9 x = −3 distance cannot be negative therefore x = 9 9.

A.

𝑎 𝑏 = sin A sin B 15 AC = sin (40) sin (79) 15 × sin (79) AC = sin (40)

AC = 22.9cm B. ∠BAC = 180o − ∠ABC − ∠ACB = 180o − 79o − 40o = 61o 1 2 1 =2

A = absin C × AB × AC × sin (61)

= 150.22 cm2

Trigonometry

273

C.

opp

sin θ = hyp ℎ

sin (61) = 15

h = 15 sin (61) h = 13.1cm shortest distance = 13.1cm 10.

A. ∠ABC = 180o − ∠BAC − ∠ACB = 180o − 125o − 40o = 15o 𝑎 𝑏 = sin A sin B 275 BC = sin (15) sin (125) 275 × sin (125) BC = sin (15)

BC = 870.36km 1

B. A = 2 absin C 1

= 2 × AC × BC × sin (40) 1

= 2 × 275 × 870.36 × sin (40) = 76925.29 km2 opp

C. sin θ = hyp sin (40) =

ℎ 275

h = 275 sin (40) h = 176.77 km shortest distance = 176.77 km 11.

A. 𝑎2 = 𝑏 2 + 𝑐 2 − 2abcosA JL2 = JK 2 + KL2 − 2(KJ)(KL)cos (110) JL2 = 52 + 32 − 2(5)(3)cos (110) JL2 = 44.2606 JL = 6.65 km 1 2 1 =2 1 =2

B. A = absin C × JK × KL × sin (110) × 5 × 3 × sin (110)

= 7.05 km2

274

Trigonometry

11.

C.

𝑎 𝑏 = sin A sin B KL JL = sin (∠KJL) sin (∠JKL) 3 6.65 = sin (∠KJL) sin (110) 3 × sin (110) sin (∠KJL) = 6.65

sin (∠KJL) = 0.4239 ∠KJL = sin −1(0.448) = 25.1o opp

sin θ = hyp ℎ

sin (∠KJL) = 5

h = 5 sin (24.1) h = 2.13 km shortest distance = 2.13 km 𝑏 2 + 𝑐 2 − 𝑎2 2𝑏𝑐 AB2 + AC2 − BC2 (∠BAC) = 2(AB)(AC)

12. A. cos A = cos

cos (∠BAC) =

432 + 432 − 372 2(43)(43)

cos (∠BAC) = 0.62979 ∠BAC = cos−1 (0.544) ∠BAC = 50.9646 ∠BAC = 51o B.

𝑎 𝑏 = sin A sin B AB BC = sin (∠ACB) sin (∠BAC) 43 sin (∠ACB)

37

= sin (51)

sin (∠ACB) =

43 × sin (51) 37

sin (∠ACB) = 0.903 ∠ACB = sin −1(0.903) ∠ACB = 64.577 ∠ACB = 64.6o 1

C. A = 2 absin C 1

= 2 × AB × AC × sin (∠BAC) 1

= 2 × 43 × 43 × sin (51) = 718.47 km2

Trigonometry

275

D.

opp

sin θ = hyp ℎ

sin (∠BAC) = AC sin (51) =

ℎ 43

h = 43 sin (51) h = 33.4 km shortest distance = 33.4 km

276

Trigonometry

30

VECTORS

Introduction A vector is a quantity that has both magnitude and direction. Vectors are applied in physics and engineering to calculate the velocity of a moving object, the force acting on an object, the pressure exerted by an object and so on.

Specific outcomes This unit introduces you to vectors. By the end of this unit, you will be able to: ❖ Represent vectors by arrows and vector column ❖ Know types of vectors ❖ Add and subtract vectors using: • Triangle law • Parallelogram law ❖ Solve position vector problems ❖ Find the magnitude of a vector ❖ Work out scalar multiplication of vectors

Vectors

277

VECTORS 30.1 VECTOR • A vector is a quantity that has both magnitude and direction. • Vectors are usually denoted by a lowercase letter such as a or b, or using arrow such as 𝑎⃗ or ⃗⃗⃗⃗⃗⃗⃗ OA. • They are represented by arrows, with the arrow length representing the magnitude (size) of the vector and the arrowhead pointing the direction of the vector.

•

•

Vectors also are represented by the column vector. The top number shows the number of units moved in the x-direction and the bottom number shows the number of units moved in the y-direction. 𝑥 a = (𝑦) 2 For example, a = ( ), means vector a moved 2 units in the x-direction and 4 units y-direction. 4 Graphically, this vector can be drawn as shown below.

30.2 TYPES OF VECTORS 1. EQUAL VECTORS • Two vectors are equal when they have the same magnitude and direction. a

a=b

b

2. ZERO VECTORS • A zero vector is a vector with a zero magnitude. • Zero vectors are also called null vectors. • They are denoted by O. • They do not have magnitude and direction.

.O 278

Vectors

3.

UNIT VECTORS • A unit vector is a vector with a magnitude of 1. • It is usually denoted by a lowercase letter with a hat. 𝑎̂

4.

NEGATIVE VECTORS • A negative vector is a vector with the same magnitude but having an opposite direction to a vector. • A negative vector is also referred to as the inverse of a vector. ⃗⃗⃗⃗⃗⃗). • For example, if ⃗⃗⃗⃗⃗⃗ AB is a vector, a negative vector of ⃗⃗⃗⃗⃗⃗ AB is ⃗⃗⃗⃗⃗⃗ BA (or −AB ⃗⃗⃗⃗⃗⃗ AB

⃗⃗⃗⃗⃗⃗ BA

5.

