Oswaal CBSE Question Bank Class 9 Mathematics, Chapterwise and Topicwise Solved Papers For 2025 Exams 9789359589589, 9359589586

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20th EDITION

I SB N SYLLABUS COVERED

YEAR 2024-25 “9789359589589”

CENTRAL BOARD OF SECONDARY EDUCATION DELHI

PUBLISHED BY OSWAAL BOOKS & LEARNING PVT. LTD.

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RESERVED

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DISC L AIM ER This book is published by Oswaal Books and Learning Pvt Ltd (“Publisher”) and is intended solely for educational use, to enable students to practice for examinations/tests and reference. The contents of this book primarily comprise a collection of questions that have been sourced from previous examination papers. Any practice questions and/or notes included by the Publisher are formulated by placing reliance on previous question papers and are in keeping with the format/pattern/guidelines applicable to such papers. The Publisher expressly disclaims any liability for the use of, or references to, any terms or terminology in the book, which may not be considered appropriate or may be considered offensive, in light of societal changes. Further, the contents of this book, including references to any persons, corporations, brands, political parties, incidents, historical events and/or terminology within the book, if any, are not intended to be offensive, and/or to hurt, insult or defame any person (whether living or dead), entity, gender, caste, religion, race, etc. and any interpretation to this effect is unintended and purely incidental. While we try to keep our publications as updated and accurate as possible, human error may creep in. We expressly disclaim liability for errors and/or omissions in the content, if any, and further disclaim any liability for any loss or damages in connection with the use of the book and reference to its contents”.

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Contents l

Latest CBSE Syllabus

6 - 11

In each chapter, for better comprehension, questions have been categorized according to the typology issued by CBSE as follows : R - Remembering, U - Understanding,

A - Analysing, AP - Applying,

C - Creating,

E - Evaluating.

Unit I : Number Systems 1. Real Numbers l Self Assessment Paper-1

1 - 25 26 - 27

2. Polynomials 3. Linear Equations in Two Variables l Artificial Intelligence l Self Assessment Paper-2

28 - 43 44 - 51 52 - 52 53 - 54



55 - 63 64 - 65 66 - 68

Unit II : Algebra

Unit III : Coordinate Geometry 4. Coordinate Geometry

l Artificial Intelligence l Self Assessment Paper-3

Unit IV : Geometry

5. Introduction to Euclid's Geometry 6. Lines and Angles 7. Triangles 8. Quadrilaterals l Artificial Intelligence 9. Circles l Artificial Intelligence l Self Assessment Paper-4

69 - 79 80 - 93 94 - 108 109 - 121 121 - 122 123 - 138 139 - 139 140 - 141



10. Areas 11. Surface Areas and Volumes l Artificial Intelligence l Self Assessment Paper-5

142 - 154 155 - 166 167 - 168 169 - 170



171 - 180 180 - 182 183 - 184

Unit V : Mensuration

Unit VI : Statistics 12. Statistics

l Artificial Intelligence l Self Assessment Paper-6

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How to use this Book Chapter Navigation Tools

Syllabus

Topicwise & Conceptwise Segregation

Revision Notes

Prescribed by CBSE

For Focused & Systematic Study

Chapter Summary Developed by Oswaal Experts

Latest Typologies of Questions Objective/SA/LA/ Case-based Questions

Mind Maps & Mnemonics

QR Codes

For better retention of concepts

For Concept Videos

Previous Years’ Board Papers

Academically Important Questions

Practice Questions

To decode the paper pattern

To Look out for Highly Expected Questions for the upcoming exams

For Better Practice

Competency based Questions

Self Assessment Papers

Commonly Made Errors & Answering Tips

To demonstrate application based knowledge of Concepts

For Unitwise Assessment

To write perfect Answers

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Preface Elevate Your Performance, Surpassing the Past Elevate Your Performance, Surpassing the Past Get ready for another epic journey through the academic wonders of the academic year 2024-2025 with your trustworthy companion—Oswaal Books! Remember last year’s triumphs? Well, buckle up because we are about to make this year even more awesome! As the legendary dancer Martha Graham once said, “Practice means to perform, repeatedly in the face of all obstacles, some act of vision, of faith, of desire.” We have taken this wisdom to heart and packed it into our brand-new Question Banks for 2024-2025. They are a magical mix of CBSE Board Updates, cool questions from the past, and specially crafted ones tailored to the Latest Typologies. Oh, and did we mention the fantastic Learning Resources that come with them?

What makes these Question Banks truly exceptional? • 100% Updated Syllabus & With Latest Questions Typologies: We have got you covered with the latest and 100% updated curriculum • Timed Revision with Topic-wise Revision Notes, Smart Mind Maps & Mnemonics: Study smart, not hard! • Extensive Practice with 1000+ Questions & SAS Questions (Sri Aurobindo Society): To give you 1000+ chances to become a champ!

• Concept Clarity with 500+ Concepts & Concept Videos: For you to learn the cool way—with videos and mind-blowing concepts



• NEP 2020 Compliance with Competency-Based Questions & Artificial Intelligence: For you to be on the cutting edge of the coolest educational trends.

If you are looking to conquer every study challenge, these Question Banks are your secret weapon. It is like having a superhero ally for your exams! So, let’s kick off this exciting journey, fill those learning gaps, and rock the year with ease and confidence. Big shoutout to our superhero team—the Oswaal Editorial Board! They’re the brains behind this incredible resource, working day and night just for you. And a massive thank you to you, our fellow Students, Parents & Teachers for your awesome inputs that make this book one-of-a-kind. Wishing you all the best, superheroes-in-the-making! Strive for greatness! Team Oswaal Books

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Syllabus MATHEMATICS

COURSE STRUCTURE Class - IX (Code No. 041) Units

Unit Name

Marks

NUMBER SYSTEMS

10

II

ALGEBRA

20

III

COORDINATE GEOMETRY

04

IV

GEOMETRY

27

V

MENSURATION

13

VI

STATISTICS

06

I

Total

80

UNIT I : NUMBER SYSTEMS 1. REAL NUMBERS

(18) Periods

1. Review of representation of natural numbers, integers, and rational numbers on the number line. Rational numbers as recurring/ terminating decimals. Operations on real numbers.



2. Examples of non-recurring/non-terminating decimals. Existence of non–rational numbers (irrational numbers) such as

2 , 3 and their representation on the number line. Explaining that every real

number is represented by a unique point on the number line and conversely, viz. every point on the number line represents a unique real number.

3. Definition of nth root of a real number.

1 1 4. Rationalization (with precise meaning) of real numbers of the type a +b x and x + y (and their



combinations) where x and y are natural number and a and b are integers.

(6)

Syllabus

5. Recall of laws of exponents with integral powers. Rational exponents with positive real bases (to be done by particular cases, allowing learner to arrive at the general laws.)

UNIT II : ALGEBRA 1. POLYNOMIALS

(26) Periods

Definition of a polynomial in one variable, with examples and counter examples. Coefficients of a polynomial, terms of a polynomial and zero polynomial. Degree of a polynomial. Constant, linear, quadratic and cubic polynomials. Monomials, binomials, trinomials. Factors and multiples. Zeros of a polynomial. Motivate and State the Remainder Theorem with examples. Statement and proof of the Factor Theorem. Factorization of ax2 + bx + c, a ≠ 0 where a, b and c are real numbers, and of cubic polynomials using the Factor Theorem.



Recall of algebraic expressions and identities. Verification of identities :



(x + y + z)2 = x2 + y2 + z2+2xy + 2yz + 2zx



(x ± y)3 = x3 ± y3 ± 3xy (x ± y)

x3 ± y3 = (x ± y) (x2

xy + y2)

x3 + y3 + z3 – 3xyz = (x + y + z) (x2 + y2 + z2 – xy – yz – zx)

and their use in factorization of polynomials

2. LINEAR EQUATIONS IN TWO VARIABLES

(16) Periods



Recall of linear equations in one variable. Introduction to the equation in two variables.



Focus on linear equations of the type ax + by + c = 0. Explain that a linear equation in two variables has infinitely many solutions and justify their being written as ordered pairs of real numbers, plotting them and showing that they lie on a line.

UNIT III : COORDINATE GEOMETRY

COORDINATE GEOMETRY



The Cartesian plane, coordinates of a point, names and terms associated with the coordinate plane, notations.

(7)

(7) Periods

Syllabus UNIT IV : GEOMETRY 1. INTRODUCTION TO EUCLID'S GEOMETRY

(7) Periods

History - Geometry in India and Euclid's geometry. Euclid's method of formalizing observed phenomenon into rigorous Mathematics with definitions, common/obvious notions, axioms/postulates and theorems. The five postulates of Euclid. Showing the relationship between axiom and theorem, for example:



(Axiom) 1. Given two distinct points, there exists one and only one line through them.



(Theorem) 2. (Prove) Two distinct lines cannot have more than one point in common.

2. LINES AND ANGLES

(15) Periods

1. (Motivate) If a ray stands on a line, then the sum of the two adjacent angles so formed is 180° and the converse.



2. (Prove) If two lines intersect, vertically opposite angles are equal.



3. (Motivate) Lines which are parallel to a given line are parallel.

3. TRIANGLES

(22) Periods

1. (Motivate) Two triangles are congruent if any two sides and the included angle of one triangle is equal to any two sides and the included angle of the other triangle (SAS Congruence).



2. (Prove) Two triangles are congruent if any two angles and the included side of one triangle is equal to any two angles and the included side of the other triangle (ASA Congruence).



3. (Motivate) Two triangles are congruent if the three sides of one triangle are equal to three sides of the other triangle (SSS Congruence).



4. (Motivate) Two right triangles are congruent if the hypotenuse and a side of one triangle are equal (respectively) to the hypotenuse and a side of the other triangle. (RHS Congruence)



5. (Prove) The angles opposite to equal sides of a triangle are equal.



6. (Motivate) The sides opposite to equal angles of a triangle are equal.

(8)

Syllabus 4. QUADRILATERALS

(13) Periods



1. (Prove) The diagonal divides a parallelogram into two congruent triangles.



2. (Motivate) In a parallelogram opposite sides are equal, and conversely.



3. (Motivate) In a parallelogram opposite angles are equal, and conversely.



4. (Motivate) A quadrilateral is a parallelogram if a pair of its opposite sides is parallel and equal.



5. (Motivate) In a parallelogram, the diagonals bisect each other and conversely.



6. (Motivate) In a triangle, the line segment joining the mid points of any two sides is parallel to the third side and in half of it and (motivate) its converse.

5. CIRCLES

(17) Periods



1. (Prove) Equal chords of a circle subtend equal angles at the center and (motivate) its converse.



2. (Motivate) The perpendicular from the center of a circle to a chord bisects the chord and conversely, the line drawn through the center of a circle to bisect a chord is perpendicular to the chord.



3. (Motivate) Equal chords of a circle (or of congruent circles) are equidistant from the center (or their respective centers) and conversely.



4. (Prove) The angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle.



5. (Motivate) Angles in the same segment of a circle are equal.



6. (Motivate) If a line segment joining two points subtends equal angle at two other points lying on the same side of the line containing the segment, the four points lie on a circle.



7. (Motivate) The sum of either of the pair of the opposite angles of a cyclic quadrilateral is 180° and its converse.

(9)

Syllabus UNIT V : MENSURATION 1. AREAS

Area of a triangle using Heron's formula (without proof)

2. SURFACE AREAS AND VOLUMES

(5) Periods (17) Periods

Surface areas and volumes of spheres (including hemispheres) and right circular cones.

UNIT VI : STATISTICS

STATISTICS



Bar graphs, histograms (with varying base lengths), and frequency polygons.

( 10 )

(15) Periods

Syllabus MATHEMATICS QUESTION PAPER DESIGN Time : 3 hrs

Max. Marks : 80

S. No.

Typology of Questions

1.

Remembering : Exhibit memory of previously learned material by recalling facts, terms, basic concepts, and answers.

Total Marks

% Weightage (approx)

43

54

19

24

18

22

80

100

Understanding : Demonstrate understanding of facts and ideas by organizing, comparing, translating, interpreting, giving descriptions, and stating main ideas 2.

Applying : Solve problems to new situations by applying acquired knowledge, facts, techniques and rules in a different way.

3.

Analysing : Examine and break information into parts by identifying motives or causes. Make inferences and find evidence to support generalizations Evaluating : Present and defend opinions by making judgments about information, validity of ideas, or quality of work based on a set of criteria. Creating : Compile information together in a different way by combining elements in a new pattern or proposing alternative solutions Total

INTERNAL ASSESSMENT Pen Paper Test and Multiple Assessment (5+5)

Portfolio Lab Practical (Lab activities to be done from the prescribed books)

20 MARKS 10 Marks

05 Marks 05 Marks

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PPositive ositive A ffirmations Affirmations An affirmation is a positive statement or phrase that individuals repeat to themselves with the intention of fostering a positive mindset, self-confidence, or personal development. The practice involves affirming desired beliefs or qualities, often in the present tense, to reinforce a positive self-image and encourage a more optimistic outlook.

“ EMBRACE YOUR UNIQUENESS, NOURISH SELF-CARE, AND PRESS FORWARD RELENTLESSLY. TO ACHIEVE FLIGHT, UNLOAD THE BAGGAGE THAT SLOWS YOUR ASCENT."

I strongly trust in my vision and work deligently towards making it a reality.

I am capable of learning and growing everyday. I am not defined by grades & test scores. My true worth lies in my character, compassion and positive impact I have on others. I celebrate my quirks, for they make me beautifully unique.

I embrace my mistakes, learn from them and use them to fuel my growth & resilience.

C

E NG A H

My voice matters and I believe in the power of my words to shape the world around me.

I am not the product of circumstances, I am the product of my own choices.

I face challenges with courage and determination.

I believe that the twists & turns of my life lead me to an extraordinary destinations.

I celebrate the achievements of others without diminishning my journey.

( 13 )

Be mindful. Be grateful. Be positive. Be true. Be kind Three things that make you special

Three people you are grateful for and why

Three simple things you are grateful for

A challenging experience that made you stronger

Three ways to inject gratitude into a current challenge

Describe the last time you did something nice for someone

A fear you have overcome

Three activities you enjoy most and why

What made you smile today?

Three things you love about your family

What is your favorite place, and why?

Three things you love most about yourself

The last time you were overcome with joy

A risk you are grateful you took and why

Three everyday items you are grateful for

Three songs that bring you joy

What skill do you have that you are grateful for and why?

One luxury you are thankful for

Describe a rejection you are grateful for

Three things about your body you are grateful for

What are you most grateful for in your daily life?

Three things you are grateful for about where you live

Three items in your home you are grateful for

Say thank you to someone

Something in nature you are grateful for

A person in your past you are grateful for

Something at school you’re grateful for

What is your proudest accomplishment?

Three things you want to manifest

Describe the last time you laughed so hard you cried

( 14 )

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( 16 )

UNIT-I

NUMBER SYSTEMS

Study Time: Maximum time: 3:15 Hrs Maximum questions: 112

CHAPTER

1



REAL NUMBERS Review of representation of natural numbers, integers and rational numbers on the number line. Rational numbers as recurring/terminating decimals. Operations on real numbers.

Syllabus

Examples of non-recurring/non-terminating decimals. Existence of non-rational numbers (irrational numbers) such as 2 , 3 and their representation on the number line. Explaining that every real number is represented by unique point on the number line and conversely viz. every point on the number line represents a unique real number and Definition of nth root of a real number. Rationalization (with precise meaning) of real numbers of the type where x and y are natural numbers and a and b are integers.

1 and a+b x

1 (and their combinations) x+ y

Recall of laws of exponents with integral powers. Rational exponents with positive real bases (to be done by particular cases, allowing learner to arrive at the general laws).

List of Topics Topic-1: Rational Numbers



Topic-1 Rational Numbers

Page No. 1

Topic-2 : Irrational Numbers 

Page No. 7

Topic-3 : nth Root of a Real Number 

Revision Notes

Page No. 9

Topic-4 : Laws of Exponents with Integral Powers Page No. 12 Topic-5 : Rationalization of Real Numbers Page No. 16

 Rational Number : A number ’r’ is called a rational number, if it can be written in the form p/q, where p and q are integers and q ≠ 0, denoted by ‘Q’. 1 3 4 2 e.g., , , , − etc. are all rational numbers. 2 4 5 3 Symbolically,

p  Q =  q , q ≠ 0 and p, q ∈ I   

 Decimal Expansion of Real Numbers : The decimal expansion of real number is used to represent a number on the number line. If the decimal expansion of a real number is either terminating or non-terminating (recurring), then the real number is called a rational number.  Cases in Rational Number: p Case 1 : When Remainder becomes Zero : Every rational number , (q ≠ 0) can be expressed as q a decimal. On dividing p by q, when the remainder becomes zero, then the decimal is called a terminating decimal.

Scan to know more about this topic

Introduction of Rational and Irrational Numbers

2

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-IX



REAL NUMBERS

3

1 e.g., = 0.5 2 On dividing 1 by 2, we get value 0.5 i.e., remainder equal to zero, so 0.5 is a terminating decimal. P  Terminating decimal : If a rational number (¹ integer) can be expressed in the form n m P Î Z, (n, m) Î W, the 2 ×5 rational number will be a terminating decimal. 5 5 5 , is terminating decimal. For Example : = 3 8 2 × 50 8  Non-terminating decimal : If in a rational number

p , prime factors of q are other than 2 and 5, then the rational q

number is Non-terminating decimal. 4 4 4 For Example : = , is non terminating decimal. 45 3 2 × 2 0 × 51 45 Case 2 : When remainder never becomes Zero — A rational number expressed in the form of

p or division of q

p by q, when remainder never becomes zero and set of digits repeat periodically then the decimal is called nonterminating recurring or repeating decimal. It is denoted by the bar over it. 1 e.g., = 0.333.... = 0.3 3 On dividing 1 by 3, we get 3 again and again in the decimal part of the quotient i.e., remainder never becomes zero, so 0.3 is a non-terminating repeating decimal.  Every integer is a rational number.  There are infinitely many rational numbers between any two given rational numbers.  If x and y are any two rational numbers, then : (i) x + y is a rational number (ii) x – y is a rational number (iii) x × y is a rational number (iv) x ÷ y is a rational number, (y ≠ 0).

Example 1 Express 0.5 in the form of

p . q

Solution: Step I : Assume the given decimal expansion as x and count the number of digits which are repeated. Let x = 0.5 or x = 0.555...... ...(i) Here, 1 digit is repeated that is, 5. Step II : Multiply both sides by 10 (because one digit is repeating)

On multiplying eqn. (i) by 10, we get 10x = 5.555..... ...(ii) Step III : Subtracting eqn. (i) from (ii), we get 10x – x = 5.555....... – 0.555...... or 9x = 5 5 x= or 9 Hence,

0.5 =

5 9

Mnemonics 1. We use R symbol. Think R as in ‘Real’. The quick way to remember real numbers is that they’re the numbers that truly exist and can be represented on number line. N : Naturally (Natural numbers) I : Involve (Integers) R : Relation (Rational Numbers) I : Insanity (Irrational Numbers) Rational Numbers These numbers can be expressed as a ratio of two integers hence the name rational numbers. Denoted by symbol ‘Q’ - as in quotient so ratio means on dividing we get quotient.

4

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-IX

SUBJECTIVE TYPE QUESTIONS Very Short Answer Type Questions (1 mark each) Q. 1. Find two rational numbers between 4 and 5. U [Board Term I, 2016] 4 5 × 5 and 5 = × 5 5 5 20 25 i.e., 4 = and 5 = 5 5 21 22 . The numbers are and 5 5 Sol.



1

[CBSE Marking Scheme, 2016]

Commonly Made Error

rational numbers, students do not convert the rational numbers into like fractions.

While inserting rational number between two rational number, students should convert the rational numbers into like fractions.

Q. 5. Write the sum of 0.3 and 0. 4 . 0.3 + 0.4 = (0.333.....) + (0.444.....) Sol.

Short Answer Type Questions-I (2 marks each)

while dividing a number by denomination of 10’s.

Q. 1. Is zero (0) a rational number ? Justify your answer. R [Board Term I, 2015] Sol. Yes, zero is a rational number. 0 0 0 Zero can be expressed as , etc, , 5 26 100

Students should remember while multiplying,

decimal is placed to the right counting the number of zeros and while dividing to the left as many zeros as are there.

Q. 3. Is

98 2

a rational number or not ?

which are in the form of integers and q ≠ 0. U



1 2 Q. 4. Insert three rational numbers between − and 3 1 − . U [Board Term I, 2016] 3 −

1 4 = − 3 12

and

−

2 8 = − 3 12

So three rational numbers are

−

5 6 7 and − . , − 12 12 12 

1

p where p and q are q 2

Q. 2. If 7x = 1, then find the decimal expansion of x.

98 98 Sol. = = 49 = 7 2 2 So, it is a rational number.

Sol.

U

= 0.777...... Let, x = 0.777....... 10x = 7.777....... or, 10x – x = (7.777.......) – (0.777.......) or, 9x = 7 7 or, x = 1 9

Students make an error while placing decimal

Answering Tip



While inserting rational number between two

Answering Tip

58 Q. 2. Find the decimal expansion of . 1000 R [Board Term I, 2015] 58 Sol. = 0.058 (Decimal point is shifted three places 1000 to the left.)  1





4 =





Commonly Made Error

1 Sol. x = 7



U [Board Term I, 2012]

0.142857 7 1.000000

\ x = 0.142857

)

–7 30 – 28 20 – 14 60 – 56 40 – 35 50 – 49 1

2

[CBSE Marking Scheme, 2012]



REAL NUMBERS

Q. 3. Express – 0.00875 in the form integers and q ¹ 0. Sol.

–0.00875 = -

p , where p and q are q

U [Board Term I, 2016]

875 100000



Hence, three rational numbers between



56 57 58 , , 77 77 77



5

5 9 and are: 7 11 1

Q. 2. Find six rational numbers between 3 and 4.

U [Board Term I, 2014] - 35 -7 = = 2 Sol. Let a = 3 and b = 4 4000 800 Here, we find six rational numbers, i.e., n = 6 [CBSE Marking Scheme, 2016] b−a 4−3 1 2157 So, d= = = Q. 4. Express in the decimal form and state n +1 6+1 7 625 1 22 whether it is terminating or not. 1st rational number = a + d = 3 + = ½ 7 7 R [Board Term I, 2012] Sol. 3.4512 2 23 2nd rational number = a + 2d = 3 + = ½ 625) 2157 1½ 7 7 –1875 3 24 2820 3rd rational number = a + 3d = 3 + = ½ 7 7 – 2500 4 25  3200 4th rational number = a + 4d = 3 + = ½ –3125 7 7 750 5 26 5th rational number = a + 5d = 3 + = ½ –625 7 7 1250 6 27 – 1250 6th rational number = a + 6d = 3 + = ½ × 7 7 2157 So, six rational numbers are: = 3.4512 (Terminating) ½ 625 22 23 24 25 26 27 , , , , and Q. 5. Find the rational numbers between 7 7 7 7 7 7 0.121221222122221 ..... and 0.141441444144441 .... in 1 1 p Q. 3. Find four rational numbers between and . the form, where p and q are integers and q ¹ 0. 6 5 q [Board Term I, 2012] U R [Board Term I, 2016] Sol. Since, LCM of 5 and 6 is 30. Sol. Two rational numbers between 0.1212212221 ... 1 1 5 5 5 25 \ = × = ½ × = and 0.1414414441 ... are 0.13 and 0.14. 30 5 150  6 6 5 13 14 1 1 6 6 5 30 i.e., and and = × = ½ × = 100 100 5 5 6 30 5 150  13 7 1 1 or, and . 2 Hence, four rational numbers between and 100 50 6 5 are : [CBSE Marking Scheme, 2016] 26 2627 2728 2829 29 ½+½+½+½ , ,, , ,and, 150 150 150 150 150 150 150  150 Short Answer Type

Questions-II

(3 marks each)

Q. 4. Express 0·328 in the form of

p , where p and q are q

5 9 and . integers and q ¹ 0. 7 11 A [Board Term I, 2012] Sol. Let x = 0·328 = 0·3282828....... ½ U [Board Term I, 2014] Sol. Since, LCM of 7 and 11 is 77. or, 10x = 3·282828..... ½ or, 1000x = 328·282828........ ½ 5 5 11 55 = \ = × 1 or, 1000x – 10x = 328·2828..... – 3·2828..... ½ 7 7 11 77 or, 990x = 325·000 325 65 9 9 7 63 or, x = = 1 = × = and 1 990 198  11 11 7 77 Q. 1. Find three rational numbers between

6

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-IX

Long Answer Type Questions (5 marks each)

or, Again, let

Q. 1. Give two rational numbers whose

(i) difference is a rational number,



(ii) sum is a rational number,

or, or, or,

(iii) product is a rational number, (iv) division is a rational number. Justify also.

R [Board Term I, 2015]

Sol. Any example and verification of example : 4 9 Let m = , n = 5 2 (i) Difference =



9 4 37 (Rational Number) 1½ − = 2 5 10

45 - 8 37 = 10 10

(ii)

Sum =

4 9 53 (Rational Number) + = 5 2 10

84+ 459 553 + =  2 510 2 10 (iii)



Product =

1½

9 4 45 (Rational Number) 1 ÷ = 2 5 8

Q. 2. Arrange in descending order 12

3

2, 4 5,

6

7

and

U [Board Term I, 2012]

3.

Sol. LCM of 3, 4, 6 and 12 is 12 3

4

6

1 3

4 12

1 4

3 12

1 6

2 12

= 2 2= 2 = 5 5= 5 = 7 7= 7

=

12

16

=

12

125

=

12

49





12

125 ,

i.e.,

4

12

5,

49 ,

12

7,

3

6

1309 + 350 990

1659 = 990 553 = 330 Q. 4. Express 0.6 +0.7 +0.47 in the form

12

2 and

3.

12

and q are integers and q ≠ 0.

3.

1 p , where p q

Ap

Sol. Let x = 0. 7 = 0.777...

…(i)

Multiply both sides by 10 10x = 7.777... Subtracting (i) from (ii),

…(ii)



9x = 7



x =

7 1½ 9 …(iii)

10x = 4.777...

...(iv)

Multiply both sides of (iv) by 10

100x = 47.777...

...(v)

Subtracting (iv) from (v), 

1+1+1+1+1  [CBSE Marking Scheme, 2012]

Q. 3. Express 1.32 + 0.35 in the form

Sol. Let, or, or,

=

Multiply both sides of (iii) by 10



16 and

integers and q ≠ 0.

100y = 35.3535........ 100y – y = (35.3535......) – (0.3535.......) 99y = 35 35 y = 2 99

119 35 \ + 1.32 + 0.35 = x + y = 90 99 119 × 11 + 35 × 10 = 990



12 = 3 3= 12 3 Descending order is



y = 0.35 = 0.353535.......

Let x = 0.47 = 0.477...

1 12



or,

100x – 10x = (132.222.......) – (13.222.....) 90x = 119 119 x = 2 90

4 9 36 18 × = = 5 2 10 5 (Rational Number) 1

(iv) Division =

or, or,

p , where p and q are q

A [Board Term I, 2012]



90x = 43



x =

1½

6 7 43 + + \ 0.6 + 0. 7 + 0.47 = 10 9 90 =

x = 1.32 = 1.32222....... 10x = 13.222.......... 100x = 132.222......

43 90 



=

54 + 70 + 43 90 167  90

2



REAL NUMBERS

7

Topic-2 Irrational Numbers Revision Notes  Irrational Number : If a number cannot be written in the form of p/q, where q ≠ 0 and p, q ∈ I, then it is called an irrational number.

e.g.,

2 , 3 , 5 , 2 + 5 , 3 − 7 , π etc. are all irrational numbers.  The decimal expansion of an irrational number is non-terminating and non-recurring.  The addition, subtraction, multiplication and division of rational and irrational number is an irrational number. i.e., If x and y are two real numbers where x is rational and y is an irrational, then (i) x + y is an irrational number. (ii) x – y is an irrational number. (iii) x × y is an irrational number. (iv) x ÷ y is an irrational number.

Example 2 Prove that Solution:

5 is an irrational number.

Let us assume

then



5 be rational number, 5 =

p q

(where p and q are integers and q ¹ 0) and p, q have no common factor other than 1. p 5 = So, q or

5 =

p2 q2

(on squaring both sides)

5q2 = p2 ...(i) 5 divides 5q2 so 5 divides p2 also. \ 5 divides p. Let p = 5a where a is an integer On squaring both sides p2 = 25a2 ...(ii) Substituting value of p2 from (i) into (ii) we get 5p2 = 25a2 q2 = 5a2 (5 divides 5a2 so 5 divides q2 as well) \ 5 divides q Thus 5 is common factor of p and q. But this contradicts the fact that p and q have no common factor other than 1. Hence, 5 is irrational number.

SUBJECTIVE TYPE QUESTIONS Very Short Answer Type Questions (1 mark each) Q. 1. Identify an irrational number among the following numbers : 0.13 , 0.1315 , 0.1315 , 0.3013001300013 ... R [Board Term I, 2014] Sol. 0.13 is a terminating number. So, it is not an irrational number.

0.1315 = 0.131515......, is repeating continuously, so it is not an irrational number.



0.1315 = 0.13151315...., is repeating continuously, so, it is not an irrational number. 0.3013001300013...., is non-terminating and nonrecurring decimal. Hence, it is an irrational number. So, 0.3013001300013... is an irrational number. 1



Q. 2. Is the product of two irrational numbers always an R irrational number ?

Sol. No, it may be rational or irrational.

1 R

Q. 3. Write the sum of 2 5 and 3 7 .

1 Sol. Sum of 2 5 and 3 7 = 2 5 + 3 7 .  Q. 4. Calculate the irrational number between 2 and 2.5.  U Sol. Irrational number =

2×2.5 =

5

Since, 5 = 2.236.... Hence, the irrational number between 2 and 2.5 is



1

5 .

( 2 + 5) . 2 5) = ( 2) + ( 5) + 2 × 2 ×

Q. 5. Simplify the number

Sol.



(

2+

2

2

2



= 2 + 5 + 2 10



= 7 + 2 10 (Irrational Number)

A

5

1

8

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-IX

Short Answer Type Questions-I (2 marks each)



Q. 1. Find any two irrational numbers between 0.1 and 0.12.

R [Board Term I, 2014]

Sol. Required two irrational numbers are : (i) 0.10100100010000 ......... and

1

(ii) 0.10200200020000 ........

1

Q. 2. Find any two irrational numbers between 0.5 and 0.55. R [Board Term I, 2012] Sol. Required two irrational numbers are : (i) 0.5101001000100001..... and

1

(ii) 0.502002000200002.... 

1 1 2 Q. 3. Find an irrational number between and , 7 7 1 when it is given that = 0.142857 7 Sol. Given,



U [Board Term I, 2012] 1 = 0.142857142857......... ½ 7 2 = 0.285714285714...... ½ 7 

\ Hence, required irrational number lies between 0.142857 ...... 0.285714 It can 0.142858 or 0.20203

Short Answer Type Questions-II (3 marks each)

So, 3 – r is rational, so

2 is also rational which

contradicts the statement that number.

2 is an irrational 1

\ 3 - 2 is irrational number.

1

Hence Proved Q. 3. Simplify: (a) 6 5 ´ 2 5 (b) 8 15 ¸ 2 3 Sol. (a) 6 5 ´ 2 5 = ( 6 ´ 2 ) 5 ´ 5 = 12 × 5 = 60

1½

(b) 8 15 ¸ 2 3 = 8 5 ´ 3 ¸ 2 3 8 5´ 3 = 4 5 2´ 3

1½

Q. 4. Insert an irrational number between Sol. One irrational number between -

2 1 and . 5 2

2 1 and 5 2

5) 2 (0.4 –0 20 20 0 -

2 = –0.4 5

1 5 9 2) 1 (0.5 Q. 1. Find three irrational numbers between and . 7 11 –0 10 U [Board Term I, 2016] 10 5 0 = 0·714285 Sol. 1 7 = 0.5 1 2 9 1 = 0·81 So, there are infinite irrational numbers between 11 2 1 5 - and Hence, three irrational numbers between and 5 2 7 one irrational number among them can be 9 can be : 0.1010010001 11 0·727227222........... 1 Long Answer Type 0·737337333........... Questions (5 marks each) 0·747447444........... 1 [CBSE Marking Scheme, 2016] Q. 1. Give an example of two irrational numbers whose: Q. 2. 2 is an irrational number. Prove that 3 - 2 is also an irrational number

Ap

Sol. 2 is an irrational number

[Board Term I, 2013] (given)

Let us assume 3 - 2 be a rational number ‘r’ So

3 - 2 = r





3–r=



We know that r is rational

1 2

(i) difference is an irrational number, (ii) sum is an irrational number, (iii) product is an irrational number, (iv) division is an irrational number. Justify also. R [Board Term I, 2015] Sol. Let two irrational numbers be : 6 and

3 ,

(i) 6 − 3 = Difference is an irrational number.

1 1



REAL NUMBERS

(ii) 6 + 3 = sum is an irrational number.

1

(iii) 6 × 3 = 18 = 3 2 = product is an irrational number.

1





7 6 is an irrational number 12

= 3

(d)

8

+

1

6

(iv) = 2 3 = division is an irrational number. Q. 2. Simplify :



+

1

2

1½



=





=

24 54 + 8 9





=

3





2 2 5 2 ´ = (By rationalizing the denominator) 2 2 2





=

8

+

1 2

45 - 3 20 + 4 5

Sol. (a)







3´3´5 - 3 2´2´5 + 4 5 3 5 - 3´2 5 + 4 5





3 5 -6 5 +4 5





7 5 -6 5





=





(b) 3 3 + 2 27 + 7 3 3 3 + 2 3´3´3 + 7 3 3 3 + 2´3 3 +7 3









1

5 is an irrational number

1

=





=





=





2´2´2´3 2´3´3´3 + 9 8 2 6 3 6 + 8 9

6 6 + 4 3 By taking LCM 3 6 +4 6 = 12





24 54 + 8 9



2 2

2

(By taking LCM)

5

5 2 is an irrational number 4

form be

= 16 3 is an irrational number



3+2

Sol. If possible let

(3 + 6 + 7 ) 3

(c)

2 2

Q. 3. Examine whether

3 3 +6 3 +7 3



3

1

+



(c)



2´2´2

1

(b) 3 3 + 2 27 + 7 3



2 3

=

45 - 3 20 + 4 5

(a)

(d)



9

1½

a . b

2 is rational or irrational.

Ap [Board Term I, 2016] be rational and let its simplest 2



1

where a and b are integers having no common factor other than 1 and b ¹ 0. a a2 Now, 2 = or, 2 = 2 [on squaring both sides] 1 b b



or, 2b2 = a2 ...(i) or, 2 divides a2 [∵ 2 divides 2b2] or, 2 divides a [∵ 2 is prime and divide a2]  1 Let a = 2c for some integer c Putting a = 2c in (i), we get 2b2 = 4c2 or, b2 = 2c2 or, 2 divides b2 [∵ 2 divides 2c2] 1 or, 2 divides b [2 is prime and 2 divides b2] Thus, 2 is a common factor of a and b. But this contradicts the fact that a and b have no common factor other than 1. The contradiction arises by assuming that 2 is



1



rational. Hence, 2 is irrational.

Topic-3 nth Root of a Real Number Revision Notes  Definition: In an = b, a and b are real numbers and n is a positive integer,. (i) a is an nth root of b. n (ii) It can also be written as b = a.

10

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-IX

(iii) It is also known as radical. Example : (i) 3 is fourth root of 81 i.e., 34 = 81 or 3 =

4

81

nd

 Square root: The “2 ” root is the square root.  Cube root: The “3rd” root is the cube root. a × a = a : Square root is used two times in a multiplication to get the original value.

 

3

a × 3 a × 3 a = a : Cube root is used three times in a multiplication to get the original value.

n n × n a × ..........  a a = a : The nth root is used n times in a multiplication to get the original value. n terms  Identities used for radicals : Identities for two positive real numbers r and s :

rs = r · s

(i)



(iii)

(

r+ s

)(

r − s =r−s

(v)

(

r− s

)

=r−2 r s+s

2

r = s

(ii)

)

(

Scan to know more about this topic

r s

)(

)

(iv) r + s r − s = r 2 − s



nth root of number

 Laws of radicals : Laws for two positive real numbers a and b : (i)

n n

a = a

(ii)

(iii)

n

a n b = n ab , (a, b > 0 be real number)

(iv)

p n

a

(v)

p m

a

p



p

= an − m

( )

m

(vi)

mn n n

a =nma

a na = b b

p n

p

a × am = an + m

1 −m (viii) a = a m

p

n = anm (vii) a Examples :

(i) 2 × 2 = 2 × 2 = 2 (ii) (iii)

( (

2+ 3

)(

5+ 7

9 (iv) = 4

) ( 2) − ( 3) 2

2− 3 =

) = ( 5) + ( 7) 2

2

2

2

= 2 − 3 = −1

+ 2 5 7 = 5 + 7 + 2 35 = 12 + 2 35

9 3 = 4 2

Example 3

Simplify :

4

81 − 8 3 216 + 15 5 32 + 225

Solution: Step I : Write the exponents in the form of powers 1 4



1 3

i.e., ( 81) − 8.( 216 ) + 15.( 32 ) + ( 225)

1 2

Step II: Factorize the radical 1



1 5

1

1

1

Here, ( 3 4 ) 4 − 8.( 6 3 ) 3 + 15( 2 5 ) 5 + (15 2 ) 2



Step III : Multiplying the powers



i.e., ( 3)



= 3 – 8 (6) + 15 (2) + 15



Step IV : Solving the expression

4×

1 4

− 8.( 6 )

3×

1 3

+ 15.( 2 )

3 – 48 + 30 + 15 = 0

48 – 48 = 0

5×

1 5

+ (15)

2×

1 2



REAL NUMBERS

11

SUBJECTIVE TYPE QUESTIONS Very Short Answer Type Questions (1 mark each)

(

)(

Q. 3. Simplify : 3 3 40 − 4 3 320 − 3 5 .

)

Q. 1. Simplify : 5 + 5 5 − 5 .

(5 + 5 )(5 − 5 ) = {52 − ( 5 ) } 2

½

= 25 – 5 ½

= 20

[CBSE Marking Scheme, 2014]

Q. 2. Write the equivalent of

12 × 8 .

Sol. 12 × 8 = 2 3 × 2 2 = 4 3 × 2 = 4 6 . Q. 3. Calculate the value of 4 28 √ 3 7 .

U

1 U

8 Sol. 4 28 ÷ 3 7 = 4 × 2 7 ÷ 3 7 = 3 2 Q. 4. If b > 0 and b = a, then find the value of 2 Sol. b = a or, a = b.

Q. 5. Simplify :

1 a . R 1

3

Sol.

72 + 800 − 18

= 36 × 2 + 400 × 2 − 9 × 2 = 6 2 + 20 2 − 3 2 = 23 2



1

Short Answer Type Questions-I (2 marks each) Q. 1. Simplify : 8 3 − 2 3 + 4 3 .



3 3 320 = 2 × 2 × 2 × 2 × 2 × 2 × 5







3 3 3 \ 3 40 − 4 320 − 5





= 3 × 23 5 − 4 × 43 5 − 3 5





= 6 3 5 − 16 3 5 − 3 5 = −113 5 .

8 3 − 2 3 + 4 3 = 3 ( 8 − 2 + 4 ) = 10 3 2 [CBSE Marking Scheme, 2015]

Q. 2. Simplify : Sol.

40 =

= 2 × 23 5 = 43 5

(

)(

50 - 98 + 162 5×5×2 − 7×7×2 + 3×3×3×3×2



=



= 5 2 − 7 2 + 9 2



= 7 2 

1 ½ ½

[CBSE Marking Scheme, 2012]

)



(

)(

U [Board Term I, 2012]

)

( 2 (17 + 8 2 )

= 5 2 15 + 3 2 + 5 2 + 2













= 85 2 + 40 × 2





= 85 2 + 80

= 5

)

1

1

Short Answer Type Questions-II (3 marks each) Q. 1. Simplify : 3 45 − 125 + 200 − 50 . U [Board Term I, 2014]

Sol. 3 45 − 125 + 200 − 50

= 9 5 − 5 5 + 10 2 − 5 2

1½



= 4 5 + 5 2 

1½ −1 6

Q. 2. Find the value of (729 ) . U [Board Term I, 2012]

50 − 98 + 162 . U [Board Term I, 2012]



Q. 4. Find the product of 5 2 3 + 2 5 + 2 .



U [Board Term I, 2015]



2 × 2 × 2 × 5 = 23 5



[CBSE Marking Scheme, 2016]

Sol.

3



U [Board Term I, 2016]



½

½ 3 3 3 3 3 3 ∴ 3 40 − 4 320 − 5 = 3 × 2 5 − 4 × 4 5 − 5 = −113 5 1  [CBSE Marking Scheme, 2012] Detailed Solution :

Sol. 5 2 3 + 2 5 + 2

72 + 800 − 18 .

3

40 = 2 3 × 5 = 2 3 5 3 320 = 4 3 5 3

Sol.

U [Board Term I, 2014]

Sol.

U [Board Term I, 2012]

Sol.

(729 ) 6 = ( 36 ) −1



−

1 6

= 3 −1



2

1 1 3  [CBSE Marking Scheme, 2012] =

Q. 3. Evaluate :

5 + 2 6 + 8 − 2 15 . U [Board Term I, 2012]

12

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-IX

Sol.

5 + 2 6 =



= = 8 − 2 15 =



=

3+2+2 6

=

(

=

3+ 2

)

2

3+ 2

(

5− 3

)

2

Q. 2. =

3+ 2+ 5− 3

½

24 54 + 8 9

1

1 7

1 4

= ( 2 × 2 × 3) × ( 2 × 3) = ( 2 )

1 1 1 + + 4 4 7

× ( 3)

9 14

= ( 2 ) × ( 3) =

14

28

9

2 × 3

Q. 3. Simplify :

U

4

11 28

11

1 1 + 4 7

1



1



1 1



16 − 6 343 + 18 243 − 196 . 3

5

U [Board Term I, 2014]



24 54 + 8 9

Sol.

1

1

Long Answer Type Questions (5 marks each) Q. 1. Simplify :

Ap

12 × 7 6

= (12 ) 4 × ( 6 ) 7

2+ 5 ½ [CBSE Marking Scheme, 2012]



4

1

7 4 Sol. 12 × 6

5− 3 1

∴ 5 + 2 6 + 8 − 2 15 = =

1

7 6 12 Evaluate

1

5 + 3 − 2 15

3 6 +4 6 12

4

16 =

4

3

343 =

3

7×7×7 =7

5

243 =

5

3×3×3×3×3 = 3

Sol.

=

4×6 9×6 + 9 8

1

=

2 6 3 6 + 9 8

1

6 6 = + 4 3

1



2×2×2×2 =2

196 = 14

4

3

1

5

\ 16 − 6 343 + 18 243 − 196 = 2 – 6 × 7 + 18 × 3 – 14 = 2 – 42 + 54 –14 = 56 –56 = 0 1

Topic-4 Laws of Exponents with Integral Powers Revision Notes  Let a > 0 be a real number and ‘r ’ and ‘s’ be rational numbers, then (i)

(ii) (ar)s = ars

r

a = ar–s, r > s as 1 (v) a–r = r a

(iii)

ar.as = ar+s

Scan to know more about this topic

(iv)

ar br = (ab)r

(vi)

a s = (ar)1/s = (a1/s)r

r

r

ar  a (vii)   = r   b b (ix) a0 = 1 Examples :

 a (viii)     b

(i) (3)4 × (3)3 = 34+3 = 37

(ii)



−r

 b =    a

r

( 4 )7 = (4)7–2 = 45 ( 4 )2

Laws of exponents



REAL NUMBERS

(iii) (3)2 × (4)2 = (12)2 1 (9) = 2 9

(iv)



–2

(v)

 3   5

−2

 5 =   3

13

2



Example 4

Find the value of

3 40 + 339 + 338 3 41 + 3 40 − 339



Solution: Step I : Taking common factor from numerator and denominator as possible we can. 338 ( 3 2 + 31 + 1) 3 40 + 339 + 338 i.e., 41 = 40 39 339 ( 3 2 + 31 − 1) 3 + 3 − 3 Step II : Shifting the common factor which in denominator and solving the expression which are in bracket.

338 − 39 ( 9 + 3 + 1) ( 9 + 3 − 1)

i.e.,

Step III : Solving the expression

i.e.,

3 −1 × 13 = 11

13 13 = = 3 × 11 33

SUBJECTIVE TYPE QUESTIONS Very Short Answer Type Questions (1 mark each) 1 Q. 1. Simplify :   2 1 Sol.   2







−2

2 +  3

−2

−2

2 +  3

3 +  4

−2

−2

3 +  4

−2

U

 2

4 2 3 = (2) +   +   2 3

9 16 + 4 9 144 + 81 + 64 289 = = 36 36

5

 7   12  .   Sol. Given,  7  ⇒    12 

1 −4

 7  ×   12 

=

3x

 7 

−4 + 3 x



⇒

3 x−1

7

2

3x–1

½

[CBSE Marking Scheme, 2012]

(i) (ab + ba)–1 5

U

+ 10

½ ½ ½

Q. 2. If a = 2 and b = 3, then find the value of :

5

=    12 

 7  ⇒ =    12  ⇒ – 4 + 3x = 5 ⇒ 3x = 9 ⇒ x = 3 1 Q. 3. If (23x – 1 + 10) ÷ 7 = 6, then find the value of x.  2

(13 + 23 + 33)–3/2 = (1 + 8 + 27)–3/2 = (36)–3/2 = [(6)2]–3/2 = 6–3 1 1 = 3 = 216 6



 7  ×   12 

1

U [Board Term I, 2012]

Sol.