CO-INITIAL VECTORS • Co-initial vectors are two or more vectors with the same initial point. • For example, vector ⃗⃗⃗⃗⃗⃗ OA and ⃗⃗⃗⃗⃗⃗ OB are co-initial vectors since they have the same starting point. ⃗⃗⃗⃗⃗⃗ OA ⃗⃗⃗⃗⃗⃗ OB

6.

COLLINEAR VECTORS • Collinear vectors are vectors that lie along the same line or parallel lines. • They can have the same or different magnitude and direction. • All vectors below are collinear vectors.

b a

7.

c

COPLANAR VECTORS • Coplanar are two or more vectors that lie in the same plane. • All vectors below are coplanar.

Vectors

279

30.3 ADDITION AND SUBTRACTION OF VECTORS • When two or more vectors are added or subtracted, the vector produced is called the resultant vector. • The resultant vector is drawn from the tail of one vector to the head of another vector. • Vectors can be added or subtracted by two methods, namely triangle law and parallelogram. 1. TRIANGLE LAW OF VECTORS • Triangle law of vector states that when the two sides of a triangle represent two vectors in magnitude and direction in the same order, then the third side of the triangle represents the magnitude and direction of the resultant vector. • In other words, if you have two vectors a and b that are represented by the two sides of a triangle, then the third vector represents the resultant vector of a and b

❖ HOW TO ADD VECTORS USING TRIANGLE LAW i. Draw one vector. ii. Translate the other vector so that its tail touches the head of the previous vector. iii. Draw the resultant vector from the tail of the first vector to the head of the second vector ❖ EXAMPLES 1. Find the sum of the following vectors using the triangle law method

2 3 2. Given the following vectors ⃗⃗⃗⃗⃗⃗⃗ AB = ( ) and ⃗⃗⃗⃗⃗⃗⃗ AC = ( ). Find the sum of vector ⃗⃗⃗⃗⃗⃗⃗ AB and ⃗⃗⃗⃗⃗⃗⃗ AC 3 2

280

Vectors

❖

SOLUTIONS 1.

2.

❖

𝑝 𝑟 Note: for column vectors a = (𝑞 ) and b = ( ) 𝑠 𝑝 𝑟 a + b = (𝑞 ) + ( ) 𝑠 𝑝+𝑟 = (𝑞 + 𝑠)

HOW TO SUBTRACT VECTORS USING TRIANGLE LAW i. Draw one vector. ii. Translate the other vector so that its tail touches the tail of the previous vector. iii. Draw a resultant vector from the head of the vector you are subtracting to the head of the vector you are subtracting from.

Vectors

281

❖

EXAMPLES 1. Find the difference of the following vectors using the triangle law method.

2 3 2. Given the following vectors ⃗⃗⃗⃗⃗⃗⃗ AB = ( ) and ⃗⃗⃗⃗⃗⃗⃗ AC = ( ). Find the ⃗⃗⃗⃗⃗⃗⃗ AB − ⃗⃗⃗⃗⃗⃗⃗ AC 3 −2

❖

SOLUTIONS

1.

282

Vectors

2.

2.

𝑝 𝑟 Note: for column vectors a = (𝑞 ) and b = ( ) 𝑠 𝑝 𝑟 a − b = (𝑞 ) − ( ) 𝑠 𝑝 𝑟 = (𝑞 𝑠 )

PARALLELOGRAM LAW OF VECTORS • Parallelogram law of vectors states that when the adjacent sides of a parallelogram represent two vectors in magnitude and direction, then their resultant is represented by the diagonal drawn from the meeting points of the vectors. • In other words, if you have two vectors a and b that are represented by the adjacent sides of the parallelogram, then their resultant vector is represented by the diagonal of the parallelogram drawn from the meeting points of the two vectors.

❖ HOW TO ADD VECTORS USING PARALLELOGRAM LAW i. Draw one vector. ii. Translate the other vector so that its tail touches the tail of the previous vector. iii. Complete the parallelogram. iv. The resultant vector is the diagonal of the parallelogram. ❖ EXAMPLES 1. Find the sum of the following vectors using the parallelogram law method.

Vectors

283

❖

SOLUTIONS

1.

❖

HOW TO SUBTRACT VECTORS USING PARALLELOGRAM LAW i. Draw a vector you are subtracting from. ii. Translate the negative vector of the vector you are subtracting and connect its tail to the tail of the previous vector. iii. Complete the parallelogram. iv. The resultant vector is the diagonal of the parallelogram

❖

EXAMPLES

1.

284

Vectors

❖

SOLUTIONS

1.

30.4 POSITION VECTOR • Position vector is a vector that starts from the origin and terminates at a point. • It links the location of a point with its origin. • In the cartesian coordinate system, the position vector is expressed as (𝑥, 𝑦). ⃗⃗ and B ⃗⃗. The position vector for vector A is (2,1) and the • The graph below shows two vectors A position vector for vector B is ( 4,3). The origin can be any point. In this example, the origin is (0,0).

Vectors

285

• •

From the previous graph, we can find the position vector AB (from point A to point B) or position vector BA (from point B to point A). To determine position vector AB, subtract the coordinates ⃗A⃗ of from ⃗B⃗. ⃗⃗⃗⃗⃗⃗ AB = ⃗B⃗ − ⃗A⃗ 4 2 =( )−( ) 3 1 6 =( ) 2 AB is (−6,2)

•

⃗⃗ ⃗⃗ from A To determine position vector BA, subtract the coordinates of B ⃗⃗⃗⃗⃗⃗ = A ⃗⃗ − B ⃗⃗ BA 4 2 =( )−( ) 1 3 6 =( ) 2 BA is (6,−2)

❖

EXAMPLES 1 1. Given that ⃗⃗⃗⃗⃗⃗ PQ = ( ) and Q is the point (4,5), find the coordinates of P. 3 3 2. The vector CD = ( ). Given that point C is (1,4), find the coordinate of point D. 2 ⃗⃗⃗⃗⃗⃗⃗ is a column vector. 3. The points L and M have coordinates (2,4) and (−3,1) respectively. Express LM

❖

SOLUTIONS 1. ⃗Q⃗ − ⃗P⃗ = ⃗⃗⃗⃗⃗⃗ PQ 𝑥 4 1 ( ) − (𝑦) = ( ) 5 3 4−x=1 x=4−1 x=3 5 − y = −3 y=5+3 y=8 P is (3,8)

286

Vectors

2.

⃗D⃗ − ⃗C⃗ = ⃗⃗⃗⃗⃗⃗ CD 𝑥 1 3 (𝑦) − ( ) = ( ) 4 2 x − 1 = −3 x = −3 + 1 x = −2 y−4=2 y=2+4 y=6 D is ( 2,6)

3.

⃗⃗⃗⃗⃗⃗⃗ ⃗⃗⃗⃗ − L ⃗⃗ L M=M 3 2 =( )−( ) 1 4 5 =( ) 3

30.5 MAGNITUDE OF A VECTOR • The magnitude of a vector is the length of a vector. • It is also called the modulus of a vector. • It is denoted by |𝑎⃗|. 𝑥 • When vector 𝑎⃗ = (𝑦), then: |𝒂 ⃗⃗| = √𝒙𝟐 + 𝒚𝟐

where |𝑎⃗| = magnitude of a vector x = x-coordinate y = y-coordinate

❖

EXAMPLES 1. What is the magnitude of vector c = (3,4)? 8 ⃗⃗| 2. Given that ⃗P⃗ = ( ), find |P 6 ⃗⃗⃗⃗⃗⃗ whose initial point A is at (1,3) and B final point B is at (5,6). 3. Find the magnitude of the vector AB

❖

SOLUTIONS 1. |c| = √𝑥 2 + 𝑦 2 = √32 + 42 = √9 + 16 = √25 =5 2.

⃗⃗| = √𝑥 2 + 𝑦 2 |P = √( 8)2 + 62 = √64 + 36 = √100 = 10

Vectors

287

3.

⃗⃗⃗⃗⃗⃗| = √(𝑥2 |AB

𝑥1 )2 + (𝑦2

= √(5 1)2 + (6 = √42 + 32 = √16 + 9 = √25 =5

𝑦1 )2 3)2

30.6 SCALAR MULTIPLICATION • A scalar is a quantity that has magnitude but no direction. • A scalar is simply a real number. • Multiplying a vector by a scalar scales a vector (makes a vector bigger or smaller). • To perform scalar multiplication of vectors, multiply a scalar by each component of a vector. • When a vector 𝑎⃗ = (x,y), then: k𝑎⃗ = k(x,y) = (kx,ky)

❖

•

When k > 0, k𝑎⃗ has the same direction as 𝑎⃗ but the magnitude changes. The magnitude is equal to the product of k and 𝑎⃗.

•

When k < 0, k𝑎⃗ has the opposite direction of 𝑎⃗ and the magnitude changes. The magnitude is equal to the product of k and 𝑎⃗.

EXAMPLES 1. Given that 𝑚 ⃗⃗⃗ = (6,2), find A. 2m 1 B. 2m C. −3𝑚

288

where 𝑎⃗ = vector k = scalar x = x-coordinate y = y-coordinate

Vectors

2.

⃗⃗ = (3) Given that P 4 ⃗⃗ A. 2P ⃗⃗ B. Magnitude of 2P

❖

SOLUTIONS 1. A. 2m = 2(6,2) = (12,4) B.

1 m 2

1

= 2(6,2) = (3,2)

C. −3𝑚 = −3(6,2) = (−18,−6) ⃗⃗ = 2 (3) 2. A. 2P 4 6 =( ) 8 ⃗⃗⃗⃗⃗| = √𝑥 2 + 𝑦 2 B. |2P = √62 + 82 = √36 + 64 = √100 = 10

Vectors

289

REVIEW QUESTIONS ⃗⃗⃗⃗⃗⃗ = (4) and B is the point (1,6), find the coordinates of A. 1. Given that AB 2 6 2. The vector YZ = ( ). Given that point Y is (2,1), find the coordinate of point Z. 2 ⃗⃗⃗⃗⃗⃗ is a column vector. 3. The points P and Q have coordinates (−3,1) and (2,4) respectively. Express PQ 4. Given that 𝑘⃗⃗ = (5,−3), find A. 3k B.