3x



 7   12   

Sol. Given,

23x–1 + 10 = 42 23x–1 = 42 – 10 23x–1 = 32 23x–1 = (2)5 3x – 1 = 5 3x = 6 x = 2

Q. 1. Find the value of (13 + 23 + 33)–3/2.

U

−4

⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒

Short Answer Type Questions-I (2 marks each)

= 4+

 7  Q. 2. Find the value of x for which    12 



2



= 6

+ 10 = 6 × 7

(ii) (aa + bb)–1 

U [Board Term I, 2012]

Sol. Given, a = 2 and b = 3. (i) (ab + ba)–1 = (23 + 32)–1 = (8 + 9)–1 = 17–1 1 = 17 (ii) (aa + bb)–1 = (22 + 33)–1 = (4 + 27)–1 = 31–1 1 = 31

½ ½ ½ ½

[CBSE Marking Scheme, 2012]

14

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-IX

 81  Q. 3. Simplify :    16 

−

5 4

 25  ×   9

−

5 2

U [Board Term I, 2012]

 81  Sol.     16

−

5 4

 25  ×   9



−

5 2

 3  4  =       2   3 =    2



−5

−

5 4

 5  2  ×     3  

 5 ×   3

5



 2  3 =   ×   3  5 25 = 5 5 32 = 3125 











−

5 2



Short Answer Type Questions-II (3 marks each)

½

−5

½



(

Q. 1. Simplify : 2 2 − 5

5

Sol.



U [Board Term I, 2012]

=        

    

1



=



Q. 5. Show that :

x a( b − c ) x

b( a − c )

a( b − c )

Sol. LHS =

Þ

1



x x b( a − c ) ab − ac

1 4 ( 54 )=

( ) ( )

1

5= 5 c

( ) ( ) bc

x x ÷ ac ba − bc x x

2

2

2+ 3



c

1

2

)

2

2 −1

2

(

– 2(2 2 )(5) + (5)2 + 3 2

( 2)

2

)

2



= 8 + 25 – 20 2 + 18 + 3 + 6 6 – 2 – 1 + 2 2 1



1 = 51 – 18 2 + 6 6 [CBSE Marking Scheme, 2012]

( x)

−

2 3

1

( xy )

y4 ÷

– (1)2 +2( 2 ) (1)

−

1 2

.



U [Board Term I, 2012] 2 − 3

(xy )

y4 ÷

 1 2 =  x   

1 − 2

= x

−

−1 3 ·y 2

2 3

.

1 y4 2

( )

÷ ( xy )

−





−1 x 3 ·y 2

1 × ( xy ) 4

=



1

1  − 2 ÷ ( xy ) 2    1 4



−1 1 1 + 2+ x 3 4 ·y 4



=



= x 12 ·y 4

2

9

9



1

)

2





y4

1

1 x 12



  Q. 3.  x   xb    a



) −(

−1



) −(

2 −1 . U [Board Term I, 2012]

=

 xb  ÷  a  = 1.  x  U [Board Term I, 2014]

 xb  ÷ a   x   

)

) + (3

1

3 3 =  5( 2 + 3)  4 =  5( 5)  4



2+ 3

+ 2(3 2 ) ( 3 ) + ( 3 )2 –

( )



1   1 1 3  4  5 ( 2 3 ) 3 + ( 33 ) 3  

(

Sol. x

1 4

  1 Sol.  5 8 3 +   

2



base of exponent same and then solve according to the law of exponents.

1 1 3  4 27 3  

2 −5

Q. 2. Simplify :

The students should be careful in making the

1 3   1  3 3 Q. 4. Simplify : 5 8 + 27   .    

(2

= 2 2



While solving the real numbers with integral

powers, the bases of numbers should be made the same to solve but students sometimes do it directly and simply reduce it to lowest term, thereby marks are not awarded to them inspite of correct answer.





1

Answering Tip

) + (3



Commonly Made Error

(xab–ac–ba+bc) ÷ (xbc–ac) x(bc–ac)÷ x(bc–ac) x0 = 1 = RHS 1 [CBSE Marking Scheme, 2014]

Þ Þ Þ

.

 xa  Sol.  b  x 

a+ b

a+b

 xb  × c  x     xb  × c  x   

b+ c

b+c

 xc  × a  x   

 xc  × a  x   

c+a



R [KVS 2019]

c+a

= (xa – b)a + b × (xb – c)b + c × (xc – a)c + a = x

a2 − b 2

= x a

2

×x

b2 − c2

×x

− b 2 + b 2 − c 2 + c 2 − a2

= x0 = 1

1

c 2 − a2



1 1



15

REAL NUMBERS 7

 5−1 × 7 2  2  5−2 × 7 3  Q. 4. Simplify :    × 3  5 × 7 −5   52 × 7 −4 

−

5 2

.

U [Board Term I, 2012] 7

Sol.



 5 −1 × 7 2  2  5 −2 × 7 3    2  × 3  5 × 7 −5   5 × 7 −4 









5 −  75 × 73  2

=  2 1 × 3 2 5 ×5  5 ×5  1  7 42  2



1  525  2

1 1

 7 42 525  2 =  21  ×  40  =  21 × 40  ½ 7  5 7  5  1

2 4 = (7 × 5 ) 2 ½ 2 = 7 × 5 = 7 × 25 ½ = 175 [CBSE Marking Scheme, 2012]



2 Q. 5. Find x, if    3

x

 3 ·   2 x

Sol.

½



7  74 × 72  2





5 2

−

2x

 2  3   ·  3 2



81 = A [Board Term I, 2012] 16 

2x



=

3 3  3  16  4  9  2  2   =   ×   ÷   25   5   81    

1

3 3    2 4  4  32  2  2  3  =  4  ×  2  ÷     5 5 3     

½

3 3    2  4  4   3  2  2  5  3  =    ×     ×       2  3     5     3  3  3  5  3   2 =   ×   ×     3  2   5   3  3 3 2 3 5 = 3 ×  3 × 3  3  5 2  2 3 33 = 3× 3 3 2  = 1



1

2 x 32 x 34 or, · = 4 1 x 2x 3 2 2 or, 2x–2x.32x–x = 34.2–4 or, x–2x = – 4 \ x = 4 1 [CBSE Marking Scheme, 2012]

Long Answer Type Questions (5 marks each) Q. 1. Simplify : 2 4 81 − 8 3 216 + 15 5 32 + 225 − 4 16

or,

( )

1

( )

1

( )

81 Q. 2. Evaluate :    16 

3  −3   9  2  5 ×   ÷    .  25   2   

Sol.  81   16 

Ap −3 4

3  −3   9  2  5   ×   ÷   25   2   

A [Board Term I, 2016]

a – b = 0 2−x − = 0 1 2 x − 2 2 x +1 2 x – 1 – (x – 2) – 2 – x – (x + 1) = 0 2 x – 1 – x + 2 – 2 – x – x – 1 = 0 1 21 – 2 – 2x – 1 = 0 2– 2x –1 = 21 -2x – 1 = 1 1 -2x = 2 x = –1 2 [CBSE Marking Scheme, 2016] 2 x −1



or, or, or, or, or, or, or,

 xb   xc   

1

= 2 34 4 − 8 6 3 3 + 15 2 5 5 + 15 − 2 4 4 2  = 2 × 3 – 8 × 6 + 15 × 2 + 15 – 2 1 = 6 – 48 + 30 + 15 – 2  1 = 51 – 50 = 1  1 [CBSE Marking Scheme, 2016, 2019 ] −3 4

½

rational number then simplify :

Sol. 2 4 81 − 8 3 216 + 15 5 32 + 225 − 4 16 1

½

Q. 4. If x is positive real number and exponents are

U [Board Term I, 2016]

( )

½

2 2 , b = x+1 and a – b = 0, find the value 2x−2 2

Q. 3. If a =

Sol.

½

-x

x −1

of x.

81 16

1

b+ c − a

 xc  × a x 

c+ a−b

 xa  × b x 

a+ b− c

U [Board Term I, 2016]

Sol.

 xb   xc   

b+c−a



 xc  × a x 

b−c =  x 

b = x

2

c+ a−b

b+c−a

− c 2 − ab + ac

 xa  × b x 

×  x c − a 

× xc

2

a+b−c

c+ a −b

− a 2 − bc + ab

×  x a − b 

× xa

2

− b 2 − ac + bc

½



1 2

1 = x = x0 = 1 [CBSE Marking Scheme, 2016] 1 b2 − c 2 − ab + ac + c 2 − a 2 − bc + ab + a 2 − b2 − ac + bc

[Board Term I, 2016]

a+b−c

16

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-IX

Topic-5 Rationalization of Real Numbers Revision Notes  Rationalization: If a given number is transformed into an equivalent form, such that the denominator is a rational number then the process is known as rationalization.  Rationalizing the denominator : If the denominator of a fraction contains a term with root (a number under a radical sign), the process of converting it to an equivalent expression with rational denominator is called as rationalizing the denominator.  List of Rationalization Factors. Term

Rationalizing Factor

1 r

r

1 r −s

r +s

1 r +s

r −s

1 r− s

r+ s

1 r+ s

r− s

Example 5 Rationalize the denominator of



7 5- 2

Solution: To rationalize the denominator, we will multiply the numerator and denominator by its conjugate to remove the radical sign from the denominator. Step I : Assume the given fraction as x and write the denominator. 7

Let x =

and denominator =

5- 2

5- 2 Step II : Find the conjugate of denominator. Here, the conjugate of denominator ( 5 - 2 ) is

Step III : Multiply the numerator and denominator of x by the conjugate of denominator and rationalize it.



x = =



7

5+ 2

×

5− 2

5+ 2

7( 5 + 2 ) ( 5 )2 − ( 2 )2

=

7( 5 + 2 ) 5−2

=

7 ( 5 + 2) 3

( 5 + 2)

SUBJECTIVE TYPE QUESTIONS Very Short Answer Type Questions (1 mark each)

Q. 1. Write the rationalizing factor of

1 . 50 

R



Sol.

1 = 50 =

1 5×5×2 1 5 2

×







So, rationalizing factor is

2 2 = 2 10 2.

1



Q. 2. Rationalize the denominator of 2

Sol.

3 3

=

2 3 3

2 . U [NCERT] 3 3 

×

3 3

2 = 1.414, then, find the value of

1

1 . 2 +1

U [Board Term I, 2012] 1 1 ( 2 − 1) = × Sol. 1 ( 2 + 1) ( 2 − 1) 2 +1











Q. 2. Taking

( 2 − 1) = ( 2 )2 − (1)2 =



2 − 1 1.414 − 1 = = 0.414 2 −1 1

1

1 2 = 1.414 and r = 3.141, evaluate +r 2

upto three places of decimal.



U [Board Term I, 2012]

1 1 2 = × (rationalizing) 2 2 2 2 1.414 = = = 0.707 1 2 2 Sol.

1 + p = 0.707 + 3.141 \ 2 = 3.848 1 [CBSE Marking Scheme, 2012] Q. 3. If x = 3 − 2 2 , find the value of Sol.

or,

2 x = ( 2 − 1)



or,

x =

or,

1 = x

2 −1

(

)



\



1 x+ = x

½ ½

=



and

y = =

1 3+2 2 × 3−2 2 3+2 2 3+2 2 = 3+2 2 9−8

1

1 3−2 2 × 3+2 2 3−2 2 3−2 2 =3−2 2 9−8

x + y+ xy = 3+2 2 + 3 – 2 2 + (3+2 2 )



½ 2

3 , then find the value of x +

Q. 2. If x = 2 +

1

(3 – 2 2 ) ½

= 6 + 9 – 8 = 7

1 x2

.

A [Board Term I, 2015]

Sol.

x = 2 + 3



1 1 = x 2+ 3 2− 3 1 × 2+ 3 2− 3

=



= 2 − 3



1 = 2 – x

\



x+

x2 +

Q. 3. If a = 2 +

3

1

1 = 4 x

1 + 2 = 16 x2

1

1 = 14 x2

1

x2 +

5 and b =



b =

½ ½

1 2 2 , find a + b . a A [Board Term I, 2016]

Sol.

2 −1+ 2 +1

= 2 2

x =

or,

2 +1 1 × 2 −1 2 +1

= 1 + 2



Sol.

A [Board Term I, 2014]

Squaring both sides, we get

x = 3 − 2 2 x = 1 + 2 − 2 2



1 . x 

A [Board Term I, 2014]

or,



x+

1 1 and y = , then find the 3−2 2 3+2 2

value of x + y+ xy.

Short Answer Type Questions-I (2 marks each) Q. 1. If

Short Answer Type Questions-II (3 marks each) Q. 1. If x =

2 3 2 3 = = 9  3×3



17

REAL NUMBERS



1 2+ 5

=

2− 5 1 × 2+ 5 2− 5

=

2− 5 = −2 + 5 −1

1

18

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-IX



(

a2 = 2 + 5

2

(

b2 = − 2 + 5



)

(

=9+4 5  ½

)

2

=9−4 5 ½ 

a2 + b2 = 9 + 4 5 + 9 − 4 5 = 18  1 [CBSE Marking Scheme, 2016]

\

3

Q. 4. If x =

1  2 – 1, then find the value of  x −  .  x



) ( 2 − 1) − 2 ( 1) ( 2 + 1) 2

2 +1 −















On comparing

2

= a + 2 b

1



2+1+ 2 2 − 2 −1+ 2 2 =a+ 2b 2 −1 4 2 = a +

2b

1

a = 0, b = 4

[CBSE Marking Scheme, 2016]

A [Board Term I, 2014]

Sol.

x =

Commonly Made Error

2 –1

or,

1 = x

1 2 +1 × 2 −1 2 +1

1

or,

1 2 +1 = = 2 +1 x 2 −1

½



Generally, students make mistake while

adding or subtracting rational numbers while rationalizing them.

Answering Tip

3

1   x −  = ( 2 − 1 − 2 − 1)3 \ x = (–2)3

½



The student should be slow and precise while working with rational numbers.

1

= – 8 Q. 5. Find a and b if 1 − 3 = a + b. 1+ 3 

Long Answer Type Questions (5 marks each)

A [Board Term I, 2016]

Sol.

a + b =

1− 3 1+ 3

1− 3 1− 3 = × 1 + 3 1 − 3



(1 − 3 ) =

\



and



=

1+3− 2 3 −2

=

4−2 3 −2

a = –2 b =

1

or, a =

3

3

b = – 2

1

Q. 6. Find the value of a and b if

2 −1

Sol.



−

2 −1 2 +1

=

3+ 8 8+ 7 ( 7 + 6) − + ( 3 )2 − ( 8 )2 ( 8 )2 − ( 7 )2 ( 7 )2 − ( 6 )2 −

A [Board Term I, 2016]

2 −1 2 +1 − =a+ 2 −1 2 +1

2 b

1

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







3 + 8 ( 8 + 7 ) ( 7 + 6) ( 6 + 5) − + − = 6−5 9−8 8−7 7−6 +

1

( 5 + 2) [ a2 – b2 = (a – b) (a + b)] 2 5−4



= 3+ 8 − 8 − 7 + 7 + 6 − 6 − 5 + 5 +2



= 3 + 2 = 5 = RHS

1 2

Q. 2. If x = 4 –

=a+ 2 b

U [Board Term I, 2014]

1 1 1 1 1 − + − + = 3 − 8 8 − 7 7 − 6 6 − 5 5 −2

3

[CBSE Marking Scheme, 2016]

2 +1

Sol. LHS

2

1−3

a + b = –2 +



1

Q. 1. Prove that : 1 1 1 1 1 − + − + = 5 3 − 8 8 − 7 7 − 6 6 − 5 5 −2

Sol.

or,

1  15 , then find the value of  x +  .  x  U [Board Term I, 2014] x = 4 – 15 4 + 15 1 1 = × x 4 − 15 4 + 15 4 + 15 1 = 16 − 15 x

1 1



1 = 4 + 15 x



1

Q. 5. If a =

2

1  2 1  x +  = ( 4 − 15 + 4 + 15 ) x = (8)2 = 64 1 2 −1 2 +1 Q. 3. If x = and y = then find the value of 2 +1 2 −1 \

x2 + 5xy + y2.

A [Board Term I, 2016]

of

2

é 2 -1 2 -1 2 + 1ù 2 +1 + ´ ê ú +5 2 - 1 úû 2 +1 2 -1 êë 2 + 1

2

a2 + ab + b2 . a2 − ab + b2 

A

Sol.

a2 + ab + b2 = (a + b)2 – ab



a2 – ab + b2 = (a – b)2 + ab





( =

=

2







= (6)2 +3



1



[CBSE Marking Scheme, 2016] 5+1 and y = 5 −1

Q. 4. If x =

5 −1 , then find the value of 5+1

x2 + y2.

U [Board Term I, 2016]

5 +1 5 −1

x =





 5 + 1 6+2 5 3+ 5 =  = x =  5 − 1 6−2 5 3− 5  

 5 − 1  y =   5 + 1

2

=

2



\











6−2 5 3− 5 = 6+2 5 3+ 5









\



2

1



2

7 − 2 10 3

 7 + 2 10 7 − 2 10  (a + b)2 – ab =  +  3 3 



1 2



½

 7 + 2 10   7 − 2 10  −    3 3 

=

187 9

½

 7 + 2 10 7 − 2 10  (a – b)2 + ab =  −  3 3 



1

(3 + 5 ) + (3 − 5 ) (3 − 5 ) (3 + 5 )

2

1

½

 7 + 2 10   7 − 2 10  +    3 3 

=

28 4



 4 10  49 − 40 + =  9  3  =

160 9 169 + = 9 9 9

½

( a + b )2 − ab a 2 + ab + b 2 \ = 2 2 ( a − b )2 + ab a − ab + b



9+5+6 5 +9+5−6 5 9−5

x + y = 7

5− 2 5− 2 × 5+ 2 5− 2

2

=



1



5 + 2 + 2 10 7 + 2 10 = 3 3





2

\

1

2

x2 + y2 =

2

2

3+ 5 3− 5 + x +y = 3− 5 3+ 5 2



5 −1 5 +1

y =



5−2

)

14 49 − 40  196 9 =   −   3  9  = 9 − 9

2

2





=

5+ 2

2

Sol.



b =

1

= 36 + 3 = 39

5+ 2 5+ 2 × 5− 2 5+ 2

a =



2 +1−2 2 + 2 +1+ 2 2 =   ++33 1 2 −1 

5− 2 . Find the value 5+ 2

5 + 2 and b = 5− 2

Sol. x2 + 5xy + y2 = (x + y)2 + 3xy



19

REAL NUMBERS

1

[CBSE Marking Scheme, 2016]

187 = 9 169 9 187 = 1 169

20

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-IX

OBJECTIVE TYPE QUESTIONS A Multiple Choice Questions Q. 1. Every rational number is (A) a natural number

A [NCERT Exemp.]

Ans. Option (C) is correct. Explanation: We know that rational and irrational numbers taken together are known as real numbers. Therefore, every real number is either a rational number or an irrational number. Hence, every rational number is a real number. Therefore, (c) is the correct answer. Q. 2. Between two rational numbers (A) there is no rational number (B) there is exactly one rational number (C) there are infinitely many rational numbers (D) there are only rational numbers and no A [NCERT Exemp.]

Ans. Option (C) is correct. Explanation: Between two rational numbers there are infinitely many rational number, for example, between 4 and 5 there are 4.1, 4.2, 4.22, 4.223............ Q. 3. Decimal representation of a rational number cannot be (A) terminating (B) non-terminating (C) non-terminating / repeating (D) non-terminating / non-repeating  A [NCERT Exemp.] Ans. Option (D) is correct. Explanation: The decimal representation of a rational number cannot be non-terminating and non-repeating. Q. 4. A rational number between 2 and 3 is 2− 3 2+ 3 (A) (B) 2 2 (C) 1.5 (D) 1.8  Ans. Option (C) is correct. Explanation: We know that

2 is

(B) 1.41421

(C) a real number

irrational numbers

Explanation: The product of any two irrational numbers is sometimes rational and sometimes irrational. Q. 6. The decimal expansion of the number (A) a finite decimal

(B) an integer (D) a whole number

(1 mark each)

A [NCERT Exemp.]

= 2 1= .4142135.... and 3 1.732050807....

We see that 1.5 is a rational number which lies between 1.4142135….. and 1.732050807…. Q. 5. The product of any two irrational numbers is (A) always an irrational number (B) always a rational number (C) always an integer (D) sometimes rational, sometimes irrational Ans. Option (D) is correct.

(C) non-terminating recurring (D) non-terminating non-recurring Ans. Option (D) is correct. Explanation: The decimal expansion of the number is 1.41421…… non- terminating non - recurring Q. 7. Which of the following is irrational? 12 4 (B) (A) 3 9 (C) 7 (D) 81 Ans. Option (C) is correct. Explanation: (A)

4 9

=

12 (B) = 3

2 which is a rational number. 3

4× 3 = 2 which is a rational number. 3

(C) 7 is an irrational number. (D) 81 = 9 which is a rational number. Q. 8. Which of the following is irrational? (A) 0.14 (B) 0. 1416 (C) 0. 1416 (D) 0.4014001400014... Ans. Option (D) is correct. Explanation: A number is irrational if and only if its decimal representation is non-terminating and non-recurring. (A) 0.14 is a terminating decimal and therefore cannot be an irrational number. (B) 0. 1416 is a non-terminating and recurring decimal and therefore cannot be irrational. (C) 0. 1416 is a non-terminating and recurring decimal and therefore cannot be irrational. (D) 0.4014001400014... is a non-terminating and non-recurring decimal and therefore is an irrational number. Q. 9. 2 3 + 3 is equal to (A) (B) 6 2 6 4 6 (C) 3 3 (D) Ans. Option (C) is correct. 3 3 Explanation: 2 3 + 3 =

A [NCERT Exemp.]



21

REAL NUMBERS

Q. 10. 10 × 15 is equal to (A) 6 5 (B) 5 6 (C) 25 (D) 10 5

Explanation: 2560.16 ´ 2560.09 = 2560.16 + 0.09 = 2560.25

Ans. Option (B) is correct. Explanation: We have 10 × 15 = 10 × 15 = 5 6 32 + 48 Q. 11. The value of is equal to 8 + 12 (A) 2 (B) 2 (C) 4 (D) 8

A [NCERT Exemp.]  Ans. Option (B) is correct. Explanation: 32 + 48 4 2 + 4 3 4( 2 + 3 ) 4 = = = = 2 8 + 12 2 2 + 2 3 2( 2 + 3 ) 2

(A) 3 3

(B) 9 3

(C) 9 2 Ans. Option (C) is correct. Explanation:

(D) 2 2

Q. 13. 2 equal −1 (A) 2 6 (B) 2–6 2

1

(C) 2 6 (D) 26  A [NCERT Exemp.] Ans. Option (C) is correct. 4 3

3

2

4

4

12

2 1/3

2 = (2 )

= (2 ) 2 3

1 4

2 1 × 4

= 23

1

= 26

Q. 14. The product 2. 2. 32 equals (A) 2 2 (B) 12 32 (C) 12 2 (D)  A [NCERT Exemp.] Ans. Option (B) is correct. Explanation: We have, 3

4

1 3

12

1 4

5 12

2. 2 32 = 2 .2 .2 = 2

1 1 5 + + 3 4 12

12 12

4

Explanation:

4

81-2

=4

is equal to 1

1 (A) (3 - 2 2 ) 2

(B)

(C) 3-2 2

(D) 3 + 2 2

3+2 2 A

Ans. Option (D) is correct. Explanation: 1

=

9− 8

1 3− 8

=

1 3− 8

×

3+ 8 3+ 8

=

7 +2 (A) 3

(B)

7 -2 3

7 +2 (C) 5

(D)

7 +2 45

3+ 8 9−8

1

4´ æ æ 1 ö 2´ 2 ö 1 4 1 = çç ÷ ÷ = = çè 9 ø ÷ 9 9 è ø

Q. 16. Value of 256 0.16 × 256 0.09 is (A) 4 (B) 16 (C) 64 (D) 256.25  A [NCERT Exemp.] Ans. Option (A) is correct.

A

Ans. Option (A) is correct. Explanation:

1 7 −2

=

1 7 −2

×

7+2 7 +2

=

7+2 7+2 = 7−4 3

Q. 19. After rationalizing the denominator of

7 3 3 −2 2

we get the denominator as (A) 13 (B) 19 (C) 5

(D) 35

A

Ans. Option (B) is correct. Explanation: 7

=2 =2

Q. 15. Value of 81−2 is 1 1 (A) (B) 9 3 1 (C) 9 (D) 81  A [NCERT Exemp.] Ans. Option (A) is correct. 1 4

1 4

Q. 18. The number obtained on rationalizing the 1 denominator of is 7 −2

Þ 3´3 2 = 9 2

Explanation:

9− 8

4´

= 3+ 2×4 = 3+ 2 2

6 ´ 27 = 2 ´ 3 ´ 3 ´ 3 ´ 3

4 3

1

Q. 17.

6 ´ 27 is

Q. 12. The value of

1

1

= 256 4 = ( 44 ) 4 = 4

A [NCERT Exemp.]

3 3−2 2 =

=

7 3 3−2 2

×

(3 3 + 2 2 ) 3 3+2 2

3 3+2 2 3 3+2 2 7( 3 3 + = 2 2 ) 7( 3 3=+ 2 2 ) 27= − 8 19 27 - 8 19

Therefore, we get the denominator as 19. Q. 20. If

2 = 1.4142... then is

2 −1 2 +1

equal to

(A) 2.4142

(B) 5.8282

(C) 0.4142

(D) 0.1718

Ans. Option (C) is correct. Explanation: After rationalisation, we get 2 - 1 = 1.4142 – 1 = 0.4142

AE

,

22

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-IX

= 5.82842......

B Assertion & Reason

\ Assertion is true.

Directions: In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Mark the correct choice as: (A) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A). (B) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A). (C) Assertion (A) is true but reason (R) is false. (D) Assertion (A) is false but reason (R) is true.

In case of Reason (R):



Q. 1. Assertion (A): Three rational numbers between and

2 5

A number whose decimal expansion terminates is a rational number is true. Both A and R are true but R is not correct explanation of A. Q. 3. Assertion (A): 3 is an irrational number. Reason (R): Square root of a positive integer which is not a perfect square is an irrational number. Ans. Option (A) is correct. Explanation: In case of Assertion (A): 3 has non terminating and non-recurring decimal

3 9 10 11 are , and . 5 20 20 20

expansion. \ Assertion is true. In case of Reason (R): Square root of a positive integer which is not perfect square has a non terminating and non-recurring decimal expansion. \ Reason is true.

Reason (R): A rational number between two 1 rational numbers p and q is ( p + q ). 2

Ans. Option (B) is correct. Explanation: In case of Assertion (A): Multiplying and dividing



2 4 8 × = 5 4 20



3 4 12 × = 5 4 20

Now, rational number are \

8 12 and . 20 20

9 10 11 , , are three rational numbers between 20 20 20

2 3 and . 5 5 \ Assertion is true In case of Reason: Now, rational number between two rational numbers =

Therefore, Both A and R are true and R the correction explanation of A.

2 3 and by 4, we get 5 5

1 ( p + q ) . where p and q are rational numbers. 2

Q. 4. Assertion (A): e is an irrational number. Reason (R): p is an irrational number. Ans. Option (B) is correct. Explanation: In case of Assertion (A): e = 2.71828... \ e is an irrational number, hence, Assertion is true. In case of Reason (R): p = 3.14159... \ p is irrational number, hence, Reason is true. Both A and R are true but R is not correct explanation of A. Q. 5. Assertion (A): Value of

5 + 2 6 is an irrational

number. Reason (R): Product of a rational number and an irrational number is irrational, when rational number is not equal to 0. Ans. Option (B) is correct.

Explanation: In case of Assertion: 5 + 2 6 =

( 2 )2 + ( 3 )2 + 2 2 3

\ Reason is true



Therefore, Both A and R are true but R is not the correction explanation of A.

= ( 2 + 3 )2 = 2 + 3

Q. 2. Assertion (A): On simplification, 3 + 2 2 gives an irrational number. Reason (R): A number whose decimal expansion is terminating is rational. Ans. Option (B) is correct. Explanation: In case of Assertion: 3 + 2 2 = 3 + 2 × (1.41421. .......) = 3 + 2.82842......

In case of Reason: Product of rational and irrational number is irrational when rational number is not equal to 0. Both A and R are true but R is not correct explanation of A. Q. 6. Assertion (A): Value of 7–2 =

1 72

=

1 49

Reason (R): According to laws of radicals a–m =

1 am



REAL NUMBERS

Ans. Option (A) is correct.

Rational number is in the form of

Explanation: In case of Assertion (A): 1

1 1 = 7 ´ 7 49

In case of Reason (R): 1

Q. 9. Assertion (A): The rationalizing factor of 2 + 3 3 is

a–m =



7

2

=

am

2 - 3 3.

satisfies the laws of radicals for positive real number Hence, Both A and R are true but R is the correct explanation of A. Q. 7. Assertion (A): Value of (32)2/5 + (–7)0 is a rational number. Reason (R): (a)0 ¹ 1; where a > 0 and a is a real number. Ans. Option (C) is correct. Explanation: In case of Assertion: (32)2/5 + (–7)0 = (25)2/5 + (–7)0 5×



Reason (R): If the product of two irrational numbers is rational then each one is called the rationalizing factor of the other. Ans. Option (A) is correct. Explanation: In case of Assertion (A): ( 2 + 3 3 )( 2 - 3 3 ) Þ 4 - 6 3 + 6 3 - 27 = 4 – 27 = –23 is a rational number \ Assertion is true In case of Reason (R): If product of two irrational is rational then they are rationalizing factor of each other. \ A and R are true and R is correct explanation of A.

2

= 2 5 +1

= 22 + 1 = 4 + 1 = 5 \ Assertion is true In case of Reason: (a)0 = 1 \ Reason is false. Q. 8. Assertion (A): Value of

3

Q. 10. Assertion (A): To rationalize the denominator of 1 , we should multiply and divide it by 3 - 4. 4+ 3

( 343)-2 is a rational number.

Reason (R): Rational number is of the form

p q

where p and q are not equal to zero. Ans. Option (A) is correct. Explanation: In case of assertion (A):

3

( 343)-2 = [(343)–2]1/3 = (343)–2/3 1 × −2 3 3

= (7 ) 2   3 ×− 

= 7  3  = 7–2 1 1 = 2 = 7 49 \ Assertion is true. In case of Reason:



Reason (R): To rationalize the denominator, we will multiply the numerator and denominator by its conjugate. Ans. Option (D) is correct. Explanation: In case of assertion (A): To rationalize the denominator 1 4 ´ 3

´

4- 3

4- 3 = 4 - 3 16 - 3

=

4- 3 13

\ Assertion is False. In case of Reason (R): To rationalize the denominator, multiply numerator and denominator by its conjugate. \ Reason (R) is True: Therefore A is False and R is true.

COMPETENCY BASED QUESTIONS A Case based MCQs

p where p and q q

are not equal to 0. Both A and R are true and R is the correct explanation of A.

7–2 =



23

Read the following passage and answer any four questions.

I. Democracy has given people a powerful right- that

(4 marks each)

is to VOTE. In India, every citizen over 18 years of age has the right to vote. Instead of enjoying it as a holiday, one must vote if he/she truly wants to contribute to the nation-building process and bring about a change.

24

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-IX

Q. 5. If r is rational and s is irrational, then which statement is false? 1 (A) r + s is irrational number. (B) r – s is rational number. (C) rs is irrational number. r (D) is irrational number. s



A survey was done in a small area in which 5 9 + 2 x − 2 x voters were men and 9 + 2x

voters were women. Q. 1. Find x, if number of men is equal to number of women. 1 (A) 6 (B) 5 (C) 8

(D) 9 5

9 + 2x − 2x =

9 + 2x

⇒ =5 9 + 2 x [ 9 + 2 x − 2 x ] ⇒ ( 9 + 2 x )2 − 2 x ( 9 + 2 x ) = 5 ⇒

9 + 2x – 5 =

2 x( 9 + 2 x )

⇒

4 + 2x =

2 x( 9 + 2 x )

Squaring both sides, (4 + 2x)2 =



(

2 x( 9 + 2 x )

⇒ 16 + 4x2 + 16x = 18x + 4x2 ⇒ 16 = 2x ⇒ x = 8 Q. 2. Which mathematical concept is used here ? (A) Number system (B) Circle (C) Area (D) Statistics Ans. Option (A) is correct. Q. 3. ap . aq =  (A) ap + q (B) ap – q pq (C) a (D) (pq)a Ans. Option (A) is correct. ap.aq = ap + q

)

2

1 1

1

Q. 4.



1 73

(A)

(D)

7 7

Ans. Option (B) is correct. 1

Ans. Option (A) is correct. Q. 3. What kind of decimal expansion it has. 1 (A) Terminating (B) non terminating (C) terminating repeating (D) non terminating repeating Ans. Option (D) is correct. Q. 4. How many rational numbers are there between 5 and 7. 1 (A) 0 (B) 1 (C) 2 (D) infinite Ans. Option (D) is correct. Q. 5. Every rational number is a .................. number.1 (A) Prime (B) Composite (C) real (D) even. Ans. Option (C) is correct.

B Subjective Based Questions

1

75

4 (C) 7

5 Q. 2. What is the value of ? 7 (A) 0.714285 (B) 1.4 (C) 1.414 (D) 1.7142

Ans. Option (C) is correct. \

Ans. Option (B) is correct. II. In a school 5 out of every 7 children participated in ‘Save wild life’ campaign organised by the school authorities. Q. 1. What fraction of the students participated in the campaign. 1 2 5 (A) (B) 7 7

=

1 

-2 7 15

2 (C) 7 15

-1



(B) 7 15



(D) 7 15

-4

teacher was, whether

1



73

1

= 75

−

1 3

=7

7 is rational or irrational

number.1

Ans. Option (A) is correct. 1 75

Read the following passage and answer the following questions: I. In a classroom activity on real numbers, the students have to give answers of some questions framed by their teacher on basis of number cards picked up by first 3 Roll numbers. Q. 1. Reena picked up 7 and question asked by

3−5 15

−2

= 7 15

Ans. 7 is an irrational number as decimal expansion of an irrational number is non-terminating and nonrecurring.



REAL NUMBERS

Q. 2. Rajiv picked up card on which it is written

Then x has a non-terminating repeating decimal expansion. 29 Q. 1. Solve 2 3 5 ´2 

15 - 10 and again teacher asked whether it is rational or irrational.

1

Ans. 15 - 10 Þ Irrational number

Ans.

Because difference of two irrational numbers is irrational. Q. 3. Saumya picked third card on which

1 45

Ans. =

45

=

1 3´3´5

Þ

1 3 5

´

was

441 2 2 ´ 57 ´ 7 2

Ans. Decimal expansion of

441 2 ´ 57 ´ 7 2 2

will be non

terminating repeating as it is not of the form p 1 n 2 ´ 5m

5 5

Q. 3. Express both of them in decimal form and say what kind of decimal expansion each has? 49 (a) 100

5.

II. Decimal Form of Rational numbers can be classified into two types. Let x be a rational number whose decimal expansion terminates.

1

is.

5 15

So, Rationalizing factor is

29 29 29 Þ = = 0.145 52 ´ 23 25 ´ 8 200

Q. 2. What will be the decimal expansion of

written. Now teacher asked students to write the 1 . rationalizing factor of 45  1 1

25

Then x can be expressed in the form

Ans. (a)

2 15 

49 49 = 2 2 = 0.49 100 2 ´5

so, it has terminating decimal expansion since p it is of the form n m 1 2 ´ 5

p where p and q

q are co-prime and the prime factorisation of q is of the form 2n5m, where n and m are non-negative and p be a rational number, such vice-versa. Let x = q

(b)

2 2 = 1 1 = 0.13 15 5 ´3

so, it has non terminating decimal expansion, p since it is not of the form n m 1 2 ´ 5

that the prime factorisation of q is not of the form 2n5m where n and m are non-negative integers,



(b)





SELF ASSESSMENT PAPER - 01 Time: 1 hour

MM: 30

UNIT-I I. Multiple Choice Questions 1.

[1×6 = 6]

1

is a/an ............. number. 2 (A) Rational (B) Irrational 6x

(C) Fractional

(D) None of these

(C) 1

(D) 4

(C) 2 + 3 5 + 7

(D) 5 5 + 7

(C) (15)20

(D) (56)20

2

2. 5 = 125 , then x = ............... (A) 2 (B) 3 3. The sum of 2 5 and 3 7 is ............. (B) 2 5 + 3 7

(A) 5 12

4. Value of 710 × 810 is ........... (A) (15)10 (B) (56)10 II. Assertion and Reason Based MCQs

[1×6 = 6]

Directions: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as. (A) Both A and R are true and R is the correct explanation of A. (B) Both A and R are true but R is NOT the correct explanation of A. (C) A is true but R is false. (D) A is false and R is True. 1. Assertion (A):

2 is an irrational number.

Reason (R): The sum of a rational number and an irrational number is an irrational number.

2. Assertion (A): A number 0.57 in the form of rational number

Reason (R): Decimal Expansion of

p 29 . is q 99

29 is non terminating and recurring. 99

III. Very Short Answer Type Questions 

[1×5 = 5]

225 1. Write the simplest form of a rational number . 1250 p 2. Represent 0.239 in the form of , where p and q are integers and q ≠ 0. q 3. If a = 3 and b = 2, then find the value of ab + ba.  4. Is

196

a rational number or not ?

2 1

5. Find 32 5 . IV. Short Answer Type Questions–I 

1. Find the Product of 5 3 (3 + 3 ) ( 5 + 3 ) .  1

1

2. If 23x × 4x = (8 ) 3 × (64 ) 6 , then find the value of x.

[2×2 = 4]



SELF ASSESSMENT PAPER

V. Short Answer Type Questions-II 

27

[3×2 = 6]

1. If x + y = ( 2 − 3 ) , then find the values of x and y. 2

2. what would be the denominator after rationalizing

7 5 3 -5 2

?



VI. Long Answer Type Question 1. If

.9 × 3 × ( 3 n

2

3

)

− n / 2 −2

3m

×2

n

− ( 27 )

3

=

[5×1 = 5]

1 , prove that m – n = 1. 27

VII. Case Based Questions

[1×4 = 4]

Attempt any four sub parts from the given questions. Read the following text and answer the following questions on the basis of the same:

To judge the preparation of class IX students on the topic ‘ Real Numbers’, a maths teacher ask the following questions. 1. Which of the following is an irrational number? (A) 3.245245 (B) 3.245

(C) 3.245245245…

(D) 3.245224522245....

(C) 0.17

(D) 0.71

2. What is the decimal form of 2/11? (A) 0.81

(B) 0.18

3. Decimal form of 2/11 is (A) terminating (C) non-terminating and non-recurring

(B) non-terminating (D) non-terminating and recurring

4. Write p/q form of 0.38 . (A) 38/99 (B) 7/99

(C) 11/18

(D) 1/18

5. If p/q form of 0.38 is m/n then find the value of m + n. (A) 137

(B) 90

(C) 140

(D) 130

qq

UNIT-II

ALGEBRA

Study Time: Maximum time: 3:15 Hrs Maximum questions: 64

CHAPTER

2

POLYNOMIALS

Definition of a polynomial in one variable, with examples and counter examples. Coefficients of a polynomials terms of a polynomial and zero polynomial. Degree of a polynomial. Constant, linear, quadratic, cubic polynomials. Monomials, binomials, trinomials. Factors and multiples. Zeroes of a polynomial. Motivate and state the Remainder Theorem with examples. Statement and proof of the Factor Theorem. Factorization of ax2 + bx + c, a ≠ 0, where a, b and c are real numbers, and of cubic polynomials using the Factor Theorem. Recall of algebraic expressions and identities. Verification of identities: (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx, (x± y)3 = x3 ± y3 ± 3xy(x ± y), x3 ± y3 = (x ± y)(x2 ∓ xy + y2), x3 + y3 + z3 – 3xyz = (x + y + z)(x2 + y2 + z2 – xy – yz – zx) and their use in factorization of polynomials.

Syllabus



Topic-1 Polynomials



Page No. 28

Topic-2 : Factor Theorem 

Revision Notes

List of Topics Topic-1: Polynomials

Page No. 33

Topic-3 : Algebraic Identities 

Page No. 36

 Polynomial : The algebraic expression in which the variables involved have only non-negative integral exponent is called ‘Polynomial’. A polynomial p(x) in one variable x is an algebraic expression in x of the form p (x) = anxn + an–1xn–1+an–2xn–2 + … + a2x2 + a1x + a0. where a0, a1, a2, …, an are real numbers and an ≠ 0. Here, a0, a1, a2, … +a are respectively the co-efficients of x0, x1, n Scan to know x2, …, xn and n is called the degree of the polynomial. more about This form of polynomial is known as the “Standard form of Polynomial”. this topic e.g., : 2x3 – 4x2 + 5x – 7 is a polynomial in one variable (x).  Constant Polynomial : A polynomial of degree zero is called a constant polynomial. 7 3 Polynomials , are constant polynomials. 5 4  Zero Polynomial : A zero polynomial is a polynomial whose whole value is zero. The coefficients of the variables are equal to zero and the constant term is zero so the degree of zero polynomial is not defined. e.g.,: 4, –

 Degree of a Polynomial : Highest power of variable in a polynomial is called the ‘degree of a polynomial’. l In polynomial of one variable, the highest power of the variable is called the degree of the polynomial. e.g., : 4x7 – 3x3 + 2x2 + 3x – 6 is a polynomial in x of degree 7.



POLYNOMIALS

29

30

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-IX

l In a polynomial of more than one variable, the sum of the powers of variables in each term is taken up and the highest sum so obtained is called the degree of the polynomial. e.g.,: 7x3 – 4x2y2 + 3x2y – 3y + 9 is a polynomial in x and y of degree 4.  Types of Polynomials : l On the basis of terms Term : In a polynomial x2 + 3x + 4, the expressions x2, 3x and 4 are called terms. e.g.,: Polynomial x2 + 3x + 7 has three terms. (I) Monomial : A polynomial of one non-zero term, is called a monomial. e.g.,: 2x, –4x2, 7x3, 10x are monomials. (II) Binomial : A polynomial of two non-zero terms, is called a binomial. e.g.,: (4x2 + 8), (7y2 – 3y), (3x – 6), (10x2 – 4) are binomials. (III) Trinomial : A polynomial of three non-zero terms, is called a trinomial. 7 e.g.,: (x2 + 2x + 4), (4x2 + x+14), (3x2+3 3 x + 3 ) are trinomials. 5 l On the basis of degree (I) Linear Polynomial : A polynomial of degree 1 is called a linear polynomial. It is expressed in the form of ax + b, where a and b are real constants and a ≠ 0. e.g.,: 3x + 6 is a linear polynomial in x. (II) Quadratic Polynomial : A polynomial of degree 2 is called a quadratic polynomial. It is expressed in the form of ax2 + bx + c, where a, b and c are real constants and a ≠ 0. e.g.,: 3x2 + 4x + 1 is a quadratic polynomial in x. (III) Cubic Polynomial : A polynomial of degree 3 is called a cubic polynomial. It is expressed in the form of ax3 + bx2 + cx + d, where a, b, c and d are real constants and a ≠ 0. e.g.,: 4y3 + 3y2 + 7y + 4 is a cubic polynomial in y.  Zeroes of a Polynomial : A polynomial is when equal to zero then all the value of x are called zeroes of the polynomial. Zero of a polynomial p(x) is a number c such that p(c) = 0. (i) ‘0’ may be a zero of a polynomial. (ii) Every linear polynomial in one variable has a unique zero of a polynomial. (iii) A non-zero constant polynomial has non zero of a polynomial. (iv) Every real number is a zero of the zero polynomial. (v) Maximum number of zeroes of a polynomial is equal to its degree. Example 1 : Find the zero of the polynomial p (x) = 3x + 1. Solution : Finding the zero of p (x), is the same as solving the equation

p (x) = 0 −1 −1 . So, Now, 3x + 1 = 0, Þ x = is the zero of the polynomial 3x + 1. 3 3

Example 1 Verify whether 3 and 0 are zeroes of polynomial x2 – 3x. Solution:

Step I : Write the given polynomial equal to p(x)



p(x) = x2 – 3x ...(i)

Step II : Putting 3 in place of x and solve it to find the value of p(3). If the value of p(3) is zero, then 3 will be zero of the given polynomial.

Putting x = 3 in (i), we get

p(3) = (3)2 – 3(3) = 9 – 9 = 0 Step III : Similarly, putting 0 in place of x and solving it. Putting x = 0 in (i), we get p(0) = 02 – 3(0) = 0 So, 3 and 0 are zeroes of the given polynomial p (x).



POLYNOMIALS

31

Mnemonics 1. Memorising by ‘SOAP’ Mnemonic

Same opposite Always Positive a3 + b3 = (a + b) (a2 – ab + b2) same opposite Always Positive

2. We can check the long division using mnemonic Dirty Monkeys Smell Bad (i) Divide the leading term of the dividend by the leading term of the divisor. Write this quotient directly above the term you just divided into. (ii) Multiply: Multiply the quotient from step by the entire divisor and write it under the dividend so the like terms are lined up. (iii) Subtract: Change the sign of the subtrahend and subtract. (iv) Bring down the next term and Repeat steps (i) to (iv).