2 k 3

C. −2𝑘 4 5. Given that ⃗A⃗ = ( ) 3 ⃗ ⃗ A. 2A ⃗⃗ B. Magnitude of 2A ⃗⃗ = ( 6), find |A ⃗⃗| 6. Given that A 8 ⃗⃗⃗⃗⃗⃗ = a and OB ⃗⃗⃗⃗⃗⃗ = b. Given that OC ⃗⃗⃗⃗⃗⃗ = a + b, draw a line indicating OC ⃗⃗⃗⃗⃗⃗. 7. In the diagram below, OA

8. Show that the points L(−2,−10), M(2,2) and N(5,11) are collinear. 9. Determine whether the points A(1,−3), B(2,−5) and C(−4,7) are collinear. AC 1 10. In the diagram below, ⃗⃗⃗⃗⃗⃗ OA = a, ⃗⃗⃗⃗⃗⃗ OB = b and = CB

2

A. Express in terms of a and/or b i. ⃗⃗⃗⃗⃗⃗ AB ii. ⃗⃗⃗⃗⃗⃗ AC iii. ⃗⃗⃗⃗⃗⃗ OC B.

290

1 Given that M is the midpoint of OC, show that ⃗⃗⃗⃗⃗⃗⃗ AM = 6 (b − 4a)

Vectors

11. In the diagram below, OABC is a parallelogram in which ⃗⃗⃗⃗⃗⃗ OA = a and ⃗⃗⃗⃗⃗⃗ AB = b. AC and OB meet at D such 1

that OD = DB. OC is produced to E, such that CE = 2 OC.

Express each of the following in terms of a and/or b ⃗⃗⃗⃗⃗⃗ A. OB ⃗⃗⃗⃗⃗⃗ B. AD ⃗⃗⃗⃗⃗⃗ C. BE 12. In the diagram below, ⃗⃗⃗⃗⃗⃗ OP = 2p, ⃗⃗⃗⃗⃗⃗ OQ = 4q and PX:XQ = 1:2

A.

B.

Express in terms of p and/or q i. ⃗⃗⃗⃗⃗⃗ PQ ii. ⃗⃗⃗⃗⃗ PX iii. ⃗⃗⃗⃗⃗⃗ OX ⃗⃗⃗⃗⃗⃗ = hOX ⃗⃗⃗⃗⃗⃗, show that CQ ⃗⃗⃗⃗⃗⃗ = 4(1 − ℎ)q − 4ℎp Given that OC 3 3

Vectors

291

SOLUTIONS 1. ⃗B⃗ − ⃗A⃗ = ⃗⃗⃗⃗⃗⃗ AB 𝑥 1 4 ( ) − (𝑦) = ( ) 6 2 1−x=4 −x=4−1 −x=3 x = −3 6−y=2 −y=2−6 − y = −4 y=4 A is ( 3,4) ⃗⃗ − Y ⃗⃗ = ⃗⃗⃗⃗⃗ 2. Z YZ 𝑥 2 6 (𝑦) − ( ) = ( ) 1 2 x−2=6 x=6+2 x=8 y − 1 = −2 y = −2 + 1 y = −1 D is (8, 1) 3.

⃗⃗⃗⃗⃗⃗ = Q ⃗⃗ − P ⃗⃗ PQ 2 3 =( )−( ) 4 1 2 ( 3) =( ) 4 1 5 =( ) 3 5 ) 3 3×5 =( ) 3 × ( 3) 15 =( ) 9 = (15,−9)

4. A. 3k = 3(

B.

2 k 3

2

=3( = (2 3

=( =

292

Vectors

5 ) 3 2 3

×5

) × ( 3) 10 3)

2

10 ( , 3

2)

5 ) 3 2×5 =( ) 2 × ( 3) 10 =( ) 6 = (−10,6)

4.

C.

−2k = −2(

5.

A.

4 ⃗⃗⃗⃗⃗⃗ 2A = 2( ) 3 2 × ( 4) =( ) 2 × ( 3) 8 =( ) 6

B.

⃗⃗⃗⃗⃗⃗| = √𝑥 2 + 𝑦 2 |2A = √( 8)2 + ( 6)2 = √64 + 36 = √100 = 10

6.

⃗⃗| = √𝑥 2 + 𝑦 2 |A = √( 6)2 + (8)2 = √36 + 64 = √100 = 10

7.

Vectors

293

8.

Find the gradient of 2 points L(−2,−10), M(2,2) 𝑦2 − 𝑦1

m= = = =

𝑥2 − 𝑥1

To find the gradient of a line refer to section 11.4

2 −(−10) 2−(−2) 2 + 10 2+2 12 4

=3

Determine whether the other point produce the same gradient M(2,2), N(5,11) m= = =

𝑦2 − 𝑦1 𝑥2 − 𝑥1 11 − 2 5−2 9 3

=3 The points L(−2,−10), M(2,2) and N(5,11) are collinear 9.

Find the gradient of 2 points A(1,−3), B(2,−5) m= = =

𝑦2 − 𝑦1 𝑥2 − 𝑥1 −5 −(−3) 2−1 −5 + 3 1

= −2 Determine whether the other point produce the same gradient B(2,−5), C(−4,7) m= = = =

𝑦2 − 𝑦1 𝑥2 − 𝑥1 7 −(−5) −4 − 2 7+5 −6 12 −6

= −2 The points A(1,−3), B(2,−5) and C(−4,7) are collinear

294

Vectors

10.

A.

i.

⃗⃗⃗⃗⃗⃗ AB = ⃗⃗⃗⃗⃗⃗ AO + ⃗⃗⃗⃗⃗⃗ OB = −a + b =b−a ⃗⃗⃗⃗⃗⃗ AC : ⃗⃗⃗⃗⃗⃗ AB = 1 : 3

ii.

1

⃗⃗⃗⃗⃗⃗ = AC

⃗⃗⃗⃗⃗⃗ AB

3 1

= (b − a) 3

⃗⃗⃗⃗⃗⃗ OC = ⃗⃗⃗⃗⃗⃗ OA + ⃗⃗⃗⃗⃗⃗ AC

ii.

1

=a+ =a+ 2𝑎

= B.

3

⃗⃗⃗⃗⃗⃗⃗ AM = ⃗⃗⃗⃗⃗⃗ AO +

= −a +

=

−

𝑎

3 3 𝑏

+

3

⃗⃗⃗⃗⃗⃗ OC

2 1 2𝑎

𝑏

( + 3) 2 3

= −a +

=−

1

(b − a)

3 𝑏

2𝑎

2𝑎 6

+

+

𝑏 6

𝑏

3 6 −4𝑎 + 𝑏 6

1

= (−4a + b) 6 1

= (b − 4a) 6

11.

A.

⃗⃗⃗⃗⃗⃗ OB = ⃗⃗⃗⃗⃗⃗ OA + ⃗⃗⃗⃗⃗⃗ AB =a+b

B.

⃗⃗⃗⃗⃗⃗ AD = ⃗⃗⃗⃗⃗⃗ AO +

1

⃗⃗⃗⃗⃗⃗ OB

2 1

= −a + (a + b) = −a + =− =

𝑏 2

𝑎 2

−

2 𝑎

+

𝑏

2 2 𝑏

+

2

𝑎 2

Vectors

295

C.

⃗⃗⃗⃗⃗⃗ BE = ⃗⃗⃗⃗⃗⃗ BC + ⃗⃗⃗⃗⃗ CE = −a + =

12.

A.

𝑏 2

i.

1 2

b

−a

⃗⃗⃗⃗⃗⃗ = PO ⃗⃗⃗⃗⃗⃗ + OQ ⃗⃗⃗⃗⃗⃗ PQ = −2p + 4q = 4q − 2p

ii. ⃗⃗⃗⃗⃗ PX : ⃗⃗⃗⃗⃗⃗ PQ = 1 : 3 ⃗⃗⃗⃗⃗ PX =

1 3 1

⃗⃗⃗⃗⃗⃗ PQ

= (4q − 2p) 3

⃗⃗⃗⃗⃗⃗ = OP ⃗⃗⃗⃗⃗⃗ + PX ⃗⃗⃗⃗⃗ iii. OX 1

= 2p + (4q − 2p) = 2p + =

4𝑝

+

3 4

3 4𝑞

3 4𝑞

−

2𝑝 3

3

= (p + q) 3

B.

⃗⃗⃗⃗⃗⃗ CQ = ⃗⃗⃗⃗⃗⃗ CO + ⃗⃗⃗⃗⃗⃗ OQ ⃗⃗⃗⃗⃗⃗ + 4p = −hOX 4𝑝

= −h( =−

3 4ℎ𝑝 3

= 4q − = 4(𝑞 = 4(1

296

Vectors

−

+

3 4ℎ𝑞

4ℎ𝑞 3

−

4𝑞

+ 4q

3

−

) + 4q

4ℎ𝑝 3

ℎ𝑞

4ℎ𝑝

ℎ

3 4ℎ𝑝

)− 3

− )q − 3

3

31

TRANSFORMATION

Introduction Transformation is the moving of an object from its original position to a new position. Objects are moved in various ways. Most objects are moved from one place to another without changing their size and shape such as airplanes and cars. Some objects move by rotating about an axis such as fun blades and doors. Some objects are reflected such as images in a mirror. Some objects are stretched such as shadow stretching due to different positions of the sun.

Specific outcomes This unit covers the transformation of objects. You will learn about different types of transformation. By the end of this unit, you will be able to: ❖ Translation: • Define translation • Perform translation of an object ❖ Reflection: • Define reflection • Perform reflection of an object ❖ Rotation: • Define rotation • Find the centre and angle of rotation • Perform rotation of an object ❖ Enlargement: • Define enlargement • Find the centre of enlargement • Find the scale factor • Perform enlargement of an object

Transformations

297

TRANSFORMATIONS 31.1 TRANSFORMATION • Transformation is the moving of an object from its original position to a new position. • An object in its initial position is called a preimage. • An object changes its position or its size into an image. • There are various types of transformation which include translation, reflection, rotation and enlargement.

31.2 TRANSLATION • Translation is a type of transformation in which every point of an object is moved in the same direction and distance. • It is also referred to as slide. • The size and shape of an object and image are the same, only the position differs.

• •

The above object moved 7 units in the x-direction and 3 units in the y-direction. The movement can be written in the rule of translation or vector column.

1. RULE OF TRANSLATION • Rule of translation is written in the form of (𝑥, 𝑦) → (𝑥 + 𝑎, 𝑦 + 𝑏), where a are units moved in the x-axis and b are units moved in the y-axis. • When a is positive, an object moves to the right. When a is negative, an object moves to the left. • When b is positive an object moves upward. When b is negative an object moves downward. • The above translation can be written as (𝑥, 𝑦) → (𝑥 + 8, 𝑦 + 3)

298

Transformations

2.

VECTOR COLUMN • • • •

𝑥 Translation in vector column is written in the form of (𝑦), where x are units moved in the x-axis and y are units moved in the y-axis. When a is positive then an object moves to the right. When a is negative an object moves to the left. When b is positive an object moves upward. When b is negative an object moves downward. 8 The above translation can be written as ( ). 3

31.3 REFLECTION • Reflection is a type of transformation in which all points of an object are flipped on a line of reflection. • A line of reflection is midway between an object and an image. • Each point of an object has an equal distance from the line of reflection as the image. • The shape and size of an image are equal to the object, but it is laterally inverted.