SUBJECTIVE TYPE QUESTIONS Very Short Answer Type Questions (1 mark each) Q. 1. What is the degree of the polynomial (x3 + 5) (4 – x5) ? Sol.

U

Degree of x3 + 5 = 3 Degree of 4 – x5 = 5







Now,



= –x8 – 5x5 + 4x3 + 20



(x3 + 5)(4 – x5) = 4x3 – x8 + 20 – 5x5 3

5

Degree of (x + 5)(4 – x ) = 8.

U

π π Sol. Coefficient of x in expression x2 + x – 7 is . 1 2 2 Q. 3. If –4 is a zero of the polynomial p(x)= x2 + 11x + k, then calculate the value of k. Sol. Given, p(x) = x2 + 11x + k Since, – 4 is a zero of polynomial p(– 4) = 0 or, (– 4)2 + 11 × (– 4) + k = 0 or, 16 – 44 + k = 0 \ k = 28

Sol. For zeroes, put

\



Therefore,

(a) x2 + x

(b) x – x3

(c) 1 + x

(d) 7x3

Sol. Linear polynomial ® 1 + x; degree = 1 3

3

Cubic polynomial ® x – x , 7x ; degree = 3

½ ½ 1

Q. 2. Find the value of the polynomial p(x) = x3 – 3x2 – 2x + 6 at x = 2 U [Board Term I, 2014] 3

1 U

p(x) = 0

x(x – 2)(x – 3) = 0, x = 0, 2, 3

Q. 1. Classify the following as linear, quadratic and R [Board Term I, 2013] cubic polynomials :

Quadratic polynomial ® x + x; degree = 2 U

1

Short Answer Type Questions-I (2 marks each)

2

Q. 4. Write the zeroes of the polynomial  p(x) = x(x – 2)(x – 3).

U

p(0) + p(2) ? Sol. Putting, x = 0 p(0) = 0 – 3 × 0 + 2 = 2 Putting, x = 2 p(2) = 22 – 3 × 2 + 2 = 4 – 6 + 2 = 0 Thus, p(0) + p(2) = 2 + 0 = 2

1

π Q. 2. In the expression x2 + x – 7, what is the 2 coefficient of x ?

Q. 5. If p(x) = x2 – 3x + 2, then what is the value of

1

2

Sol. Given, p(x) = x – 3x – 2x + 6

½

p( 2 ) = ( 2 )3 − 3( 2 )2 − 2( 2 ) + 6 ½



Then,



= 2 2 − 6 − 2 2 + 6



= 0 ½

½

32

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-IX

Q. 3. If y = 2 and y = 0 are the zeroes of the polynomial f(y) = 2y3 – 5y2 + ay + b, find the value of a and b. Ap

2

49  1  1  1 f   =   − 5   + 7 = 1  3  3  3 9



and



49 −59  1 Thus, f (2) – f(–1) + f   = 1 − 13 + =  3 9 9

[Board Term I, 2016]

3

Sol. Given, f(y) = 2y – 5y2 + ay + b \ f(2) = 2(2)3 – 5(2)2 + a(2) + b = 0 or, 16 – 20 + 2a + b = 0 1 or, 2a + b = 4 ...(i) and f(0) = b = 0 From (i), 2a + 0 = 4 or, a = 2 \ a = 2, b = 0 1 [CBSE Marking Scheme, 2016]

Short Answer Type Questions-II (3 marks each) Q. 1. If f(x) = 3x + 5, evaluate f(7) – f(5). Sol. Given, f(x) = 3x + 5 \ f(7) = 3 × 7 + 5 = 26 and f(5) = 3 × 5 + 5 = 20 \ f(7) – f(5) = 26 – 20 = 6

U

1 1 1

1

Q. 2. Find the value of the polynomial x2 – 3x + 6 at (i) x = 2 , Sol. Given,

U

(ii) x = 3. p(x) = x2 – 3x + 6



(i) When



Then, p( 2 ) = ( 2 )2 − 3 × 2 + 6



= 2 – 3 2 + 6



= 8 – 3 2



(ii) When



Then,



= 9 – 9 + 6

x = 2 1



2

x = 3 p(3) = 32 – 3 × 3 + 6

= 6 2 Q. 2. If f(x) = x3 – 3x2 + 3x – 4, find f(2) + f(– 2) + f(0). 4 3 2 Q. 3. If f(x) = x – 4x + 3x – 2x + 1, then check whether U [Board Term I, 2016] f(0) × f(– 1) = f(2) is true or not. Sol. Given, f(x) = x3 – 3x2 + 3x – 4 U [Board Term I, 2016] \ f(2) = (2)3 – 3(2)2 + 3(2) – 4 4 3 Sol. Given, f(x) = x – 4x + 3x2 – 2x + 1 = 8 – 12 + 6 – 4 f(2) = – 2 1 \ f(0) = 1 1 4 3 2 and f(– 2) = (– 2)3 – 3(– 2)2 + 3(– 2) – 4 and f(– 1) = (– 1) – 4(– 1) + 3(– 1) – 2(– 1) + 1 = – 8 – 12 – 6 – 4 = 1 + 4 + 3 + 2 + 1 = 11 f(– 2) = – 30 and f(2) = (2)4 – 4(2)3 + 3(2)2 – 2(2) + 1 and f(0) = – 4 1 = 16 – 32 + 12 – 4 + 1 \ f(2) + f(– 2) + f(0) = – 2 – 30 – 4 = – 36 1 Q. 3. If f(x) = 5x2 – 4x + 5, find f(1) + f(– 1) + f(0).

= 29 – 36 = – 7

U [Board Term I, 2016] Sol. Given, f(x) = 5x2 – 4x + 5 \ f(1) = 5 – 4 + 5 = 6 1 and f(– 1) = 5(– 1)2 – 4(– 1) + 5 = 5 + 4 + 5 = 14 and f(0) = 5 1 \ f(1) + f(– 1) + f(0) = 6 + 14 + 5 = 25 1

\ f(0) × f(– 1) = 11

...(i)

f(2) = –7

...(ii)

Long Answer Type Questions (5 marks each)

From (i) and (ii) we get

(11) ¹ (–7)

[CBSE Marking Scheme, 2016]





3

f(x) = x2 – 5x + 7

1



Then,

f(2) = 22 – 5 × 2 + 7 = 1

1



and

f(–1) = (– 1)2 – 5 (–1) + 7 = 13 1

Students commit errors while substituting and calculating values.

Answering Tip

U [Board Term I, 2014]

Sol. Given,

2

\ f(0) × f(– 1) ¹ f(2) (Not true)

Commonly Made Error

Q. 1. If f(x) = x2 – 5x + 7, evaluate f(2) – f(–1) + f  1  .

1



Practice to substitute the values in polynomial properly.



POLYNOMIALS

33

Topic-2 Factor Theorem Revision Notes  If p(x) is a polynomial of degree x ³ 1 and a is any real number, then (i) Graph of linear equation is a straight line, while graph of quadratic equation is a parabola. (ii) Degree of polynomial = Number of zeroes of polynomial. (iii) If remainder r(x) = 0, then g(x) is a factor of p(x). (iv) (x + a) is a factor of polynomial p(x), if p(– a) = 0. (v) (x – a) is a factor of polynomial p(x), if p( a) = 0. (vi) (x – a)(x – b) is a factor of polynomial p(x), if p(a) = 0 and p(b) = 0. (vii) (ax + b) is a factor of polynomial p(x), if p(– b/a) = 0. (viii) (ax – b) is a factor of polynomial p(x), if p(b/a) = 0.  Factorization of a Polynomial :  By Splitting the Middle Term : Let a quadratic polynomial be x2 + lx + m, where l and m are constants. Factorize the polynomial by splitting the middle term lx as ax + bx, so that ab = m. Then, x2 + lx + m = x2 + ax + bx + ab = x(x + a) + b(x + a). = (x + a) (x + b).  By using Factor Theorem :

Scan to know more about this topic

Factor theorem

Consider a quadratic polynomial ax2 + bx + c, where a, b and c are constants. It has two factors (x – a) and (x – b). \ ax2 + bx + c = a(x – a)(x – b) or, ax2 + bx + c = ax2– a(a + b)x + aab −b c On equating the coefficient of x and constant term, we get a + b = and ab = . a a On simplifying, we get the value of a and b. Example 1. Factorize 6x2 + 17x + 5 by splitting the middle term, and by using the Factor Theorem. Solution : (i) By Splitting the Middle Term : If we find the two numbers a and b, such that a + b = 17 and ab = 6 × 5 = 30, then we can get the factors. Factors of 30 are 1 and 30, 2 and 15, 3 and 10, 5 and 6, of these pairs, 2 and 15 will give us a + b = 17. So, 6x2 + 17x + 5 = 6x2 + (2 + 15)x + 5 = 6x2 + 2x + 15x + 5 = 2x(3x + 1) + 5(3x + 1) = (2x + 5)(3x + 1) (ii) By Factor Theorem : 17 5  6x2 + 17x + 5 = 6  x 2 + x +  = 6p(x).   6 6 5 If a and b are the zeroes of p(x), then, new line 6x2 + 17x + 5 = 6(x – a)(x – b), and ab = . 6 Then, see some possibilities for a and b. 1 1 5 5 They could be ± , ± , ± , ± , ± 1. 2 3 3 2 1 17  1  5  1 Now, p   = +   + ≠ 0 ,  2 4 6  2 6  −1  But, p   = 0. So,  3

1   x +  is a factor of p(x). 3

5  Similarly, we will get  x +  as a factor of p(x).  2

34

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-IX

1  5  \ 6x2 + 17x + 5 = 6  x +   x +   3  2  3x + 1   2 x + 5  = 6  = (3x + 1)(2x + 5).  3   2  In this example, the use of the splitting method appears more efficient.

Example 2 Factorize the cubic polynomial x3 + 6x2 + 11x + 6. Solution: Step I : Consider the given cubic polynomial as p(x) and find the constant term p(x) = x3 + 6x2 + 11x + 6 Here, constant term = 6 Step II : Find all the factors of constant term of p(x). All possible factor of 6 are ± 1, ± 2, ± 3 and ± 6. Step III : Check at which factor, p(x) is zero by trial method and get one factor of p(x). At x = – 1 p( – 1) = (– 1)3 + 6(– 1)2 + 11(– 1) + 6 = – 1 + 6 – 11 + 6

= – 12 + 12 = 0 So, (x + 1) is a factor of p(x). Step IV : Now, write p(x) as the product of this factor and a quadratic polynomial. On dividing p(x) by (x + 1), we get quotient x2 + 5x + 6 So, p(x) = (x + 1)(x2 + 5x + 6) Step V : Now, use splitting method or factor theorem to find the factor of p(x). Now, by splitting the middle term of quotient, we get p(x) = (x + 1)[x2 + 3x + 2x + 6] = (x + 1)[x(x + 3) + 2(x + 3)] = (x + 1)(x + 2)(x + 3) Hence, the factors of given polynomial are (x + 1), (x + 2) and (x + 3).

SUBJECTIVE TYPE QUESTIONS Very Short Answer Type Questions (1 mark each)

 1 Q. 1. If f(x) be a polynomial such that f  −  = 0, then  3

R calculate one factor of f(x). 1   f  −  = 0 Sol. Since,  3 1 \ − is a zero of polynomial f(x). 3 1 So, x + or 3x + 1 is a factor of f(x). 1 3 Q. 2. Find the value of m, if x + 4 is a factor of the polynomial x2 + 3x + m. U [Board Term I, 2015] Sol. Given, x + 4 is a factor of x2 + 3x + m = p(x) \ p(– 4) = 0 or, 16 – 12 + m = 0 or, m = – 4 1 Q. 3. Find the value of k, if 2x – 1 is a factor of the polynomial 6x2 + kx – 2. U [Board Term I, 2015]

Sol. Since, 2x – 1 is a factor of p(x) = 6x2 + kx – 2

Thus,



or, or,

 1 p  = 0  2 1 1 6. + k . − 2 = 0 4 2 k = 1 1 [CBSE Marking Scheme, 2015]

Q. 4. Factorize : 6 – x – x2 Sol.

U [Board Term I, 2016] 2

6 – x – x = 6 – 3x + 2x – x2

= 3(2 – x) + x(2 – x) = (2 – x)(3 + x)

1

[CBSE Marking Scheme, 2016]

Commonly Made Error

While splitting the middle term in factorisation, the students commit an error in putting signs because in multiplication two negative signs give positive product while adding two negative numbers, the result is negative.

Answering Tip

Students should be careful with signs and do

ample practice so that they do not make errors, while splitting the middle term.

3 3 U [Board Term I, 2016] Q. 5. Factorize : 8y – 125x  3 Sol. 8y – 125x3 = (2y)3 – (5x)3 = (2y – 5x)(4y2 + 10xy + 25x2) 1 [CBSE Marking Scheme, 2016]



POLYNOMIALS

Short Answer Type Questions-I (2 marks each) Q. 1. Find the value of k, so that polynomial x3 + 3x2 – kx – 3 has one factor as x + 3. U [Board Term I, 2016] f(x) = x3 + 3x2 – kx – 3

Sol. Let

Since, (x + 3) is a factor of f(x). Then,

f(– 3) = 0 3

1

Q. 5. Find the value of ‘k’ if (x – 1) is a factor of p(x) = 2x2 + kx + 2 .

Sol. Given, p(x) = 2x + kx + 2 Since, (x – 1) is a factor of p(x), then p(1) = 0. \ 2 (1)2 + k(1) +

2 = 0

½



or,

2 = 0

½



or,

3k – 3 = 0

or, k = 1 1 Q. 2. Find the value of k, if x – 2 is a factor of f(x) = x2 + kx + 2k. U [Board Term I, 2016] Sol. Given, (x – 2) is a factor of f(x). f(2) = 0

1

2



Þ (2) + k(2) + 2k = 0



Þ

4 + 2k + 2k = 0



Þ

4 + 4k = 0



Þ

k = – 1

1

[CBSE Marking Scheme, 2016]

Commonly Made Error

Sometimes students are confused between Remainder Theorem and Factor Theorem.

Answering Tip

2+k+

k = –2 – 2

½

Short Answer Type Questions-II (3 marks each)

or, – 27 + 27 + 3k – 3 = 0

\

½



or, (– 3) + 3(– 3) – k(–3) – 3 = 0



U [Board Term I, 2014]

2

2

or,

35

Understand the concepts of both the theorems

Q. 1. Factorize : x3 – 3x2 – 9x – 5

Sol. Let p(x) = x3 – 3x2 – 9x – 5, Since, p (– 1) = – 1 – 3 + 9 – 5 = 0 Therefore, (x + 1) is a factor of x3 – 3x2 – 9x – 5. ½ \ (x3 – 3x2 – 9x – 5) = (x + 1)(x2 – 4x – 5) 1 Now, x2 – 4x – 5 = x2 – 5x + x – 5 = x(x – 5) + 1(x – 5) x2 – 4x – 5 = (x + 1)(x – 5) 1½ \ p(x) = (x + 1)(x – 5)(x + 1) [CBSE Marking Scheme, 2014] Q. 2. Factorize : 2y3 + y2 – 2y – 1 U [NCERT] Sol. Let p(y) = 2y3 + y2 – 2y – 1 Q p(– 1) = 2(– 1)3 + (– 1)2 – 2(– 1) – 1 = – 2 + 1 + 2 – 1 = 0 2 \ (y + 1) is a factor of given polynomial. \ 2y3 + y2 – 2y – 1 = (y + 1)(2y2 – y – 1) Now, 2y2 – y – 1 = 2y2 – 2y + y – 1 = 2y(y – 1) + 1(y – 1) = (2y + 1)(y – 1) p(y) = (y + 1)(2y + 1)(y – 1) 1

Long Answer Type Questions (5 marks each)

and do adequate practice to solve problems based on them.

Q. 3. Find the value of ‘a‘ for which (x – 1) is a factor of the polynomial a2x3 – 4ax + 4a – 1. U [Board Term I, 2016] Sol. Let f(x) = a2x3 – 4ax + 4a – 1 Since, (x – 1) is a factor of f(x) Then, f(1) = 0 1 Þ a2(1)3 – 4a(1) + 4a – 1 = 0 Þ a2 – 4a + 4a – 1 = 0 Þ a2 – 1 = 0 Þ a = ± 1 1 [CBSE Marking Scheme, 2016] Q. 4. For what value of k, is the polynomial p(x) = 2x3 – U kx2 + 3x+ 10 exactly divisible by (x + 2) ? [Board Term I, 2014] Sol. Since, (x + 2) is a factor of p(x). ½ Thus, p(–2) = 0 or, 2(–2)3 – k(–2)2 + 3(–2) +10 = 0 1 or, – 16 – 4k – 6 + 10 = 0 k = – 3  ½

U [Board Term I, 2014], [NCERT]

Q. 1. Using factor theorem, show that (m – n), (n – p) and (p – m) are factors of m(n2 – p2) + n(p2 – m2) + p(m2 – n2) U [Board Term I, 2015] Sol. Let f = m(n2 – p2) + n(p2 – m2) + p(m2 – n2) \ f(m) = n = n(n2 – p2) + n(p2 – n2) + p(n2 – n2) = n(n2 – p2) – n(n2 – p2) + 0 = 0 So, m – n is a factor of f. Similarly, f(n = p) = 0 & f(p = m)= 0 \ (m – n), (n – p) and (p – m) are factors of f. 5 [CBSE Marking Scheme, 2015] Q. 2. Factorize : 9x3 – 3x2 – 5x – 1 U [Board Term I, 2016] Sol. Let

p(x) = 9x3 – 3x2 – 5x – 1



Factors of 1 = ± 1





p(1) = 9 – 3 – 5 – 1 = 0

1

36

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-IX

i.e., x – 1 is a factor of p(x).



= (x – 1)(9x2 + 6x + 1)

1





= (x – 1)(3x + 1)2

1



\

p(x) = (x – 1)(3x + 1)(3x + 1)

x – 1)x3 – 23x2 + 142x – 120



\ 9x3 – 3x2 – 5x – 1 = 9x2(x – 1) + 6x(x – 1) + 1(x – 1)



x2 – 22x + 120



1

x3 – x2 (–) (+)

1

[CBSE Marking Scheme, 2016]



Q. 3. Factorize : x3 – 23x2 + 142x – 120. U [KVS, 2019] Sol. x3 – 23x2 + 142x – 120 Put, x = 1 Then, x3 – 23x2 + 142x – 120 = (1)3 – 23(1)2 + 142(1) – 120 = 1 – 23 + 142 – 120 = 143 – 143 = 0 \ Remainder = 0 \ (x – 1) is a factor of x3 – 23x2 + 142x – 120.

– 22x2 + 142x – 120 – 22x2 + 22x (+) (–)

120x – 120 120x – 120 (–) (+) 0







= (x – 1)(x2 – 12x – 10x + 120)



= (x – 1)[x(x – 12) – 10(x – 12)]



\ x3 – 23x2 + 142x – 120 = (x – 1)(x – 10)(x – 12)

3

2

2

x – 23x + 142x – 120 = (x – 1)(x – 22x + 120)

Topic-3 Algebraic Identities Revision Notes  Algebraic Identities : An algebraic identity is an algebraic equation that is true for all values of the variables occurring in it.

 Some useful algebraic identities:

(i)

(x + y)2 = x2 + 2xy + y2

(ii)

(x – y)2 = x2 – 2xy + y2

(iii)

x2 – y2 = (x + y)(x – y)

(iv) (v) (vi) (vii)

Scan to know more about this topic

(x + a)(x + b) = x2 + (a + b)x + ab (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx (x + y)3 = x3 + y3 + 3xy(x + y) (x – y)3 = x3 – y3 – 3xy(x – y)

(viii) x3 + y3 + z3 – 3xyz = (x + y + z)(x2 + y2 + z2 – xy – yz– zx) (ix) (x)



x3 + y3 = (x + y)(x2 – xy + y2) x3 – y3 = (x – y)(x2 + xy + y2)

Example 1. Factorize the following :

(i) x2 + 4xy + 4y2

(ii) x3 – 8

(iii) 4x2 – 12xy + 9y2

(iv) x3 + 8y3 + 6x2y + 12xy2



Solution :



(i) x2 + 4xy + 4y2 = (x)2 + 2 × (x) × (2y) + (2y)2 = (x + 2y)2 = (x + 2y)(x + 2y)



(ii) x3 – 8 = x3 – 23 = (x – 2)(x2 + 2x + 22) = (x – 2)(x2 + 2x + 4)











(iv) x3 + 8y3 + 6x2y + 12xy2





= (x)3 + (2y)3 + 3(x)(2y)[x + 2y]





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

(iii) 4x2 – 12xy + 9y2 = (2x)2 – 2(2x)(3y) + (3y)2 = (2x – 3y)2 = (2x – 3y)(2x – 3y)

Algebraic identities

5



POLYNOMIALS

37

Example 3 Evaluate (102)3 by using suitable identities. Solution: Step I : Express the given number without power as the sum or difference of two numbers Given number without power is 102. Since, it is greater than 100, so it can be written as 100 + 2 \ (102)3 = (100 + 2)3 Step II : Compare the expression with (x + y)3. On comparing (100 + 2)3 with (x + y)3, we get x = 100 and y = 2



Step III : Use the identity (x + y)3 = x3 + y3 + 3xy (x + y) to expand it. (102)3 = (100 + 2)3 = (100)3 + (2)3 + 3(100)(2)(100 + 2) Step IV : Simplify the above expression. (102)3 = 1000000 + 8 + 60000 + 1200 = 1000000 + 8 + 61200 = 1061208

SUBJECTIVE TYPE QUESTIONS Very Short Answer Type Questions (1 mark each)

2 Q. 1. Write the coefficient of x in the expansion of 3 (x – 2) . A Sol. (x – 2)3 = (x)3 – (2)3 – 3 × x × 2(x – 2) = x3 – 8 – 6x2 + 12x Coefficient of x2 in the expansion of (x – 2)3 = – 6 1 1 1 = 4, then calculate the value of x2 + 2 . Q. 2. If x + x x  U 2 1 1     1 Sol. x 2 + 2 =  x +  − 2( x )   x x x

= (4)2 – 2 = 16 – 2 = 14 1 833 + 17 3 Q. 3. Calculate the value of . 832 − 83 × 17 + 17 2 Ap [NCERT Exemp.]





(

3





)





= (2a + 2b)[(2a)2 + (2b)2 – (2a) × (2b)] 1





[ a3 + b3 = (a + b)(a2 + b2 – ab)]

= 2(a + b) × 4(a2 + b2 – ab) = 8(a + b)(a2 + b2 – ab) 1 3 3 Q. 4. Factorize : 8x – (2x – y) U [Board Term I, 2015] Sol. 8x3 – (2x – y)3 = (2x)3 – (2x – y)3 = [2x – (2x – y)][(2x)2 + (2x – y)2 + 2x(2x – y)] [Since, (a3 – b3) = (a – b)(a2 + b2 + ab)] = y[4x2 + 4x2 + y2 – 4xy + 4x2 – 2xy]

Short Answer Type Questions-I (2 marks each)



= a2 + b2 + c2 + 2ab + 2bc + 2ca ( 2x + (–y) + z) = (2x)2 + (–y)2 + z2 + 2(2x)(–y)  + 2(–y)(z) + 2(z)(2x) = 4x2 + y2 + z2 – 4xy – 2yz + 4xz



2

2  1 Q. 2. Expand :  x − y 3 3  3

2  1 Sol.  x − y 3 3 

x 3 8 y 3 2 xy  x 2 y  − −  −  27 27 3 3 3 

x 3 8 y 3 2 x 2 y 4 xy 2 − − + 1 9 9 27 27 Q. 3. Factorize : 8a3 + 8b3 U [Board Term I, 2014] Sol. 8a3 + 8b3 = (2a)3 + (2b)3

)

Q. 1. Expand by using identity (2x – y + z)2. R [Board Term I, 2014] Sol. By using the identity, (a + b + c)2



=

= y[12x2 + y2 – 6xy] 2 [CBSE Marking Scheme, 2015] 3 3 2 2 Q. 5. Factorize : 64a – 27b – 144a b + 108ab U [Board Term I, 2014], [NCERT] Sol. 64a3 – 27b3 – 144a2b + 108ab2



= (4a)3 – (3b)3 – 3 × (4a)2 × (3b) + 3 × (4a) × (3b)2 ½



= (4a)3 – (3b)3 – 3 × 4a × 3b(4a – 3b)

½



= (4a – 3b)3 

1

3



1

=

( 83 + 17 ) 83 2 − 83 × 17 + 17 2 833 + 17 3 Sol. 2 2 = 83 − 83 × 17 + 17 83 2 − 83 × 17 + 17 2 [ a3 + b3 = (a+ b)(a2 – ab + b2)] = 83 + 17 = 100 1

(

3

1 2 1 2y  1  2  =  x −  y − 3 × x × y  x −   3  3  3 3 3 3

U [Board Term I, 2013]

Commonly Made Error

While factorizing, the students factorize once

and leave the answer without checking that few expressions can be further factorized.

38

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-IX

Now,

Answering Tip

whether an expression can be further factorized otherwise they tend to miss one step and thereby their marks.

Sol.

3 Q. 1. Using a suitable identity, find (98) . A [Board Term I, 2016]

(98)3 = (100 – 2)3 = (100)3 – (2)3 – 3 × 100 × 2(100 – 2) 1 = 1000000 – 8 – 600 × (100 – 2) = 1000000 – 8 – 60000 + 1200 1 = 1000000 – 58808 = 941192 1 Q. 2. Evaluate : ( 2 + 3 )2 + ( 5 − 2 )2 U [Board Term I, 2013] Sol. ( 2 + 3 )2 + ( 5 − 2 )2 = ( 2 )2 + ( 3 )2 + 2 × 2 × 3 + ( 5 )2 + ( 2 )2 − 2 × 5 × 2 1 = 2 + 3 + 2 6 + 5 + 2 – 2 10 1



1 Q. 3. Simplify : (2a + 3b)3 – (2a – 3b)3 U [Board Term I, 2015] Sol. Let (2a + 3b)3 – (2a – 3b)3 = x3 – y3, where 2a + 3b =x and 2a – 3b = y = [(2a + 3b) – (2a – 3b)][(2a + 3b)2 + (2a + 3b)











x3 +

1 = ± 10 x

1  2 1 1   1 =  x +   x − x· + 2  3 x  x x  x 

Sometimes students take whole cube of z2 +

1

1 z2

= 14, which leads to wrong calculation.

2

= 6b[(4a + 12ab + 9b2) + (4a 2 – 9b2)







= 6b(12a2 + 9b2)

Answering Tip

+ (4a2 – 12ab + 9b2)] 2



1

Students should use the identity 3

2

1 1   3 1   z + z3  =  z + z  − 3  z + z       

= 6b × 3 × (4a + 3b )

= 18b(4a2 + 3b2) 1 1 1 3 2 Q. 4. If x + 2 = 7. Find the value of x + 3 , taking x x 1 only the positive value of x + . x U [Board Term I, 2016] 2 1 1 1 2  Sol.  x +  = x + x 2 + 2·x· x x 1 2 = x + 2 +2 x 1   2 = 7 + 2 ∵ x + x 2 = 7    = 9 1 1    x +  = ± 3 x



x+

1

(2a – 3b) + (2a – 3b)2] 1







1 = 98 + 2 = 100 x 

Commonly Made Error

2





1 = 98 x2

 2 1  = ± (10 )  x + 2 − 1  x   = ± (10)(98 – 1) = ± 10 × 97 = ± 970 1 [CBSE Marking Scheme, 2016]

= 12 + 2 6 − 2 10 2( 6 + 6 − 10 ) =

= (x –y)(x + xy + y )

1  2 1 1   x + 2 − x·  x x x

2

or,  x +





x2 +



Sol.

2

1  = x + x3 

1  1   =  x +   x 2 + 2 − 1    x x = (3)(7 – 1) = 3 × 6 = 18 1 1 1 2 3 Q. 5. If x + 2 = 98 , then find value of x + 3 . x x U [Board Term I, 2016]

Students should be particular and check

Short Answer Type Questions-II (3 marks each)



x3 +

1 x+ = 3 [on taking +ve value] 1 x



Long Answer Type Questions (5 marks each) 1 1 4 = 2, find x + 4 . x x U [Board Term I, 2015, Set-2] Sol. On squaring both sides 2 1  2 x − 1   =2 x 1 1 x 2 + 2 − 2 x × = 4 x x 1 2 x + 2 = 4 + 2 = 6 x Q. 1. If x −

2





 2 1  4 1  x + 2  =  x + 4  + 2 x x

1



POLYNOMIALS

1 +2 1 4 x 1 Þ 36 – 2 = x 4 + 4 1 x 1 x 4 + 4 = 34 Þ 1 x Q. 2. Prove that (a2 – b2)3 + (b2 – c2)3 + (c2 – a2)3 = 3(a + b)(b + c)(c + a)(a – b)(b – c)(c – a). U [Board Term I, 2016] Þ (6)2 = x 4 +

2

2

2

2

2

Sol. Let x = a – b , y = b – c , z = c – a



x + y + z = a2 – b2 + b2 – c2 + c2 – a2

Now,



= 0

1

x + y + z = 0

1

x3 + y3 + z3 = 3xyz

1



2

Sol. RHS =

i.e., (a2 – b2)3 + (b2 – c2)3 + (c2 – a2)3

2

= 3(a – b2)(b2 – c2)(c2 – a2)

1



= 3(a + b)(a – b)(b + c)(b – c)(c + a)(c – a)



= 3(a + b)(b + c)(c + a)(a – b)(b – c)(c – a) 1 1 3 3 3 Q. 3. Prove that x + y + z – 3xyz = (x + y + z) 2 [(x – y)2 + (y – z)2 + (z – x)2] A [Board Term I, 2015, Set-1] [NCERT]



1 (x + y + z)[(x – y)2 + (y – z)2 + (z – x)2] 1 2

1 = (x + y + z) [x2 + y2 – 2xy + y2 + z2 – 2yz + z2 + x2 – 2zx] 2 1 = (x + y + z)[2x2 + 2y2 + 2z2 – 2xy – 2yz – 2zx] 2 2 1 = (x + y + z). 2[x2 + y2 + z2 – xy – yz – zx] 2 = x3 + y3 + z3 – 3xyz [Using identity x3 + y3 + z3 – 3xyz = (x + y + z) (x2 + y2 + z2 – xy – yz – zx)] 2 = LHS Q. 4. Factorize : x12 – y12. U [KVS, 2019]

Sol. x12 – y12

= (x6)2 – (y6)2



= (x6 – y6)(x6 + y6)



= {(x3)2 – (y3)2}(x6 + y6) = (x3 – y3)(x3 + y3)(x6 + y6) = {(x)3 – (y)3}{(x)3 + (y)3}(x6 + y6) = (x – y)(x2 + xy+ y2)(x + y)(x2 – xy+ y2)(x6 + y6) = (x – y)(x + y)(x2 + xy + y2)(x2 – xy + y2)  {(x2)3 + (y2)3} = (x – y)(x+ y)(x2 + xy + y2)(x2 – xy+ y2)(x2 + y2)  (x4 – x2y2 + y4) 5

OBJECTIVE TYPE QUESTIONS A Multiple Choice Questions

Q. 1. Which one of the following is a polynomial ? x2 2 (A) − 2 (B) 2 x − 1 2 x 3 2

x + (C)

3x 2 x

(D)

x −1 x +1

A [NCERT Exemp.]  Sol. Option (C) is correct. Explanation : It is a polynomial because degree of polynomial is whole number. 3x 3 / 2 = x2 + 1/ 2 x 3 1 − 2

= x 2 + 3x 2 = x 2 + 3x

3 −1 2

= x 2 + 3x

Q. 2. 2 is a polynomial of degree (A) 2 (B) 0 1 (C) 1 (D) 2 A [NCERT Exemp.]  Sol. Option (B) is correct.

39

Explanation :

(1 mark each)

2 = 2x 0

Because exponent of x is 0. Q. 3. Degree of the zero polynomial is (A) 0 (B) 1 (C) any natural number (D) not defined  A [NCERT Exemp.] Sol. Option (D) is correct. Explanation : Because in zero polynomial the coefficient of any variable is zero. So, we will not be able to determine the degree of the polynomial. Q. 4. If p ( x ) = x 2 − 2 2 x + 1, then p(2 2) is equal to (A) 0 (B) 1 (C) 4 2  Sol. Option (B) is correct. Explanation :

(D) 8 2 +1 A [NCERT Exemp.]

On putting x = 2 2 in given equation. p( x ) = x 2 − 2 2x + 1 p(2 2= ) (2 2 )2 − 2 2 × 2 2 + 1 = 8−8+1= 9−8 = 1

Q. 5. One of the factors of (25x2 – 1) + (1 + 5x)2 is (A) 5 + x (B) 5 – x (C) 5x – 1 (D) 10x Ans. Option (D) is correct. Explanation: (25x2 – 1) + (1 + 5x)2

40

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-IX

[(5x)2 – (1)2] + (1 + 5x)2 [Using identity a2 – b2 = (a + b)(a – b)] [(1 + 5x)(5x – 1)] + (1 + 5x)2 (1 + 5x)[5x – 1 + 1 + 5x] (1 + 5x)(10x) One of the factors = 10x Q. 6. The factorisation of 4x2 + 8x + 3 is (A) (x + 1) (x + 3) (B) (2x + 1) (2x + 3) (C) (2x + 2) (2x + 5) (D) (2x – 1) (2x – 3) Ans. Option (B) is correct. Explanation: 4x2 + 8x + 3 (By splitting the middle term) 4x2 + 6x + 2x + 3 2x(2x + 3) + 1(2x + 3) (2x + 1) (2x + 3) Q. 7. Factorize 12a2b – 6ab2 (A) 6ab(2a – b) (B) 2ab(6a – 3b) (C) 3ab(4a – 2b) (D) 6a(2ab – b) Ans. Option (A) is correct. Explanation: 12a2b – 6ab2 = 6ab(2a – b) [By taking common] Q. 8. If (x + 2) is a factor of g(x) = 3x2 + x – k, then value of k will be (A) 12 (B) 8 (C) 10 (D) 14 Ans. Option (C) is correct. Explanation: Since, x + 2 is a factor of g(x) then g(–2) = 0 g(–2) = 3(–2)2 + (–2) – k = 0 = 12 – 2 – k = 0 k = 10 Q. 9. Which of the following is a factor of (x + y)3– (x3+ y3)? (A) x2 + y2+ 2xy (B) x2 + y2 – xy 2 (C) xy (D) 3xy  [NCERT Exemp.] [(U) KVS 2019] Ans. Option (D) is correct. Explanation :

(x + y ) − ( x3 + y3 ) 3

[Using identity a3 + b 3 = ( a + b )( a 2 + b 2 − 2ab )]

(

)

3 = ( x + y ) − ( x + y ) x 2 + y 2 − xy  2 = ( x + y ) ( x + y ) − x 2 + y 2 − xy    [Using identity ( a + b )2 = a 2 + b 2 + 2ab]

(

)

= ( x + y )  x 2 + y 2 + 2xy − x 2 − y 2 + xy  =

(x + y )(3xy )

One of the factor of given polynomial is 3xy. x y Q. 10. If + = −1 ( x , y ≠ 0 ) the value of x3– y3 is y x

(A) 1 (B) −1 1 (C) 0 (D) 2

 Ans. Option (C) is correct.

[NCERT Exemp.]

Explanation : x y + = −1 y x x2 + y2 = −1 xy

 

x2 + y2 = − xy

x 2 + y 2 + xy = 0

...(i)

(

x 3 − y 3 = ( x − y ) x 2 + y 2 + xy

 

)

x 3 − y 3 = ( x − y )( 0 ) [ Using equation(i)]







  

x3 − y3 = 0

2 Q. 11. If 49 x −

( b)

2

1  1  =  7 x +   7 x −  then the value  2  2

of b is 1 (A) 0 (B) 2 1 1 (C) (D) 4 2  Ans. Option (C) is correct. Explanation :

49x 2 −

( ) b

2

[NCERT Exemp.]

2  2 1  = ( 7 x ) −     2   

 Using identity a 2 − b 2 = ( a + b )( a − b )  49x 2 −

( b)

2

=49x 2 −

1 4

On comparing both the sides 2 1 − b = − 4

( )

On multiplying both sides by –1 2 1 b = 4 1 b= 4

( )

Q. 12. If a + b + c = 0, then a3+ b3+ c3 is equal to (A) 0 (B) abc (C) 3abc (D) 2abc  [NCERT Exemp.] Ans. Option (C) is correct. Explanation : a + b + c = 0 a + b = – c . ..(i) On cubing both the sides

(a + b) = ( −c ) 3 3 3 a + b + 3ab ( a + b ) =− ( c) 3

3

a3 + b 3 + 3ab ( −c ) =( −c )

3

By using equation (i)



POLYNOMIALS

a3 + b 3 − 3abc =− ( c)  

3

a3 + b 3 + c 3 = 3abc

B Assertion & Reason

Directions: In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Mark the correct choice as: (A) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A). (B) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A). (C) Assertion (A) is true but reason (R) is false. (D) Assertion (A) is false but reason (R) is true. Q. 1. Assertion (A): A polynomial may have more than one zero. Reason (R): Every real number is zero of the zero polynomial. Ans. Option (B) is correct. Explanation: In case of Assertion (A): Maximum number of zeros of polynomial is equal to its degree therefore polynomial may have more than one zero. In case of Reason (R): In zero polynomial 0(x – k) where k is a real number. For determining the zero of the given polynomial put x – k = 0; x = k, where k can be any real number. Q. 2. Assertion (A): Degree of the polynomial 4x4 + 0x3 + 0x5 + 5x + 7 is 4. Reason (R): The highest power of the variable is 4, so the degree of polynomial is 4. Ans. Option (A) is correct. Explanation: In case of assertion (A): Highest power of variable x is 4. So, degree of the polynomial is 4 Q. 3. Assertion (A): The equation x2 + 5x + 6 = (x + p)(x + q) where p = 3, q = 2 Reason (R): Roots of the polynomial x2 + 5x – 6 are 3, 2 Ans. Option (C) is correct. Explanation: x2 + 5x + 6 = (x + p)(x + q) Þ x2 + 3x + 2x + 6 = (x + p)(x + q) Þ (x + 3)(x + 2) = (x + p)(x + q) Comparing the component,

Here, p = 3, q = 2 Hence, Assertion is true In reason x = 3 and 2 are the roots of x2 + 5x + 6 then, p(3) = 0 and p(2) = 0 px = x2 + 5x + 6 x = 3 x = 2 p(3) = (3)2 + 5 × 3 + 6 p(2) = (2)2 + 5 × 2 + 6 p(3) = 9 + 15 + 6 p(2) = 4 + 10 + 6 p(3) = 30 p(2) = 20 \ p(3) ¹ 0 \ p(2) ¹ 0 Since, 3, 2 are not roots of x2 + 5x + 6 Hence, reason is false. Q. 4. Assertion (A): If x + 2 is a factor of P(x) = 2x2 + 3x – k then value of k = 14 Reason (R): (x – a) is a factor of polynomial P(x), if P(a) = 0. Ans. Option (A) is correct. Explanation: In case of assertion (A): Since, x – 2 is a factor of P(x) then P(2) = 0 P(2) = 2 × (2)2 + 3 × 2 – k = 0 8 + 6 – k = 0 k = 14 \ Assertion is true and R is correct explanation of A. Q. 5. Assertion (A): (3x + 4y)3 = 27x3 + 64y3 + 36xy(3x + 4y) Reason (R): (x + y)3 = x3 + y3 + 3xy(x + y) Ans. Option (A) is correct. Explanation: In case of Assertion (A): (3x + 4y)3 = (3x)3 + (4y)3 + 3 × 3x × 4y(3x + 4y) Þ 27x3 + 64y3 + 36xy(3x + y) Hence, Assertion is true and identity used in this is (a + b)3 = a3 + b3 + 3ab(a + b) therefore reason is also true and correct explanation of Assertion. Q. 6. Assertion (A): Value of (111)3 is 1367631 Reason (R): x3 + y3 = (x + y)(x2 – xy + y2) Ans. Option (B) is correct. Explanation: In case of assertion (A): (111)3 = (100 + 11)3 = (100)3 + (11)3 + 3(100)2 × 11 + 3(100)(11)2 = 1367631 \ Assertion is True. In case of Reason (R) : Identity used is true but not for Assertion. \ Both (A) and (R) are true but (R) is not correct explanation of (A).

COMPETENCY BASED QUESTIONS A Case based MCQs

Read the following passage and answer any four questions. I. National Association For The Blind (NAB) aimed to empower and well-inform visually challenged population of our country, thus enabling them to lead a life of dignity and productivity.

41

(4 marks each)

42

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-IX

1  Ravi donated `  x 3 +  to NAB. When his cousin x3   asks to tell the amount donated by him, he just gave the below hint. 1 x + = 10 x Answer the following questions : Q. 1. (x + a)(x + b) = x2 + ................ x + ab (A) a + b (B) ab a (C) a – b (D) b

1

Ans. Option (A) is correct. Q. 2. (x – y)3 = 1 (A) x3 – y3 – 3xy (B) x3 – y3 – 3xy(x – y) (C) x3 – y3 – 3xy(x + y) (D) x3 – y3 Ans. Option (B) is correct. Q. 3. Which mathematical concept is involved in the above situation ? 1 (A) Polynomial (B) Circle (C) Lines and angles (D) Triangle Ans. Option (A) is correct. Q. 4. Find the amount donated by Ravi. 1 (A) ` 1000 (B) ` 850 (C) ` 970 (D) ` 900 Ans. Option (C) is correct. Explanation: 1 x+ = 10 x By cubing both sides we get 3

1ö æ 3 ç x + x ÷ = (10) è ø 1 1 1 3 x + 3 + 3x  x +  = 1000 x x x 1 3 (x + 3 ) + 3(10) = 1000 x 1 (x3 + 3 ) = 1000 – 30 = 970 1 x 1 Q. 5. Find the amount donated by Ravi if x + = 7. 1 x (A) ` 522 (B) ` 422 (C) ` 222 (D) ` 322 Ans. Option (D) is correct. Explanation: 1 x+ =7 x By cubing both sides we get 3

1ö æ 3 ç x + x ÷ = (7) è ø 1 1 1 3 x + 3 + 3x (x + )= 343 x x x 1 (x3 + 3 ) + 3(7) = 343 x 1 (x3 + 3 ) = 343 – 21 = 322. x

II. Beti Bachao, Beti Padhao (BBBP) is a personal campaign of the Government of India that aims to generate awareness and improve the efficiency of welfare services intended for girls.

1

In a school, a group of (x + y) teachers, (x2 + y2) girls and (x3 + y3) boys organised a campaign on Beti Bachao, Beti Padhao. Q. 1. Which mathematical concept is used here? 1 (A) Linear equations (B) Triangles (C) Polynomials (D) Area Ans. Option (C) is correct. Q. 2. Which is the correct identity? 1 2 2 2 (A) (a + b) = a + b – 2ab (B) (a + b)2 = a2 + b2 + 2ab (C) (a + b)2 = a2 – b2 – 2ab (D) All are correct. Ans. Option (B) is correct. Q. 3. (x – y)3 =  1 (A) (x2 – y2 – 3xy(x – y)) (B) (x3 – y3 – 3xy(x – y)) (C) (x3 – y3 – 2xy(x – y)) (D) (x3 – y3 – 3xyx – y) Ans. Option (B) is correct. Q. 4. If in the group, there are 10 teachers and 58 girls, then what is the number of boys? 1 (A) 300 (B) 360 (C) 350 (D) 370 Ans. Option (D) is correct. Explanation: No. of teachers = x + y = 10 By squaring both sides Þ (x + y)2 = (10)2 2 Þ x + y2 + 2xy = 100 [Since (a + b)2 = a2 + b2 + 2ab] No. of students = (x2 + y2) = 58 Þ 58 + 2xy = 100 Þ 2xy = 100 – 58 Þ 2xy = 42 42 Þ xy = 2 Þ xy = 21½ Now, since (x + y)3 = [x3 + y3 + 3xy(x + y)] Þ (10)3 = [x3 + y3 + 3 × 21(10)] Þ 1000 = (x3 + y3 + 630) Þ 1000 – 630 = (x3 + y3) Þ (x3 + y3) = 370½ Q. 5. Using part (iv), find (x2 – y2) if x – y = 23. 1 (A) 200 (B) 330 (C) 120 (D) 230



POLYNOMIALS

Ans. Option (D) is correct. Explanation: Given x – y = 23 Also, x + y = 10 x2 – y2 = (x + y)(x – y)½ = 10 × 23 = 230½

(4x – 3)(3x + 5) 1 Share of Ashish and Amit are either (3x + 5) and (4x – 3) or (4x – 3) and (3x + 5) respectively. Q. 3. Find value of x if their shares are equal 1 Ans. According to question if their share are equal. 4x – 3 = 3x + 5 4x – 3x = 5 + 3 x = 8 1

B Subjective Based Questions

43

II. Two students in class of IX named Ria and Ravya were assigned a polynomial by their maths teacher. The polynomial was p(x) = x2 – 5x + 6.

Read the following Passage and answer the questions.



They were asked to express this polynomial as product of factors. Both applied factorisation by splitting the middle term and got different answer.