❖

HOW TO PERFORM A REFLECTION 1. REFLECTION OVER THE X-AXIS • When the line of reflection is the x-axis, the x-coordinates remain the same but ycoordinates acquire the opposite sign.

Transformations

299

2.

300

REFLECTION OVER THE Y-AXIS • When the line of reflection is the y-axis, the y-coordinates remain the same but the x-coordinates acquire the opposite sign.

Transformations

3.

REFLECTION OVER Y = X • When the line of reflection is y = x, the x-coordinates and y-coordinates are interchanged.

31.4 ROTATION • Rotation is a type of transformation in which an object is rotated about a fixed point. • A fixed point at which rotation occurs is called the centre of rotation. • The direction of rotation can either be clockwise or anticlockwise. • The amount of rotation made by an object is called the angle of rotation. • The shape and size of an image are equal to the object but turned in a different direction.

•

For any rotation, find the centre of rotation, the direction of rotation and the amount of rotation

Transformations

301

1.

HOW TO FIND THE CENTRE OF ROTATION i. Join any two corresponding points with a straight line. ii. Draw a perpendicular bisector of each line. iii. The point of intersection is the centre of rotation.

2.

DIRECTION OF ROTATION • The direction of rotation is either clockwise or anticlockwise. • When the direction of rotation is clockwise the angle is described as negative. • When the direction of rotation is anticlockwise the angle is described as positive.

3.

ANGLE OF ROTATION • The angle of rotation is obtained by measuring the angle between two corresponding lines at the centre of rotation. • The sign of the angle depends on the direction of rotation. • When the rotation is clockwise the sign is negative. • When the rotation is anticlockwise the sign is positive.

31.5 ENLARGEMENT • Enlargement is a type of transformation in which an object is enlarged. • An image has the same shape as an object but with a different size and position. • For an object to be enlarged, a centre of enlargement and a scale factor is needed.

302

Transformations

❖

CENTRE OF ENLARGEMENT • The centre of enlargement is a fixed point where enlargement occurs. • To find the centre of enlargement, join corresponding points of an image and object. • The point of intersection is the centre of intersection.

❖

SCALE FACTOR • A scale factor is a scaling number of an object into an image. • It describes the size of enlargement of an object into an image. • A scale factor is found by the following methods. Scale factor =

distance from a point of an image from the centre of enlargement distance of corresponding point of an object from the centre of enlargement

Or Scale factor =

❖

distance between two points on an image distance between corresponding points on an object

EXAMPLES

scale factor =

8 4

=2

scale factor =

2 1

=2 Transformations

303

REVIEW QUESTIONS 1. A is a point (4,2). Find the coordinates of the image of point A under A. a reflection in the line y = x 3 B. a translation with vector T = ( ) 3 2. The diagram below shows two triangles A and B. Describe fully the transformation which maps A into B

3. In the diagram below triangle P is mapped onto triangle Q by a single transformation. Describe fully the transformation.

304

Transformations

4. The diagram below shows two shapes X and Y. Describe fully the transformation which maps X into Y.

5. In the diagram below triangle J is mapped onto triangle K by a single transformation. Describe fully the transformation.

6. In the diagram below, triangle A is mapped onto triangle B by transformation X followed by transformation Y. Describe fully the transformations X and Y

Transformations

305

7. Triangle P has vertices (2,2), (3,1) and (3,2). Triangle Q has vertices (−2,2), (−3,1) and (−3,2) A. Using a scale of 2cm to represent 1 unit on each axis, draw axes for values of x and y in the range −4 ≤ x ≤ 4 and −4 ≤ y ≤ 6. Draw and label triangles P and Q. B. Describe fully a single transformation that maps triangle P onto Q. C. Triangle R is the image of triangle P after a rotation of 180o about the origin. Draw triangle R. D. Triangle P is mapped onto S with coordinates (1,−2), (2,−2) and (2,−3). Describe fully a single transformation that maps triangle P onto triangle S. 1 0 E. A transformation with matrix ( ) maps triangle P onto triangle T. Draw and label triangle T and 0 2.5 name this transformation. 8. The diagram below shows triangle A, B, C, D and E

A. Triangle A is mapped onto triangle B by a single transformation. Describe fully this transformation. B. Triangle A is mapped onto triangle C by an enlargement. Find the i. centre of enlargement ii. scale factor of an enlargement C. Triangle A is mapped onto triangle D by a single transformation. Find i. matrix representing this transformation ii. the area scale factor of the transformation D. Triangle A is mapped onto triangle E by a single transformation. Describe fully this transformation.

306

Transformations

SOLUTIONS 1. A.

2. 3. 4. 5. 6.

Coordinates of image (−4,3) 4 3 B. A + T = ( ) + ( ) 2 3 4 3 =( ) 2+3 1 =( ) 5 Coordinates of image (1,5) 6 Translation ( ) 2 Reflection in the line y = x Rotation 90o about point (−1,2) Enlargement, scale factor 3, centre of enlargement (0,0) X is reflection over y-axis 7 Y is translation ( ) 3

7. A.

307

7.

Reflection over the y-axis 1 D. Translation ( ) 4 E. Stretch transformation B.

8. A. Rotation 90o about point (0,−2) B. i. Centre of enlargement (0,0) ii. Scale factor = = C.

i.

distance between two points on image distance between corresponding two points on object 4 2

=2 1 0 T(E) = ( ) 0 2

ii. Area scale factor (asf) =

area of image area of object 1

Area of image = 2 bh 1

=2×2×2 =2

1

Area of object = 2 bh 1

=2×2×1 =1

2

Area scale factor = 1 = 1 D.

308

6

Shear, invariant line y-axis, shear factor = 3 = 2

32

STATISTICS

Introduction The world is full of random information. To make sense, data has to be collected, analysed and interpreted. Statistics is a field of mathematics that is responsible for this role. Statistics is the study of collection, analysis, presentation and interpretation of data. Statistics is important in our everyday lives. To come up with a better plan, correct data has to be collected and interpreted. Weather forecast and news reports you watch or read use statistics. Statistics is used in almost all fields.

Specific outcomes The unit introduces you to statistics. By the end of this unit, you will be able to: ❖ Define statistics ❖ Describe qualitative and quantitative data ❖ Represent data using: • Bar graph • Histogram • Pie chart • Line graph ❖ Measure the following central tendencies: • Mode • Median • Mean ❖ Calculate the standard deviation ❖ Construct and represent a cumulative frequency graphically

Statistics

309

STATISTICS 32.1 STATISTICS • Statistics is the study of collection, analysis, presentation and interpretation of data. 32.2 DATA • Data are a collection of information on a particular subject. • Data can be numbers or words that have been observed or recorded. ❖ TYPES OF DATA 1. QUALITATIVE DATA • Qualitative data is a collection of information that cannot be expressed as numbers. • The information is descriptive. • For example, “it’s cloudy this week”. “He wore a blue T-shirt”. • The above examples cannot be expressed as numbers. The first example is a qualitative data describing the weather. The second example is a qualitative data describing the colour of a T-shirt. 2. QUANTITATIVE DATA • Quantitative data is a collection of information that can be expressed as numbers. • For example, it has rained twice this week. He is 1.6 metres tall. • The above examples can be expressed as numbers. The first example is a quantitative data describing the number it rained in a week. The second example is a quantitative data describing the height of a man.

32.3 DATA REPRESENTATION • Data representation is a way of displaying data in an organised manner. • Data is usually displayed in the form of graphs. ❖ TYPES OF DATA REPRESENTATION 1. BAR GRAPH • A bar graph is used to represent and compare data in discrete categories. • The x-axis consists of categories of information while the y-axis is the frequency of each category. • The height of each bar indicates the amount of information in a category. • The bars have equal width and do not touch each other. ❖ EXAMPLE 1. The following table shows the favourite sports of 200 students of a certain school. Represent the data as a bar graph. Types of sports Number of students

310

Statistics

Volleyball Football

60 90

Netball Basketball

20 30

❖

SOLUTION

2.

HISTOGRAM • Histogram is used to represent and compare continuous data. • Some data such as weight in a particular population produces a range of values. For example, in a class 10 students weigh between 40kg to 60kg, 15 students weigh 60kg to 80kg. • Histogram is similar to bar graph but it displays data in ranges instead of discrete categories. • The x-axis represents the range of data while the y-axis is the frequency of data. • The height of bars indicates the amount of information in each range. • The bars have equal width and are connected to each other. ❖ EXAMPLE 1. The table below shows the mass of female students in a class. Draw a histogram to represent these results. Mass (kg)

Frequency

20 < m ≤ 30

1

30 < m ≤ 40

4

40 < m ≤ 50

8

50 < m ≤ 60

12

60 < m ≤ 70

6

70 < m ≤ 80

2

80 < m ≤ 100

1

Statistics

311

❖

SOLUTION

3.

PIE CHART • Pie chart display data in a circular path. • It is used to indicate the relative sizes of the whole data. • Each slice is proportional to the quantity of a category and represents a portion of the whole. ❖ HOW TO DRAW A PIE CHART 1. Draw a circle that represents the entire data. 2. Calculate what proportion of 360o each category corresponds to using the following. Angular size =

frequency total

× 360o

3. Draw a slice corresponding to the angular size. 4. Check that the total degrees of different slices add up to 360o. ❖ EXAMPLE 1. A football team plays 36 matches in a season. The results are as follows. Win

Draw

lose

26

7

3

A. Calculate the angle one match represents in a pie chart. B. Calculate the angle of each category i. Win ii. Draw iii. Lose C. Draw a pie chart showing the proportion of each category.

312

Statistics

❖

SOLUTIONS 1. A. Angular size of one match = =

frequency total 1 36

× 360o

× 360o

= 10o frequency

B. i. Angular size of win = =

total 26 36

× 360o

× 360o

= 260o frequency

ii. Angular size of draw =

total

=

7 36

× 360o

× 360o

= 70o iii. Angular size of lose = =

frequency total 3 36

× 360o

× 360o

= 30o C.

4.

LINE GRAPH • A line graph is used to represent data that changes with time. • The frequencies are plotted using points and straight lines connecting these points. ❖ EXAMPLE • The following table shows the average monthly rainfall in Lusaka. Represent the following data in a line graph. Months Rainfall (inch)

Jan Feb Mar Apr May Jun 10 8 5 2 0 0

Jul 0

Aug Sep 0 0

Oct 1

Nov Dec 4 6 Statistics

313

❖ SOLUTIONS

32.4 MEASURES OF CENTRAL TENDENCY • The measure of central tendency is simply the measure of the average set of data. • They indicate the main tendency of the set of data. • For example, if the set of data is 2, 2, 2, 2, 3, 2 it can be seen that the main tendency of data is 2 or close to 2. • The common measures of central tendency are mode, mean and median. 1. MODE • Mode is a data value that occurs most frequently. • It is found by counting how many times each data value occurs. • The value that occurs the most is the mode. • Examples: Find the mode of the following A. 2, 2, 2, 2, 3, 2 B. 2, 1, 3, 4, 2, 3, 3, 4, 8, 1, 9, 3, 2 C. 4, 4, 4, 4, 4, • Solutions: A. 2 B. 3 C. 4