Riya’s answer : (x – 3)(x – 2)



Ravya’s answer : (x + 3)(x – 2)

I. Two brothers Ashish and Amit wanted to start a business together. They decided to share their amount depending upon the variable expenditure. The amount of two partners is given by the expression 12x2 + 11x – 15, which is the product of their individual share factors. On the basis of above information answer the following questions. [CBSE SAS] Q. 1. Find the total Expenditure of Ashish and Amit when x = Rs 100 1 Ans. Total Expenditure = 12x2 + 11x – 15 Put x = 100 = 12 × (100)2 + 11 × 100 – 15 = 120000 + 1100 – 15 = Rs. 121,085 1 Q. 2. Find individual share factor of Ashish and Amit in terms of x.2 Ans. Total Amount = 12x2 + 11x – 15 By splitting the middle term, we get 12x2 + 11x – 15

On basis of above information answer the following questions: Q. 1. Find out whose answer is correct and show factorisation 2 Ans. p(x) = x2 – 5x + 6 = x2 – 3x – 2x + 6 1 = x(x – 3) – 2(x – 3) = (x – 3)(x – 2) 1 Q. 2. Find the value of P(–1) 1 Ans. p(x) = x2 – 5x + 6 p(–1) = (–1)2 – 5 × – 1 + 6 1 = 1 + 5 + 6 = 12 1 So, Riya answered correctly. Q. 3. Which identity is used in the above situation. 1

12x2 + 20x – 9x – 15 4x(3x + 5) – 3(3x + 5)

Ans. Identity used in above situation is (x + a)(x + b) = x2 + (a + b)x + ab





1







1

CHAPTER

3

Syllabus

Study Time: Maximum time: 3:15 Hrs Maximum questions: 23

LINEAR EQUATIONS IN TWO VARIABLES

Recall of linear equations in one variable. Introduction to the equation in two variables. Focus on linear equations of the type ax + by + c = 0. Explain that a linear equation in two variables has infinitely many solutions and justify their being written as ordered pairs of real numbers. Plotting them and showing that they lie on a line.

Revision Notes  Linear equations in one variable are x + 1 = 0, x +

2 = 0 etc. we know that such equations have

unique solution and solution of these type of equations can be represented on number line.

Scan to know more about this topic

 Linear equation in two variables :

An equation of the form ax + by + c = 0, where a, b and c are real numbers, such that a and b are both non zero, is called a linear equation in two variables.



e.g., x + y = 16, p + 4q = 7, 3 =



All are linear equations in two variables.

Linear equations in Two Variables

7x − y and 2l + m = 3

 Solution of linear equation in two variables :

Any pair of values of x and y which satisfies the equation ax + by + c = 0, is called its solution. This solution can be written as an ordered pair (x, y), first writing value of x and then value of y.

 Linear equation in two variables has infinitely many solutions.

For finding the solution of linear equation in two variables (i.e., ax + by + c = 0), we use following steps :



Step 1: Write the given equation in two variables, if not present.



Step 2: Put an arbitrary value (for convenience put x = 0 or y = 0) of x (or y) in the given equation and then it reduces into linear equation of one variable, which gives a unique solution. Thus, we get one pair of solution of given equation.



Step 3: Repeat step 2 for another arbitrary value of x (or y) and get another pair of solution of given equation.

Example 1

Find four different solutions of the equation 2x + y = 7. Solution: Step I : Write the given linear equation. Given, linear equation in two variables is 2x + y = 7 ...(i) Step II : Put an arbitrary value of x (or y) in the



given equation and find corresponding value of y (or x). or, On putting x = 0 in eq. (i), we get 2(0) + y = 7 or, y = 7 So, (0, 7) is a solution of the given equation or, On putting y = 0 in eq. (i), we get 2x + 0 = 7



LINEAR EQUATIONS IN TWO VARIABLES

45

46

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-IX



or,



So, (



Step III : Repeat step 2 for other solutions. or, On putting x = 1 in eq. (i), we get 2(1) + y = 7 or, y = 5 So, (1, 5) is also a solution of the given equation.

x =

7 2

7 , 0) is also a solution of the given equation. 2

or, On putting y = 1 in eq. (i), we get 2x + 1 = 7 or, 2x = 6 or, x = 3 So, (3, 1) is also a solution of the given equation. Step IV : Write all the solutions. 7 (0, 7), ( , 0), (1, 5) and (3, 1) are four solutions of 2 the given equation.

Mnemonics 1. Solving equations “Don’t Call Me After Midnight” (i) Distribute (multiply term outside parentheses by what’s inside) x + 4(2 + 3x) = 21 (ii) Combine like terms x + 8 + 12x = 21 (iii) Move variable x + 12x (iv) Add or subtract 21 – 8 (v) Multiply or divide 13x = 13 x = 13 = 1 13

SUBJECTIVE TYPE QUESTIONS Very Short Answer Type Questions (1 mark each) Q. 1. Find the value of k, if x = 2, y = 1 is a solution of the equation 2x + 3y = k. OR Find two different solutions of the equation 3x + y = 19. A [KVS 2019]  Sol.  x = 2 and y = 1 is a solution of the equation 2x + 3y = k \ 2 × 2 + 3 × 1 = k \ k = 7 1 OR 3x + y = 19 When x = 1 y = 19 – 3x = 19 – 3(1) = 16 When x = 5 y = 19 – 3(5) = 19 – 15 = 4 Hence, solutions are (1, 16) and (5, 4). 1 Q. 2. In – 2y + 3x = 14, express y in terms of x. R [Board Term II, 2017] Sol.

– 2y + 3x = 14 ½ 3x – 14 = 2y 3x − 14 y = ½ 2 [CBSE Marking Scheme, 2017]

Q. 3. Express

x − 3 y = 7 in the form of ax + by + c = 0. 4 R [Board Term II, 2017]



x − 3 y = 7 4



x − 12 y = 7 4



x – 12y = 28



x – 12y – 28 = 0

Sol.

½

½

[CBSE Marking Scheme, 2017] Q. 4. Total number of legs in a herd of goats and hens is 40. Represent this in the form of linear equation of two variables. A [Board Term II, 2014] Sol. Let the number of goats and hens in herd are x & y respectively then, 

Þ

4x + 2y = 40 2x + y = 20

1

Q. 5. An equation of the form ax + by + c = 0 will be a linear equation in two variables, when a ¹ 0, b ¹ 0. Is it True or False ? R Sol. True. 1



LINEAR EQUATIONS IN TWO VARIABLES

Short Answer Type Questions-I (2 marks each) Q. 1. Find three solutions of linear equation 7x – 5y = 35 in two variables. U [Board Term II, 2014] 7 x − 35 1 Sol. So, when y = 5 x

5

0

10

y

0

–7

7

1 Q. 2. Check which of the following is (are) the solution(s) of the equation 3y – 2x = 1. (i) (4, 3)

(ii)

(2

2, 3 2

)

U [Board Term II, 2011] Sol. (i) Put x = 4 and y = 3, then 3y – 2x = 3(3) – 2(4) = 1 So, (4, 3) is the solution of the equation. 1



(ii) Again put

x = 2 2



and

y = 3 2 , then,



47

Short Answer Type Questions-II (3 marks each) Q. 1. ABCD is a square. Co-ordinates of A and C are (–1, –1) and (1, 1) respectively. Write the coordinates of B and D. Also write the equations of all the sides of square. A [Board Term II, 2014]

Sol. Given, A(–1, –1) and C(1, 1) Then, B(1, –1) and D(–1, 1) Also, equations of sides of square are, AB : y = –1 BC : x = 1 CD : y = 1 DA : x = –1

1 ½ ½ ½ ½

3y – 2x = 3( 3 2 ) − 2( 2 2 )

= 5 2 ≠ 1 So, ( 2 2 , 3 2 ) is not a solution of the given



1

equation.

Q. 3. Express y in terms of x from the equation 3x + 2y = 8 and check whether the point (4, – 2) lies on the line. U [NCERT] Sol. 2y = 8 – 3x 8 − 3x or, y = 1 2 For x = 4 ½ 8−3×4 y = 2 8 − 12 −4 = = =–2 2 2 \ (4, – 2) lies on the line. ½ Q. 4. Find the point at which the equation 3x – 2y = 6 meets the x-axis. U [NCERT] Sol. On x-axis, y co-ordinate is zero. So, put y = 0 in 3x – 2y = 6, we get 1 3x – 0 = 6 6 or, x = = 2 ½ 3 \ 3x – 2y = 6 meets the x-axis at (2, 0).



½

Q. 5. If the point (2k – 3, k + 2) lies on the graph of the equation 2x + 3y + 15 = 0, find value of k.

U [Board Term II, 2011]

Sol. Putting x = 2k – 3, y = k + 2 in 2x + 3y + 15 = 0, we get 2(2k – 3) + 3(k + 2) + 15 = 0

½



½

or, 4k – 6 + 3k + 6 + 15 = 0

or,

−15 k = 7

1

Q. 2. For what value of k, the linear equation 2x + ky = 8 has x = 2 and y = 1 as its solution ? If x = 4, then find the value of y. U [Board Term II, KVS 2014] Sol. The linear equation is 2x + ky = 8 At x = 2, y = 1, 2(2) + k(1) = 8 1 or, 4 + k = 8 \ k = 4 1 If x = 4, then or, 2(4) + 4y = 8 or, 8 + 4y = 8 or, 4y = 0 \ \ y = 0 1

Q. 3. For what value of p; x = 2, y = 3 is a solution of (p + 1) x – (2p + 3)y – 1 = 0 and write the equation. U [Board Term II, 2013] Sol. Given equation is (p + 1)x – (2p + 3)y – 1 = 0 ...(i) If x = 2, y = 3 is the solution of the equation (i), then (p + 1)2 – (2p + 3)3 – 1 = 0 or, 2p + 2 – 6p – 9 – 1 = 0 or, – 4p – 8 = 0 or, p = – 2 2 Put the value of p in equation (i), then –x + y – 1 = 0 or, x – y + 1 = 0, is the required equation. 1

48

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-IX

or,

Commonly Made Error

While opening the bracket the students tend to forget all the terms inside the bracket with one outside and to change the sign too.

Students have to be careful when they multiply a term with another and be particular about signs.

Q. 4. Find the equations of any two lines passing through the point (– 1, 2). How many such lines R [Board Term II, 2017] can be here ? Sol. Equation of two lines passing through the point (– 1, 2) are x + y = 1 1 2x + y = 0 1 Infinite lines can pass through the point (– 1, 2). 1 [CBSE Marking Scheme 2017]

Long Answer Type Questions (5 marks each) Q. 1. The auto rickshaw fare in a city is charged ` 10 for first kilometre and @ ` 4 per kilometre for subsequent distance covered. Write the linear equation to express the above statement and find 3 solutions to the equations. A [Board Term II, KVS 2016]

Sol. Total distance covered = x km. Total fare = ` y Fare for the first kilometre = ` 10 Subsequent distance = (x – 1) km

1

(i) If the temperature is 30°C, what is the temperature in Fahrenheit? (ii) It the temperature is 95°F, what is the temperature in Celsius ? (iii) If the temperature is 0°C, what is the temperature in Fahrenheit and if the temperature is 0°F, what is the temperature in Celsius ? (iv) Is there a temperature which is numerically the same in both Fahrenheit and Celsius ? If yes, find it. A + U [KVS, 2019] 9 Sol. F = C + 32 5

(i)

(ii)

\ Fare for the subsequent distance = ` 4(x – 1) 1½ According to question, y = 10 + 4(x – 1) or, y = 10 + 4x – 4 or, y = 4x + 6 \ Required linear equation y = 4x + 6 where x ³ 1 x

1

2

3

y

10

14

18

2½

0 10 15 25 15 10 (ii) 14 + y = 25 y = 25 – 14 2½ y = 11 years Hence, Akhil’s age = 11 years Q. 3. Given below a linear equation that converts Fahrenheit to Celsius and vice-versa. 9 F = C + 32 5

Answering Tip

y = 25 – x x y

2½

Q. 2. A student Amit of class IX is unable to write in his examination, due to fracture in his arm. Akhil a student of class VI writes for him. The sum of their ages is 25 years. A [Board Term II, 2013] (i) Write a linear equation for the above situation and find three solutions to the equations. (ii) Find the age of Akhil, when age of Amit is 14 years. Sol. Let Age of Amit = x years Age of Akhil = y years (i) According to the question the linear equation for the above situation is x + y = 25

(iii) if, then, If, then, or,

C = 30° 9 F = × 30 + 32 5 F = 54 + 32 = 86 30°C = 86°F F = 95° 9 95° = × C + 32 5 95 – 32 =

9 ×C 5

63 × 5 =C 9 C = 35 95°F = 35°C C = 0 9 F = C + 32 5 F =



– 32 =



C =

1

9 × 0 + 32 5

F = 32 0°C = 32°F F = 0 9 F = C + 32 5 0 =

1

9 C + 32 5 9 C 5 5 × −32 9

1



LINEAR EQUATIONS IN TWO VARIABLES

=

−160 9

−160 \ 0°F = °C 9 (iv) Let



9x = 32 5 5x − 9 x = 32 5



−4 x = 32 5

1

°C = °F = x 9C F = + 32 5 9x x = + 32 5

x−

32 × 5 = – 40° 4

Þ

x = −

\

– 40°C = – 40°F

OBJECTIVE TYPE QUESTIONS

Q. 1. The linear equation 2x – 5y = 7 has (A) a unique solution (B) two solutions (C) infinitely many solutions U (D) no solution Ans. Option (C) is correct. Explanation: In given equation 2x − 5y = 7, for every value of x, we get a corresponding value of y and vice-versa; therefore, the linear equation has infinitely many solutions.

Q. 3. If (2, 0) is a solution of the linear equation 2x + 3y = k, then the value of k is (A) 4 (B) 6 A (C) 5 (D) 2 Ans. Option (A) is correct. Explanation: Putx = 2, y = 0 in the equation 2x + 3y = k 2 (2) + 3 (0) = k 4 + 0 = k Hence, k  = 4 Q. 4. Any solution of the linear equation : 2x + 0y + 9 = 0 in two variables is of the form 9  9   (A)  − , m  (B)  n,  2 2     9 (C)  0,   2  Ans. Option (A) is correct.

(D) (–9, 0)

A

1

(1 mark each)

0 Explanation: 2 x + 0 y + 9 = 2x + 9 = 0 2 x = −9 9 x= − 2

A Multiple Choice Questions

Q. 2. The equation 2x + 5y = 7 has a unique solution, if x, y are (A) natural numbers (B) positive real numbers (C) real numbers U (D) rational numbers. Ans. Option (A) is correct. Explanation: In natural numbers, there is only one pair, i.e., (1, 1) which satisfies the given equation but in positive real numbers, and rational numbers, there are many pairs to satisfy the given linear equation.

49

 9  And y can be any real number. Hence,  − , m   2  is the required form of solution of the given linear equation.

B Assertion & Reason

Directions: In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Mark the correct choice as: (A) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A). (B) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A). (C) Assertion (A) is true but reason (R) is false. (D) Assertion (A) is false but reason (R) is true. Q. 1. Assertion (A): The point of the form (a, –a) lies on the line x + y = 0. Reason (R): Any point which satisfies the euqation ax + by + c = 0 is the solution of the equation. Ans. Option (A) is correct. Explanation: In case of Assertion (A): Let x = a and y = –a x + y = a + (–a) Þ a – a = 0 \ Assertion is true. \ In case of Reason (R): Any pair (x, y) satisfies ax + by + c = 0 then (x, y) is the solution. \ Reason is true. Both (A) and (R) are true and (R) is the correct explanation of (A). Q. 2. Assertion (A): x + 2 y

2 = 0 is a linear equation. 9

Reason (R): A linear equation in two variables is of the form ax + by + c = 0 Ans. Option (A) is correct.

50

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-IX

Explanation: In case of Assertion (A): 2 x + 2y - = 0 9

\ Assertion is true. In case of Reason (R): A linear equation in two variables is of the form ax + by + c = 0 which a ¹ 0 and b ¹ 0. Hence, Both (A) and (R) are true and (R) is the correct explanation of (A).

9 x + 18 y - 2 =0 9 9x + 18y – 2 = 0

COMPETENCY BASED QUESTIONS A Case based MCQs

Read the following passage and answer any four questions of the following : I. Prime Minister’s National Relief Fund (also called PMNRF in short) is the fund raised to provide support for people affected by natural and manmade disasters. Natural disasters that are covered under this include flood, cyclone, earthquake etc. Man-made disasters that are included are major accidents, acid attacks, riots, etc.

Two friends Sita and Gita, together contributed ` 200 towards Prime Minister’s Relief Fund. Q. 1. Which out of the following is not the linear equation in two variables ? (A) 2x = 3 (B) 4 = 5x – 4y

(C) x2 + x = 1 (D) x − 2 y = 3

(4 marks each)

Q. 4. If both contributed equally, then how much is contributed by each? (A) ` 50, ` 150 (B) ` 100, ` 100 (C) ` 50, ` 50 (D) ` 120, ` 120 Ans. Option (C) is correct. Explanation: If x = y then x + x = 200½ 2x = 200 200 x = = 100 ½ 2 Thus, each contributed is ` 100. Q. 5. Which is the standard form of linear equation x=–5? (A) x + 5 = 0 (B) 1.x – 5 = 0 (C) x + 0.y + 5 = 0 (D) 1.x + 0.y = 5 Ans. Option (C) is correct. Explanation: Since, x = –5 Þ x + 5 = 0 Thus, standard form of x = – 5 is 1.x + 0.y + 5 = 0. 1 II. Rainwater harvesting system is a technology that collects and stores rainwater for human use. Anup decided to do rainwater harvesting. He collected rainwater in the underground tank at the rate of 30 cm3/sec.

Ans. Option (C) is correct. Explanation: x2 + x = 1 is not linear as highest power is 2. Also, it is an equation in one variable. Thus, it is not a linear equation in two variables.1 Q. 2. How to represent the above situation in linear equations in two variables ? (A) 2x + y = 200 (B) x + y = 200 (C) 200x = y (D) 200 + x = y Ans. Option (B) is correct. Explanation: Here, x represents Sita’s contribution and y represents Gita’s contribution. 1 Q. 3. If Sita contributed ` 76, then how much was contributed by Gita ? (A) ` 120 (B) ` 123 (C) ` 124 (D) ` 125 Ans. Option (C) is correct. Explanation: If x = 76 then 76 + y = 200

½



y = 200 – 76



y = 124½

Q. 1. What will be the equation formed if volume of water collected in x seconds is taken as y cm3 ?1 (A) 30x = y (B) x = 30y (C) 30 – x = y (D) 30 + y = x Ans. Option (A) is correct. Q. 2. What is the type of solution of the equation formed? 1 (A) a unique solution (B) only two solutions (C) no solution (D) infinitely many solutions



LINEAR EQUATIONS IN TWO VARIABLES

Ans. Option (D) is correct. Explanation: Because for every value of x, there is a corresponding value of y and vice versa. 1 Q. 3. Write the equation in standard form. 1 (A) 30x – y + 0 = 0 (B) 30x + y + 0 = 0 (C) 30x – y – 0 = 0 (D) 30x – y = 0 Ans. Option (A) is correct. Explanation: Standard form of a linear equation in two variables is ax + by + c = 0. 1 Q. 4. How much water will be collected in 60 sec ? 1 (A) 1500 cm3 (B) 2 cm3 3 (C) 1800 cm (D) 1 cm3 Ans. Option (C) is correct. Explanation: Since, y = 30x If x = 60, then, y = 30 × 60 = 1800 Required volume is 1800 cm3 1 Q. 5. How much time will it take to collect water in 900 1 cm3 ? (A) 20 sec (B) 50 sec (C) 40 sec (D) 30 sec Ans. Option (D) is correct. Explanation: Since, y = 30x If y = 900, then, 900 = 30x

x = 900 = 30 30

Required time is 30 sec.

1

Q. 1. Write an equation to express the above situation. 1 Ans. y is directly proportional to x. Þ y = cx where c is a constant. 1 Q. 2. Find acceleration produced in the body, if force applied is 10 newton. Ans. Given, Force applied = 10 newtons c = 10 Since, y = cx Þ 10 = 10x Þ x = 1 Thus, acceleration produced = 1 m/sec2 1 Q. 3. Find force applied, if acceleration produced is 2.5m/sec2. Ans. Given, acceleration produced = 2.5 m/sec2 y = cx,

Read the following passage and answer the following questions: I. The force applied on a body is directly proportional to the acceleration produced in the body. Assume x be acceleration produced in the body and y be the force. Take constant as 10. (S.I unit of force is Newton and S.I. unit of acceleration is m/sec2.

y = 10 × 2.5

= 25 newton

1

II. Temperature is a measure of the warmth or coldness of an object or substance with reference to some standard value. In countries like USA and Canada, temperature is measured in Fahrenheit, whereas in countries like India, it is measured in Celsius. An equation that converts Fahrenheit to Celsius is:

B Cased Based Subjective Questions

51

æ 9ö F = ç ÷ C + 32 è 5ø

On reading the above information, some queries come in Aman’s mind. Help him to resolve his following queries:

Answer the following : Q. 1. How many variables are there in the equation, æ 9ö 1 F = ç ÷ C + 32 ? è 5ø  Ans. There are two variables, C and F. 1 Q. 2. If the temperature is 30°C, what is the temperature in Fahrenheit ?  1 Ans.

9 5

F =   30 + 32

= 54 + 32 = 86°F 1 Q. 3. What is a temperature which is numerically the same in both Fahrenheit and Celsius ? 2

9 5

Ans.

F =   x + 32



x =   x + 32



9 5

5x = 9x + 160 5x – 9x = 160 – 4x = 160 x = – 40

2

52

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-IX

Artificial Intelligence PARAMETERS Chapter Covered

DESCRIPTION Chapter 4: Linear Equations in Two Variables

AI CONCEPTS INTEGRATED

Name of the book Mathematics, Class 9, NCERT Subject and Artificial Understanding the concept of Linear equations in two variables Intelligence Integrated Introductory video:

Learning Objectives

Time Required Classroom Arrangement Material Required Pre-Preparation Activities Previous Knowledge Methodology

Learning Outcomes

Follow up Activities Reflections

Geogebra AI Applications How google map and Uber app are interconnected. To understand concept of Linear equation of 2 variables

Autodraw

To understand concept of Framing of equations in two variables To find the solutions Represent graphically linear equation in two variable on graph paper. Students can practice of graph in Geogebra of different equations. 5 periods of 40 minutes each Flexible Pen, paper, Black Board chalk, Graph paper Laptops and Internet connection Students will asked to arrange graph paper one day prior· Students are given an idea of linear equations in one variable and its graph representation in class VIII Activity 1. Teacher will assign the task in groups to the students. Each group will create word problems by their own based on Linear equation in two variables and then they will plot the graph on graph paper. Activity II: Practice Activity Google map, Uber App Ask students to apply their understanding of Linear equations in two variables in solving ex 4.1 to ex 4.3. Activity III The points of intersection of two graphs represent common solutions to both equations.

Activity IV: At the End of the chapter students were asked to gather a information on Uber or OLA App. How AI works in these apps. How these app calculating Distance And total Fare. Students will understand concept of Linear equation of 2 variables Students will able to frame of equations in two variables. They will able to find the solutions. Students will able to Represent linear equation in two variable on graph paper. Students will able to draw graph of different linear equations in two variables in Geogebra Students will give the presentation in groups and teacher will give their inputs. In this way teacher will evaluate students’ task in groups. Ask the students to explore more AI based applications app. Where linear equations has used. Explore different app like Arogya app (predicting infectious people) Swiggy app. Zomato App.







SELF ASSESSMENT PAPER - 02 Time: 1 hour

MM: 30

UNIT-II I. Multiple Choice Questions

 1. The linear equation 3x – 4y = 9 has (A) a unique solution (B) two solutions (C) infinitely many solution s (D) no solution

[1 × 6 = 6]

2. The equation x = 5, in two variables can be written as (A) 1.x + 1.y = 5 (B) 1.x + 0.y = 5 (C) 0.x + 1.y = 5

(D) 0.x + 0.y = 5

3. x = 5, y = 2 is a solution of the linear equation. (A) x + 2y = 7 (B) 5x + 2y = 7

(C) x + y = 7

(D) 5x + y = 7

4. The equation of x-axis is of the form (A) x = 0 (B) y = 0

(C) x + y = 0

(D) x = y

II. Assertion and Reason Based MCQs Directions: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as. (A) Both A and R are true and R is the correct explanation of A. (B) Both A and R are true but R is NOT the correct explanation of A. (C) A is true but R is false. (D) A is false and R is True. 1. Assertion (A): (4, 3) is a solution of 3y – 2x = 1. Reason (R): Any pair of values of x and y which satisfies the equation ax + by + c = 0 is the solution of the equation. 2. Assertion (A): A linear equation in two variables has infinitely many solutions. Reason (R): An equation x + 2y = 6 has solutions (2, 2) and (6, 0). III. Very Short Answer type Questions 1 + 5x + 7 a polynomial? Justify your answer. 1. Is 5x −2



[1 × 5 = 5]

2. Write the coefficient of x2 in (x – 1) (3x – 4). 3. Write the degree of the given polynomial 4 – y2. 4. The graph of the linear equation x + 2y = 7 passes through the point (0, 7). Is this statement True or False? Justify your answer. 5. Solve for x if

5x =x–2 6

IV. Short Answer Type questions–I  1. If f(x) = 4x + 7, evaluate f(5) – f(2). 2. Find the remainder when x3 + x2 + x + 5 is divided by x + 1, using remainder theorem. 3. Find the value of k, if x – 4 is a factor of p(x) = x2 – kx + 2k. 1 1 4. Find the value of x 2 + 2 , if x − = 5 . x x

[2 × 5 = 10)

5. If x = 5 and y = 2 is the solution of the linear equation x + 2y + k = 0, find the value of k. V. Short Answer Type questions-II



2 1. Show that the points A(3, 0), B(0, – 2) and C(6, 2) lie on the graph of the linear equation x − y = 2 . 3 2. If x + y + z = 12 and x2 + y2 + z2 = 50, then find xy + yz + zx and x3 + y3 + z3 –3xyz.

[3 × 2 = 6]

54

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-IX

VI. Long Answer Type questions  1. Ravi took a cab to go to his office. The cab fare is as follows: For the first kilometre, the fare is Rs 50 and for the subsequent distance it is Rs 10 per kilometre. Take the distance covered as x km and total fare as Rs y. (i) Form a linear equation in two variables (ii) If Ravi covered 7 km, then how much fare he has to pay.

[5 × 1 = 5]

II. Case Based Questions V From the following graph, ABCD is a rectangle.

[1 × 4 = 4]



3 2

Give the answer of the following questions : 1. Equation of the side AB is (A) x = – 1 (B) y = – 4 2. The area of rectangle ABCD is (A) 16 sq. units (B) 15 sq. units 3. The area of DABC is (A) 6 sq. units (B) 3 sq. units 4. Equation of the side CD is (A) y = 0 (B) y = – 4

(C) y = 0

(D) x = 1

(C) 20 sq. units

(D) 12 sq. units

(C) 7 sq. units

(D) 8 sq. units

(C) x +1

(D) x +4 = 0

qq

UNIT-III

CO-ORDINATE GEOMETRY

Study Time: Maximum time: 2:30 Hrs Maximum questions: 31

CHAPTER

4

Syllabus

CO-ORDINATE GEOMETRY The Cartesian plane, co-ordinates of a point, names and terms associated with the coordinate plane, notations..



List of Topics Topic-1: Cartesian System



Topic-1 Cartesian System

Page No. 55

Topic-2 : Co-ordinate Plane 

Page No. 58

Revision Notes  Cartesian System : The system by which we can describe the position of a point in a plane is called Cartesian System.  The horizontal line XOX’ is called the X–axis and vertical line YOY’ is called Y–axis.

Scan to know more about this topic

Introduction to Coordinate Geometry



    

    

The point where XOX’ and YOY’ intersect is called the origin, and is denoted by O. Location of a point P in cartesian system, written in the form of ordered pair say P(a, b) in above figure. a is the length of perpendicular of P(a, b) from Y-axis and is called abscissa of P. b is the length of perpendicular of P(a, b) from X-axis and is called ordinate of P. Positive numbers lie on the directions OX and OY, are called the positive directions of the X-axis and the Y-axis, respectively. Similarly, OX’ and OY’ are called the negative directions of the X-axis and the Y-axis respectively. How to write the co-ordinates of a point : x – co-ordinate (or abscissa) = perpendicular distance of a point from Y-axis. y – co-ordinate (or ordinate) = perpendicular distance of a point from X-axis. If abscissa of a point is x and ordinate is y, then the co-ordinates of the point are (x, y). The abscissa of every point on Y-axis is zero.

56

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-IX



CO-ORDINATE GEOMETRY

57

 The ordinate of every point on X-axis is zero.  Co-ordinate of a point on X-axis are of the form (x, 0).  Co-ordinate of a point on Y-axis are of the form of (0, y).  X-axis and Y-axis intersect at origin, represented by O and its co-ordinates are (0, 0). Example. On which axes do the given points lie. A(0, 2), B(– 3, 0), C(0, – 3), D(0, 4), E(6, 0), F(3, 0). Solution : On X-axis : B(–3, 0), E(6, 0), F(3, 0) On Y-axis : A(0, 2), C(0, – 3), D(0, 4)

Example 1

Write the quadrant in which each of the following points lie :

(i) (– 2, – 3), (ii) (3, – 4), (iii) (– 1, 2) Sol. If both (abscissa and ordinate) co-ordinates are positive, then point lie in I quadrant. If abscissa is negative and ordinate is positive, then point lies in II quadrant. If both co-ordinates are negative then point lies in III quadrant. If abscissa is positive and ordinate is negative, then point lies in IV quadrant.

Hence, (i) the point (– 2, – 3) lies in III quadrant because its both co-ordinates are negative. (ii) The point (3, – 4) lies in IV quadrant because its x-co-ordinate is positive and y-co-ordinate is negative. (iii) The point (– 1, 2) lies in II quadrant because its x-co-ordinate is negative and y-co-ordinate is positive.

SUBJECTIVE TYPE QUESTIONS Very Short Answer Type Questions (1 mark each) Q. 1. Find one solution of y – 5 = 0 in a cartesian plane. 

R [Board Term II, 2017]

Sol.

y – 5 = 0

½

y = 5

½







\ (0, 5), (1, 5), (2, 5) any one

[CBSE Marking Scheme, 2017] Q. 2. In which quadrants the points P(2, – 3) and Q(–3, 2) lie ? R [Board Term-I, 2015] Sol. P(2, – 3) and Q(–3, 2) lie in IV and II quadrants respectively. 1 Q. 3. If (a, b) = (0, – 2). Find the value of a and b.

R [Board Term I, 2016]

Sol. or,

(a, b) = (0, – 2) a = 0, b = – 2

1

Q. 4. The points P(a, b) lies in the IV quadrant. Find which of a or b is greater ? R [Board Term I, 2016] Sol. Since, P(a, b) lies in IV quadrant \ a > 0, and b < 0 \ a > b 1 [CBSE Marking Scheme, 2016] Q. 5. Find the reflection of the points (–3, –2) in y-axis.

A [Board Term-I, 2015]

Sol. (3, –2)

1 [CBSE Marking Scheme, 2015]

Short Answer Type Questions-I (2 marks each) Q. 1. Write ordinates of following points : R (3, 4), (4, 0), (0, 4), (5, –3) R [Board Term I, 2015] Sol. 4, 0, 4, –3.

2 [CBSE Marking Scheme, 2015]

Q. 2. Find distances of points C(– 3, – 2) and D(5, 2) from x-axis and y-axis. R [Board Term I 2016] Sol. C(– 3, – 2), distance from X-axis = 2 units  ½ distance from Y-axis = 3 units ½ D(5, 2), distance from X-axis = 2 units ½ distance from Y-axis = 5 units ½

Short Answer Type Questions-II (3 marks each) Q. 1. In which quadrant or on which axis does each of the following points lie  R (– 5, 3), (4, – 3), (5, 0), (6, 6), (– 5, – 4) ? Sol. In point (– 5, 3), x < 0 and y > 0 \ Point (– 5, 3) lies in II quadrant. ½ In point ( 4, – 3), x > 0 and y < 0 \ Point (4, – 3) lies in IV quadrant. ½ In point (5, 0), x > 0 and y = 0. \ Point (5, 0) lies on X-axis ½ In point (6, 6), x > 0 and y > 0 \ Point (6, 6) lies in I quadrant. ½ In point (– 5, – 4), x < 0 and y < 0. \ Point (– 5, – 4) lies in III quadrant 1

58

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-IX

Q. 2. Write the co-ordinates of each point P, Q, R, S, T and O from the figure given below. U [NCERT Exemplar]

Sol.

½ ½ ½ ½ ½ ½

Co-ordinates of P = (1, 1) Co-ordinates of Q = (– 3, 0) Co-ordinates of R = (– 2, – 3) Co-ordinates of S = (2, 1) Co-ordinates of T = (4, – 2) Co-ordinates of O = (0, 0)

2 Q. 2. See figure and write the following : (i) The co-ordinate of B. (ii) The point identified by the coordinates (–3, –2). (iii) The abscissa of the point D. (iv) The ordinate of the point C.

Long Answer Type Questions (5 marks each) Q. 1. Write the quadrant in which each of the following points lie : (i) (– 3, – 5)

(ii) (2, – 5)

(iii) (– 3, 5)

Also, verify by locating them on the cartesian plane. 

A [NCERT Exemplar]

Sol. (i) (– 3, – 5) lies in III quadrant, as x < 0 and y < 0.

(ii) (2, – 5) lies in IV quadrant, as x > 0 and y < 0.



(iii) (– 3, 5) lies in II quadrant, as x < 0 and y > 0.



½+½+½

Verification : The points (– 3, – 5), (2, – 5) and (– 3, 5) are plotted as shown in figure :



Result is verified : (i) (– 3, – 5) lies in III quadrant.



(ii) (2, – 5) lies in IV quadrant.



(iii) (– 3, 5) lies in II quadrant.

1½

A [Board Term I, 2014] Sol. (i) The co-ordinates of B = (–2, 3) 1 (ii) E is the point which is identified by the co-ordinates (–3, –2) 1 (iii) The co-ordinates of the point D is (6, 2) 1½ \ Abscissa is 6. (iv) The co-ordinates of the point C is (3, –1) 1½ \ Ordinate is –1.

Topic-2 Co-ordinate Plane Revision Notes  When two mutually perpendicular lines (one horizontal and other vertical) intersect each other they locate the position of a point or an object in a plane.  The plane is called the co-ordinate plane and the lines are called the co-ordinate axes.  The coordinate axes divide the plane into four parts called quadrants (one-fourth part) numbered I, II, III and IV anti-clockwise from OX.



CO-ORDINATE GEOMETRY

59

Example 2

If the perpendicular distance of a point p from the x-axis is 5 units and the foot of the perpendicular lies in the negative direction of x-axis, then the ycoordinate of p are? Solution: The perpendicular distance of a point from the x-axis gives y-coordinate of that point. The foot of



the perpendicular lies on the negative direction of x-axis, so perpendicular distance can be measure in II and III quadrant. So, the point p has y-coordinate = 5 or –5.

SUBJECTIVE TYPE QUESTIONS Very Short Answer Type Questions (1 mark each) Q. 1. From the figure given below find the co-ordinates of point Q. R

Q. 2. In the co-ordinate plane, draw a square of side 3 units, taking origin as one vertex. Also, write the co-ordinates of its vertices. R [Board Term I, 2015] Sol.

O

Sol. The co-ordinate of point Q = (– 3, – 3.5)

1

Short Answer Type Questions-I (2 marks each)

Vertices are (0, 0), (3, 0), (3, 3), & (0, 3). 2 Q. 3. In which quadrant or on which axis do the points (–2, – 4), (2, 4), (0, –2) and (4, – 6) lie ? A [Board Term I, 2014]

Sol. The given points are : (–2, – 4) = A

Q. 1. Which of the following point lie



(2, 4) = B

(i) on X-axis ? (ii) on Y-axis ? A(0, 2), B(5, 6), C(23, 0), D(0, 23), E(0, 4), F(6, 0), G(3, 0)



(0, –2) = C



(4, – 6) = D



R [Board Term-I, 2016]

Sol. (i) on X-axis = C, F, G 1 (ii) on Y-axis = A, D, E 1 [CBSE Marking Scheme, 2016]



The point A will lie in III quadrant.



The point B will lie in I quadrant.



The point C will lie on Y-axis (i.e., x = 0)



The point D will lie in IV quadrant.

1

60

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-IX

Short Answer Type Questions-II (3 marks each)



Q. 1. Solve the equation 2x + 1 = x – 3 and represent the solution(s) on (i) The number line (ii) The Cartesian plane. U [KVS 2019] Sol. (i) 2x + 1 = x – 3 Þ 2x – x = – 3 – 1 Þ x = – 4 1 (i) number line –8 –7 –6 –5 –4 –3 –2 –1 (ii) Cartesian plane

0

1

2

3

1

Since, DABC is an equilateral triangle with side 2a units, therefore



AB = BC = CA = 2a units



OA =



=

5



4

OB =

3



2

=

1 –1

1 (2a) = a units 2

AB2 − OA 2 ( 2 a )2 − a 2



2

3

4

= 4 a2 − a2

5 X

–2

=

–3 x =–4



–4 –5



Y'

1

Long Answer Type Questions (5 marks each) Q. 1. In the given figure, DABC and DADC are equilateral triangles on common base AC, each side of triangles being 2a units. Vertices A and C lies on X-axis, vertices B and D lies on Y-axis. O is the mid-point of AC and BD. Find the co-ordinates of the point B. A [NCERT Exemplar]

Sol.





1



1 –3 –2 –1 0

1 AC 2

Now, in right DAOB

Y

X' –7 –6 –5 –4

1

O is the mid-point of AC, then

1

3a 2 = a 3 units

Thus, co-ordinates of B are (0, a 3 )

1 1

Q. 2. If the coordinates of a point A are (–2, 9) which can also be expressed as (1 + x, y2) and y > 0, then find in which quadrant do the following points lie : P(y, x), Q(2, x), R(x2, y – 1), S(2x, – 3y) Sol. Here, A(–2, 9) can also be expressed as (1 + x, y2) Þ (–2, 9) = (1 + x, y2) where y > 0 \ 1 + x = –2, y2 = 9 1 x = – 3, y = 9 y = 3 ( y > 0) Now, P(y, x) = (3, –3), it lies in IV quadrant 1 Q(2, x) = (2, –3), it lies in IV quadrant 1 R(x2, y – 1) = [(–3)2, (3 – 1)] = (9, 2), it lies in I quadrant 5(2x, –3y) = [2 × (–3), (–3 × 3)] = (–6, –9), it lies in III quadrant 1 Q. 3. (i) Find values of a and b, if two ordered pairs (a – 3, –6) and (4, a + b) are equal. (ii) Find distances of point (a, b) obtained from x-axis and y-axis. (iii) Find in which quadrant they lie. Sol. (i) Here, two ordered pairs are equal \ a – 3 = 4; a + b = 6 a = 4 + 3; 7 + b = – 6 (substituting value of 'a') a = 7; b = –6 – 7 b = –13 2 Hence a = 7 and b = –13 (ii) Distances of point (7, –13) from x-axis is, 13 units in negative direction and from y-axis it is 7 in positive direction. 2 (iii) Clearly point (7, –13) lie in IV quadrant. 1



CO-ORDINATE GEOMETRY

OBJECTIVE TYPE QUESTIONS A Multiple Choice Questions Q. 1. Point (–3, 5) lies in the (A) first quadrant (B) second quadrant R (C) third quadrant (D) fourth quadrant Ans. Option (B) is correct. Explanation: In point (–3, 5), x-coordinate is negative and y-coordinate is positive. So, the point lies in second quadrant. Q. 2. Signs of the abscissa and ordinate of a point in the second quadrant are respectively (A) +, + (B) –, – R (C) –, + (D) +, – Ans. Option (C) is correct. Explanation: In second quadrant, x-axis is negative and y-axis is positive. So, sign of abscissa of a point is negative and sign of ordinate of a point is positive. Q. 3. Point (0, –7) lies (A) on the x–axis (B) in the second quadrant (C) on the y-axis R (D) in the fourth quadrant Ans. Option (C) is correct. Explanation: In the point (0, –7), x-coordinate is zero, So, it lies on y-axis and y-coordinate is negative so, the point lies on y-axis in the negative direction. Q. 4. The point at which two coordinate axes meet is called the ............... . (A) abscissa (B) ordinate R (C) origin (D) quadrant Ans. Option (C) is correct. Explanation: It is the point at which both the axes; i.e., x-axis and y-axis meet and its co-ordinates are (0, 0). Q. 5. Points (1, –1), (2, –2), (4, –5), (–3, –4) (A) lie in II quadrant (B) lie in III quadrant (C) lie in IV quadrant (D) do not lie in the same quadrant Ans. Option (D) is correct. Explanation: In points (1, –1), (2, –2) and (4, –5) x-coordinate is positive and y-coordinate is negative. So, they all lie in IVth quadrant. In point (–3, –4), x-coordinate and y-coordinate both are negative, so it lies in III quadrant. So, all the points do not lie in same quadrant. Q. 6. If the coordinates of the two points are P (–2, 3) and Q (–3, 5), then (abscissa of P) – (abscissa of Q) is (A) –5 (B) 1 (C) –1 A (D) –2

61

(1 mark each)

Ans. Option (B) is correct. Explanation: Abscissa of P = –2 Abscissa of Q = –3 (Abscissa of P) – (Abscissa of Q) = (–2) – (–3) Þ –2 + 3 = 1 Q. 7. If P (5, 1), Q (8, 0), R (0, 4), S (0, 5) and O (0, 0) are plotted on the graph paper, then the point(s) on the x-axis are (A) P and R (B) R and S (C) Only Q A (D) Q and O Ans. Option (D) is correct. Explanation: A point lies on x-axis, if its y-coordinate is zero. Here, Q(8, 0) and O(0, 0) has y-coordinate zero. Q. 8. Abscissa of a point is positive in (A) I and II quadrants (B) I and IV quadrants (C) I quadrant only (D) II quadrant only R Ans. Option (B) is correct. Explanation: Abscissa of a point is positive in I and IV quadrants.

B Assertion & Reason Directions: In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Mark the correct choice as: (A) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A). (B) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A). (C) Assertion (A) is true but reason (R) is false. (D) Assertion (A) is false but reason (R) is true. Q. 1. Assertion (A): The point P(–3, 0) lies on x-axis. Reason (R): Every point on x-axis is of the form (x, 0). Ans. Option (A) is correct. Explanation: In case of assertion (A): The point P(–3, 0) lies on x-axis as ordinate is 0. \ Assertion is true. In case of reason (R): Every point on x-axis has ordinate 0(zero). \ Reason is true. Therefore, Both A and R are true and R is the correct Explanation of A. Q. 2. Assertion (A): The point P(–6, –4) lies in the quadrant III. Reason (R): The signs of points in quadrants I, II, III, IV are respectively (+, +), (–, +), (–, –), (+, –). Ans. Option (A) is correct. Explanation: In case of assertion (A): P(–6, –4) lies in III quadrant as the points having negative signs lies in quadrant III.

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Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-IX

In case of reason (R): This statement is fact which is always true. Therefore, Both A and R are true and R is the correct Explanation of A. Q. 3. Assertion (A): Point (–5, –4) lie in quadrant II. Reason (R): Point lying in quadrant I has x > 0 and y > 0. Ans. Option (D) is correct. Explanation: In case of assertion (A): In point (–5, –4) x < 0 and y < 0 \ (–5, –4) lies in III quadrant. \ Assertion is false In case of reason (R):

Any point lying in I quadrant has both the co-ordinates positive. \ Reason is true. Q. 4. Assertion (A): Point (0, 5) lies on y-axis. Reason (R): Every point on y-axis is of the form (0, y). Ans. Option (A) is correct. Explanation: In case of assertion (A): Point (0, 5) lies on y-axis as x-coordinate is zero. In case of reason (R): Every point on y-axis has x-coordinate zero is fact. Therefore, Both A and R are true and R is the correct Explanation of A.

COMPETENCY BASED QUESTIONS A Case based MCQs

Read the following passage and answer any four questions of the following : I. Five friends playing a game in which they are standing at different positions, P, S, T, R and Q.

Rohan is watching them playing. Few questions came to Rohan's mind while watching the game. Give answer to his questions by looking at the figure. Q. 1. What are the co-ordinates of P? 1 (A) (– 1, 1) (B) (1, – 1) (C) (1, 1) (D) (– 1, – 1) Ans. Option (C) is correct. Explanation: Co-ordinates of P(1, 1)1 Q. 2. Name the point whose y-co-ordinate is zero : 1 (A) P (B) Q (C) R (D) S Ans. Option (B) is correct. Explanation: As point Q lies on x-axis. Therefore, its y-co-ordinate is zero.1 Q. 3. Name the polygon formed on joining all these five points in an order. 1 (A) Quadrilateral (B) Hexagon (C) Pentagon (D) Triangle

(4 marks each)

Ans. Option (C) is correct. Explanation: As polygon formed on joining five points will be pentagon.1 Q. 4. Name the point lying in the third quadrant. 1 (A) R (B) P (C) Q (D) T Ans. Option (A) is correct. Explanation: In third quadrant, both x-co-ordinate and y-co-ordinate are negative.1 Q. 5. (x, y) = (y, x), if 1 (A) x > y (B) x b. Reason (R): Point (0, 0) lies on y-axis. III. Very Short Answer Type Questions

[1 × 5 = 5]

Observe the points plotted in the figure and give answer the following questions :

1. The co-ordinates of the point P is (A) (1, 1)

(B) (– 1 , 3)

(C) (1 , 3)

(D) (0, 0)

(C) (– 2 , – 3)

(D) (3, 1)

2. The coordinates of the point S is (A) (2, 1)

(B) (4, – 2)



SELF ASSESSMENT PAPER

67

3. The coordinates of the point T is (A) (4, – 2)

(B) (4 , 2)

(C) (– 2, 4)

(D) (2, 4)

(C) 4

(D) – 4

(C) 5

(D) – 5

4. The abscissa of P- abscissa of S is (A) 2

(B) – 2

5. The ordinate of P + ordinate of S is : (A) 4

(B) – 4

IV. Short Answer Type questions–I

[2 × 2 = 4]

1. From the given figure, write the following : (i) Find the coordinates of P. (ii) The abscissa of the point Q. (iii) The ordinates of the point R. (iv) The points whose abscissa is O.

2. Locate and write the coordinates of a point : (i) Above X-axis lying on Y-axis at a distance of 3 units from origin. (ii) Below X-axis lying on Y-axis at a distance of 5 units from origin. V. Short Answer Type questions-II [3 × 2 = 6] 1. In which quadrant or on which axis does each of the following points lie (– 4, 2), (5, – 4), (6, 0), (8, 8), (– 1, – 3)? 2. Find the coordinates of the point : (i) Whose abscissa is 4 and ordinate is 2. (ii) Whose abscissa is 5 and which lies on X-axis. (iii) Whose ordinate is 7 and which lies on Y-axis. VI. Long Answer Type questions

[5 × 1 = 5]

1.



[CBSE SAS]

In a cartesian plane given below. A child is standing at a certain Point P and his mother is standing at a point O. (i) What are the co-ordinates of mother?

1

(ii) What are the co-ordinates of child?

1

(iii) In which Quadrant a child is standing?

1

(iv) What is the distance between the child and his mother?

2

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Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-IX

VII. Case Based (Attempt any 4 parts)

[1 × 4 = 4]

Read the following text and answer the following questions on the basis of the same:

Priya draws a kite on the graph paper. She has following points: (0, 4), (3, 0), (–3, 0), (0, –4). A

1. Name the closed figure obtained.

(A) Triangle (B) Circle 2. What is ordinate of the point (–3, 0)?

(C) Pentagon

(D) Quadrilateral

(A) 0

(C) 2

(D) 3

(C) – 4

(D) 3

(C) (2, 3)

(D) (0, – 4)

(B) 1

3. What is abscissa of the point (3, 0)?

(A) 0

(B) – 3

4. What is the mirror image of (0, 3)? (A) (0, 4)

(B) (0, –3)

5. Name the point where x-co-ordinate is positive and y-co-ordinate is zero. (A) (0, 4)

(B) (3, 0)

(C) (–3, 0)

(D) (0, 0)

qq

UNIT-IV

GEOMETRY

CHAPTER

5

Syllabus

Study Time: Maximum time: 2:30 Hrs Maximum questions: 40

INTRODUCTION TO EUCLID’S GEOMETRY

History : Geometry in India and Euclid’s geometry. Euclid’s method of formalizing observed phenomenon into rigorous Mathematics with definitions, common/obvious notions, axioms/postulates and theorems. The five postulates of Euclid. Showing the relationship between axiom and theorem, for example : (Axiom) 1. Given two distinct points, there exists one and only one line through them. (Theorem) 2. (Prove) Two distinct lines cannot have more than one point in common.

Topic-1 Euclid’s Geometry



List of Topics Topic-1: Euclid’s Geometry



Revision Notes

Page No. 69

Topic-2 : Euclid’s Postulates 

Page No. 73

 Axiom : Axioms are the assumptions which are obvious universal truths. They are not proved.  Euclid’s Axioms :



Things which are equal to the same thing are equal to one another.       e.g., If AB = PQ and PQ = XY , then AB = XY .



Scan to know more about this topic

If equals are added to equals, the wholes are equal.

e.g., If m∠1 = m∠2, then m∠1 + m∠3 = m∠2 + m∠3.



If equals are subtracted from equals, the remainders are equal.

e.g., If m∠1 = m∠2, then m∠1 – m∠3 = m∠2 – m∠3.



Things which coincide with one another are equal to one another.     e.g., If AB coincides with XY , such that A falls on X and B falls on Y, then AB = XY



The whole is greater than the part.

e.g., If m∠1 = m∠2 + m∠3, then m∠1 > m∠2 & m∠1 > m∠3.



Things which are double of the same thing are equal to one another.

e.g., If a = 2c and b = 2c, then a = b.



Things which are halves of the same thing are equal to one another.

c c e.g., If a = and b = , then a = b 2 2

Euclid Geometry

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INTRODUCTION TO EUCLID’S GEOMETRY

71

Mnemonics Postulates are true assumptions specific to Geometry (PG-Post Graduate) Axioms are true assumption, not specifically linked to geometry (Requires no proof) (AP–Andhra Pradesh).

SUBJECTIVE TYPE QUESTIONS Very Short Answer Type Questions (1 mark each) Q. 1. Explain when a system of axioms is called consistent.

R [Board Term I, 2015]

Sol. A system of axioms is called consistent, when it is impossible to deduce from these axioms, a statement that contradicts any axiom or previously proved statement. 1 [CBSE Marking Scheme, 2015] Q. 2. Express in variables the things which are double R of the same thing are equal. Sol. Let, First thing = x = 2a Second thing = y = 2a then, x = y 1 R Q. 3. What does a theorem require ? Sol. Theorem requires a proof. 1

Short Answer Type Questions-I (2 marks each) Q. 1. In the given figure AC = DC, CB = CE, show that AB = DE. R [Board Term I, 2012]

AB = AD AC = AD AB = AC 1 Things which are equal to the same thing are equal to one another. 1 [CBSE Marking Scheme, 2012] Q. 3. Solve the equation x + 4 = 10 and state Euclid’s axiom used. R [Board Term I, 2016] Sol. x + 4 = 10 x + 4 – 4 = 10 – 4 1 Þ x = 6 If equals are subtracted from equals, the remainders are equal. 1 [CBSE Marking Scheme, 2016] Q. 4. In a triangle PQR, X and Y are the points on PQ and QR respectively. If PQ = QR and QX = QY, show that PX = RY.  R [Board Term I, 2016] Sol. Here

PQ = QR



QX = QY

1

Write Euclid’s axiom to support this. Sol. AC = DC  (Given) CB = CE Adding, AC + CB = DC + CE ½ AB = DE ½ If equals are added to equals, the wholes are equal. 1 Q. 2. In the given figure, we have AB = AD and AC = AD. Prove that AB = AC. State the Euclid’s axiom to support this. R [Board Term I, 2012] Sol.



If equals are subtracted from equals, the remainders are also equal. We have

PQ – QX = QR – QY PX = RY

1

Q. 5. P and Q are the centres of two intersecting circles. Prove that PQ = QR = PR. U [Board Term I, 2016; 2013; 2012]

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Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-IX

Sol. In a circle having centre at P, we have

PR = PQ = radius

½

In a circle having centre at Q, we have

QR = QP = radius

½

Euclid’s first axiom : Things which are equal to the same thing are equal to one another. ½ ∴

½

PR = PQ = QR.

Short Answer Type Questions-II (3 marks each) Q. 1. In the given figure, if ∠1 = ∠3, ∠2 = ∠4 and ∠3 = ∠4, write the relation between ∠1 and ∠2, using Euclid’s axiom.

Sol. Given, ∠ABC = ∠ACB or, ∠1 + ∠4 = ∠2 + ∠3 1 or, ∠1 + ∠4 – ∠4 = ∠2 + ∠3 – ∠3  (As, ∠3 = ∠4) 1 If equals are subtracted from equals then remainders are also equal. \ ∠1 = ∠2. 1 Q. 4. Prove that every line segment has one and only one mid-point. U [KVS 2019] Sol. Suppose C and D are two mid-points of the line segment AB. A

\

C

D

AC = CB =

1 AB 2

AD = DB =

1 AB 2

3 4

1



2

R [NCERT Exemplar]

Sol. Here, ∠3 = ∠4 and ∠1 = ∠3 and ∠2 = ∠4. Euclid’s first axiom says, the things which are equal to same thing are equal to one another. 2 Given, ∠3 = ∠4 and ∠1 = ∠3 \ ∠1 = ∠4 Again given ∠2 = ∠4 and ∠1 = ∠4 (Proved above) \ ∠1 = ∠2. 1 Q. 2. In a triangle ABC, X and Y are the points on AB and BC such that BX = BY and AB = BC. Show that AX = CY. State the Euclid’s Axiom used. A [Board Term I, 2015]

\

B

\ AC = AD [Things which are equal to the same thing are equal to one another] \ This is possible only when C and D coincide each other. Hence, every line segment has one and only one midpoint. Hence Proved 3

Long Answer Type Questions (5 marks each) Q. 1. In the figure we have Ð1 = Ð3 and Ð2 = Ð4. Show that ÐA = ÐC. State which axiom you used here. Also give two more axioms other than the axioms used in the above situation.

Sol.

AB = BC (given) BX = BY (given) 1 If equals are subtracted from equals, then remainders are also equal. 1 AB – BX = BC – BY \ AX = CY 1 Q. 3. In the given figure, we have ∠ABC = ∠ACB, ∠3 = ∠4. Show that ∠1 = ∠2. R [NCERT Exemplar]



U [Board Term I, 2016] [NCERT Exemplar]

Sol. Since, ∠1 = ∠3 and ∠2 = ∠4, therefore adding both equations

∠1 + ∠2 = ∠3 + ∠4

Þ

∠BAD = ∠BCD

Þ

∠A = ∠C

½ 1

If equals are added to equal, the wholes are equal 1 Two more axioms : (i) Things which are equal to the same thing are equal to one another.  1 e.g., if AB = PQ and PQ = XY , then AB = XY  ½

(ii) If equals are subtracted from equals, the remainders are equal. 

e.g., if

m∠1 = m∠2 then mÐ1 – m∠3 = m∠2 – m∠3

1



INTRODUCTION TO EUCLID’S GEOMETRY

= Q. 2. In the fig., if OX

1 1 = XY , PX XZ and OX = PX, 2 2

Show that XY = XZ. State which axiom you used here. Also give two more axioms other than the axiom used in the above situation.

73

Two more axioms : (i) Things which coincide with one another are equal to one another.  1 e.g., if AB coincide with XY , such that A falls on X and B falls on Y, then AB = XY . (ii) The whole is greater than the part

1

e.g., if mÐ1 = mÐ2 + mÐ3, then mÐ1 > mÐ2 and mÐ1 > mÐ3. Q. 3. Using Euclid’s axiom, compare length AD and AF. State which axiom you used here. Also give two more axioms other than the axiom used in the above situation. U [Board Term I, 2016] [NCERT Exemplar]

Sol. Here

OX =

1 XY 2 

½



PX =

1 XZ 2 

1



Also \

OX = PX (Given) 1 1 XY = XZ 2 2 

½

XY = XZ

1

Thing which are equal to half of same thing are equal to one another.

 R [Board Term I, 2016] Sol. AD is part of AF \ AD < AF As whole is greater than the part. 2 Two more axioms : (i) If equals are added two equals, the wholes are equal. 1½ e.g., if mÐ1 = mÐ2, then mÐ1 + mÐ3 = mÐ2 + mÐ3 (ii) If equals are subtracted from equals, the remainders are equal. 1½ e.g., if mÐ1 = mÐ2, then mÐ1 – mÐ3 = mÐ2 – mÐ3

Topic-2 Euclid’s Postulates Revision Notes  Postulates : The basic facts which are taken for granted, without proof and which are specific to geometry are called postulates.  Plane : A plane is a surface such that the line obtained by joining any two points in it will be entirely in the plane.  Incidence Axioms on lines : (i) A line contains infinitely many points. (ii) Through a given point A, (infinite) lines can be drawn.

A

(iii) One and only one line can be drawn to pass through two given points A and B.

 Collinear points : Three or more points are said to be collinear, if there is a straight line which passes through all of them.

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Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-IX

Figure I Figure II In figure I; A, B, C are collinear points, while in figure II; P, Q, R are non-collinear points.  Intersecting lines : Two lines which cut at one point are said to be intersecting lines. The point P common to two given line segments AB and CD is called their point of intersection.

 Concurrent lines : Three or more lines intersecting at a same point are said to be concurrent. l

m

P

n

 Parallel lines : Two lines l and m in a plane are said to be parallel, if they have no point in common then we write l l || m. m

The distance between two parallel lines always remains the same.  Two distinct lines cannot have more than one point in common.  Parallel Line Axiom : If l is a line and P is a point not on the line l, there is one and only one line (say m) which passes through P and parallel to l.

 If two lines l and m are both parallel to the same line n, they will also be parallel to each other. l m n

 If l, m, n are lines in the same plane such that l intersects m and n || m, then l also intersects n. l

m n

 If l and m are intersecting lines, l || p and q || m, then p and q also intersect. l

p m q

 If line segments AB, AC, AD and AE are parallel to a line l, then points A, B, C, D and E will be collinear.  Betweenness : Point B is said to lie between the two points A and C, if : (i) Points A, B and C are collinear, and (ii) AB + BC = AC.



INTRODUCTION TO EUCLID’S GEOMETRY

75

 Mid-point of a Line-segment : For a given line segment AB, a point M is said to be the mid-point of AB, if : (i) M is an interior point of AB, and (ii) AM = MB.  Euclid’s Five Postulates : Postulate 1 : A straight line can be drawn from any one point to another point.

Postulate 2 : A terminated line can be produced indefinitely i.e., ‘A line segment can be extended on either side to form a line’.

Postulate 3 : A circle can be drawn with any centre and any radius.

Postulate 4 : All right angles are equal to one another.

If ∠XYZ = 90° and ∠PQR = 90°, then ∠XYZ = ∠PQR.  [congruent angles] Postulate 5 : If a straight line falling on two straight lines makes the interior angles on the same side of it taken together less then two right angles, then the two straight lines, if produced indefinitely, meet on that side on which the sum of angles is less than two right angles.  Non-Euclidean geometries : All the attempts to prove the Euclid’s fifth postulate using the first 4 postulates failed. But they led to the discovery of several other geometries, called non-Euclidean geometries.  Theorems : Theorems are statements which are proved using definitions, axioms, previously proved statements and deductive reasoning.

SUBJECTIVE TYPE QUESTIONS Very Short Answer Type Questions (1 mark each) Q. 1. How many lines can be passed through two distinct points ? R Sol. Only one line passes through two distinct points.

Q. 3. Give any one example of a geometrical line from your surroundings. A Sol. Meeting place of two walls.

1

Q. 4. What is a surface ?

R

Sol. A surface is that which has length and breadth. y

B

1



A

Q. 2. What is a straight line ?

Surface breadth (y) x length (x)

R

Sol. Two planes intersect each other to form a straight line. 1





1

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Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-IX

Q. 5. How can we identify parallel lines ? R Sol. Lines are parallel if they do not intersect on being extended. For example :



1 Lines A and B are parallel lines or they have no common point.

Short Answer Type Questions-I (2 marks each) Q. 1. How many planes can be made to pass through : (i) Three collinear points. (ii) Three non-collinear points. R [Board Term I, 2012] Sol. (i) Infinite, if they are collinear. (ii) Only one, if they are non-collinear. 1+1 Q. 2. Show that of all the line segments drawn from a given point to a line, not on it, the perpendicular line segment is the shortest. U [Board Term I, 2011]



(ii) How would you rewrite Euclid’s fifth postulate so that it would be easier to understand ? A [NCERT] Sol. (i) Since, it is true for things in any part of universe so, this is a universal truth. 1 (ii) If the sum of the co-interior angles made by a transversal intersect two straight lines at distinct points is less than 180°, then the lines cannot be parallel. 1 Q. 5. Consider two ‘postulates’ given below : (i) Given any two distinct points A and B, there exists a third point C which is in between A and B. (ii) There exist at least three points that are not on the same line. Do these postulates contain any undefined term ? Are these postulates consistent ? Do they follow with Euclid’s postulates ? Explain. U [NCERT] Sol. There are two undefined terms, line and point. They are consistent, because they deal with two different situations. (i) Says that given two points A and B, there is a point C, lying on the line which is in between them. 1 (ii) Says that given A and B, we can take C not lying on the line passing through A and B. These ‘Postulates’ do not go with Euclid’s postulates. 1

Long Answer Type Questions (5 marks each)

Sol. Let AB be perpendicular to a line l and AP is any other line segment. In right ∆ABP, ∠B > ∠P, (Q ∠B = 90°) 1 or, AP > AB or AB < AP. 1 Q. 3. In figure, AE = DF, E is the mid-point of AB and F is the mid-point of DC. Using an Euclid’s axiom, show that AB = CD. A [Board Term I, 2011, 2010]

Sol. AB = 2AE (E is the mid-point of AB) ½ CD = 2DF  (F is the mid-point of CD) ½ Also, AE = DF (Given) Therefore, AB = CD (Things which are double of the same thing are equal to one another) 1 Q. 4. (i) Why is Axiom 5, in the list of Euclid’s axioms, considered a ‘universal truth’ ? (Note that the question is not about the 5th postulate.)

Q. 1. A square is a polygon made up of five line segments, out of which, length of three line segments are equal to the length of fourth of one and all its angles are right angles” Define the term used in this definition which have been highlighted/ underlined. R [Board Term I, 2016] Sol. Polygon : A simple closed figure made up of three or more line segments. 1½ Line Segment : Part of a line with two end points. 1 Angle : A figure formed by two rays with a common initial point.  1½ Right angle : Angle whose measure is 90°. 1 Q. 2. Two salesmen make equal sales during the month of June. In July, each salesmen doubles his sale of the month of June. Compare their sales in July. State which axiom you use here. Also give two more axioms other than the axiom used in the above situation. R [Board Term I, 2015] Sol. Their sales in July will also be equal as things which are double of the same thing are equal to one another.  1½ Two other axioms : (i) The whole is greater than the part. 1½ (ii) Things which are halves of the same thing are equal to one another. 2 Q. 3. It is known that if a + b = 10, then a + b – c = 10 – c. Write the Euclid’s axiom that best illustrates this statement. Also give two more axioms other than the axiom used in the above situation. R [Board Term I, 2015]



INTRODUCTION TO EUCLID’S GEOMETRY

Sol. Axiom : If equals are subtracted from equals, the remainders are equal.  2 Two more axioms : (i) Things which are halves of the same thing are equal to one other. 1½ (ii) The whole is greater than the part. 1½

Commonly Made Error

77

Answering Tips

To prove anything the student must be familiar





with Axioms, postulates and properties of lines, triangles, circle etc. Students should memorize the Euclid’s Axioms by heart.

Sometimes students make mistakes or find difficulty in proving statements and equations.

OBJECTIVE TYPE QUESTIONS A Multiple Choice Questions Q. 1. The number of dimension, a point has (A) 0. (B) 1. (C) 2. (D) 3.  R [NCERT Exemp.] Ans. Option (A) is correct. Explanation: According to Euclid, a point is that which has no part,that is, no length, no breadth and no height. Therefore, it has no dimension. Q. 2. A pyramid is a solid figure, the base of which is (A) only a triangle. (B) only a square. (C) only a rectangle. (D) any polygon.  Ans. Option (D) is correct.

R [NCERT Exemp.]

Explanation: A pyramid is a solid figure, the base of which is any polygon. Q. 3. It is known that if x + y = 10 then x + y + z = 10 + z. The Euclid’s axiom that illustrates this statement is (A) First Axiom. (B) Second Axiom. (C) Third Axiom. (D) Fourth Axiom.  R [NCERT Exemp.] Ans. Option (B) is correct. Explanation: If x + y = 10 and then x + y + z = 10 + z. The Euclid’s second axiom states that if equals are added to equals, the wholes are equal. Q. 4. Which of the following needs a proof? (A) Theorem (B) Axiom (C) Definition (D) Postulate  R [NCERT Exemp.] Ans. Option (A) is correct. Explanation: Theorem needs a proof. Q. 5. ‘Lines are parallel if they do not intersect’ is stated in the form of (A) an axiom. (B) a definition. (C) a postulate. (D) a proof.  R [NCERT Exemp. Ex. 5.1, Q. 22, Page 47] Ans. Option (B) is correct.

(1 mark each)

Explanation: ‘Lines are parallel if they do not intersect’ is the form of a definition. Q. 6. Attempts to prove Euclid’s fifth postulate using the other postulates and axioms led to the discovery of several other geometries. State whether the given statement is true or false? Justify your answer.  R [NCERT Exemp. Ex. 5.2, Q. 9, Page 49] Ans. True, these geometries are far different from Euclidean geometry and are called non-Euclidean geometries. Q. 7. The basic facts which are taken for granted, without proof and which are specific to geometry are called (A) Axiom (B) Postulates (C) Theorem (D) Definition Ans. Option (B) is correct. Explanation: Postulates means basic facts without proof. Q. 8. Three or more lines intersecting at a same point are said to be (A) Parallel lines (B) Intersecting lines (C) Concurrent lines (D) Straight line Ans. Option (C) is correct.

B Assertion & Reason Directions: In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Mark the correct choice as: (A) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A). (B) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A). (C) Assertion (A) is true but reason (R) is false. (D) Assertion (A) is false but reason (R) is true. Q. 1. Assertion (A): According to Euclid’s 1st Axiom”Things which are equal to the same thing are also equal to one another.” Reason (R): If AB = PQ and PQ = XY, then AB = XY. Ans. Option (A) is correct.

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Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-IX

Explanation: In case of Assertion (A): Euclid’s 1st Axiom is given which is an assumption and is universal truth. \ Assertion is correct. In case of Reason (R): It is correct explanation of Euclid’s 1st Axiom. \ Reason is true. Hence, Assertion and Reason are true and R is correct explanation of A. Q. 2. Assertion (A): According to Euclid’s Axiom, when equals are added to equals, wholes are equal. Reason (R): If Rita and Rivi are of same age that is 10 years then after 6 years also they will have the same age. Ans. Option (A) is correct. Explanation: In case of Assertion (A): Euclid’s Second Axiom is given which is an assumption and is universal truth. \ Assertion is correct. In case of Reason (R): If Rita and Rivi are 10 years old. After 6 years 10 + 6 = 10 + 6 16 = 16

\ Reason is true. Therefore Assertion and Reason are correct and Reason is correct explanation of Assertion. Q. 3. Assertion (A): Through two distinct points there can be only one line that can be drawn. Reason (R): From these two points we can draw only one line.



A

B

Ans. Option (A) is correct. Q. 4. Assertion (A): If ‘l’ is a line and P is a point not on the line l, there is one and only one line (say m) which passes through P and parallel to l.



Reason (R): If two lines ‘l’ and ‘m’ are both parallel to the same line n, they will also be parallel to each other.

Ans. Option (B) is correct. Explanation: Assertion is true and Reason is also true but Assertion is saying about only two lines whereas Reason is saying about more than two lines.

COMPETENCY BASED QUESTIONS A Case based MCQs

Read the following passage and answer any four questions of the following : I. Two salesmen make equal sales during the month of June. In July each salesmen doubles his sale of the month of June. Q. 1. What will their sales be in July? (A) Double (B) Half (C) Equal (D) One third Ans. Option (C) is correct. Q. 2. Which axiom is used here? (A) Things which are halves of same thing are equal to one another. (B) All right angles are equal to one another (C) Things which coincide are equal (D) None of the above Ans. Option (D) is correct. Q. 3. Euclid’s stated that all right angles are equal to each other in the form of (A) An axiom (B) a definition (C) a postulate (D) a proof Ans. Option (C) is correct. Q. 4. What does a theorem require ? (A) Meaning (B) Conclusion (C) Theory

(D) Proof

Ans. Option (D) is correct. Q. 5. Euclid’s divided his famous treatise ‘The Elements‘ into



(A) 13 chapters

(C) 11 chapters

(4 marks each) (B) 12 chapters (D) 9 chapters

Ans. Option (A) is correct. II. If a point C be the midpoint of a line segment AB, then the relationship among AB, BC and AC can be explained by:

Q. 1. Relationship Between BC and AC is (A) AC =

1 BC XY 2

(C) AC = BC

(B) AC =

1 AB XY 2

(D) All of these

Ans. Option (D) is correct. Explanation:

AC =



1 AB XY 2 ( C is midpoint of AB) 1 AB XY 2



BC =

\

AC = BC

Q Thing’s which are halves of the same thing are equal to one another.

Q. 2. Which Euclid Axiom’s state the required result. (A) All right angles are equal to one another. (B) Things which are equal to the same thing are equal to one another. (C) Things which are halves of the same thing are equal to one another.



INTRODUCTION TO EUCLID’S GEOMETRY

79

Q. 1. How many straight lines can be drawn from A to

(D) None of the above

C?

Ans. Option (C) is correct.

Q. 3. Euclid’s belong to the category.

Ans. One and only one line can be drawn from A to C. 1

(A) Babylonia (C) Greece

Q. 2. State the Euclid Axiom which states the required

(B) Egypt (D) India

result.

Ans. Option (C) is correct.

Ans. According to Euclid’s Postulate,

Q. 4. Here, A, B and C are

“A straight line can be drawn from any point to any other point.” 1



(A) Points of intersection (B) Collinear Points

(C) Non Collinear Points (D) None of these

Q. 3. Give one more Postulate.

Ans. Option (B) is correct.

Ans. Another Postulate, “A circle can be drawn with any centre and any radius.” 2

Q. 5. For given line segment AB, C is midpoint of AB, if

(C) C is not lying on AB

II. Rohan’s sister has two daughter’s of same age. Both of them have equal number of dolls. Rohan on his birthday plans to give both of them same number of dolls.

(D) None of the above

Q. 1. How many dolls will each one of them have after



(A) C’ is an exterior point of AB

(B) C is an interior point of AB

Rohan’s birthday?

Ans. Option (B) is correct.

Ans. Both of them will have equal number of dolls.

Case based Subjective B Questions

1

Q. 2. Which Euclid’s axiom is used to answer this question? Ans. Euclid’s Axiom 2, “If equals are added to equals, then the whole are equal, is used in this question. 1



Read the following passage and answer the following questions: I. Three light house towers are located at points A, B and C on the section of a national forest to protect animals from hunters by the forest department as shown in figure.



Q. 3. Write one more Euclid’s axiom? Ans. According to Euclid’s Axiom 3, “If equals are subtracted from equals, then the remainders are equal. 2





Study Time: Maximum time: 3:30 Hrs Maximum questions: 45

CHAPTER

6

Syllabus

LINES AND ANGLES (Motivate) If a ray stands on a line, then the sum of the two adjacent angles so formed is 180° and the converse. (Prove) If two lines intersect, vertically opposite angles are equal.\ (Motivate) Lines which are parallel to a given line are parallel.

Topic-1 Types of Angles Revision Notes



List of Topics Topic-1: Types of Angles



Page No. 80

Topic-2 : Exterior Angle and Intersecting Lines Page No. 86

 Line : Line is a collection of points which has only length neither breadth nor thickness.  Line Segment : A line with two end points.  Ray : A part of line with one end point.  Angle : An angle is formed when two rays originate from the same end point. The rays making an angle are called the arms and the end point is called the vertex.  Types of Angles : (i) Acute Angle : 0° < x < 90° An angle whose measure is more than 0° but less than 90° is called an Acute angle.

Scan to know more about this topic

(ii) Right Angle : x = 90° An angle whose measure is 90°, is called a right angle. Lines and Angles

(iii) Obtuse Angle : 90° < x < 180° An angle whose measure is more than 90° but less than 180° is called an obtuse angle.



LINES AND ANGLES

81

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Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-IX

(iv) Straight Angle : x = 180° An angle whose measure is 180° is called a straight angle.

(v) Reflex Angle : 180° < x < 360° An angle whose measure is more than 180° but less than 360° is called a reflex angle.

(vi) Complete Angle : x = 360° An angle whose measure is 360° is called a complete angle.

 Complementary Angles : Two angles whose sum is 90° are called complementary angles. e.g., Complement of 30° angle is 60° angle.  Supplementary Angles : Two angles whose sum is 180° are called supplementary angles. e.g., Supplement of 70° angle is 110° angle.  Adjacent Angles : Two angles are called adjacent angles, if : (i) they have the same vertex, (ii) they have a common arm, and (iii) uncommon arms on opposite side of the common arm. In the figure, ∠AOP and ∠BOP are adjacent angles.  Vertically Opposite Angles : When two straight lines intersect each other four angles are formed. The pair of angles which lie on the opposite sides of the point of intersection are called vertically opposite angles.

In figure, ∠AOC and ∠BOD are vertically opposite angles and ∠AOD and ∠BOC are also vertically opposite angles. Vertically opposite angles are always equal.  Linear Pair of Angles : Two adjacent angles are said to form a linear pair of angles, if their non-common arms are two opposite rays. OR When the sum of two adjacent angles is 180°, then they are called linear pair of angles.

In figure, ∠AOC and ∠BOC form a linear pair of angles. Examples: Find x in the figure given below



LINES AND ANGLES

Solution :





6x + 3x = 180°  9x = 180° x = 180° = 20° 9

83

[Linear pair]

Mnemonics 1. To draw an angle with the help of protractor we use the mnemonic ‘LIRO’—left inner right outer to draw angle on left side of a line we use inner scale of protractor to draw from right side we use outer scale. 2. Easy way to learn supplementary and complementary angles.

Supplementary = 180°

Complementary = 90°

(It makes 8) (It makes 9)

Example 1

In the given figure, two straight lines PQ and RS intersect each other at O. If ÐPOT = 75°, find the values of a, b, c.

Solution: Step-I : Identify the straight line and use the suitable property to find the value of b. Here, ROS is a straight line. So by property that sum of all angles on a straight line is 180°, we get ÐROP + ÐPOT + ÐTOS = 180° 4b + 75° + 3b = 180° or, 7b + 75° = 180° or, 7b = 180° – 75 ° = 105°



or,

b =

105° = 15° 7

a + 2c = 180° ...(i) Step-II : Use the property which gives relation for a and 4b find the value of a. Since, vertically opposite angles are equal \ a = 4b or, a = 4 × 15° or, a = 60° ...(ii) Step-III : Substitute value of a in eq. (i) to find the value of c. From eqn. (i) we get a + 2c = 180° or, 60° + 2c = 180° or, 2c = 180° – 60° = 120° 120° or, c = 2 = 60° Hence, a = 60°, b = 15°, c = 60°

SUBJECTIVE TYPE QUESTIONS Very Short Answer Type Questions (1 mark each) Q. 1. Two supplementary angles are in ratio 2 : 7. Find the measures of angles.  U [Board Term I, 2016] Sol. 2x + 7x = 180° Þ x = 20° So, the angles are 2x = 2 × 20° = 40° 7x = 7 × 20° = 140° So, two angles are 40° and 140°. 1 [CBSE Marking Scheme, 2016] Q. 2. What is the measure of an angle which is complement of itself ? R Sol. Let the angle be x, then

or,

Angle x = Complement of x x = 90° – x 2x = 90°. 90° x = 2

or x = 45° 1 Q. 3. Two angles measure (30° – a) and (125° + 2a). If each one is the supplement of the other, then find the value of a. A Sol. Angles (30° – a) and (125° + 2a) are supplementary of each other, then 30° – a + 125° + 2a = 180° or, a = 180° – 155° = 25°. 1

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Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-IX

Q. 4. Write the angle which is one-fifth of its complement. R Sol. Let the angle be x, then 1 By given condition, x = (90° –x) 5 or, 5x = 90° – x or, x = 15°. 1 Q. 5. In the figure below, AOB is a straight line. Calculate the measure of ∠COD. C

U

Q. 4. In the given figure, ∠AOB : ∠BOC = 2 : 3. If ∠AOC = 75°, then find the measure of ∠AOB and ∠BOC. 

D 2x – 20° 60° O

Sol. a + b = 180° (Linear pair) ½ a – b = 80° (Given) Adding, 2a = 260° ½ or, a = 130° ½ and b = 180° – a = 180° – 130° = 50°. ½ [CBSE Marking Scheme, 2012]

U [Board Term I, 2012]

x+ 20° A

B

Sol. ∵ x + 20° + 2x – 20° + 60° = 180° (∵ Straight line makes an angle of 180°) or, 3x = 180° – 60° = 120° or, x = 40° Thus, ∠COD = 2x – 20° = 80° – 20° = 60°. 1

Short Answer Type Questions-I (2 marks each) Q. 1. If (3x – 15°) and (x + 5°) are complementary angles, find the angles. R [Board Term I, 2015] Sol. (3x – 15°) + (x + 5°) = 90° or 4x = 90° + 10° = 100° 100° or x = 4 or, x = 25° 1st angle = 3x – 15 = 3(25) – 15 = 60° 2nd angle = (x + 5) = 25 + 5 = 30° Angles are 60° and 30°. 2 Q. 2. In the figure, lines XY and MN intersect at O. If ∠POY = 90° and a : b = 2 : 3, find the value of c.

U [Board Term I, 2012]

Sol. Let, ∠a = 2x and ∠b = 3x ½ Then, ∠a + ∠b = 90° or, 2x + 3x = 90° ∴ x = 18° ½ a = 2 × 18° = 36° b = 3 × 18° = 54° ∴ c = 180° – b = 180° – 54° = 126°. 1 [CBSE Marking Scheme, 2012] Q. 3. In the given figure, ∠AOC and ∠BOC form a line AB. If a – b = 80°, find the values of a and b.

U [Board Term I, 2012]

Sol. Let, ∠AOB = 2x and ∠BOC = 3x ∠AOB + ∠BOC = ∠AOC or, 2x + 3x = 75° 75° or, x = = 15° 5 ∴ and

∠AOB = 2x = 2 × 15° = 30° 1 ∠BOC = 3x = 3 × 15 = 45°. 1 [CBSE Marking Scheme, 2012]

Q. 5. In figure, ∠DOB = 87° and ∠COA = 82°. If ∠BOA = 35°, then find ∠COB and ∠COD.

U [Board Term I, 2012]

∠COA = 82°

Sol. Given, or,

∠COB + ∠BOA = 82°

or,

∠COB + 35° = 82°

(∵ ∠BOA = 35°)

or,

∠COB = 82° – 35° = 47°

Similarly,

∠DOB = ∠DOC + ∠COB

or,

87° = ∠DOC + 47°

or,

∠DOC = 87° – 47° = 40°.



1

1

[CBSE Marking Scheme, 2012]

Short Answer Type Questions-II (3 marks each) Q. 1. In the figure PQ and RS intersect each other at point O. If ÐPOR : ÐROQ = 2 : 3, Find ÐPOR and ÐROQ.



LINES AND ANGLES

85

ÐAOC + ÐAOD = ÐAOD + ÐBOD [Each equal to 180°] \ ÐAOC = ÐBOD Similarly ÐAOD = ÐBOC Hence Proved 1 [CBSE Marking Scheme, 2016] Q. 4. Prove that bisectors of pair of vertically opposite angles are in the same straight line. A [Board Term I, 2016] Sol. Given : Two lines AB and CD intersect at point O. Also, OM and ON are the bisectors of ÐAOC and ÐBOD respectively.



U [Board Term I, 2016]

Sol. ÐPOR = 2x = ÐROQ = 3x ÐPOR + ÐROQ = 180° (straight line angle) or, 2x + 3x = 180° x = 36° \ ÐPOR = 2x = 2 × 36° = 72° ÐROQ = 3x = 3 × 36° = 108° 3 [CBSE Marking Scheme, 2016] Q. 2. In the given figure, lines AB, CD and EF meet at O. Find the value of x, hence find all the three indicated angles.

Sol.

1





To prove : MON is a straight line. Prove : Since, the sum of all the angles around a point O is 360°, we have ÐAOC + ÐBOC + ÐBOD + ÐAOD = 360° or, 2ÐMOC + 2ÐBOC + 2ÐBON = 360° 1 [∵ ÐBOC = ÐAOD ( vertically : opp. angles OM is bisector of ÐAOC, ON is bisector of ÐBOD] or, ÐMOC + ÐBOC + ÐBON = 180° or, ÐMON = 180° \ ÐMON is a straight angle Hence, MON is a straight line. 1 Q. 5. In the figure, if x + y = w + z, then prove that AOB is a straight line.

U [Board Term I, 2016]

ÐCOF = 2x (vertically opposite angles)

\ or,

3x + 2x + 5x = 180 (straight line angle) 10x = 180 Þ x =18



ÐAOC = 3x = 3 × 18 = 54



ÐBOF = 5x = 5 × 18 = 90



ÐDOE = 2x = 2 × 18 = 36 3 [CBSE Marking Scheme, 2016]

Q. 3. Prove that if two lines intersect, vertically opposite angles are equal.  U [Board Term I, 2016] [NCERT Exemp.] Sol. Given : Two lines AB and CD intersect at a point O.

R [Board Term I, 2015] [NCERT] Sol. x + y + w + z = 360° or, 2(x + y) = 360° (∵ x + y = w + z) or, x + y = 180° ∴ AOB is a straight line. 3

Long Answer Type Questions (5 marks each) Q. 1. In figure, m and n are two plane mirrors perpendicular to each other. Show that incident ray CA is parallel to reflected ray BD.

1 To Prove (i) ÐAOC = ÐBOD (ii) ÐAOD = ÐBOC Proof \ Ray OA stand on line CD \ ÐAOC + ÐAOD = 180° [Linear pair] ...(i) Again ray OD stand on line AB 1 \ ÐAOD + ÐBOD = 180° [Linear pair]...(ii) from eqn. (i) and (ii),



A

[Board Term I, 2016, NCERT Exemplar]

86

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-IX

Sol. 84° + 2x = 180° (linear pair) or, 2x = 96° 2 or, x = 48° y + 75° = 2x (VOA) or, y = 2 × 48° – 75° = 96° – 75° = 21° 1 75° + 21° + z = 180° (straight line angles) 2 z = 180° – 96° z = 84°

Sol. Let normals at A and B meet at P.

1 As mirrors are perpendicular to each other therefore BP || OA and AP || OB. So BP ^ PA, i.e., ÐBPA = 90° 1 Therefore Ð3 + Ð2 = 90° ...(i) (Angle sum property) Also Ð1 = Ð2 and Ð4 = Ð3  1 (Angle of incidence = Angle of reflection) Therefore Ð1 + Ð4 = 90° ...(ii) [from (i)] Adding (i) and (ii), we have Ð1 + Ð2 + Ð3 + Ð4 = 180° i.e., ÐCAB + ÐDBA = 180° Hence, CA || BD 2 Q. 2. In the given figure, lines AB and CD intersect each other at O. Find the values of x, y and z.



Q. 3. AB, CD and EF are three concurrent lines passing through the point O such that OF bisect ÐBOD. If ÐBOF = 35°, find ÐBOC and ÐAOD. Sol.

AB, CD and EF are three concurrent lines passing through the point O. OF bisects ÐBOD Þ ÐBOF = ÐFOD = 35° 1½ But ÐBOD = ÐBOF + ÐFOD = 35° + 35° = 70° 1½ ÐBOD + ÐAOD = 180° (Linear pair) ÐAOD = 180° – 70° = 110° ÐBOC = ÐAOD (Vertically opposite angles are equal) \ ÐBOC = 110° 2 Hence, ÐAOD = 110° and ÐBOC = 110°.

U [Board Term I, 2014]

Topic-2 Exterior Angle and Intersecting Lines Revision Notes  Intersecting Lines : Two lines are said to be intersecting when the perpendicular distance between the two lines is not same everywhere. They meet at one point.  Non-Intersecting lines : Two lines are said to be non-intersecting lines when the perpendicular distance between them is same every where. They do not meet. If these lines are in the same plane these are known as parallel lines.

THEOREM  Theorem 1 : Statement : Lines which are parallel to the same line are parallel to each other.  Theorem 2 : Statement : If a side of a triangle is produced, then

the exterior angle so formed is equal to the sum of the two interior opposite angles. Given : A Triangle ABC with interior angles x, y and z, and exterior angle ‘e’. To Prove :



LINES AND ANGLES

87

Proof : In the figure above : x + y + z = 180°

...(i) (Angle Sum Property) e + z = 180° ...(ii) (Linear Pair) Comparing equations (i) and (ii), x + y + z = e + z therefore, x + y = e, Hence Proved.

Example 2

In given figure find value of x?

Solution:

In DABC Exterior Hence,

ÐA = 50° ÐACE = 127° ÐA + ÐB = 127° (Exterior angle property) 50° + ÐB = 127° x = 127° – 50° x = 77°

SUBJECTIVE TYPE QUESTIONS Very Short Answer Type Questions (1 mark each) Q. 1. What will be measure of an angle which is five times its complement? Sol. Let the angle be x Complement of x = (90 – x) \ x = 5(90 – x) \ x = 450° – 5x 6x = 450° x = 75° 1 Q. 2. What will be measure of an angle whose supplement is one-fourth of itself? Sol. Let the angle be x 1 Supplement of x = x 4 1 \ x + x = 180° 4 4x + x = 180° 4 180° ´ 4 x = 5 x = 144° 1 Q. 3. In the figure given below, find the value of y. U [Board Term I, 2015, 2012]

ÐCOF = 5y Now AB is a straight line \ ÐAOC + ÐCOF + ÐFOB = 180° (Straight line angle) 5y + 5y + 2y = 180° 12y = 180° 180 y = = 15° 1 12 Q. 4. In the figure given below, Find ÐACE?

Sol. In DABC ÐACE = exterior angle \ ÐACE = ÐBAC + ÐABC ( exterior angle is equal to sum of interior opposite angles) ÐACE = 42° + 56° = 98° 1 Q. 5. What can you say about lines l and n if l||m and m||n in figure given below?

Sol. l||m and m||n (given) \ l||n Because, Lines which are parallel to the same line are parallel to each other. Sol.

ÐCOF = ÐEOD (Vertically opposite angles are equal)

88

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-IX

Short Answer Type Questions-I (2 marks each)

Short Answer Type Questions-II (3 marks each)

Q. 1. Two supplementary angles are in the ratio 3 : 7, R find the angles. Sol. Let the two supplementary angles be 3x and 7x, then 3x + 7x = 180° 10x = 180° 180 x = = 180° 1 10 Hence, the angles are 3 × 18 = 54° 7 × 18 = 126°

1

Q. 2. In figure given below, calculate the value of angle U [CBSE Term-I, 2015, 2012] q. [CBSE SAS]

Sol. AB is a straight line \

x + 50° + 90° = 180° (straight line angles)



x = 180 – 140°



x = 40°

Now,

1

Q. 1. In the fig. given below, An exterior angle of a triangle is 130° and the two interior opposite angles are equal. Find each of these angles.

Sol. In a triangle ABC x + x = 130° (exterior angle is equal to the sum of interior opposite angles) 2x = 130° x = 65° 1 \ ÐA = ÐB = 65° ÐACB + ÐACD = 180° (Linear pair) ÐACB + 130° = 180° 1 ÐACB = 50° Hence, ÐABC = 65°; ÐBAC = 65° and ÐACB = 50° 1 Q. 2. In given figure PO ^ AB, if x : y : z = 1 : 3 : 5, then find the degree measure of x, y and z.

x = q



(Vertically opposite angles are equal)

\

q = 40°

1

Q. 3. In figure, Prove that ÐAOB + ÐBOC + COD + ÐDOA = 360°

U [Board Term I, 2014, 2012]

A [Board Term I, 2012]

Sol. Extend AO to E, E

\ ÐAOB + ÐBOE = 180° (Linear Pair) ...(i) ½ ÐEOC + ÐCOD + ÐDOA = 180° (Adjacent angles) ...(ii) ½ Adding (i) and (ii) we get ÐAOB + ÐBOE + ÐEOC +ÐCOD + ÐDOA = 180° + 180° ÐAOB + ÐBOC + ÐCOD + ÐDOA = 360° 1 Hence Proved [CBSE Marking Scheme, 2012]

Sol. So, Let \

OP ^ AB ½ ÐPOA = 90° x = ÐPOQ = a y = ÐQOR = 3a z = ÐROA = 5a ½ a + 3a + 5a = 90° 9a = 90° a = 10° ½ x = 10° ½ y = 3 × 10 = 30° ½ z = 5 × 10 = 50° ½ [CBSE Marking Scheme, 2012, 2014]

Q. 3. In the figure given below ÐABD is an exterior angle of DABC.

Sol.

ÐABD = ÐBAC + ÐACB (exterior angle of a triangle is equal to sum of interior opposite angles) 2 3x + 2x + 1 = 2x2 + 1 + 3x + 6



LINES AND ANGLES

3x + 2x – 2x – 3x = 7 – 1 x2 – x – 6 = 0 (x – 3) (x + 2) = 0 x = 3 or –2 But angle cannot be in negative \ x = 3 2 ÐABC + ÐABD = 180° (Linear pair) ÐABC = 180° – (3 × 32 + 2 × 3 + 1) ÐABC = 180° – 34° \ ÐABC = 146° 1 Q. 4. In the figure AB ||CD, EF ^ CD and ÐGFC = 130°. Find x, y and z. U [Board Term I, 2013, 2012] 2

2

89

and, ÐAPD + ÐCPD = 100° ½ Let ÐCPD = x ½ \ ÐBPC = 146° – x ½ and ÐAPD = 100° – x ½ ÐAPD + ÐCPD + ÐBPC = 180° (straight line angles) 1 100° – x + x + 146° – x = 180° 246° – x = 180° x = 66° 1½ Hence, ÐCPD = 66° Q. 2. In the figure given below, if AB and CD are straight lines intersecting at O and OE is perpendicular to CD, find the value of x and y.