A. 2 occurs the most: 5 times B. 3 occurs the most: 4 times C. 4 is the only data value, ∴ is the mode

2. MEDIAN • Median is a data value in the middle when all the data values are arranged in ascending or descending order. • How to find the median: i. Rearrange data values in ascending or descending order. ii. Count all the data values. iii. Find the data value in the middle. iv. When there are two middle values add them and divide by 2. 314

Statistics

•

Examples Find the median of the following A. 3, 10, 7, 5, 9 B. 3, 10, 7, 5, 9, 6

•

Solutions A. 3, 5, 7, 9, 10 median = 7 B.

Rearrange data values in ascending or descending order. Find the data value in the middle.

3, 5, 6, 7, 9, 10 median =

Rearrange data values in ascending or descending order. Find the data value in the middle. The middle values are 6 and 7. Therefore, add then divide them by 2.

6+7 2

= 6.5

3.

MEAN • Mean is the average of the data values. • It is denoted by 𝑥̅ . • Mean is calculated by summing all values and divide by the number of values • Formula: 𝑥̅ =

sum of all values number of values

•

Examples: Find the mean of the following A. 3, 10, 7, 5, 9, 6, 2 B. 8, 15, 1, 25, 12

•

Solutions A. 𝑥̅ = =

sum of all values number of values 3 + 10 + 7 + 5 + 9 + 6 + 2 7

=6 B. 𝑥̅ = =

sum of all values number of values 8 + 15 + 1 + 24 + 12 5

= 12

Statistics

315

32.5 CUMULATIVE FREQUENCY • Cumulative frequency is the running total of all frequencies. • It is used to determine how many data values lie above or below a particular value in a data set. • It can be calculated by adding each frequency to the total frequencies of all data values before it in the data set. • The last value for the cumulative frequency will always be equal to the total number of data values. • A cumulative frequency diagram is drawn by plotting the upper-class boundary. ❖ EXAMPLES 1. The table below shows how long 100 took to complete a puzzle. Time (min) 0 < t ≤ 10 10 < t ≤ 15 15 < t ≤ 20 20 < t ≤ 25 25 < t ≤ 30

Frequency 15 35 30 15 5

A. Construct a cumulative frequency table. B. Draw a cumulative frequency diagram. 2. Speeding is one of the major risk factors for high accident rates. The table below shows the speed and the number of accidents associated. Speed (km/h) 60 < x ≤ 70 70 < x ≤ 80 80 < x ≤ 90 90 < x ≤ 100 100 < x ≤ 110 110 < x ≤ 120 120 < x ≤ 130 130 < x ≤ 140 140 < x ≤ 150 150 < x ≤ 160

Frequency 3 2 6 40 50 30 15 12 3 2

A. Construct a cumulative frequency table. B. Draw a cumulative frequency diagram. ❖ SOLUTIONS 1. A. Time (min) ≤ 10 ≤ 15 ≤ 20 ≤ 25 ≤ 30

316

Statistics

Cumulative frequency 15 50 80 95 100

B.

2.

A.

Speed (km/h) ≤ 70 ≤ 80 ≤ 90 ≤ 100 ≤ 110 ≤ 120 ≤ 130 ≤ 140 ≤ 150 ≤ 160

Frequency 3 5 11 51 101 131 146 158 161 163

Statistics

317

B.

❖

MEDIAN FROM CUMULATIVE FREQUENCY DIAGRAM • To find the median from a cumulative frequency diagram, draw a line from the middle value of the cumulative frequency to the graph. • The corresponding value is the median.

❖

LOWER QUARTILE • The lower quartile is defined as the median of the lower half of the data. • It is one-quarter of the data when all data values are arranged in order. • It is denoted by Q1.

❖

UPPER QUARTILE • The upper quartile is defined as the median of the upper half of the data. • It is three-quarters of the data when all data values are arranged in order. • It is denoted by Q3.

❖

INTERQUARTILE • Interquartile is the difference between the upper quartile and lower quartile. • It shows how the data is spread.

318

Statistics

❖

EXAMPLE 1. The following data shows the mass of 60 students Mass (kg) 40 < m ≤ 45 45 < m ≤ 50 50 < m ≤ 55 55 < m ≤ 60 60 < m ≤ 65 65 < m ≤ 70 70 < m ≤ 75 75 < m ≤ 80 A. B. C. D. E. F.

Frequency 3 5 8 12 14 9 7 2

Construct a cumulative frequency table. Draw a cumulative frequency diagram. Find the median. Find the lower quartile. Find the upper quartile. Find the interquartile.

Statistics

319

❖

SOLUTION 1. A.

Mass (kg) ≤ 45 ≤ 50 ≤ 55 ≤ 60 ≤ 65 ≤ 70 ≤ 75 ≤ 80

B.

C. Median = 61 kg D. Q1 = 55 kg

320

E.

Q3 = 66.5 kg

F.

Interquartile = Q3 − Q1 = 66.5 − 55 = 11.5 kg

Statistics

Frequency 3 8 16 28 42 51 58 60

REVIEW QUESTIONS 1. The table below shows the amount of money spent by 100 learners at a school on a particular day. Amount of kwacha Frequency

0