E

Sol. According to figure, Sol. \ \ Exterior

EF ^ CD ÐCFE = 90° 90 + z = ÐCFG z = 130° – 90 = 40° ÐFEG = 90° ( FE ^ AB) ÐBGF = Ðz + ÐFEG ( exterior angle property) x = 40° + 90° x = 130° 1 x + y = 180° (Linear pair) 1 130° + y = 180° y = 50° [CBSE Marking Scheme, 2012, 2013]

Commonly Made Error

Students make a mistake in identifying and

AB is a straight line \ ÐAOD + ÐDOE + ÐEOB = 180°

6x = 90°



x = 15°

Now,

1

x + 90° + 5x = 180° 1

ÐAOC = ÐDOB (vertically opposite angles)



ÐAOC = ÐDOE + ÐEOB



y = 90 + 5 × 15



y = 165°

1

Hence, x = 15° and y = 165° 2 Q. 3. In the given figure, the side QR of ÐPQR is produced to a point S. If the bisectors of ÐPQR and ÐPRS meet at point T, then prove that ÐQTR = 1 ÐPQR. 2

applying exterior angle property.

Answering Tip

Students should be well versed with exterior angle property.

Long Answer Type Questions (5 marks each) Q. 1. In the figure given below ÐAPC = 100° and ÐBPD = 146°. Find ÐCPD ?

Sol. According to figure ÐBPC + ÐCPD = 146°

½

Sol. In DPQR, QR is produced to point S, \ ÐPRS = ÐP + ÐPQR (exterior angle property) 1 1 1 So, ÐPRS = ÐP + ÐPQR 1 2 2 2 1 ÐTRS = ÐP + ÐTQR ...(i) 2 ( QT and RT are bisectors of ÐPQR and ÐPRS respectively) 1 Now in DQTR, ÐTRS = ÐT + ÐTQR ...(ii) (exterior angle property of a triangle) From (i) and (ii) we have

90

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-IX

1 ÐP + ÐTQR = ÐTQR + ÐT 2 1 Þ ÐP = ÐT 2

1

Hence,



1 ÐQPR = ÐQTR 2 OR 1 ÐQTR = ÐQPR 2



Proved

OBJECTIVE TYPE QUESTIONS A Multiple Choice Questions Q. 1. In the given figure, POQ is a line. The value of x is. (A) 20° (B) 25° A (C) 30° (D) 35°

Ans. Option (A) is correct. Explanation: 40° + 4 x + 3x = 180° ( POQ is a straight line) 7 x = 180° - 40° 7 x = 140° 140° x= 7 x = 20° Q. 2. An exterior angle of a triangle is 105° and its two interior opposite angles are equal. Each of these equal angles is 1 1 (A) 37 ° (B) 52 ° 2 2 1 (C) 72 ° 2

(D) 75°

A

Ans. Option (B) is correct. Explanation: Let each interiors be x° Exterior ∠s = 105° Sum of interior opp. ∠s = Exterior ∠s x + x = 105° 2 x = 105° 105° x= 2



= 52.5° or 52

1

(1 mark each)

(A) 20° (B) 30° A (C) 10° (D) 40° Ans. Option (C) is correct. Explanation: Angles (55° + 3a) and (115° – 2a) are supplement of each other, Then, 55° + 3a + 115° – 2a = 180° a + 170° = 180° a = 10° Q. 5. In the adjoining figure, AOB is a straight line. If x : y : z = 4 : 5 : 6, then y = ? (A) 60° (B) 80° (C) 48° (D) 72°

Ans. Option (A) is correct. Explanation: Let p bet the common ratio. \ x = 4p, y = 5p, 6z = 6p x + y + z = 180° [angles on a straight line] 4p + 5p + 6p = 180° 15p = 180° 180° p = = 12 15 Now y = 5 × 12 = 60° Q. 6. In the given figure, AOB is a straight line. If ÐAOC = 4x° and ÐBOC = 5x° then find ÐAOC. (A) 40° (B) 60° (C) 80° (D) 100°

1 ° 2

Q. 3. Write the complementary angle of 65°. (A) 65° (B) 25° R (C) 75° (D) 35° Ans. Option (B) is correct. Explanation: Complementary angle of 65° = 90 – 65° = 25 ° ( Sum of complementary angles is 90°) Q. 4. Two angles measure (55° + 3a) and (115° – 2a). If each is supplement of the other, then calculate the value of a.

Ans. Option (C) is correct. Explanation: ÐAOC + ÐBOC = 180° [Angles on a straight line] 4x + 5x = 180° 9x = 180° 180° x = 9

x = 20° ÐAOC = 4x = 4 × 20° = 80°



LINES AND ANGLES

Q. 7. In the given figure, AOB is a straight line, if ÐAOC = (3x + 10)° and ÐBOC = (4x – 26)°, find ÐBOC. (A) 96° (B) 86° (C) 76° (D) 106°

Ans. Option (B) is correct. Explanation: ÐAOC + ÐBOC = 180° [Angles on a straight line] (3x + 10)° + (4x – 26°) = 180° 3x + 10° + 4x – 26° = 180° 3x + 4x = 180° + 16° 7x = 196° 196° x = 7 x = 28° ÐBOC = (4x – 26)° = (4 × 28 – 26) = 86° Q. 8. In the given figure, AOB is a straight line. If ÐAOC = (3x – 10)°, ÐCOD = 50° and ÐBOD = (x + 20)° then find ÐAOC. (A) 40° (B) 60° (C) 80° (D) 50°

Ans. Option (C) is correct. Explanation: ÐAOC + ÐCOD + ÐDOB = 180° [Angles on a straight line] 3x – 10° + 50° + x + 20° = 180° 4x + 60° = 180° 4x = 180° – 60° 4x = 120° 120 x = = 30° 4 ÐAOC = (3x – 10)° = (3 × 30 – 10) = 90 – 10 = 80°

B Assertion & Reason Directions: In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Mark the correct choice as: (A) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A). (B) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).

91

(C) Assertion (A) is true but reason (R) is false. (D) Assertion (A) is false but reason (R) is true. Q. 1. Assertion (A): If angles 'a' and 'b' form a linear pair of angles and a = 40° then b = 150°. Reason (R): Sum of linear pair of angles is always 180°. Ans. Option (D) is correct. Explanation: In case of Assertion (A): 'a' and 'b' are linear pair (given) So a + b = 180° But 40° + 150° = 190° ¹ 180° \ Assertion is false. In case of Reason (R): The sum of linear pair of angles is always 180°. \ Reason is true. Q. 2. Assertion (A): An angle is 14° more than its complementary angle, then angle is 52°. Reason (R): Two angles are said to be supplementary if their sum is 180°. Ans. Option (B) is correct. Explanation: In case of assertion (A): Let the angle be x° Complement of x° = (90 – x)° Since, the difference is 14° \ x – (90° – x) = 14° Þ x – 90° + x = 14° Þ 2x = 14° + 90° 2x = 104° x = 52° \ Assertion is true. In case of reason (R): Sum of supplementary angles is 180°. \ Reason is also true. But Reason is not correct explanation of Assertion. Q. 3. Assertion (A): Sum of the pair of angles is 120° and 60° is supplementary. Reason (R): Two angles, the sum of whose is 90° are called supplementary angles. Ans. Option (C) is correct. Explanation: In case of assertion (A): Sum of 120° and 60° is 120° + 60° = 180° which is supplementary. \ Assertion is true. In case of reason (R): Two angles whose sum is 90° are called complementary angles. \ Reason is false. Q. 4. Assertion (A): An exterior angle of a triangle is 110° and its two interior angles are equal. Therefore each of these equal angle is 55°. Reason (R): Sum of two interior opposite angles is equal to the exterior angle. Ans. Option (A) is correct. Explanation: In case of assertion (A): Given, interior angles are equal Let the interior angles be x. x + x = 110° 2x = 110° x = 55° \ Assertion is true.

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Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-IX

In case of reason (R): We know that the exterior angle is equal to the sum of two interior opposite angles.

\ Reason is also true. Hence, Both A and R are true and R is the correct Explanation of A.

COMPETENCY BASED QUESTIONS A Case based MCQs

Read the following passage and answer any four questions of the following :

I. BSE stands for a disease called Bovine Spongiform Encephalopathy. “Bovine” means that the disease affects cows, “spongiform” refers to the way the brain from a sick cow looks spongy under a microscope, and “encephalopathy” indicates that it is a disease of the brain. This disease is commonly called “mad cow disease.”

A farmer has a field ABCD formed by two pair of parallel roads as shown below in which l || m and p || q. His four cows suffering from BSE. Thus, he tied them at four corners of the field ABCD.

Q. 1. If ∠BAC = 30°, find ∠CAD. (A) 60° (B) 90° (C) 30° (D) 20° Ans. Option (C) is correct. Explanation: AC is diagonal. \ ÐBAC = ÐCAD ( Diagonals of a parallelogram bisect one of its angles) Thus ÐCAO = 30° 1 Q. 2. ∠ABC + ∠BCD = 180° as : (A) Corresponding angles are supplementary. (B) Alternate interior angles are supplementary. (C) Alternate exterior angles are supplementary. (D) Angles on the same side of a parallelogram are supplementary.

Ans. Option (D) is correct.

(4 marks each) 1

Q. 3. If cow at C and cow at D is 2 km apart, then what is the distance between cow at A and cow at B? (A) 1 km (B) 2 km (C) 3 km (D) 4 km Ans. Option (B) is correct. Explanation: Since, p||q and l||m thus, ABCD is a parallelogram. Also, since opposite sides of a parallelogram are equal. So, AB = CD½ Given, distance between cow at C and cow at D = CD = 2 km ⇒ AB = 2 km Hence, distance between cow at A and cow at B is 2 km. ½ Q. 4. If ∠B = 45°, then ∠D = _________ . (A) 50° (B) 40° (C) 45° (D) 55° Ans. Option (C) is correct. Explanation: Since, ∠B = 45 ⇒ ∠D = 45° (opposite angles of a parallelogram are equal) 1 Q. 5. If we join BD such that BD meet AC at O and ∠BOC = 30°, then what is the measure of ∠AOD?  (A) 90° (B) 60° (C) 45° (D) 30° Ans. Option (D) is correct. Explanation: ∠BOC = ∠AOD = 30° (vertically opposite angles are equal)  1 II. A plane mirror is a mirror with a flat reflective surface.

An incident ray is a ray of light that strikes a surface. The reflected ray corresponding to a given incident ray, is the ray that represents the light reflected by the surface. In figure, m and n are two plane mirrors perpendicular to each other.



LINES AND ANGLES

Q. 1. Parallel lines

(A) meet at a point (B) never meet (C) sometimes meet (D) meet at right angle Ans. Option (B) is correct. Q. 2. Which statement is incorrect ? (A) Corresponding angles formed at corresponding corners. (B) Alternate interior angles are equal (C) Angles on the same side of the transversal are complementary (D) Vertically opposite angles are equal Ans. Option (C) is correct. Q. 3. Incident ray CA is : (A) parallel to AB (B) perpendicular to BD (C) parallel to BD (D) perpendicular to AO Ans. Option (C) is correct. Q. 4. ∠BOA = (A) 270°°

straight road cuts the other at 30°. Now using the given information, answer the following questions.

Q. 1. Find the measure of ÐBOD. Ans. \

ÐBOD = ÐAOC (Vertical opposite angles) ÐAOC = 30° (given) ÐBOD = 30° 1

Q. 2. Find the measure of ÐAOD Ans.

ÐAOC + ÐAOD = 180° 30° + ÐAOD = 180° ÐAOD = 150°

(Linear pair) 1

Q. 3. Which property is used in this case. Ans. Intersecting lines property is used which states, that, "Two lines are said to be intersecting when the perpendicular distance between the two lines is not same everywhere. They meet at one point." 2 II. To Protect poor people from cold weather, Ramlal, has given his land to make a shelter home for them. In the given figure sides PQ and RQ of DPQR are produced to point S and T respectively. If ÐPQT = 110° and ÐSPR = 135°, answer the following questions.

1

Ans. Option (B) is correct. 1 Explanation: m and n are two plane mirrors perpendicular to each other So, AO is perpendicular to BO \ ∠BOA = 90° Q. 5. If BO = 3 cm, AB = 5 cm then, AO =  (A) 3 cm (B) 2 cm (C) 4 cm (D) 6 cm Ans. Option (C) is correct.  1 Explanation: m and n are two plane mirrors perpendicular to each other. So, AO is perpendicular to BO. Thus, triangle AOB is a right-angled triangle. OA2 + OB2 = AB2 OA2 + 32 = 52 OA2 + 9 = 25 OA2 = 25 – 9 = 16 OA = 4 cm

Q. 1. Find ÐQPR. Ans. ÐSPR +ÐQPR = 180° 135° + ÐQPR = 180° \ ÐQPR = 45°

(Linear pair) 1

Q. 2. Find ÐPRQ. Ans. In DPQR, ÐPQT is an exterior angle \ ÐPQT = ÐQPR + ÐPRQ (exterior angle proeprty) 110° = 45° + ÐPRQ 110 – 45° = ÐPRQ \ ÐPRQ = 65° 1

B Case based Subjective Questions



Read the following passage and answer the following questions: I. Harry was going on a road trip with his father. They were travelling on a straight road. After riding for some distance, they reach a cross road where one



93

Q. 3. Which property is used over here.



Ans. In this case exterior angle property is used which states that, "An exterior angle of a triangle is equal to the sum of interior opposite angles." 2



Study Time: Max. Time: 3 Hrs Max. Questions: 42

CHAPTER

7



Syllabus

TRIANGLES (Motivate) Two triangles are congruent if any two sides and the included angle of one triangle are equal to any two sides and the included angle of the other triangle (SAS Congruence).

(Prove) Two triangles are congruent if any two angles and the included side of one triangle is equal to any two angles and the included side of the other triangle (ASA Congruence). (Motivate) Two triangles are congruent if the three sides of one triangle are equal to three sides of the other triangle (SSS Congruence). (Motivate) Two right triangles are congruent if the hypotenuse and a side of one triangle are equal (respectively) to the hypotenuse and a side of the other triangle (RHS Congruence). (Prove) The angles opposite to equal sides of a triangle are equal. (Motivate) The sides opposite to equal angles of a triangle are equal.

Criteria for Congruence of

Topic-1 Triangles

List of Topics Topic-1: Criteria for Congruence of Triangles Page No. 94 Topic-2 : Properties of Triangles 

Page No. 100

Revision Notes  Congruence of Triangle : The geometrical figures of same shape and size are congruent to each other i.e., two triangles DABC and DPQR are congruent if and only if their corresponding sides and the corresponding angles are equal.

If two triangles DABC and DPQR are congruent under the correspondence A ¾® P, B ¾® Q and C ¾® R, then symbolically it is expressed as

DABC @ DPQR



 SAS Congruence Rule : Two triangles are congruent if two sides and the included angle of one triangle are equal to the sides and the included angle of the other triangle.



 ASA Congruence Rule : Two triangles are congruent if two angles and the included side of one triangle are equal to two angles and the included side of the other triangle.



 AAS Congruence Rule : Two triangles are congruent if any two pairs of angles and one pair of corresponding sides are equal.

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Triangles Part-I



TRIANGLES

95

96

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-IX

 SSS Congruence Rule : If three sides of a triangle are equal to the three sides of another triangle, then the two triangles are congruent.



 RHS Congruence Rule : If in two right triangles, the hypotenuse and one side of a triangle are equal to the hypotenuse and one side of the other triangle, then the two triangles are congruent.

Example 1 Q. In an isosceles DPQR with PQ = PR, S and T are points on QR such that QT = RS show that PS = PT. Solution: Step I : Read the question carefully and write the given conditions.

DPQR is an isosceles triangle in which PQ = PR ...(i) S and T are points on QR such that QT = RS ...(ii) Step II : Apply the theorems related to given conditions to find other information.

Since PQ = PR So ÐR = ÐQ ...(iii) [Property of isosceles triangle] From eqn. (ii) we have QT = RS Þ QT – ST = RS – ST [Subtracting ST from both sides] Þ QS = RT ...(iv) Step III : Apply the suitable congruence rule in two triangles. In DPQS and DPRT, PQ = PR [From eqn. (i)] ÐQ = ÐR [From eqn. (iii)] and QS = RT [From eqn. (iv)] \ DPQS @ ÐPRT (By SAS congruence rule) Step IV : Apply the property CPCT i.e., corresponding part of congruent triangles, to get the required result. As DPQS @ DPRT Then PS = PT (CPCT)

Mnemonics SSS has filled the form with SAS and AAS. Concept: Congruency of a triangle Interpretation Side Side Side SSS (3 sides are equal) Side Angle Side SAS (2 sides and included angles) Angle Angle Side AAS (2 angles and corresponding side)

SUBJECTIVE TYPE QUESTIONS Very Short Answer Type Questions (1 mark each) Q. 1. Write ASA congruence rule for two triangles.

R [Board Term I, 2015]

Sol. ASA congruence : Two triangles are congruent, if two angles and the included side of one triangle are equal to two angles and the included side of other triangle. 1 Q. 2. In DABC and DDEF, AB = DE, ÐA = ÐD. What will be the condition in which the two triangles will be congruent by SAS axiom ? R Sol. Since AB = DE, ÐA = ÐD and DABC @ DDEF by SAS.

Therefore

AC = DF.

1

Q. 3. What do we call a triangle if the angles are in the ratio 5 : 3 : 7 ? U Sol. Let the angles of triangle are 5x, 3x and 7x, then 5x + 3x + 7x = 180° or, 15x = 180° Thus, x = 12° \ Angles are 60°, 36°, 84° ∵ Each angle is less than 90° \ The triangle is an acute angled triangle.

1



TRIANGLES

It can also be a scalene triangle, since all angles are of different measure. Hence, corresponding sides are unequal. Q. 4. DABC @ DPQR, AB = PQ. Which statement has been followed in this ? A Sol. If two triangles are congruent, then one side of a triangle is equal to the corresponding side of the other triangle. Hence,

AB = PQ

AB = DC (Given) ÐABD = ÐCDB (Given) BD = BD (Common) DABD @ DCDB. Hence Proved 1

Short Answer Type Questions-I (2 marks each)

(CPCT) ½ Hence Proved Q. 2. In the figure below, the diagonal AC of quadrilateral ABCD bisects ÐBAD and ÐBCD. Prove that BC = CD. A [Board Term I, 2011, 2010]

Sol. In DADC and DABC, AC bisects ÐA and ÐC So ÐDAC = ÐBAC ½ ÐDCA = ÐBCA (Given) ½ AC = AC (Common) Hence, DADC @ DABC (By ASA rule) ½ \ CD = BC.  (CPCT) ½ Hence Proved.

Commonly Made Error

Q. 1. In the figure, DABC and DDBC are two isosceles triangles on the same base BC. Prove that ÐABD = ÐACD.

ÐABD = ÐACD

1

Q. 5. In the figure, if AB = DC, ÐABD = ÐCDB, which congruence rule would you apply to prove DABD @ DCDB ? R

Sol. By SAS.

\

97

Students make an error in putting the condition

AAS or ASA rule as both denote angle side or angle side angle.

Answering Tip

A [Board Term I, 2014]



ASA congruence rule states that two angles and

included side of one triangle are equal and AAS congruence rule states that if any two pairs of angles and one pair of corresponding sides are equal.

Q. 3. In the figure, OA = OB and OD = OC. Show that : (i) DAOD @ DBOC, (ii) AD || BC.

Sol.



Join AD.



In DABD and DACD,



AB = AC (Given)



BD = CD (Given)

AD = AD (Common) 1 By using SSS Congruence Rule, DABD @ DACD ½

U [Board Term I, 2011] Sol. (i) In DAOD and DBOC, OA = OB (Given) OD = OC (Given) ÐAOD = ÐBOC (Vertically opposite angles) So, by SAS criteria, DAOD @ DBOC 1 (ii) ÐCBO = ÐDAO (CPCT) AD and BC are two lines intersected by AB and ÐCBA = ÐDAB (Proved above) and they form a pair of alternate angles.

\

AD || BC.

Hence Proved 1

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Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-IX

Q. 4. In figure ÐB = ÐE, BD = CE and Ð1 = Ð2. Show DABC @ DAED. U [Board Term I, 2011]

Sol. Let, ÐDAC be Ð3 Ð1 = Ð2 (Given) ½ By adding Ð3 on both sides Ð1 + Ð3 = Ð2 + Ð3 ÐBAC = ÐEAD ...(i) ½ Given that, BD = CE By adding DC on both sides BD + DC = CE + DC ½ or, BC = DE ...(ii) ÐB = ÐE (Given) ...(iii) From (i), (ii) and (iii), we get DABC @ DAED. (By AAS rule) ½ Hence Proved

Short Answer Type Questions-II (3 marks each)

Sol. In DBMR and DDNR, BM = DN (Given) \ ÐN = ÐM (each right angle) 1 ÐNDR = ÐMBR (Alternate angle) \ DBMR @ DDNR (ASA Congruency) BR = DR or, R is mid-point of BD 1 \ AC bisects BD. Hence Proved 1 [CBSE Marking Scheme, 2016]

Commonly Made Error

Students get confused between the condition of

congruency AAS and ASA and put the incorrect condition.

Answering Tip

To prove the triangle students should be clear

in mind about all the properties of triangles and which condition of congruency will be applied in which situation.

Q. 3. In the given figure AD = BC and BD = AC. Prove that, ÐADB = ÐBCA and ÐDAB = ÐCBA.

Q. 1. In the figure, if AF = CD and ÐAFE = ÐCDE, prove that EF = ED.



U [Board Term I, 2016]

Sol. In DAFE and DCDE, AF = CD (Given) \ ÐAFE = ÐCDE (Given) ÐE = ÐE (Common) \ ÐFAE = ÐDCE (Third angle of D) \ DAFE @ DCDE. (ASA) Hence EF = ED (CPCT) Proved. [CBSE Marking Scheme, 2016] 3 Q. 2. In the figure, BM and DN are both perpendicular to AC and BM = DN. Prove that AC bisects BD.



U [Board Term I, 2016]

A [Board Term I, 2014, 2012] Sol. Proof : In DABD and DBAC, AD = BC (Given) BD = AC (Given) AB = AB (Common side) 2 DABD @ DBAC By SSS congruence or, ÐADB = ÐBCA (CPCT) 1 and ÐDAB = ÐCBA (CPCT)  Hence Proved. Q. 4. In figure, PQRS is a square and SRT is an equilateral triangle. Prove that : (i) PT = QT (ii) ÐTQR = 15° U [Board Term I, 2014; 2012, 2011, 2010]



Sol. PQRS is a square. 

TRIANGLES

(Given)

A

B

(i) SRT is an equilateral triangle.(Given) \ ÐPSR = 90°, ÐTSR = 60° ½ or, ÐPSR + ÐTSR = 150°. Similarly, ÐQRT = 150° In DPST and DQRT, we have PS = QR ÐPST = ÐQRT = 150° and ST = RT ½ By SAS, DPST @ DQRT or, PT = QT  (CPCT) Hence Proved. (ii) In DTQR, QR = RT (Square and equilateral D on same base) ½ or, ÐTQR = ÐQTR = x ½ \ x + x + ÐQRT = 180° or, 2x + 150° = 180° or, 2x = 30° \ x = 15°. or, ÐTQR = 15°. Hence Proved. 1 Q. 5. In a right angled triangle, if one acute angle is double the other, then prove that the hypotenuse is double the smallest side. A [Board Term I, 2016] Sol. DABC is right angled at B.

Let ÐCAB = x and ÐACB = 2x produce CB to D so that BD = BC, join AD 1 In DABC and DABD, we have AB = AB (Common) BC = BD (By Construction) and ÐABC = ÐABD = 90° 1 \ DABC @ DABD (SAS) \ AC = AD and ÐCAB = ÐDAB (CPCT) Now, ÐCAD = x + x = 2x° = ÐACB Then AD = CD or AD = 2BC or AC = 2BC 1 \ Hypotenuse AC is double the smallest side BC. Hence Proved [CBSE Marking Scheme, 2016]

Q. 6. In an isosceles triangle ABC with AB = AC, D and E are points on BC such that BE = CD (as given figure). Show that AD = AE U + A [KVS 2019]

99

D

E

C

Sol. Step I : Read the question carefully and write the given conditions.

DABC is an isosceles triangle in which AB = AC ...(i) D and E are points on BC such that BE = CD ...(ii) Step II : Apply the theorems related to given conditions to find other information. Since, AB = AC So, ÐB = ÐC ...(iii) [Property of isosceles triangle] From eqn. (ii) we have BE = CD Þ BE – DE = CD – DE [Subtracting DE from both sides] Þ BD = CE ...(iv) 1 Step III : Apply the suitable congruence rule in two triangles. In DABD and DACE, AB = AC ÐB = ÐC [From eqn. (iii)] and BD = CE [From eqn. (iv)] \ DABD @ DACE (By SAS congruence rule) 1 Step IV : Apply the property CPCT i.e., corresponding part of congruence triangles, to get the required result. As DABD @ DACE Then AD = AE (CPCT) 1

Long Answer Type Questions (5 marks each) Q. 1. In the given figure, if AC = BC, ÐDCA = ÐECB and ÐDBC = ÐEAC, then prove that BD = AE.

U [Board Term I, 2014; 2012]

Sol. Given, ÐDCA = ÐECB Adding ÐDCE to both sides, we get ÐDCA + ÐDCE = ÐECB + ÐDCE or, ÐECA = ÐDCB

1½

100

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-IX

In DACE and DBCD, AC = BC (Given) ÐECA = ÐDCB (Proved) ÐEAC = ÐDBC (Given) \ DACE @ DBCD (By ASA cong.) 2 \ BD = AE. (CPCT) Proved. 1½ [CBSE Marking Scheme, 2012] Q. 2. ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that (i) D is the mid-point of AC (ii) MD ^ AC  1 (iii) CM = MA = AB U [KVS 2019] 2 Sol. Given : ABC is a triangle right angle at C and M is the mid point of side AB.

(ii) DM || BC \ ÐACB = ÐADM  (Corresponding angles) Hence, MD ^ AC 1 (iii) In DADM and DDMC ÐADM = ÐMDC = 90° (Proved) AD = DC(Proved) MD = MD (Common) \ DADM @ DCDM (SAS Congruence rule) \ AM = CM (CPCT) 1 \ CM = AM = AB 2 2  Hence Proved. Q. 3. In the figure, OA = OB, OC = OD and ÐAOB = ÐCOD. Prove that AC = BD.

A

D

M

C

B

To prove (i) D is the mid point of AC (ii) MD ^ AC

1 AB 2 Proof (i) M is the mid point of side AB and MD || BC \ D will be mid point of side AC (by converse of mid point theorem)  2 (iii)

CM = MA =

U [Board Term I, 2016] Sol. ÐAOB = ÐCOD (Given) ÐAOB – ÐCOB = ÐCOD – ÐCOB 1 ÐAOC = ÐBOD In DAOC and DBOD 1 AO = OB (Given) ½ OC = OD (Given) ½ ÐAOC = ÐBOD (Proved above) \ DAOC @ DBOD (SAS Criterion) 1 AC = BD (CPCT) Hence Proved 1

Topic-2 Properties of Triangles Revision Notes  A triangle is isosceles if its any two sides are equal. Here, we will discuss some properties related to isosceles triangle. (i) Angles opposite to equal sides of a triangle are equal. In figure, ÐB = ÐC

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(ii) The sides opposite to equal angles of a triangle are equal. Properties of In figure, AB = AC Triangles  In an isosceles triangle, bisector of the vertical angle of a triangle bisect the base.  The medians of an equilateral triangle are equal in length.  A point equidistant from two intersecting lines lies on the bisector of the angles formed by the two lines.



TRIANGLES

101

Example 2

AB is a line segment. C and D are points on opposite side of AB such that each of them is equidistant from the points A and B. Show that line CD is the perpendicular bisector of AB.

Solution: Step I : Read the question carefully and write all given conditions. AB is a line segment. C and D are points on opposite sides of AB such that CA = CB ...(i) and DA = DB ...(ii) CD intersects AB at point O. Step II : Find what is given to show. CD is the perpendicular bisector of AB. Step III : Show that DCAD and DCBD are congruent, further use CPCT to find relation between angles. In DCAD and DCBD CA = CB [from eqn. (i)] AD = BD [from eqn. (ii)] and CD = CD [Common side]

\ DCAD º DCBD [By SSS Congruence rule] or, ÐACD = ÐBCD [By CPCT] or ÐACO = ÐBCO ...(iii) Step IV : Show that DCAO and DCBO are congruent and further use CPCT to find relation between angles and sides. In DCAO and DCBO CA = CB [from eqn. (i)] ÐACO = ÐBCO [from eqn. (iii)] and CO = CO [Common side] \ DCAO º DCBO [By SAS Congruence rule] or, AO = BO ...(iv) [By CPCT] and ÐAOC = ÐBOC ...(v) [By CPCT] Step V : Since, AB is a line segment, so we use the property of linear pair and find the measure of ÐAOC or ÐBOC. AB is a line segment. So, ÐAOC + ÐBOC = 180° or, ÐAOC + ÐAOC = 180° [from eqn. (v)] or, 2ÐAOC = 180° 180° or, ÐAOC = 2 or, ÐAOC = 90° AO = BO [From eqn. (iv)] ÐBOC = ÐAOC = 90° each Hence, CD is the perpendicular bisector of AB.

SUBJECTIVE TYPE QUESTIONS Very Short Answer Type Questions (1 mark each) Q. 1. DABC is an isosceles right angled triangle in which ÐA = 90°. Calculate ÐB. U Sol. AB = AC (given) \ ÐC = ÐB (Angle opposite to equal side)

\ \

PE = PE DPEQ @ DPER

(Common) (by ASA)

PQ = PR. (CPCT) 1 Hence Proved.

Short Answer Type Questions-I (2 marks each)

ÐA + ÐB + ÐC = 180°

[Angle sum property] or, 90° + ÐB + ÐB = 180° or, 2ÐB = 90° \ ÐB = 45° 1 Q. 2. In DPQR, PE is the perpendicular bisector of ÐQPR, then prove that PQ = PR. U Sol. In DPEQ and DPER, ÐPEQ = ÐPER = 90° ÐQPE = ÐRPE (Given)

Q. 1. ABC is an isosceles triangle with AB = AC. Draw AP ^ BC. Show that ÐB = ÐC. A [Board Term I, 2011] Sol.

In DABP and DACP, AB = AC

(Given) ½

102

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-IX



AP = AP (Common) ÐAPB = ÐAPC = 90°, (AP ^ BC) ½

\

DABP @ DACP (By RHS rule) ½

or, ÐB = ÐC. (CPCT) Hence Proved ½ Q. 2. PS is an altitude of an isosceles triangle PQR in which PQ = PR. Show that PS bisects ÐP. A [Board Term I, 2011] Sol. In DPQS and DPRS,

By RHS rule, or,

PQ = PR (Given) ½ PS = PS (Common) ÐPSQ = ÐPSR = 90° (PS is altitude) ½ DPQS @ DPRS ÐQPS = ÐRPS

\ PS bisects ÐP.

½ (CPCT) ½

Q. 1. In a triangle ABC, X and Y are the points on AB and 1 1 BC respectively. If BX = AB and BY = BC 2 2 and AB = BC. Show that BX = BY. A [Board Term-I, 2015]

Sol.

or,



But But



A [NCERT, Board Term I, 2014, 2012]

Hence Proved

Short Answer Type Questions-II (3 marks each)



or, ÐA = ÐB ....(2) ½ (Angles opp. to equal sides) From (1) and (2), we have ÐA = ÐB = ÐC ½ Also, ÐA + ÐB + ÐC = 180°, (Angle sum property) \ ÐA + ÐA + ÐA = 180° or, 3ÐA = 180° or, ÐA = 60° \ ÐA = ÐB = ÐC = 60°. 1 Thus, each angle of an equilateral triangle is 60°. Hence Proved. Q. 3. Triangle ABC is an isosceles triangle such that AB = AC. Side BA is produced to D, such that AD = AB. Show that ÐBCD is a right angle.

1 AB = BC (Given) 1 1 AB = BC 2 2 1 AB = BX ...(i) ½ 2

Sol.

In DABC, AB = AC or, Ð1 = Ð2 ...(1) Angles opp. to equal sides are equal. In DADC, AB = AD \ AC = AD ½ In DBCD, Ð3 = Ð4 ...(2) ½ Ð1 + (Ð2 + Ð3) + Ð4 = 180° ½ or, Ð2 + Ð2 + Ð3 + Ð3 = 180° ½ or, 2(Ð2 + Ð3) = 180° or, Ð2 + Ð3 = 90° ½ or, ÐBCD is a right angle. Hence Proved ½ [CBSE Marking Scheme, 2014, 2012] Q. 4. In the figure, ABC is an isosceles triangle in which AB = AC and LM is parallel to BC. If ÐA = 50°, find ÐLMC.

1 BC = BY 2

...(ii) ½ From (i) and (ii), BX = BY Hence Proved 1 [CBSE Marking Scheme, 2011, 2015] Q. 2. Prove that each angle of an equilateral triangle is 60°. U [Board Term I, 2012, NCERT] Sol. Let DABC be an equilateral triangle, so that AB = AC = BC. Now, AB = AC ½ or, ÐB = ÐC ....(1) ½ (Angles opp. to equal sides) CB = CA

U [Board Term I, 2016] Sol. In DABC, AB = AC ½ \ ÐABC = ÐACB = q or, ÐB = ÐC = q ÐA + ÐB + ÐC = 180°



TRIANGLES

50° + q + q = 180° 2q = 180° – 50° ½ = 130° q = 65° \ ÐB = ÐC = 65° LM || BC (given) 1 \ ÐLMC + ÐBCM = 180° [∵ ÐLMC and ÐBCM are co-interior angles] ÐLMC + 65° = 180° ÐLMC = 180° – 65° ÐLMC = 115° 1 Q. 5. In the given figure, AB = AC and BE and CF are bisectors of ÐB and ÐC respectively. Prove that DEBC @ DFCB.

103

AB = AC (Given) 1 \ ÐABC = ÐACB (Angles opp. to equal sides) ÐABC + ÐACB = 137°, (ext. angle) 137ϒ \ ÐABC = ÐACB = = 68.5° 1 2 CH = CB (Given) or, ÐCBA = ÐCHB = 68.5° 1 \ ÐHCB = 180° – 137° = 43° 1 ÐCHK = ÐHCB = 43°. (Alternate angles) 1 [CBSE Marking Scheme, 2013, 2012] Q. 2. In the given figure, ABCD and BPQ are straight lines. If BP = BC and DQ is parallel to CP prove that : (i) CP = CD (ii) DP bisects ÐCDQ

U [Board Term I, 2016] Sol. AB = AC (Given) \ ÐABC = ÐACB ...(i) BE and CF are the bisectors of ÐB and ÐC 1 1 \ ÐEBC = ∠ ABC = ∠ACB 2 2 = ÐFCB ÐEBC = ÐFCB ...(ii) In DBEC and DCFB ÐABC = ÐACB \ ÐECB = ÐFBC ÐEBC = ÐFCB (Proved) BC = BC (Common) \ DBEC @ DCFB (By ASA) DEBC @ DFCB Hence Proved 3 [CBSE Marking Scheme, 2016]

U [Board Term I, 2016]

Sol. \ \

BP = BC (Given) ÐBCP = ÐBPC = y (Angle opp. to equal sides) 1 Ð1 = y (Corresponding angles)

Long Answer Type Questions (5 marks each) Q. 1. In figure, AB = AC, CH = CB and HK||BC. If ÐCAX = 137°, then find ÐCHK. U [Board Term I, 2013, 2012]

Sol.

ÐXAK + ÐKAH = 180° (Linear pair) ÐKAH = 180° – 137° = 43° (∵ ÐCAX = ÐXAK = 137°, Given)

1 4x = 2y (external angle)1 Ð2 = x (Alternate angles) y = x + Ð3 4x ( 4x = 2y, = y) 2 2x = x + Ð3 or, x = Ð3 1 or, CP = CD (side opp. to equal angles) Also, we have Ð2 = x, x = Ð3 1 \ DP bisects ÐCDQ Hence Proved [CBSE Marking Scheme 2016]

Q. 3.  ABCD is a square and ABE is an equilateral triangle 1 outside the square prove that ÐACE = ÐABE. 2 

U [Board Term I, 2016]

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Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-IX

Sol. ∵ DABE is an equilateral triangle \ ÐABE = ÐBEA = ÐEAB = 60°

1

1 Also or,

ÐBAC = 45° BE = BC Ð1 = Ð2

1

\ Ð1 + Ð2 = 180° – (60° + 90°) = 180° – 150° = 30° 1 or, 2Ð1 = 30° or, Ð1 = 15° or, Ð3 = 45° – 15° = 30° \ ÐACE = 30°, ÐABE = 60° 1 \ ÐACE = ∠ABE Hence Proved 1 2 [CBSE Marking Scheme, 2016]

OBJECTIVE TYPE QUESTIONS

(1 mark each) F

A

A Multiple Choice Questions Q. 1. Which of the following is not a criterion for congruence of triangles? (A) SAS (B) ASA (C) SSA (D) SSS  R [NCERT Exemp.] Ans. Option (C) is correct. Explanation: Criterion for congruency are SAS (side angle side), ASA (angle side angle), SSS (side side side), AAS (angle-angle side). Q. 2. If AB = QR, BC = PR and CA = PQ, then (A) ∆ ABC ≅ ∆ PQR (B) ∆ CBA ≅ ∆ PRQ (C) ∆ BAC ≅ ∆ RPQ (D) ∆ PQR ≅ ∆ BCA  A [NCERT Exemp.] Ans. Option (B) is correct. Explanation: AB = QR & BC = PR Þ ∠B = ∠R ...(i) BC = PR & CA = PQ Þ ∠C = ∠P …(ii) CA = PQ & AB = QR Þ ∠A = ∠Q …(iii) From equations (i), (ii) and (iii), we get ∆CBA ≅ ∆PRQ. Hence, the option (b) is correct. Q. 3. In triangles ABC and PQR, AB = AC, ∠C = ∠P and ∠B = ∠Q. The two triangles are (A) isosceles but not congruent (B) isosceles and congruent (C) congruent but not isosceles (D) neither congruent nor isosceles  A [NCERT Exemp.] Ans. Option (A) is correct. Explanation: In ∆ ABC, AB = AC [Given] ∠C = ∠B [Angles opposite to equal sides are equal] So, ∆ABC is an isosceles triangle But it is given that ∠B = ∠Q ∠C = ∠P Therefore, ∠P = ∠Q [Since, ∠C = ∠B] \ QR = PR [Sides opposite to equal angles are equal] So, ∆PQR is also an isosceles triangle.

C

B

E

D

Therefore, both triangles are isosceles but not congruent. As we know that AAA is not a criterion for congruence of triangles. Q. 4. In triangles ABC and DEF, AB = FD and ÐA = ÐD. The two triangles will be congruent by SAS axiom if (A) BC = EF (B) AC = DE A (C) AC = EF (D) BC = DE Ans. Option (B) is correct. Explanation: Given, in ∆ABC and ∆DEF, AB = FD, ÐA = ÐD We know that, two triangles will be congruent by SAS axiom if two sides and the included angle between them are equal.

Q. 5. In ∆ABC, AB = AC and ÐB = 50°. Then ÐC is equal to (A) 40° (B) 50° (C) 80° (D) 130°  A [NCERT Exemp.] Ans. Option (B) is correct. Explanation : Given that AB = AC and ÐB = 50° A

50° B

C

In ∆ABC, AB =AC (given) ÐC = ÐB (angles opposites to equal sides are equal) ÐC = 50°



TRIANGLES

Q. 6. In ∆ABC, BC = AB and ÐB = 80°. Then ÐA is equal to (A) 80° (B) 40° (C) 50° (D) 100° 

A [NCERT Exemp.]

Ans. Option (C) is correct. Explanation : In ∆ABC, BC = AB, ÐB = 80° ÐC =ÐA (angles opposite to equal sides are equal) ÐA + ÐB +ÐC = 180° (sum of all angles of triangle) ÐA+80°+ÐA = 180°

2ÐA = 180° – 80° 100° ∠A = 2 ∠A = 50°

   

Q. 7. It is given that ∆ ABC @ ∆ FDE and AB = 5 cm, ÐB = 40° and ÐA = 80°. Then which of the following is true? (A) DF = 5 cm, ÐF = 60°

105

Q. 1. Assertion (A): Each angle of an equilateral triangle is 60°. Reason (R): Angles opposite to equal sides of a triangle are equal. Ans. Option (A) is correct. Explanation: In case of assertion (A): We know that each angle of an equilateral triangle is 60°. \ Assertion is true. In case of reason (R): Angles opposite to equal sides are equal. \ Reason is true. Hence, Both A and R are true and R is the correct explanation of A. Q. 2. Assertion (A): In DABC and DPQR, AB = PQ, AC = PR and ÐBAC = ÐQPR, DABC @ DPQR. Reason (R): Both the triangles are congruent by SSS congruence. Ans. Option (C) is correct. Explanation: In case of assertion (A):

(B) DF = 5 cm, ÐE = 60° (C) DE = 5 cm, ÐE = 60°

(D) DE = 5 cm, ÐD = 40° A [NCERT Exemp.]

Ans. Option (B) is correct. Explanation : Given, ∆ ABC @ ∆ FDE and AB = 5 cm, ÐB = 40°, ÐA = 80° Since,

In DABC and DPQR, AB = PQ (given) AC = PR (given) ÐBAC = ÐQPR \ DABC @ DPQR (By SAS Rule) \ Assertion is true. In case of reason (R): The reason is false as the triangles are congruent by SAS and not SSS.



∆FDE @ ∆ ABC F

A 80°

40° B

C

D

E

∴

DF = AB



DF = 5 cm

and

ÐE = ÐC



ÐE = ÐC = 180° – (ÐA + ÐB)



[by CPCT]

Q. 3. Assertion (A): DABC and DDBC are two isosceles triangles on the same base BC. Then ÐABD = ÐACD.

[by CPCT]

[By angle sum property of a ΔABC]



ÐE = 180° – (80° + 40°)



ÐE = 60°

B Assertion & Reason Directions: In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Mark the correct choice as: (A) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A). (B) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A). (C) Assertion (A) is true but reason (R) is false. (D) Assertion (A) is false but reason (R) is true.



Reason (R): The angles opposite to equal sides of a triangle are equal. Ans. Option (A) is correct. Explanation: In case of assertion (A): AB = AC (given) Þ ÐABC = ÐACB ...(i) DC = DB (given) Þ ÐDBC = ÐDCB ...(ii) Subtracting (ii) from (i) Þ ÐABC – ÐDBC = ÐACB – ÐDCB \ ÐABD = ÐACD In case of reason (R): We know that the angles opposite to equal sides of a triangle are equal.

106

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-IX

Explanation: In case of assertion (A): In DABD, AB + BD > AD ...(i) Similarly in DADC, AC + CD > AD ...(ii) Adding (i) and (ii), we get AB + AC + BD + CD > AD + AD Þ AB + AC + BD + DC > 2AD Þ AB + AC + BC > 2AD In case of reason (R): We know that the sum of two sides is greater than the third side. Hence, Both A and R are true and R is the correct Explanation of A.

Hence, Both A and R are true and R is the correct Explanation of A. Q. 4. Assertion (A): If AD is a median of DABC, then AB + AC + BC > 2AD.



Reason (R): In a triangle the sum of two sides is greater than the third side. Ans. Option (A) is correct.

COMPETENCY BASED QUESTIONS Ans. Option (B) is correct.

A Case based MCQs

Read the following passage and answer any four questions of the following : I. A children‘s park is in the shape of isosceles triangle say PQR with PQ = PR, S and T are points on QR such that QT = RS. P

Q

S

T

R

Q. 1. Which rule is applied to prove that congruency of DPQS and DPRT. (A) SSS (B) SAS (C) AAS (D) RHS Ans. Option (B) is correct. Explanation: In DPQS and DPRT PQ = PR 

(Given)



(Given)



QS = TR  ÐPQR =ÐPRQ

(corresponding angles of an isosceles D)

By SAS congurency



(4 marks each)

DPQS @ DPRT

Explanation: An isosceles D has 2 sides equal 1 Q. 4. If PQ = 6 cm and QR = 7 cm, then perimeter of DPQR is (A) 19 cm (B) 20 cm (C) 13 cm (D) 18 cm Ans. Option (A) is correct. Explanation: Perimeter = sum of all 3 sides 1 PQ = PR = 6 cm, QR = 7 cm So P = (6 + 6 + 7) cm = 19 cm 1 Q. 5. If ÐQPR = 80°, find ÐPQR. (A) 20° (B) 100° (C) 50° (D) 40° Ans. Option (C) is correct. Explanation: let ÐQ = ÐR = x and ÐP = 80° In DPQR, ÐP + ÐQ + ÐR = 180° (Angle sum property of D) 80° + x + x = 180° 2x = 180° – 80

(A) Height



(C) Heron‘s formula (D) Highest

that BD = DC.

1

Q. 3. An isosceles triangle has  (A) 3 sides equal



(B) 2 sides equal



(C) None of these sides equal (D) All angles equal

1

with AD as the perpendicular bisector of BC such

Ans. Option (B) is correct.



100° 2

II. Neeraj has a plot in the shape of a triangle said ABC

(B) Hypotenuse

Explanation: ‘H‘ stands for hypotenuse

x =

= 50°

Q. 2. In RHS rule ‘H‘ stands for

2x = 100°







Q. 1. Which rule is applied to prove the congruency of DABD and DACD? (A) RHS (B) SAS (C) SSS (D) AAS Ans. Option (B) is correct. Explanation: In DABD and DACD BD = DC (given) ÐADB = ÐADC = 90° (AD is altitude) AD = AD (common) \ DABD @ DACD (by SAS) Q. 2. If ÐABD = 50°, then the value of ÐBAD. (A) 60° (B) 50° (C) 40° (D) 100° Ans. Option (C) is correct. Explanation: Given ÐABD = 50° and ÐADB = 90° In DADB ÐBAD + ÐABD + ÐADB = 180° (angle sum property) ÐBAD + 50° + 90° = 180° ÐBAD + 140° = 180° ÐBAD = 180° – 140° ÐBAD = 40° Q. 3. In DADC find ÐACD. (A) 55° (B) 20° (C) 60° (D) 50° Ans. Option (D) is correct. Explanation: ÐBAD = ÐCAD = 40° (AD is the perpendicular bisector) In DADC ÐACD + ÐADC + ÐCAD = 180° (angle sum property) ÐACD + 90° + 40° = 180° ÐACD + 130° = 180° ÐACD = 180° – 130° = 50° Q. 4. If AB = 10 cm and BD = 6 cm then find perimeter of DABD. (A) 50 cm (B) 24 cm (C) 30 cm (D) 20 cm Ans. Option (B) is correct. Explanation: DABD is a right angled triangle AB = 10 cm, BD = 6 cm. Using Pythagoras Theorem (AB)2 = (AD)2 + (BD)2 (10)2 = (AD)2 + (6)2 100 = (AD)2 + 36 (AD)2 = 100 – 36 = 64 AD = 8 cm Now, perimeter = AB + AD + BD = (10 + 8 + 6) cm = 24 cm Q. 5. Find perimeter of DABC (A) 25 cm (B) 50 cm (C) 32 cm (D) 24 cm

TRIANGLES

107

Ans. Option (C) is correct. Explanation: As DABD @ DACD \ AB = AC (Corresponding parts of congruent triangles) AC = 10 cm BC = BD + DC Since BD = DC = 6 cm (given) \ BC = (6 + 6) cm = 12 cm Perimeter of DABC = AB + BC + AC = (10 + 12 + 10) cm = 32 cm

B Case based Subjective Questions





Read the following passage and answer the following questions:

I. As shown above: In Rajesh Village there was a big pole PC. This pole was tied with a strong wire of 10 m length. Once there was a big spark on this pole, thus wires got damaged very badly. Any small fault was usually repaired with the help of a rope which normal board electricians were carrying on bicycles. This time electricians need a staircase of 10 m, so that it can reach at point P on the pole and this should make 60° with line AC. Q. 1. In DPAC and DPBC which side is common? Ans. As shown in figure side PC is common. 1 Q. 2. Find the value of Ðx? Ans. In DPAC ÐA = 60°, ÐPCA = 90° (given) ÐAPC + ÐPAC + ÐPCA = 180° (Angle sum property) Þ x + 60° + 90° = 180° x = 30° 1 Q. 3. In figure, DPAC and DPBC are congruent due to which criteria? Ans. In DPAC and DPBC 2 PC = PC (Common) ÐPCA = ÐPCB = 90° AP = BP (given) ( Length of Rope = Length of Stairs) \ DPAC @ DPBC (By RHS case) Hence, DPAC and DPBC are congruent due to 'RHS' criteria.

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Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-IX

50° + 95° + ÐDCA = 180° So, ÐDCA = 35° 1 Q. 2. Find ÐBAC and ÐABC? Ans. ÐDCA = ÐBAC (CPCT) ½ \ ÐBAC = 35° ÐADC = ÐABC = 95° (CPCT) \ ÐABC = 95° ½ Q. 3. Which rule is applied to prove the congruency? Is AB = CD? Ans. In DABC and DADC BC = AD (given) ÐDAC = ÐBCA (given) AC = AC (common) 1½ (a) \ DABC @ DACD (SAS Criteria) (b) AB = CD (CPCT) ½

II. Raj is having a quadrilateral open space in his plot. He divided the land into two parts by drawing the boundary AC and AD = BC. ÐDAC = ÐBCA = 50° and ÐADC = 95°

Q. 1. In DACD find ÐDCA. Ans. ÐDAC = 50° and ÐADC = 95° (given) In DACD ÐDAC + ÐADC + ÐDCA = 180° (Angle sum property)







Study Time: Max. Time: 2:30 Hrs Max. Questions: 35

CHAPTER

8



Syllabus



QUADRILATERALS ((Prove) The diagonal divides a parallelogram into two congruent triangles. (Motivate) In a parallelogram opposite sides are equal, and parallel and conversely. (Motivate) In a parallelogram opposite angles are equal, and conversely.

(Motivate) A quadrilateral is a parallelogram if a pair of its opposite sides is parallel and equal. (Motivate) In a parallelogram, the diagonals bisect each other and conversely. (Motivate) In a triangle, the line segment joining the mid-points of any two sides is parallel to the third side and is half of it and (motivate) its converse.

Topic-1 Properties of a Parallelogram

List of Topics Topic-1: Properties of a Parallelogram Page No. 109 Topic-2 : Mid-point Theorem 

Page No. 114

Revision Notes � Opposite sides of a parallelogram are parallel. � Opposite sides of a parallelogram are equal. � Opposite angles of a parallelogram are equal. � Consecutive angles (conjoined angles) of a parallelogram are supplementary. � A diagonal of a parallelogram divides it into two congruent triangles.

Scan to know more about this topic

� Diagonals of a parallelogram bisect each other. � If each pair of opposite sides of a quadrilateral is equal and parallel then it is a parallelogram. � If in a quadrilateral each pair of opposite angles is equal, then it is a parallelogram.

Properties of a Parallelogram

� If the diagonals of a quadrilateral bisect each other, then it is a parallelogram. If one pair of opposite side is equal and parallel then it is a parallelogram. Square, rectangle and rhombus are all parallelograms. Kite and trapezium are not parallelograms. A square is a rectangle. A square is a rhombus. A parallelogram is a trapezium. ¾¾ Every rectangle is a parallelogram; therefore, it has all the properties of a parallelogram. Additional properties of a rectangle are :  All the interior angles of a rectangle are right angles.  The diagonals of a rectangle are equal. � ¾¾ ¾¾ ¾¾ ¾¾ ¾¾

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Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-IX



QUADRILATERALS

111

¾¾ Every rhombus is a parallelogram; therefore, it has all the properties of a parallelogram. Additional properties of a rhombus are :  All the sides of a rhombus are equal.  Diagonals of a rhombus bisect at right angles. ¾¾ Every square is a parallelogram; therefore, it has all the properties of a parallelogram. Additional properties of a square are :  All sides are equal.  All angles are equal to 90°.  Diagonals are equal.  Diagonals bisect each other at right angle.





Diagonals bisect the angles of vertex.

SUBJECTIVE TYPE QUESTIONS Very Short Answer Type Questions (1 mark each) Q. 1. Two consecutive angles of a parallelogram are in the ratio 1 : 3, then what will be the smaller angles ?

or

½

∠1 + ∠2 = 90°

\

½

x + 3x = 180° ( consecutive angles are supplementary)

or,

4x = 180°

or,

1 1 ∠ADC + ∠BCD = 90° 2 2

U [Board Term II, 2011]

Sol. Let the consecutive angles be x and 3x.



or,

∠1 + ∠2 + ∠DOC = 180°

x = 45°

\



Smaller angle = x

= 45°

½

Q. 2. Diagonals AC and BD of a parallelogram ABCD intersect each other at O. If OA = 3 cm and OD = 2 cm, determine the lengths of AC and BD. 

[NCERT Exemp. Ex. 8.2, Q. 1, Page 75] Sol. We know that the diagonals of a parallelogram bisect each other.



Therefore,



AC = 2 × OA = 2 × 3 cm = 6 cm



BD = 2 × OD = 2 × 2 cm = 4 cm

Therefore, AC = 6 cm and BD = 4 cm. [2×1=2] Q. 3. Diagonals of a parallelogram are perpendicular to each other. Is this statement true? Give reason for your answer.  [NCERT Exemp. Ex. 8.2, Q. 2, Page 75] Sol. This statement is not true. We know that the diagonals of a parallelogram bisect each other but need not be perpendicular. [2×1=2]

Short Answer Type Questions-I (2 marks each) Q. 1. In a parallelogram, show that the angle bisectors of two adjacent angles intersect at right angle. 

U [Board Term II, 2011]

Sol.

∠ADC + ∠BCD = 180°

In ∆ODC,

½

( consecutive angles are supplementary)

∴

∠DOC = 90° Hence Proved [1]

Q. 2. The angle between the two altitudes of a parallelogram through the vertex of an obtuse angle is 50°. Find the angles of a parallelogram. U [Board Term II, 2011] Sol. AM ⊥ DC, AN ⊥ BC In quadrilateral AMCN, ∠MAN + ∠M + ∠C + ∠N = 360° or ∠ MAN + 90° + ∠C + 90° = 360° ∴ ∠MAN + ∠C = 180° or, 50° + ∠C = 180° or, ∠C = 130° 1

In parallelogram, ∠A = ∠C = 130° ∠B = ∠D = 180° – 130° = 50°

1

Q. 3. Two opposite angles of a parallelogram are (3x – 2)° and (63 – 2x)°. Find all the angles of a parallelogram. 

U [Board Term II, 2011]



Sol. Since, opposite angles of a parallelogram are equal. ∴ 3x – 2 = 63 – 2x or 5x = 65 or, x = 13° 1 Angles of the parallelogram are : (39 – 2)°, (180 – 37)°, (63 – 26)° and (180 – 37)°



i.e., 37°, 143°, 37° and 143°.

1

112

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-IX

Q. 4. In a parallelogram ABCD if ∠A = 115°, find ∠B, ∠C and ∠D. U [KVS, 2019] C D Sol.

115° A B In a||gm, Consecutive angles are supplementary \ ∠A + ∠B = 180° 115° + ∠B = 180° ∠B = 180° – 115° \ ∠B = 65° ∠A = ∠C = 115° 1  (Opposite angles are equal in ||gm) ∠B = ∠D = 65° 1

Short Answer Type Questions-II (3 marks each) Q. 1. Diagonal AC of a parallelogram ABCD bisects ∠A. Show that : (i) it bisects ∠C also (ii) ABCD is a rhombus U [Board Term II , 2012, NCERT]

Answering Tip

Students should learn the properties of parallelogram thoroughly.

Q. 2. PQRS is a parallelogram and PL and RM are perpendiculars drawn from the vertices P and R of the parallelogram on diagonal SQ. Show that (i) ∆PQL @ ∆RMS (ii) PL = RM U [KVS 2014, NCERT] Sol. (i) In ∆RSM and ∆PQL,

½ ∠RSM = ∠PQL  (Alternate interior angles) ∠M = ∠L  (90° each) SR = PQ 1  (Parallel sides of parallelogram are equal) By AAS, DRSM ≅ ∆PQL 1

(ii)

PL = RM 

(CPCT) ½ Hence Proved

Q. 3. In the given parallelogram ABCD, two points P and Q are taken on the diagonal BD such that DP = BQ. Show that :

Sol.

(i) AB || CD, AD || BC AC is transversal So, Ð1 = Ð3 ...(i) Ð2 = Ð4 (Alt. interior angles) But Ð1 = Ð2 ...(ii) (Given, diagonal AC bisects ÐA) Ð3 = Ð4 (On comparing eqn. (i) and eqn. (ii)) \ AC bisects ÐC, Proved 1½ (ii) As, Ð1 = Ð2 = Ð3 = Ð4 Hence, Ð1 = Ð4 \ AB = BC (sides opp. to equal angles) 1 Hence, ABCD is a rhombus (in a parallelogram if one pair of adjacent sides are equal then it is a rhombus). ½ [CBSE Marking Scheme, 2012]

Commonly Made Error

Diagonal of a parallelogram bisects its vertex

angle. Students make a mistake that the two bisected angles are equal but it is not so, actually Ð1 ¹ Ð2, in case of rhombus but in case of rectangle they are equal.

(i) ∆APD ≅ ∆CQB (ii) ∆AQB ≅ ∆CPD

(iii) APCQ is a parallelogram

A

[Board Term II, 2012, NCERT]

Sol. (i) In ∆APD and ∆CQB

AD = BC



PD = QB

(Opp. sides of a parallelogram)

∠ADP = ∠CBQ



(Given)



(Alt. Angles)



So, ∆APD ≅ ∆CQB



or,



(ii) In ∆AQB and ∆CPD

AP = CQ (it is given)

(SAS Rule) (CPCT) 1

AB = DC (opp. sides of parallelogram) BQ = DP (given) and ∠ABQ = ∠PDC (Alt. angles) ∴ ∆AQB ≅ ∆CPD (SAS Rule) or, AQ = CP (CPCT) 1

(iii) In quad. APCQ,



AP = CQ and AQ = CP

\ APCQ is a ||

gm

(Proved above) Hence Proved. 1



QUADRILATERALS

∴ ABED is a parallelogram. Proved. 1 So, AD = BE and AD || BE. ...(i) (ii) In quadrilateral BCFE, BC = EF and BC || EF i.e., one pair of opposite sides are equal and parallel. ∴ BCFE is a parallelogram. Proved. 1 So, CF = BE and CF || BE. ....(ii) (iii) From equations (i) and (ii), we get AD = CF and AD || CF \ ACFD is a parallelogram. So, AC = DF and AC || DF Proved. 1 (iv) In ∆ABC and ∆DEF, AB = DE (Given) BC = EF (Given) and AC = DF (Proved above in part (iii))

Long Answer Type Questions (5 marks each) Q. 1. In the given figure, ABC is an isosceles triangle in which AB = AC, AD bisects the exterior angle PAC and CD || AB. Show that :

(i) ∠DAC = ∠BCA, and (ii) ABCD is a parallelogram. U [Board Term II , 2012] [NCERT] Sol. (i) DABC is an isosceles triangle. So, ∠ABC = ∠BCA ½ ∠PAC = ∠ABC + ∠BCA ½ (Q Sum of two interior opposite angles is equal to exterior angle) = 2∠BCA ...(i) 1 AD bisects ∠PAC So, ∠PAC = 2∠DAC ...(ii) From (i) and (ii), ∠BCA = ∠DAC 1



(ii) ∠BCA = ∠DAC (Proved above) These are alternate angles when lines BC and AD are intersected by AC 1 \ BC || AD, Also, BA || CD (Given) ∴ ABCD is a parallelogram. Hence Proved 1 Q. 2. If DABC and DDEF are two triangles such that AB, BC are respectively equal and parallel to DE, EF then show that : (i) quadrilateral ABED is a parallelogram. (ii) quadrilateral BCFE is a parallelogram. (iii) AC = DF (iv) DABC ≅ DDEF. A [Board Term II, 2012] [NCERT] Sol. Given : Two triangles ABC and DEF, such that

1 AB = DE and AB || DE Also, BC = EF and BC || EF Proof : (i) In a quadrilateral ABED, AB = DE and AB || DE i.e., one pair of opposite sides are equal and parallel.

113



So by SSS rule DABC ≅ DDEF.

Hence Proved 1

Q. 3. ABCD is a trapezium in which AB || CD and AD = BC, Show that : (i) ÐA = ÐB

(ii) ÐC = ÐD

(iii) DABC @ DBAD A [Board Term II 2012] [NCERT]

Sol.

(i) Through C draw CE || DA

\ AECD is a parallelogram

So,

ÐA + Ð2 = 180°

...(i)

(Consecutive interior angles) AD = CE ( AECD is a ||gm) AD = BC (Given) BC = CE

and \ Now in DBCE, BC = CE (Proved above) 1 or, Ð1 = Ð2 Also, ÐABC + Ð1 = 180° (Linear Pair) ...(ii) From (i) and (ii), ÐA = ÐB Hence Proved (ii) Now, ÐA + ÐD = 180° ...(iii) ÐB + ÐC = 180° ...(iv) (Consecutive interior angles) (on comparing eqn. (iii) and eqn. (iv))

ÐA + ÐD = ÐB + ÐC 1 So, ÐC = ÐD (Q ÐA = ÐB) 1 (iii) In D's ABC and BAD AB = BA (Common) ÐB = ÐA (Proved above) 1 BC = AD (Given) So, DABC ≅ DBAD (By SAS) 1 Hence Proved [CBSE Marking Scheme 2012]

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Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-IX

Topic-2 Mid-Point Theorem Revision Notes ¾¾ Mid-Point Theorem : The line segment joining the mid-points of any two sides of a triangle is parallel to the third side and equal to half of it. ¾¾ Converse of mid-point theorem : The line drawn through the mid-point of one side of a triangle parallel to the another side, bisects the third side. ¾¾ If there are three or more parallel lines and the intercepts made by them on a transversal are equal, then the corresponding intercepts on any other transversal are also equal.

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Mid point

SUBJECTIVE TYPE QUESTIONS Very Short Answer Type Questions (1 mark each) Q. 1. In an equilateral triangle ABC, D and E are the mid-points of sides AB and AC respectively, then find the length of DE.

R [Board Term II, 2011, NCERT]

Sol.

Short Answer Type Questions-II (3 marks each) Q. 1. ABCD is a quadrilateral in which P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. Show that PQRS is a parallelogram. A [NCERT][Board Term II , 2012] Sol. Join diagonal AC. In ∆ADC, S is the mid-point of AD. R is the mid-point of DC.



Since, D and E are mid-points of sides AB and AC 1 respectively. So, by mid-point theorem, DE = BC. 2 1 Q. 2. D, E, F are the mid-points of sides BC, CA and AB of DABC. If perimeter of DABC is 12·8 cm, then find perimeter of DDEF. R [Board Term II, 2011] A Sol.

E

F

B



D

½

C





∴

SR || AC 1 and SR = AC (By mid-point theorem) 2 ...(i) 1 Similarly in ∆BAC, PQ || AC 1 and PQ = AC ...(ii) 1 2 From (i) and (ii), SR || PQ and SR = PQ ½ ∴ PQRS is a parallelogram. Hence Proved ½ Q. 2. D, E and F are the mid-points of sides PQ, QR and RP respectively of an equilateral ∆PQR. Show that ∆DEF is also an equilateral triangle.

Given, perimeter of DABC = 12·8 cm

U [Board Term II, 2012]

1 \ Perimeter of DDEF = (perimeter of DABC) 2

Sol.

12·8 cm = 2 =6·4 cm

½



QUADRILATERALS



DE || PR and DE =

115

1 PR 2

(By mid-point theorem) 1 Similarly, EF || PQ and EF = PQ 2

DF || QR and DF =

1 QR 2

1½



As



So,



or, ∆DEF is an equilateral triangle. Hence Proved ½

Perimeter of DEF = DE + EF + DF 1 = (AC +AB + BC) 2

PQ = QR = PR ( DPQR is an equilateral triangle) DE = EF = DF 1

Q. 3. Prove that the quadrilateral formed by joining the mid-points of the consecutive sides of a rectangle is a rhombus.  A [Board Term II , 2012] OR

=

1

1 1 (7·8 + 6 + 7·2) = (21) 2 2 1

= 10·5 cm

Q. 5. In the figure, ABCD is a parallelogram and E is the mid-point of side BC. DE and AB on producing meet at F. Prove that AF = 2AB.

ABCD is a rectangle and P, Q, R and S are mid points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus. [NCERT]

U [Board Term II , 2012] Sol. In ∆DCE and ∆FBE, ½ ∠DCE = ∠FBE (alternate interior angles) CE = BE (Given) ∠DEC = ∠BEF (VOA) \ ∆DCE ≅ ∆FBE (ASA) or, DC = FB (CPCT) 1 DC = AB (opp. sides of ||gm) \ AB = FB ½ or, AF = AB + BF = AB + AB = 2AB Hence Proved. 1

Sol.

In DABC, P and Q are mid points of AB and BC respectively. 1 \ PQ = AC and PQ || AC ... (i) 2



RS =

1 AC and RS || AC ...(ii) 2



Similarly,



\ PQ || RS, PQ = RS =

1 AC 2

\ PQRS is a parallelogram 1 Also, AD= BC or, AS = BQ In DAPS and DBPQ, AP = BP, AS = BQ and ÐA = ÐB = 90° 1 \ DAPS @ DBPQ (SAS Congnency) Hence PS = PQ ...(iii) (CPCT) \ PQRS is a parallelogram with PQ = PS i.e., PQRS is a rhombus. Hence Proved. 1 [CBSE Marking Scheme 2012] Q. 4. In ∆ABC, D, E and F are the mid-points of sides AB, BC and CA. If AB = 6 cm, BC = 7·2 cm and AC = 7·8 cm, find the perimeter of ∆DEF. R [Board Term II, 2012]

Sol.



1 DE = AC 2

Long Answer Type Questions (5 marks each) Q. 1. In the figure, ABCD is a parallelogram. E and F are the mid-points of sides AB and CD respectively. Show that the line segments AF and EC trisect the diagonal BD. U [Board Term II , 2013; 2012]

Sol. According to the question, E and F are the midpoints of sides AB and CD. 1 ∴ AE = AB 2



EF =

1 AB 2



DF =

1 BC (By Mid-point theorem) 2

1

and

CF =

1 CD 2

In the parallelogram opposite sides are equal, so AB = CD 1 1 AB = CD 2 2

1½

116

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-IX

∴ Again,

AE = CF AB || CD 1 1 AB || CD 2 2

So, AE || FC 1 Hence, AECF is a parallelogram. In ∆ABP, E is the mid-point of AB and EQ || AP. ∴ Q is the mid-point of BP. 1 (By converse of mid-point theorem) Similarly, P is the mid-point of DQ. \ DP = PQ = QB ∴ Line segments AF and EC trisect the diagonal BD. Hence Proved 1½ Q. 2. In given fig., AD is the median of ∆ABC. E is the mid-point of AD. DG || BF. Prove that AC = 3AF.

AC = AF + FG + GC = 3AF Hence Proved 2 Q. 3. Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other. A [Board Term II, 2012] [NCERT] OR P, Q, R, S are the mid-points of the sides AB, BC, CD and DA respectively of the quadrilateral ABCD. Prove that PR and QS bisect each other.  [Board Term II, 2012] Sol. Const. : Join SP, PQ, QR, RS and AC



U [Board Term II, 2012] Sol. In DADG, E is the mid-point of AD and EF || DG. \ F is the mid-point of AG (converse of mid-point theorem). 1 So, AF = FG ...(i) In DCBF, BF || DG, D is the mid-point of BC. So, G is the mid-point of FC. 1 \ FG = GC ...(ii) 1 From (i) and (ii), AF = FG = GC

Proof : In ∆DAC,



RS || AC



RS =

and

1 AC 2

(Mid-point theorem) ...(i) 1

In ∆ABC,



PQ || AC

1 and PQ = AC 2 ...(ii)

(Mid-point theorem) 1

From (i) and (ii), we get



RS || PQ and RS = PQ

1½



or, PQRS is a parallelogram.



Since, diagonals of a parallelogram bisect each other.



∴ PR and QS bisect each other.

OBJECTIVE TYPE QUESTIONS

1½

(1 mark each)

A Multiple Choice Questions Q. 1. A diagonal of a rectangle is inclined to one side of the rectangle at 25°. The acute angle between the diagonals is (A) 55° (B) 50° (C) 40° (D) 25°  A [NCERT Exemp.]

Ans. Option (B) is correct. Explanation: ABCD is a rectangle in which diagonal AC is inclined to one side, AB of the rectangle at an angle of 25°.

Now, AC = BD \ or 

[Diagonals of a rectangle are equal] 1 1 AC = BD 2 2   OA = OB

In triangle AOB, we have     OA = OB



QUADRILATERALS

∠BAO = 25° \  ∠OBA = (Angles opposite to equal sides)

By angle sum property, we have So, ∠OBA + ∠AOB + ∠BAO = 180°             

25° + 25° + ∠AOB = 180°

117

Ans. Option (C) is correct. Explanation: AD is parallel to BC and AC cuts them.

∠DAC = ∠ACB [Alt. int. angles] ∠DAC = 32° ⇒ ∠ACB = ∠DAC = 32°

∠AOB = 180° − 50= ° 130°

We know that, ÐAOB and ÐAOD form linear pair. ∠AOB + ∠AOD = 180° So, 130° + ∠AOD = 180°

       ∠AOD= 180° − 130°= 50° Therefore, the acute angle between the diagonal is 50°. Q. 2. If bisectors of ÐA and ÐB of a quadrilateral ABCD intersect each other at P, of ÐB and ÐC at Q, of ÐC and ÐD at R and of ÐD and ÐA at S, then PQRS is a (A) rectangle (B) rhombus (C) parallelogram (D) quadrilateral whose opposite angles are supplementary A [NCERT Exemp.] Ans. Option (D) is correct. Explanation:

According to figure, In DASD, DS bisects ÐD and AS bisects ÐA, 1 1 \ ÐDAS + ÐADS = ÐA + ÐD 2 2

=

1 (ÐA + ÐD) 2



=

1 × 180° 2

( Consecutive angles of a||gm are supplementary) So, ÐDAS + ÐADS = 90° Also ÐDAS ++ ÐADS + ÐDSA = 180° (Angle sum property) 90° + ÐDSA = 180° So, ÐDSA = 90° ÐPSR = ÐDSA (Vertically opposite angles) \ ÐPSR = 90° Similarly, ÐSPQ, ÐPQR and ÐQRS = 90° Hence, PQRS is a quadrilateral whose opposite angles are supplementary. Q. 3. The diagonals AC and BD of a parallelogram ABCD intersect each other at the point O. If ÐDAC = 32° and ÐAOB = 70°, then ÐDBC is equal to (A) 24° (B) 86° (C) 38° (D) 32°  A [NCERT Exemp.]



In triangle BOC, CO is produced to A Therefore, Ext. ÐBOA = ÐOCB + ÐOBC [By exterior angle theorem] ° 32° + ∠OBC       70= ∠OBC = 70° − 32° = 38° ∠OBC = 38° ⇒ ∠DBC = Q. 4. In the following figure, ABCD and AEFG are two parallelograms. If ÐC = 60°, then ÐGFE is

(A) 30° (B) 60° (C) 90° (D) 120° Ans. Option (B) is correct. Explanation: In parallelogram ABCD ÐA = ÐC = 60° ( Opposite angles of a parallelogram are equal) Similarly, In parallelogram AEFG ÐA = ÐF (opposite angles of a||gm) \ ÐF = 60° Q. 5. The quadrilateral formed by joining the midpoints of the sides of a quadrilateral PQRS, taken in order, is a rectangle, if : (A) PQRS is a rectangle (B) PQRS is a parallelogram (C) diagonals of PQRS are perpendicular (D) diagonals of PQRS are equal A [NCERT Exemp.] Ans. Option (C) is correct. Explanation: The perpendicular.

diagonals

of

PQRS

are

Q. 6. In fig. S is the mid-point of PQ and ST || QR then PT is equal to

118

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-IX

(A) SQ (C) TR

(B) PS (D) QR

Ans. Option (C) is correct. Explanation: The line drawn through the midpoint of one side of a triangle parallel to another side, bisects the third side. Therefore PT = TR Q. 7. In fig. D and E are the mid-points of AB and AC respectively. The length of DE is.

B Assertion & Reason Directions: In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Mark the correct choice as: (A) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A). (B) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A). (C) Assertion (A) is true but reason (R) is false. (D) Assertion (A) is false but reason (R) is true.



(A) 8.2 cm (C) 4.9 cm

(B) 4.1 cm (D) 5.1 cm

Ans. Option (B) is correct. Explanation:

DE =

1 BC 2

(According to mid-point theorem)

DE =

1 ´ 8.2 2

= 4.1 cm Q. 8. In fig. PQRS is a rectangle X and Y are mid-points of PS and PQ respectively. The length of XY is



(A) 4 cm (C) 2.5 cm

(B) 5 cm (D) 2 cm

Ans. Option (C) is correct. Explanation: In rectangle PQRS

ÐR = 90°

\ DSRQ is a right angled triangle Now, According to pythagoras theorem

QS2 = SR2 + RQ2



= 42 + 32



= 16 + 9



= 25



QS = 5 cm

X and Y are mid-points of RS and PQ \

XY =

Q. 1. Assertion (A): A parallelogram consists of two congruent triangles. Reason (R): Diagonal of a parallelogram divides it into two congruent triangles. Ans. Option (A) is correct. Explanation: According to properties of a parallelogram a diagonal of a parallelogram divides it into two congruent triangles therefore parallelogram has two congruent triangles. Hence, A and R are true and R is correct explanation of A. Q. 2. Assertion (A): Two opposite angles of a parallelograms are (3x – 8)° and (60 – x)°. The measure of one of the angle is 43°. Reason (R): Opposite angles of a parallelogram are supplementary. Ans. Option (C) is correct. Explanation: In case of assertion (A): Opposite angles of parallelogram are equal \ 3x – 8 = 60 – x 3x + x = 60 + 8 4x = 68 x = 17° By substituting 'x' we get (3x – 8)° = 43° (60 – x)° = 43° Hence, Assertion is true. In case of reason (R): According to properties of parallelogram, opposite angles of a parallelogram are equal. \ Reason is false. Q. 3. Assertion (A): ABCD and PQRC are rectangles and Q is a midpoint of AC, then DP = PC.

1 QS 2 (By mid-point theorem)



XY =

1 XS 2

= 2.5 cm



Reason (R): The line segment joining the midpoint of any two sides of a triangle is parallel to third side and is equal to half of it.



QUADRILATERALS

Ans. Option (B) is correct. Explanation: In case of assertion (A): Q is a midpoint of AC So, P is also a midpoint of DC (According to converse of midpoint theorem) \ Assertion is true. In case of reason (R): It is a midpoint theorem \ Reason is true but not correct explanation of Assertion. Q. 4. Assertion (A): In DABC, median AD is produced to X such that AD = DX. Then ABXC is a parallelogram. Reason (R): Diagonals AX and BC bisect each other at right angles.

Ans. Option (C) is correct. Explanation: In case of assertion (A): ABXC is a ||gm  In a parallelogram Diagonals bisects each other. \ Assertion is true. In case of reason (R): Diagonals of a ||gm do not bisect each other at right angles. \ Reason is false. Hence, Assertion is true but Reason is false.

COMPETENCY BASED QUESTIONS A Case based MCQs I. Read the following passage and answer any four questions of the following : Maths teacher of class 9th gave students coloured paper in the shape of quadrilateral and then ask the students to make parallelogram from it by using paper folding.

119

(4 marks each)

Q. 3. If ∠RSP = 50°, then ∠SPQ =  (A) 130° (B) 120° (C) 110° (D) 100° Ans. Option (A) is correct. 1 Explanation: Adjacent angles of a parallelogram are supplementary. Thus, ∠RSP + ∠SPQ = 180° 50° + ∠SPQ = 180° ∠SPQ = 180° – 50° = 130° Q. 4. If SP = 3 cm, then RQ =  (A) 4 cm

(B) 2 cm

(C) 3 cm

(D) 5 cm

Ans. Option (C) is correct. Q. 5. Which statement parallelogram ?

1 is

incorrect

about

the

(A) Consecutive angles are supplementary (B) Opposites sides are parallel Q. 1. How can a parallelogram be formed by using paper folding ? (A) By joining any two vertices

(D) None of the above 1

Q. 2. If ∠RSP = 30°, then ∠RQP =  (A) 150°

(B) 80°

(C) 50°

(D) 30°

Ans. Option (D) is correct.

(D) Diagonals are equal in length. Ans. Option (D) is correct.

(B) By joining one pair of opposite vertices. (C) By joining mid points of sides of a quadrilateral Ans. Option (C) is correct.

(C) Diagonal bisects each other

Q. 1. How many types of quadrilaterals can be possible? (A) 3

1

1

II. Read the following passage and answer any four questions of the following : During maths lab activity, teacher gives four sticks of lengths 6 cm, 6 cm, 4 cm and 4 cm to each student to make different types of quadrilateral.

(B) 4

(C) 5 (D) 6 Ans. Option (A) is correct.

120

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-IX

Q. 2. Write the name of quadrilateral that can be formed with these sticks.

B Case based Subjective Questions

(A) Kite, rectangle, rhombus (B) Parallelogram, rectangle , trapezium



Read the following passage and answer the following questions:



I. ABCD is an area in the shape of rhombus in which ∠ADC = 120°. Samay and Tarun lived at D and C and their school located at O.

(C) Kite, rectangle, parallelogram (D) Square, rectangle, kite Ans. Option (C) is correct. Explanation: Opposite sides of a parallelogram and a rectangle are equal. Also, adjacent sides of kite are equal. Thus, kite, rectangle and parallelogram can be formed with the sticks of lengths 6 cm, 6 cm, 4 cm and 4 cm. Hence, option (C) is correct. 1 Q. 3. Which statement is incorrect ? (A) Opposite sides of a parallelogram are equal (B) A kite is not a parallelogram (C) Diagonals of a parallelogram bisect each other (D) A trapezium is a parallelogram.



Q. 1. Measure of ∠DCB is :

Ans. Option (D) is correct.

Answer the following questions: Sol. ABCD is a rhombus and adjacent angles of a rhombus are supplementary.

Explanation: A trapezium has only one pair of parallel sides.

Thus,

Thus, it cannot be a parallelogram.

1

⇒

Q. 4. A student formed a rectangle with these sticks. What is the length of the diagonal of the rectangle formed by the student ?

⇒ 

(A) 6 13 cm

(B)



3 13 cm

∠CDA + ∠DCB = 180°

½

120° + ∠DCB = 180° ∠DCB = 180° – 120° = 60° ½

Q. 2. Who can reach school early? Sol. ∠CDO = 60° and ∠DCO = 30°

13 cm (C) (D) 2 13 cm Ans. Option (D) is correct. Explanation: All the angles of a rectangle are right angle. So, diagonal of a rectangle divides it into two right angled triangles. Thus,

∠CDO > ∠DCO

Since, ⇒ 

CO > DO. (side opposite to greater angle is greater.)

⇒ Samay will reach school early.

1

Q. 3. Measure of ∠CDO and ∠DCO are ______ and ______ respectively. Sol. Diagonal of a rhombus bisects the angles passing from. So,





6 + 4 = l 36 + 16 = l2 l2 = 52 2

2

2

l = 2 13 cm



1

Q. 5. A diagonal of a parallelogram divides it into two _______ triangles. (A) Similar

(B) Congruent

(C) Equilateral

(D) Right angled

Ans. Option (B) is correct.1

∠CDO = =

1 ∠CDA 2 1 (120)° = 60° 1 2

∠ADC + ∠DCB = 180° 120° + ∠DCB = 180°

⇒

∠DCB = 180° – 120° = 60°



∠DCO =



=

1 ∠DCB 2 1 (60°) = 30° 1 2





QUADRILATERALS

II. Harish makes a poster in the shape of a parallelogram on the topic SAVE ELECTRICITY for an inter school competition as shown in the follow figure.

Q. 1. If AB = (2y – 3) and CD = 5 cm then what is the value of y ? Sol. AB = CD (opposite sides of a||gm are equal)

121

⇒ 2y – 3 = 5 ½ ⇒ 2y = 8 ⇒ y = 4 ½ Q. 2. Which mathematical concept is used here? Sol. Properties of a parallelogram. 1 Q. 3. If ∠A = (4x + 3)° and ∠D = (5x – 3)°, then find the measure of ∠B Sol. Since, ABCD is a parallelogram. ∠A + ∠D = 180° (consecutive of a||gm are supplementary) \ (4x + 3)° + (5x – 3)° = 180°1 9x = 180° x = 20° ∠D = (5x – 3)° = 97° ∠D = ∠B (opposite angles of a parallelogram are equal) Thus, ∠B = 97°1

Artificial Intelligence PARAMETERS

DESCRIPTION

Chapter Covered

Chapter 8: Quadrilaterals

Name of the book

Mathematics, Text book for Class 9

AI CONCEPTS INTEGRATED

Subject and Artificial To understand the concept of Quadrilaterals using AI tools. Intelligence Integrated Learning Objectives

● ●

To understand the concept of Quadrilaterals. - Quadrilaterals - Angle sum property of a quadrilateral - Types of Quadrilaterals Visualization of Quadrilaterals using AI tool Autodraw. Autodraw

Time Required

2 periods of 40 minutes each.

Classroom Arrangement

Flexible

Material Required

Scrapbook, paper, pencil, scale, scissor, eraser, White Board. Laptops/ Desktops and Internet connection.

Pre-Preparation Activities

Play the game Ask the students to collect the sticks: Thirteen sticks of 6cm, one stick of 8 cm, 7 cm and 12cm each in length. Using the sticks, construct: ● Quadrilateral with two pairs of parallel sides. ● Quadrilateral with four right angles. ● Quadrilateral with four congruent sides. ● Quadrilateral with exactly one pair of parallel sides.

Previous Knowledge

● The students are asked to recall the knowledge of parallel lines, perpendicular lines, triangles, rectangle, and square. ● List the objects of quadrilateral shapes they faced in daily - life. ● Some questions will be asked related to quadrilaterals.

122

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-IX

Methodology

● Ask students to go to www.kahoot.it Teacher will create the quiz on www.create.kahoot.it Ask the students to play the quiz, this will access previous knowledge about the topic. The Teacher would get results instantly and they can assess the students. The students will be divided into four groups. The teacher will provide each group with materials needed, each group will do the activity Activity 1. ● Ask students to take a scrap book and cut into two congruent triangles and join them in such a way that they form a 4 sided closed figure. ● As they got the idea of a four sided closed figure introduced the term quadrilateral. ● Make perfect quadrilaterals with the help of AI tool Autodraw and ask students to do the same thing. Activity, ask the students to go to http://autodraw.com. Once they land on this website, ask the students to select the first icon from the left side toolbar. This icon activates the AI element of the tool. Now ask the student to draw any quadrilateral shape ● Ask students to go to https://ncase.me/loopy/ once they land on this activity, introduce the concept of similar factors and dissimilar factors of various forms of quadrilateral with the parallelogram .And then explain all properties of parallelogram and interrelation of parallelogram with many forms of quadrilateral. ● Ask students to solve the exercises

Learning Outcomes

Upon completion of the lesson , they are able ● To identify the all types of Quadrilateral ● To understand angle sum property of Quadrilateral ● Properties of parallelogram ● Properties of other quadrilaterals ● Visualization of interconnection of various quadrilaterals with parallelogram using AI tool loopy. ● Visualization of Quadrilaterals using AI tools- Autodraw

Follow up Activities

Make a table with the headings - know, want to know and learn about the quadrilaterals using fontjoy- an AI element. ● Students will be divided into four groups. Students will construct their own desired structures such as houses, schools and churches applying the idea of quadrilateral around The World. The students output will be assessed using constructed rubrics. ● Students would create their own quiz on www.create.kahoot.it and play with their friends.







Autodraw.com

Study Time: Max. Time: 3:30 Hrs Max. Questions: 43

CHAPTER

9



Syllabus

CIRCLES (Prove) Equal chords of a circle subtend equal angles at the centre and (motivate) its converse. (Motivate) The perpendicular from the centre of a circle to a chord bisects the chord and conversely, the line drawn through the centre of a circle to bisect a chord is perpendicular to the chord.

(Motivate) Equal chords of a circle (or of congruent circles) are equidistant from the centre (or their respective centres) and conversely. (Prove) The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle. (Motivate) Angles in the same segment of a circle are equal. (Motivate) If a line segment joining two points subtends equal angles at two other points lying on the same side of the line containing the segment, the four points lie on a circle. (Motivate) The sum of either of the pair of the opposite angles of a cyclic quadrilateral is 180° and its converse.

Topic-1 Basic Properties of Circles

List of Topics Topic-1: Basic Properties of Circles Page No. 123 Topic-2 : Cyclic Quadrilaterals 

Page No. 129

Revision Notes

 In a circle, equal chords subtend equal angles at the centre.  The chords corresponding to congruent arcs are equal.  If two arcs of a circle (or of congruent circles) are congruent, then the corresponding chords are equal.  If two chords of a circle (or of congruent circles) are equal, then their corresponding arcs (minor, major or semicircular) are congruent. Scan to know  An infinite number of circles can be drawn through a given point P. more about  An infinite number of circles can be drawn through the two given points. this topic  Perpendicular bisectors of two chords of a circle intersect each other at the centre of the circle.  The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle. Circle Theorems  Angles in the same segment of a circle are equal.  An angle in a semi-circle is a right angle.  The arc of a circle subtending a right angle at any point of the circle in its alternate segment is a semi-circle.

124

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-IX



CIRCLES

125

Example 1 Q. Two circles of radii 10 cm and 8 cm intersect and the length of the common chord is 12 cm. Find distance between their centres. Solution:

Step IV : Apply Pythagoras theorem in DOLP.

Step I : Draw figure as per given information. Let O and O' be the centres of the circles of radii 10 cm and 8 cm, respectively. Let PQ be their common chord.

or,

OL =

OP 2 - PL2



=

(10)2 − (6)2



Step II : Write the given information. Given, OP = 10 cm, O'P = 8 cm and PQ = 12 cm. Find OO'. Step III : For finding OO', first find PL. PL =



1 PQ = 6 cm 2

( Perpendicular from the centre of a circle to a chord bisects the chord)

In right angle DOLP, we have OP2 = OL2 + PL2

= 64 = 8 cm Step V : Apply Pythagoras theorem in DO'LP. In right angle DO'LP, we have O'P2 = PL2 + O'L2

O'L =



=

(OP )2 - ( PL)2 82 − 62

64 − 36 = 28

=



= 5.29 cm. \ OO' = OL + O'L = 8 + 5.29 = 13.29 cm



SUBJECTIVE TYPE QUESTIONS Very Short Answer Type Questions (1 mark each) Q. 1. In the given figure, O is the centre of the circle and PA = PB. Find ∠OPA.

R [Board Term II, 2013]

Sol. Line drawn through the centre of circle to bisect a chord is perpendicular to the chord. Given,

PA = PB ; OP ^ AB

Hence, ∠OPA = 90° 1 Q. 2. In the given figure, A, B, C and D are the points on a circle such that ∠ACB = 40° and ∠DAB = 60°, the measure of ∠DBA is ....................... . U [Board Term II, 2012]

Sol. ∠ACB = ∠ADB [Angles in the same segment] ½ ∴ ∠ADB = 40° Now, in ∆ADB, ∠ADB + ∠DBA + ∠BAD = 180° or, 40° + ∠DBA + 60° = 180° or, ∠DBA = 80° ½ Q. 3. In the given figure, AD || BC and ∠BCA = 40°. The measure of ∠DBC is equal to ................... . R [Board Term II, 2012]

126

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-IX

Sol. ∠BDA = ∠BCA = 40° (Angles in the same segment) ½ Now, since AD || BC, ∠DBC = ∠BDA (Alternate interior angles) ∴ ∠DBC = 40° ½ Q. 4. In the given figure, A, B, C and D are four points of a circle. AC and BD intersect at a point E such that ∠BEC = 130° and ∠ECD = 20°. Find ∠BAC.  [Board Term II, 2012, KVS 2017, 2019, NCERT] D A E 130°



U [KVS 2014]

20° C

B

Sol. 

Sol. In ∆ADB, By angle sum property ∠ABD + ∠ADB + ∠BAD = 180° ½ ∴ 50° + ∠ADB + 60° = 180° ∴ ∠ADB = 180° – (50° + 60°) = 70° 1 ∴ ∠ACB = ∠ADB = 70° (Q angles in the same segment of a circle are equal) ½ Q. 2. In the given figure, O is the centre of the circle and chord AC and BD intersect at P such that ∠APB = 120° and ∠PBC = 15°, find the value of ∠ADB.

∠BEC = ∠EDC + ∠DCE

(Exterior angle property) 130° = ∠EDC + 20° ∠EDC = 110° = ∠BDC ∠BAC = ∠BDC = 110° 1  (Angles in the same segment) Q. 5. In the given figure, ∠ACP = 40° and ∠BPD = 120°, then ∠CBD = ....................... . U [Board Term II, 2012]

ÐADB = ÐACB = 40° [Q Angles in the same segment are equal] ½ Now, in DDPB ÐDPB + ÐDBP + ÐPDB = 180° (Angle sum property) or, 120° + ÐDBP + 40° = 180° or, ÐDBP = 180 – (120° + 40°) or, ÐDBP = 20° \ ÐCBD = ÐPBD = 20° ½

Sol. In ∆PCB, ∠PCB + ∠PBC = ∠APB (exterior angle of a D is equal to the sum of two Interior opposite angles) 1 ∠PCB + 15° = 120° ∴ ∠PCB = 105° or, ∠ACB = 105° \ ∠ADB = ∠ACB = 105° [Angle in same segment] 1 Q. 3. If a line intersects two concentric circles with common centre O, at A, B, C and D. Prove that AB = CD. A [NCERT]  [Board Term II, 2016; KVS 2014; 2012]

Sol.

Short Answer Type Questions-I (2 marks each) Q. 1. In the figure, if ∠DAB = 60°, ∠ABD = 50°, then find ∠ACB.

U [KVS 2016]

Sol. Draw OP perpendicular to XY from the centre to a chord bisecting it. OP ⊥ to chord BC. or, BP = PC ...(i) 1 Similarly, AP = PD ...(ii) Subtracting eqn. (i) from eqn. (ii), we get ½ AP – BP = PD – PC or AB = CD ½  Hence Proved Q. 4. Prove that ‘‘equal chords of a circle subtend equal angles at the centres.’’ A [NCERT, KVS 2016, 2012] Sol. Given, AB and CD are the chords of a circle with centre at O such that AB = CD



CIRCLES

To Prove : ∠AOB = ∠COD Proof : In ∆AOB and ∆COD, AO = CO (Radii of same circle) AB = CD (Given) BO = DO (Radii of same circle) ∆AOB @ ∆COD (SSS) ∴ ∠AOB = ∠COD (CPCT) 2 Hence Proved. Q. 5. A chord of length 10 cm is at a distance of 12 cm from the centre of a circle. Find the radius of the circle. U [Board Term II, 2016] Sol. Given, AB = 10 cm ON = 12 cm

Also, ON ⊥ AB and AN = BN ( Perpendicular drawn from the centre of the circle to chord of circle bisects the chord) 1 In ∆ONB, OB2 = ON2 + NB2 (By pythagoras theorem) ∴ OB2 = 122 + 52 (Q BN = 5 cm) = 144 + 25 = 169 ∴ OB = 13 cm 1 Hence, the radius of the circle is 13 cm.

127

Q. 2. If O is the circumcentre of a ∆ABC and OD ⊥ BC, then prove that ∠BOD = ∠BAC. U [Board Term II, 2013]

Sol. Given : OD ⊥ BC Proof : In ∆OBD and ∆OCD, OB = OC (Radii) 1 OD = OD (Common) ∠ODB = ∠ODC (Each 90°) \ ∆OBD ≅ ∠OCD (RHS rule) So, ∠BOD = ∠COD (CPCT) 1 and, ∠BOC = 2∠BOD But , ∠BOC = 2∠BAC  (angle at centre is twice the  angle at the circumference) From (i) and (ii) ∠BOD = ∠BAC Hence Proved 1 Q. 3. In the given figure, AB and CD are two parallel chords of a circle with centre O and radius 5 cm such that AB = 8 cm and CD = 6 cm. If OP is perpendicular to AB and OQ is perpendicular to CD, determine the length of PQ. U [Board Term II, 2016]

Short Answer Type Questions-II (3 marks each) Q. 1. In the given figure, AB and CD are two chords of a circle with centre O such that MP = NP. If OM ⊥ AB and ON ⊥ DC, show that AB = CD. U [Board Term II, 2014]

Sol. Construction : Join OP. Proof : In ∆OMP and ∆ONP, ∠OMP = ∠ONP = 90° (Given) OP = OP (Common) 1 MP = NP (Given) ∴ ∆OMP @ ∆ONP (RHS) ∴ OM = ON (CPCT) 1 ∴ AB = CD  (Chords equidistant from the centre are equal) 1 Hence Proved

Sol. Construction : Join OA and OC. Proof : Since, perpendicular from centre of the circle to the chord bisects the chord.



∴



and,

In ∆OAP, or, ∴

1 AB = 4 cm 2 1 CQ = QD = CD = 3 cm 2 AP = PB =

1

OP2 = OA2 – AP2 (Pythagoras theorem) OP2 = 52 – 42 = 25 – 16 = 9 OP = 3 cm ½

128

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-IX

In ∆OCQ, ∴ ∴

OQ2 = OC2 – CQ2 (Pythagoras theorem) = 52 – 32 = 25 – 9 = 16 OQ = 4 cm ½ PQ = OP + OQ = 3 + 4 = 7 cm 1 [CBSE Marking Scheme, 2016]

Q. 4. If two equal chords of a circle intersect within a circle, prove that the line segment joining the point of intersection to the centre makes equal angles with the chords. U [NCERT][Board Term II, 2016]

Sol. Let AB and CD be two equal chords intersecting at P. Let O be the centre of the circle. Construction: Draw OM ⊥ CD and ON ⊥ AB. In ∆OMP and ∆ONP, ∠OMP = ∠ONP = 90° (By construction) OM = ON

∴ OA bisects the chord CP, (Perpendicular from the centre to the chord bisects the chord) 1 ∴ AP = CP 2 or, CP = 2AP ...(i) 1 Similarly, O'B ⊥ PD 1 ∴ BP = PD 2 or, PD = 2BP ...(ii) CD = CP + DP = 2AP + 2BP [from (i) and (ii)] = 2(AP + BP) ...(iii) 1 = 2(AB) In quadrilateral ABO'O, OA = O'B (Two lines ⊥ to same parallel line CD) AB || OO' (Given) ∴ ABO'O is ||gm and AB = OO'. (Opp. sides of parallelogram are equal) ∴ CD = 2AB = 2OO' Hence Proved 1 [CBSE Marking Scheme, 2015]

Long Answer Type Questions (5 marks each) Q. 1. In the given figure, AB is a diameter of the circle with centre O. If AC and BD are perpendiculars on a line PQ and BD meets the circle at E, then prove that AC = ED. U [Board Term II, 2013]

1 (equal chords are equidistant from the centre) OP = OP (Common) ∴ ∆OPM @ ∆OPN (R.H.S.) ∴ ∠OPM = ∠OPN (CPCT) 2 Hence Proved [CBSE Marking Scheme, 2016] Q. 5. Two circles whose centres are O and O' intersect at P. Through P, a line parallel to OO', intersecting the circles at C and D is drawn as shown. Prove that CD = 2OO' A [Board Term II, 2015, 2014, NCERT Exemplar]

Sol. Construction : Draw OA and O'B perpendicular to CD from O and O' respectively.

Sol. Proof : ∠AEB = 90°  (Angle in semi-circle) ∠AEB + ∠AED = 180° (Linear pair) 1 \ ∠AED = 90° ∠EAC + ∠ACD + ∠CDE + ∠AED = 360° (Sum of angles of a quad.) or, ∠EAC + 90° + 90° + 90° = 360° 1 or, ∠EAC = 360° – 270° 1 = 90° 1 Hence, each angle of quadrilateral is 90°. ∴ EACD is a rectangle. ∴ AC = ED. Hence Proved. 1 Q. 2. PQ and RS are two parallel chords of a circle whose centre is O and radius is 10 cm. If PQ = 16 cm and RS = 12 cm, find the distance between PQ and RS when they lie, (i) On the same side of centre O. (ii) On the opposite sides of centre O. Sol. Given,



Proof :

OA ⊥ CD



A [Board Term II, 2015]

OP = OR = 10 cm (Radii of same circle) PQ = 16 cm, RS = 12 cm



CIRCLES

Draw OL ^ PQ and OM ^ RS. 1 Since, perpendicular from the centre to the chord bisects the chord. 1 \ PL = LQ = PQ = 8 cm 2

RM = MS =

1 RS = 6 cm 2

In right triangle OLP, OP2 = OL2 + PL2 (By Pythagoras theorem) 100 = OL2 + 64 OL = 100 − 64 = 36 1

Q. 3. A circular park of radius 20 m is situated in a village. Three girls Rita, Sita and Gita are sitting at equal distance on its boundary each having a toy telephone in their hands to talk to each other. Find the length of the string of each phone. (There is no slack in the string). A OR A circular park of radius 20 m is situated in a colony. Three boys Ankur, Syed and Dayd are sitting at equal distance on its boundary each having a toy telephone in his hands to talk each other. Find the length of the string of each phone. 

OL = 6 cm In right triangle OMR, OR2 = OM2 + RM2 (By Pythagoras theorem) 100 = OM2 + 36

OM = 100 - 36 = 64

OM = 8 cm (i) If PQ and RS lie on same side of centre O.

Distance between PQ and RS = LM = OM – OL = 8 – 6 = 2 cm (ii) If PQ and RS lie on opposite sides of centre O

1

129

A [NCERT]

Sol.

1

Here, A, B and C are the three points where three girls/boys are sitting. DABC is an equilateral triangle. In an equilateral triangle, the circumcentre is the point of intersection of median. ½ ∴ O divides AD in the ratio 2 : 1.  ½ Hence, if AO = 20 m ½ then, OD = 10 m Also, median is same as the altitude for an equilateral triangle. ½ In right triangle ∆ODC, OC2 = OD2 + DC2 or, 202 = 102 + DC2 or, DC2 = 400 – 100 = 300 ½ or,

DC = 10 3 m

or, BC = 2DC Perpendicular drawn from the centre to the chord bisects the chord. Distance between PQ and RS = LM = OL + OM = 6 + 8 cm = 14 cm

1

= 20 3 m 1

Length of the string of each phone = 20 3 m

= 20 × 1.732 1

= 34.64 m (Approx)

Topic-2 Cyclic Quadrilaterals Revision Notes  If a line segment joining two points subtends equal angles at two other points lying on the same side of the line segment, the four points are concyclic, i.e., lie on the same circle.  If the sum of any pair of opposite angles of a quadrilateral is 180°, then the quadrilateral is cyclic.  Any exterior angle of a cyclic quadrilateral is equal to the interior opposite angle.  Concentric Circles : Circles with a common centre are called concentric circles.

Scan to know more about this topic

Cyclic Quadrilateral

½

130

   

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-IX

The degree measure of a semi-circle is 180°. The degree measure of a circle is 360°. The degree measure of a major arc is (360°– q), where q is the degree measure of the corresponding minor arc. Area of a circle = πr2 sq. units

Example 2 Prove that the quadrilateral formed (if possible) by the internal angle bisectors of any quadrilateral is cyclic.

1 = 180° – (ÐP + ÐQ) 2

...(i)

[PB and QD are bisectors of ÐP and ÐQ, respectively] Similarly, ÐBCD = ÐRCS

Solution: Step I : Draw a figure according to given information. Let PQRS be a quadrilateral in which the angle bisectors PB, QD, RD and SB of internal angles P, Q, R and S, respectively form a quadrilateral ABCD Step II : Write the proving statement. ABCD is a cyclic quadrilateral i.e., ÐA + ÐC = 180° or, ÐB + ÐD = 180° Step III : For proving the result, first find the angles ÐBAD and ÐBCD. Since, ÐBAD = ÐPAQ (Vertically opposite angles) Also, ÐPAQ = 180° – (ÐAPQ + ÐAQP) ( In ∆PAQ, using angle sum property i.e., ÐPAQ + ÐAPQ + ÐAQP = 180°)



= 180° – (ÐCRS + ÐRSC)





= 180° –

1 (ÐR + ÐS) 2

...(ii)

Step IV : Adding the results obtained in step III and further use the property of a quadrilateral, which prove the required results. On adding eqs (i) and (ii), we get 1 ÐBAD + ÐBCD = 180° – (ÐP + ÐQ) + 180° 2 –

1 (ÐR + ÐS) 2

1 = 360° – (ÐP + ÐQ + ÐR + ÐS) 2 1 = 360° – × 360° 2

(sum of angles of a quadrilateral is 360°) = 360° – 180°

= 180° \ ABCD is a cyclic quadrilateral because sum of a pair of opposite angles of quadrilateral ABCD is 180°. Hence Proved

SUBJECTIVE TYPE QUESTIONS Very Short Answer Type Questions (1 mark each) Q. 1. In the given figure, if ∠POR is 110°, then find the value of ∠PQR.

R [Board Term II, 2012]

Sol. Reflex angle POR = 360° – 110° = 250° ½ \ By degree measure theorem, 1 ÐPQR = (reflex angle POR) 2 =

1 (250°) 2

= 125°

½

Q. 2. The sum of the opposite angles of a cyclic quadrilateral is : ............. R [Board Term II, 2011] Sol. 180°.

1



CIRCLES

Q. 3. In the given figure, quadrilateral PQRS is cyclic. If

131

Sol.

ÐP = 80°, then ÐR is equal to ................. . R [Board Term I, 2011]





Sol. Since, quadrilateral PQRS is cyclic \ ÐP + ÐR = 180° ½ (sum of opposite angles of a cyclic quadrilateral) or, 80° + ÐR = 180° or, ÐR = 100° ½

Short Answer Type Questions-I (2 marks each) Q. 1. In the given figure, find the value of x. R [Board Term II, 2012]

Sol. In a cyclic quadrilateral, ÐA + ÐC = 180° (opposite angles of cyclic quadrilateral are supplementary) or, 2x + 4° + 4x – 64° = 180° or, 6x – 60° = 180° or, 6x = 180° + 60° = 240° 1 240ϒ or, x= 6 ∴ x = 40° 1 Q. 2. ABCD is a cyclic quadrilateral in which AC and BD are its diagonals. If ∠DBC = 55° and ∠BAC = 45°, find ∠BCD. R [Board Term II, 2012]

Since, sum of the opposite pairs of angles in a cyclic quadrilateral is 180°. Hence, ∠B + ∠D = 180° or, ∠B = 180° – 70° = 110° 1 Again, AB || CD and AD is its transversal, so ∠A + ∠D = 180° (Consecutive interior angles) ∠A + 70° = 180° or, ∠A = 180° – 70° = 110° ½ and ∠A + ∠C = 180° or, 110° + ∠C = 180° ∴ ∠C = 180° – 110 = 70° ½ Q. 4. If diagonals of a cyclic quadrilateral are diameters of the circle through the opposite vertices of the quadrilateral, prove that the quadrilateral is a rectangle. R [Board Term II 2017] Sol. ABCD is a cyclic quadrilateral in which AC and BD are diameters

Since, AC is diameter \ ÐABC = ÐADC = 90° (Angles of semicircle) 1 BD is the diameter \ ÐBAD = ÐBCD = 90° (Angles of semicircle) Since, all angles of quadrilateral ABCD are 90°. 1 \ ABCD is a rectangle. Hence Proved. Q. 5. In the given figure, O is the centre of the circle and BA = AC. If ÐABC = 50°, find ÐBOC and ÐBDC.  A [Board Term II, 2017]

Sol.





∠BAC = ∠BDC = 45°, (Angles in the same segment) ½

Now, in DDBC, ∠DBC + ∠BCD + ∠CDB = 180° (Angle sum property) ½ or, 55° + ∠BCD + 45° = 180° or, ∠BCD = 80° 1 Q. 3 ABCD is a cyclic quadrilateral in which AB || CD. If ∠D = 70°, find all the remaining angles.

R [Board Term II, 2012]

Sol. (i) \

AB = AC (Given) ÐABC = ÐACB = 50° (Isosceles D property) By angle sum property of a triangle ÐBAC = 180° – ÐABC – ÐACB ÐBAC = 180° – 50° – 50° = 80° \ ÐBOC = 2ÐBAC 1 (Angle at the centre is twice the angle at the circumference)

132

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-IX

= 2 × 80° = 160° (ii) ÐBDC + ÐBAC = 180° (Opp. angles of cyclic quadrilateral) \ ÐBDC = 180° – 80° = 100° 1

Short Answer Type Questions-II (3 marks each)

\ Ð1 + Ð3 = 180° [By (i) and (ii)] Ð2 + Ð4 = 180° [By (i) and (iii)] 1 As these are opposite angles of a quadrilateral. \ BCED is a cyclic quadrilateral. 1 [CBSE Marking Scheme 2012] Q. 3. In the given figure, PQ = QR = RS and ÐPQR = 128°. Find ÐPTQ, ÐPTS and ÐROS.

Q. 1. In the given figure, AB is a chord equal to the radius of the given circle with centre O. Find the values of a and b.

U [NCERT]

U [Board Term II, 2014]

Sol. OB = OA (radii of circle) OA = OB = AB (given) \ DOAB is an equilateral triangle. \ ÐAOB = 60° 1 (angles of equilateral D are 60° each) \ a + ÐAOB = 180° (Linear pair) \ a + 60° = 180° \ a = 120° Reflex angle BOD = 2ÐBCD 1 (angle subtended by an arc at the centre is twice at the circumference) 360° – a = 2b 360° – 120° = 2b 2b = 240° b = 120° 1 [CBSE Marking Scheme, 2014] Q. 2. If a line is drawn parallel to the base of an isosceles triangle to intersect its equal sides, prove that the quadrilateral so formed is cyclic. U [Board Term II , 2012, NCERT Exemplar]

Sol.

A

Sol. Given, In DPQR

PQ = QR = RS, ÐPQR = 128°



Ð1 = Ð2 =

(180° − 128°) 2

52° = = 26° 2  \ 

1

ÐPTQ = ÐQRP = 26° (Angle in same segment)

(ii)

ÐPTQ = ÐQTR

= ÐRTS = 26° (\ equal chords make equal angle at circumference or centre) And

ÐPTS = ÐPTQ + Ð4 + Ð3

= 26° + 26° + 26°

E

D 1

1

4

= 78° (iii)

2 B

3

C

Given, DE || BC AB = AC or, Ð2 = Ð3 ...(i) (Angles opposite to equal sides to are always equal) Since, DE is parallel to BC, the consecutive interior angles as supplementary Ð1 + Ð2 = 180° ...(ii) and Ð3 + Ð4 = 180° ...(iii)

1

ÐROS = 2ÐRTS



(Angle subtended by an arc at the centre



is twice at circumference)

= 2 × 26° = 52°

1

Long Answer Type Questions (5 marks each) Q. 1. In the given figure, PQ is the diameter of the circle. If ∠PQR = 65°, ∠QPT = 60°, then find the measure of : (i) ∠QPR (ii) ∠PRS (iii) ∠PSR (iv) ∠PQT. U [Board Term II, 2012]



CIRCLES

1 ÐPOT 2 (Angle subtended by an arc at the centre is twice the angle at the remaining circle) 1 = × 150° 2 1 = × 150° 2 \ ÐPUT = 75° 1 [CBSE Marking Scheme, 2015] Q. 3. A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc. U [Board Term II, 2012] Sol. According to the question, OA = AB = OB 1

Sol. (i) ∠QRP = 90° (Angle in the semi-circle)  1 ∴ ∠QPR = 25° (By angle sum property) (ii) ∠QPS = ∠QPR + ∠RPS = 25° + 25° = 50° ∠QRS = 180° – 50° = 130° (PQRS is a cyclic quad.) 1 ∠PRS = 130° – ∠QRP 1 = 130° – 90° = 40° (iii) ∠PSR = 180° – 65° = 115°  (PQRS is a cyclic quad. & ∠Q = 65°) 1 (iv) ∠PTQ = 90° ∴ ∠PQT = 180° – (90° + 60°) = 30°1 Q. 2. In the given figure ÐSPQ = 45°, ÐPOT = 150° and O is the centre of circle. Find the measures of ÐRQT, ÐRTQ and ÐPUT.

ÐPUT =

U [Board Term II, 2015]

Sol. In the given figure ÐPOT + reflex ÐPOT = 360° 150° + reflex ÐPOT = 360° reflex ÐPOT = 210° 1 reflex ÐPOT = 2ÐPST (Angle subtended by arc at the centre is twice at circumference) 210° = 2ÐPST \ ÐPST = 105° 1 ÐPQT + ÐPST = 180° (Opposite angles of cyclic quadrilateral PQTS are supplementary) ÐPQT = 180° – 105° = 75° ½ ÐRQT + ÐPQT = 180° (Linear pair) ÐRQT = 180° – 75° = 105° ½ ÐRTQ = ÐSPQ (Exterior angle of a cyclic quadrilateral is equal to interior opposite angle) = 45° 1

∴ DOAB is an equilateral triangle ∠AOB = 60° 1 1 ∠ACB = ∠AOB 1 2 (Angle subtended by an arc at the circumference is half of the angle at the centre of circle) 1 ÐACB = × 60° 2 ∠ACB = 30° 1 ∠ACB + ∠ADB = 180° (Opposite angles of cyclic quadrilateral are supplementary) or, ∠ADB = 180° – ∠ACB ∠ADB = 180° – 30° = 150° 1

Commonly Made Error

Students fail to identify which angle is subtended by the chord from which arc.

Answering Tip

A chord has its both points on the circle.

OBJECTIVE TYPE QUESTIONS A Multiple Choice Questions

133

Q. 1. AD is a diameter of a circle and AB is a chord. If AD = 34 cm, AB = 30 cm, the distance of AB from the centre of the circle is

(1 mark each)

(A) 17 cm (B) 15 cm (C) 4 cm (D) 8 cm A [NCERT Exemp.] Ans. Option (D) is correct.

134

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-IX

(A) 30° (B) 60° (C) 90° (D) 45°  A [NCERT Exemp.] Ans. Option (D) is correct. Explanation: As AOB is a diameter of the circle,

Explanation:

[Angles in a semi-circle is 90o] ∠C=90 o Now, AC = BC ∠A=∠B [Angles opposite to equal sides of triangle are equal] Using angle-sum property of a triangle, we have

Draw OP ⊥ AB We know that the perpendicular from the centre of the circle to the chord bisects the chord.

1 1 Therefore, AP = ×AB= ×30=15 cm 2 2 1 Radius = OA = ×34=17 cm 2 In right DOPA, we have

∠A + ∠B + ∠C =180o ∠A + ∠A + 90o = 180o   2∠A=180 o − 90 o ∠ A=



90 o = 45o 2

Q. 4. In the given figure, BC is a diameter of the circle and ÐBAO = 60°. Then ÐADC is equal to A

2 2  OP = OA − AP

60°

= 17 2 − 15 2

B

= 289 − 225 = 64 = 8 cm Q. 2. If AB = 12 cm, BC = 16 cm and AB is perpendicular to BC, then the radius of the circle passing through the points A, O and C is (A) 6 cm (B) 8 cm (C) 10 cm (D) 12 cm  A [NCERT Exemp.] Ans. Option (C) is correct. Explanation:

.

O

A

C

12 cm

16

cm

O

C

D (A) 30° (B) 45° (C) 60° (D) 120°  A [NCERT Exemp.] Ans. Option (C) is correct. Explanation: In DOAB, OA = OB [Radii of the same circle] ∠BAO \   ∠ABO = [Angles opposite to equal sides are equal] So, [Given] ∠BAO = 60o    ∠ABO = Now, ∠ADC = ∠ABC = 60o [Angles in the same segment of a circle are equal.] 60o Therefore, ∠ADC = Q. 5. In the given figure, O is the centre of a circle and ÐBOD = 150°, then the value of x is

B

AB is perpendicular to BC, therefore, ABC is right angle triangle. In right angle triangle ABC, we have AC = AB 2+BC 2

(By Pythagoras Theorem)

= 12 2+16 2 = 144+256  

= 400=20

Therefore, AC = 20 cm (Diameter of circle) 1 Radius = ×20=10 cm 2 Q. 3. In the given figure, if AOB is a diameter of the circle and AC = BC, then ÐCAB is equal to :

(A) 105° (B) 115° (C) 100° (D) 110° Ans. Option (A) is correct. Explanation: ÐBOD = 150° 1 \ ÐBCD = ´ 150° = 75° 2 (Angle at centre is twice the angle at the circumference) Now, ABCD is a cyclic quadrilateral ÐBCD + ÐBAD = 180° (\ Sum of opposite angles of a quadrilateral) 75° + ÐBAD = 180° ÐBAD = 105°





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Q. 6. In the given figure, O is the centre of a circle and diameter AB bisects the chord CD at a point P such that CP = PD = 10 cm and PB = 6 cm, then the radius of the circle is

(A) 12.3 cm (B) 11.3 cm (C) 13 cm (D) 10.3 cm Ans. Option (B) is correct. Explanation: Construction: Join OC and OD Let the radius of the circle be r cm, then OC = OD = OB = r cm, OP = (r – 6) cm But CP = PD = 10 cm (given) Since, AB bisects the chord CD at a point P, then ÐOPD = 90° \ In right triangle DOPD, OD2 = OP2 + PD2 (Using Pythagoras theorem)

r2 = (r – 6)2 + (10)2

Þ

r2 = r2 + 36 – 12r + 100

Þ 12r = 136 Þ r = 11.3 cm Q. 7. In the given figure, O is the centre of a circle, ÐAOB = 50° and ÐBDC = 120°, then ÐOBC is:

(A) 60° (B) 45° (C) 40° (D) 70° Ans. Option (A) is correct. Explanation: Here, ÐBAC = 30° (Given) Then, ÐBOC = 2 × ÐBAC = 2 × 30° = 60°  OB = OC = Radius \ ÐOBC = ÐOCB = x (Angle opposite to equal side) Now, in DOBC, ÐBOC + ÐOBC + ÐOCB = 180° Þ 60° + x + x = 180° 2x = 120° Þ x = 60°

B Assertion & Reason

Directions: In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Mark the correct choice as: (A) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A). (B) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A). (C) Assertion (A) is true but reason (R) is false. (D) Assertion (A) is false but reason (R) is true. Q. 1. Assertion (A): In the given figure, if ÐABC = 20°, then ÐAOC is 40°.

(A) 45° (B) 60° (C) 35° (D) 90° Ans. Option (C) is correct. Explanation:  ÐAOB = 50° (Given) 1 1 \ ÐDCB = ÐAOB = × 50° 2 2 = 25° Now, in DDBC, ÐBDC + ÐDCB + ÐDBC = 180° (Angle sum property) Þ 120° + 25° + ÐDBC = 180° Þ ÐDBC = 180° – (120° + 25°) = 180° – 145° = 35° Q. 8. In the given figure, O is the centre of a circle and ÐBAC = 30°, then ÐOBC is:

O A 20°

B

C



Reason (R): Angle subtended by an are of a circle at the centre is double the angle subtended by it at any on the remaining part of the circle. Ans. Option (A) is correct. Explanation: In case of Assertion (A): Arc AC subtends ÐAOC at the centre O and ÐABC at a point B on the remaining part of the circle. \ ÐAOC = 2ÐABC ( Angle subtended by an arc at the centre of the circle is double the angle subtended by it at any other part of the circle.) Hence, ÐAOC = 2 × 20° = 40°

136

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-IX

\ Assertion is True In case of Reason (R): It is a theorem of circle which is a correct explanation of Assertion. \ Reason is True Q. 2. Assertion (A): In the given figure, if OA = 5 cm, AB = 8 cm and OD is perpendicular to AB, then CD is equal to 2 cm.



Reason (R): Perpendiculars bisectors of two chords of a circle intersect each other at the centre of the circle. Ans. Option (B) is correct. Explanation: In case of Assertion (A): The perpendicular drawn from the centre to a chord bisects the chord, 1 1 \ AC = AB = ´ 8 = 4 cm 2 2 In right angle triangle ÐAOC \ OC2 = OA2 – AC2 = 52 – 42 = 25 – 16 \ OC = 9 OC = 3 cm Now, CD = OD – OC = 5 – 3 = 2 cm \ Assertion is true. In case of Reason (R): Reason is true but not correct explanation of Assertion. Q. 3. Assertion (A): In given figure ABCD is a cyclic quadrilateral in which ÐA = (2x + 40)° and ÐC = (3x + 20)°, then the value of x is 24° A (2x + 40)° B

D



Reason (R): Opposite angles of a cyclic quadrilateral are equal. Ans. Option (C) is correct. Explanation: In case of Assertion (A): ÐA + ÐC = 180° \ ÐAOC = 2ÐABC ( Sum of opposite angles of a cyclic quadrilateral is 180°.) So, 2x + 40° + 3x + 20° = 180° 5x = 180° – 60° x = 24° \ Assertion is true In case of Reason (R): Opposite angles of a cylic quadrilateral are not equal. There sum is 180° \ Reason is false. Q. 4. Assertion (A): In the given figure, BOC is a diameter of a circle and ÐB = 45°, then ÐC is 25°.

Reason (R): An angle in a semi-circle is a right angle. Ans. Option (D) is correct. Explanation: In case of Assertion (A):

ÐA = 90° ( An angle in a semi circle is a right angle) ÐB = 45° (given) Then, ÐA + ÐB + ÐC = 180° (Angle sum property) 90° + 45° + ÐC = 180° ÐC = 180° – 135° ÐC = 45° \ Assertion is false In case of Reason (R): Angle in a semicircle is right angle. \ Reason is true Hence, Assertion is false but Reason is true.

(3x + 20)° C

COMPETENCY BASED QUESTIONS A Case based MCQs

Read the following passage and answer any four questions of the following : I. Rohan draws a circle of radius 10 cm with the help of compass and scale. He also draws two chords, AB

(4 marks each)

and CD in such a way that AB and CD are 6 cm and 8 cm from the centre O. Now, he has some doubts that are given below. Help him out by answering these questions:



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137

Q. 5. Which statement is not true ?  (A) Equal chords of a circle subtend equal angles at the centre (B) The perpendicular from the centre of a circle to a chord bisects the chord. (C) Angles in the same segment of a circle are equal. (D) The sum of each pair of opposite angles of a cyclic quadrilateral is 90°. Ans. Option (D) is correct. Explanation: The sum of each pair of opposite angles of a cyclic quadrilateral is 180°. 1 II. Three STD booth are placed at A, B and C in the figure and these are operated by handicapped persons. These three booth are equidistant from each other as shown in figure given below. [CBSE SAS] A

Q. 1. What is the length of AB ? (A) 12 cm (B) 11 cm (C) 16 cm (D) 8 cm

Ans. Option (C) is correct. Explanation: Length of AB = 16 cm In DAOP h2 = p2 + b2 (Pythagoras theorem) 102 = 62 + b2 100 = 36 + b2 b = 64 = 8 cm AB = 8 + 8 (Perpendicular drawn from the centre of a circle to the chord bisects the chord) = 16 cm 1 Q. 2. What is the length of CD ? (A) 10 cm (B) 12 cm (C) 16 cm (D) 21 cm Ans. Option (B) is correct. Explanation: Length of CD = 12 cm In DQOC h2 = p2 + b2 (Pythagoras theorem) (10)2 = 82 + b2 b2 = 100 – 64 = 36 b = 6 cm CD = 6 + 6 = 12 cm 1 Q. 3. A circle divides the plane, on which it lies, in _________ parts  (A) 1 (B) 2 (C) 3 (D) 4 Ans. Option (C) is correct. Explanation: 3 parts (inside, outside and on the circle) 1 Q. 4. A quadrilateral is called cyclic if all the four vertices of it lie on a _________ (A) Circle (B) Quadrilateral (C) Pentagon (D) Triangle Ans. Option (A) is correct.

1

O

B

C

Q. 1. Value of ÐBAC will be? (A) 30° (B) 60° (C) 90° (D) 100° Ans. Option (B) is correct. Explanation: As AB = BC = CA (given) So, DABC is an equilateral triangle \ ÐBAC = 60° ( All angles of an equilateral triangle are equal) Q. 2. What is the value of ÐBOC? (A) 60° (B) 120° (C) 180° (D) 200° Ans. Option (B) is correct. Explanation: ÐBOC = 2ÐBAC = 120° 1 ( Angle subtended by an arc at the centre is double the angle subtended by it any point on the remaining part of the circle.) Q. 3. If AB = 8 cm, value of BC + CA will be? (A) 16 cm (B) 10 cm (C) 8 cm (D) 18 cm Ans. Option (A) is correct. Explanation: DABC is an equilateral triangle \ AB = BC = CA = 8 cm 1 Hence, BC + CA = 16 cm Q. 4. What will be the value of ÐOBC? (A) 60° (B) 120° (C) 30° (D) 150° Ans. Option (C) is correct. Explanation: OB = OC (equal radii) \ ÐOBC = ÐOCB = x° (Angle opposite to equal side are equal.)

138

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-IX

In DOBC ÐOBC + ÐOCB + ÐBOC = 180° x + x + 120° = 180° 2x = 60° \ ÐOBC = 30°

1

Q. 5. Which angle will be equal to ÐOBC?  (A) ÐABC (B) ÐACB (C) ÐBOC (D) ÐOCB Ans. Option (D) is correct. Explanation: OB = OC (equal radii) \ ÐOBC = ÐOCB 1 (Angle opposite to equal side)

B Case based Subjective Questions

Read the following passage and answer the following questions: I. Four Friends Rima, Mohan, Sohan and Sita are sitting on the circumference of a circular park. Their locations are marked by points A, P, Q and R. Rohit joins them and sits at the centre of the circular park, so he is equidistant from all the other friends. His position is marked as O. They are sitting in such a way that ÐPQR = 110°. [CBSE SAS] Q

ÐOPR + ÐORP + ÐPOR = 180° x + x + 140° = 180° ( ÐPOR = 2ÐPAR) 2x = 40° x = 20° Hence, ÐOPR = 20° 1 II. A farmer has a circular garden as shown in the figure given below. He has different type of trees, plants and flower plants in his garden. In the garden, there are two mango trees A and B at a distance of AB = 10 m. Similarly has two Ashok trees at the same distance of 10m as shown at C and D. AB subtends ÐAOB = 120° at the centre O. The perpendicular distance of AC from centre is 5m. the radius of the circle is 13m.



P

R

O

A Q. 1. What is measure of reflex ÐPOR? Ans. Reflex ÐPOR = 2ÐPQR ( Angle subtended by an arc at the centre is double the angle subtended by it in the remaining part of the circle.) \ Reflex ÐPOR = 2 × 110° = 220° Q. 2. What is the measure of ÐPAR? Ans. ÐPAR + ÐPQR = 180° 1 ( Sum of opposite angles of cyclic quadrilateral is 180°) \ ÐPAR = 180° – 110° = 70° 1 Q. 3. Find ÐOPR? Ans. In DOPR OP = OR (equal radii) \ ÐOPR = ÐORP = x° (Angle opposite to equal sides are equal.)

A

13 P

C

10 cm

B

cm

5 cm

120° O

10 cm

D

Q. 1. What is the value of ÐADB? Ans. ÐAOB = 2ÐADB ( An angle subtended by an arc at centre is double the angle subtended by it any point on circumference.) 1 or, ÐADB = ÐAOB 2 =

1 ´ 120° 2

\ ÐADB = 60° 1 Q. 2. What is the value of ÐACB ? Ans. ÐACB = ÐADB (Angle in same segment are equal.) \ ÐACB = 60° Q. 3. What is the distance between mango tree A and Ashok tree C? Ans. OP = 5 cm (Given) OA = 13 cm (Radii) OP ^ AC (Given) \ In right triangle AOP OA2 = OP2 + AP2 (Pythagoras theorem) 132 = 52 + AP2

AP =

169 - 25

= 12 cm Now AC = 2AP 1 \ perpendicular drawn from the centre of a circle to the chord bisects the chord \ AC = 2 cm 1



CIRCLES

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Artificial Intelligence PARAMETERS

DESCRIPTION

Chapter Covered

Chapter 10: Circles

Name of the book Subject and Artificial Intelligence Integrated Learning Objectives

Mathematics, Class 9 NCERT Understanding the concept of Circles and integrating art and math with artificial intelligence ● To understand the concept of Circles ● To discover the relationship between angles subtended by the equal chords at the center. ● To prove the theorem using triangle properties. ● To apply the theorem in solving problems.

Time Required Classroom Arrangement Material Required

2 periods of 40 minutes each Flexible Pen, paper, Black Board chalk, scissors, glue, cardboard, geometry box, Laptops/ desktops and Internet connection. Students will be asked to recall the circle shape, draw it and its related terms radius, diameter, chord etc. using https://www.autodraw.com/ and then see how many objects can you find showing the circle’s terms they already know

AI CONCEPTS INTEGRATED

AutoDraw

Pre-Preparation Activity

Previous knowledge Methodology

Learning Outcomes

Follow up Activities

Reflections

Introduce the angle subtended by a chord at a point in a circle. Ask the children to draw a circle and any two chords and measure the angles made by them at the center. Activity 1: Ask students to paste a white paper on the cardboard and draw a circle with center O on this paper. Now make two equal chords on a circle using compasses. Joining end points of both the chords they will get 2 triangles. Trace one of the triangles on the tracing paper. Place this obtained triangle on the other triangle such that the chords overlap. They will observe that both triangles completely overlap. Thus both triangles are congruent and angle subtended by both angles are equal. Thus equal chords subtend equal angles at the center. Activity 2: Proof of theorem will be explained to the students. Activity 3: Ask students whether the converse is also true. Find by following the above procedure but this time they will make equal angles not equal chords. Activity 4: Ask students to apply their understanding of theorem to attempt questions of Exercise 10.2. ● The students will understand the concept of Circles. ● The students will discover the relationship between angles subtended by the chords at the centre. ● The students will be able to prove the theorem using triangle properties. ● The students will apply the theorem in solving problems. Activity: Ask students to go to https://goart.fotor.com/Create a beautiful art by uploading a photo of a circle showing angles made by equal chords at the centre and applying different art styles with this AI image generator. It uses an algorithm inspired by the human brain. It uses the stylistic elements of one image to draw the content of another. Activity Ask students to make a model showing the above theorem. Ask them to present to small groups. Let them assess how correct they are in their presentations Teacher will see how well the students are able to capture the concept and gain confidence while presenting their model. Teacher will discuss with students● How do you like the AI tool? ● Do you know any other tool that you can use in your concept? ● Try using the tools at home.







AutoDraw

Google Maps

Goart

SELF ASSESSMENT PAPER - 04 Time: 1 hour

MM: 30

UNIT-IV I. Multiple Choice Questions 1. ABCD is a rhombus such that ÐACB = 60°. Then ÐADB is (A) 40° (B) 30° (C) 60° 2. If in the given figure ÐABC = 50°, then ÐADC = ?

(A) 40° (B) 30° (C) 50° 3. Find the supplement of 45° (A) 45° (B) 145° (C) 135° 4. A triangle whose all three sides are unequal is called (A) Scalene D (B) Isosceles D (C) Equilateral D

[1×6 = 6] (D) 45°

(D) 20° (D) 125° (D) Right D

II. Assertion and Reason Based MCQs Directions: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as. (A) Both A and R are true and R is the correct explanation of A. (B) Both A and R are true but R is NOT the correct explanation of A. (C) A is true but R is false. (D) A is false and R is True. 1. Assertion (A): According to Euclid's 1st Axiom "Things which are equal to the same thing are also equal to one another." Reason (R): If AB = PQ and PQ = XY, then AB = XY. 2. Assertion (A): In a cyclic quadrilateral ABCD, ÐA – ÐC = 60°, then the smaller of the two is 60°. Reason (R): Opposite angles of cyclic quadrilateral are equal. III. Very Short Answer Type questions 1. Solve the equation x + 5 = 12 and state Euclid‘s axiom used. 2. Write the complementary angle of 25°. 3. Three angles of a quadrilateral are 70°, 85° and 90°. Find the fourth angle. 4. In the given figure, if ∠DAB = 55°, ∠ABD = 45°, then find ∠ACB.

5. Calculate the value of x is the figure given below :

[1×5= 5]



SELF ASSESSMENT PAPER

141

IV. Short Answer Type questions–I [2×2=4] 1. If angles of a quadrilateral are is the ratio 2 : 5 : 6 : 7. Find the measure of all the angles of a quadrilateral. 2. In a DABC, ∠A + ∠B = 70° and ∠B + ∠C = 150°. Find the angles. V. Short Answer Type questions -II [3×2 = 6] 1. The diagonals AC and BD of a parallelogram, (AC, BD) intersect each other at O. If ∠DAC = 42° and ∠AOB = 80°, find ∠DBC.  2. In a right angled triangle, prove that the hypotenuse is the longest side. VI. Long Answer Type questions [5×1 = 5] 1. If two equal chords of a circle intersect within the circle, prove that the segment of one chord are equal to corresponding segments of the other chord.

II. Case Study Based Questions (Attempt Any 4 parts) V In the given figure, PQ = QR = RS and ∠PQR = 132°. Give the answer of the following questions :

1. ∠QPR is equal to (A) 48° (B) 24° 2. What is the value of ∠PTQ ? (A) 96° (B) 66° 3. The value of ∠PTS is (A) 48° (B) 132° 4. The measure of ∠ROS is (A) 48° (B) 96° 5. Which of the following relation is true ? (A) OR > OS (B) OR < OS

(C) 72°

(D) 90°

(C) 24°

(D) 48°

(C) 66°

(D) 72°

(C) 90°

(D) 42°

(C) OR = OS

(D) None of these

[1×4 = 4]

qq

UNIT-V

MENSURATION

Study Time: Max. Time: 2:30 Hrs Max. Questions: 38

CHAPTER

10 Syllabus

AREAS

Area of a triangle using Heron‘s formula (without proof)



List of Topics Topic-1: Area of Triangle Page No. 142

Topic-1 Area of Triangle

Topic-2 : Heron's Formula Page No. 146

Revision Notes  Parts of a Triangle In ∆ABC, there are : (i) three vertices, namely A, B and C. (ii) three angles, namely ∠A, ∠B and ∠C. (iii) three sides, namely AB, BC and CA.

 Area =

Scan to know more about this topic

1 × base × corresponding height. 2

 For an equilateral triangle of side ‘a’. (i) Area =

3 2 a 4

(ii) Perimeter = 3a 3 (iii) Altitude = a 2 Example: Find the area of an equilateral triangle with side 9 cm. 3 3 Solution: Area of an equilateral triangle = × a2 = × (9)2 4 4 81 3 = cm2 4  For an isosceles triangle with length of two equal sides as ‘a’ and base ‘b’. b (i) Area = 4

4 a2 − b2

(ii) Perimeter = 2a+ b

Area of Triangle



AREAS

143

144

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-IX

1 (iii) Altitude = 2

4 a2 − b2

 For right angled triangle, with ‘a’ and ‘b’ are the sides that includes the right angle.

1

(i) Area = × a × b 2 a2 + b2 )

(ii) Perimeter = (a + b + (iii) Altitude = a

Example 1 The base of a right triangle is 15 cm and its hypotenuse is 25 cm, then calculate its area. Solution: Step I : We find the height (perpendicular) of the right angled triangle by using pythagoras theorem.

AB2 = AC2 – BC2 = (25)2 – (15)2 = 625 – 225 = 400 AB = 20 cm Step II : Now we find area of triangle by using 1 Area = × base × height 2 1 é ù ê A = 2 ´ b ´ hú ë û

=

1 × 15 × 20 2

= 150 cm2

SUBJECTIVE TYPE QUESTIONS Very Short Answer Type Questions (1 mark each) Q. 1. The sides of a triangular plot are in the ratio 4 : 5 : 6 and its perimeter is 150 cm, then find its sides. U

or, 2x2 = 25 × 2 or, x = 5 cm. 1 Q. 3. What is the area of DABC in which AB = BC = 4 cm and ∠B = 90° ? U Sol.

Sol. Let sides are 4x, 5x and 6x Then,

Perimeter = 4x + 5x + 6x = 150

or, 15x = 150 or, x = 10 \ Sides are 40 cm, 50 cm and 60 cm. 1 Q. 2. Calculate the side of an isosceles right triangle of hypotenuse 5 2 cm.

U

Sol.





Area of DABC =

1 ×4×4 2

= 8 cm2. 1 Q. 4. The base of a right triangle is 6 cm and hypotenuse is 10 cm. What will be its area ? U [NCERT Exemplar]

Sol.

In right triangle ABC.

(

x2 + x2 = 5 2

)

2



AREAS

AB =

AC 2 − BC 2 = 10 2 − 6 2

=

100 − 36 = 64 = 8 cm .



1 Area of right triangle = × base × height 2 1 = × 6×8 2 48 = = 24 cm2 1 2 Q. 5. Find out the area of an isosceles triangle whose base is ‘a’ and equal sides are of length ‘b’. A Sol. Let 'b' be the equal sides length and 'a' be the base.

145

By Pythagoras theorem, AC2 = AB2+ BC2 = a2 + a2 = 2a2 or, AC = 2 a unit Perimeter of ∆ABC = AB + BC + CA = a + a +



= 2a + = a(2 +

2a

2a 2 ) unit

2

Long Answer Type Questions (5 marks each) Q. 1. From a point in the interior of an equilateral triangle, perpendiculars are drawn on the three sides. The length of the perpendiculars are 14 cm, 10 cm and 6 cm. Find the area of the triangle.  Sol. Join OA, OB and OC.

2



x=

 a b2 −   =  2

A [NCERT Exemplar]

4b2 − a2 2

1 × base × height 2 1 = ×a×x 2 Area of triangle ABC =

1 4b 2 − a2 ×a× 2 2 a 2 2 = 4 b − a (units)2 4 =

1

Short Answer Type Questions-I (2 marks each) Q. 1. If the area of an equilateral triangle is 81 3 cm . U [Board Term I, 2012] Find its perimeter. 2

Sol. Area of an equilateral triangle =

3 2 a 4

½

3 2 ∴ 1 a = 81 3 4 or, a2 = 81 × 4 a = 9 × 2 = 18 cm Perimeter of equilateral triangle = 3a = 3 × 18 = 54 cm. ½ [CBSE Marking Scheme, 2012] Q. 2. For an isosceles right angled triangle having each of equal sides a, find the perimeter. A Sol. In right angled ∆ABC,



Let, sides of equilateral triangle be ‘a’ cm. 1 ar∆OAB = × AB × OP 2



1 × a × 14 2

=

= 7a cm2 ...(i) ½ 1 × AC × OR 2

ar∆OAC =



1 ×a×6 2

=

= 3a cm2

...(ii) ½

1 × BC × OQ 2

ar∆OBC =

=

½

1 × a × 10 2

= 5a cm2 ...(iii) ½

Adding, (i), (ii) and (iii), we get

ar∆OAB + ar∆OAC + ar∆OBC = (7a + 3a + 5a) cm2 = 15a cm2 ∴ Area of equilateral DABC = 15a cm2

1

146

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-IX

or,

3 2 a = 15a 4 a =

or,

15 × 4 3 × = 20 3 cm 3 3

1

∴

Area =



=

3 × 20 3 4

(

)

2

3 × 400 × 3 = 300 3 cm2. 1 4

Topic-2 Heron’s Formula Revision Notes  Heron's Formula

Scan to know more about this topic

Consider a triangle with sides a, b and c Let AB = c, BC = a and CA = b So, Its perimeter = a + b + c a+b+c Semi-perimeter, s = 2

Heron's Formula

Area of triangle = | s( s − a )( s − b )( s − c ) |

This formula is known as ‘Heron’s formula’. This formula is applicable to all type of triangles whether it is a right triangle or an isosceles or an equilateral triangle.

Example 2

The side of a triangle are 12 cm, 16 cm, and 20 cm. Find its area. Solution: Step I : We find the semi-perimeter of a triangle by using Heron's formula

s = s =

a+b+c 2



Step II : Now we find the area of triangle by using Heron's formula



Area = | s( s - a )( s - b )( s - c )|

=

24( 24 - 12 )( 24 - 16 )( 24 - 20 )

=

24 ´ 12 ´ 8 ´ 4

= 96 cm2

12 + 16 + 20 = 24 cm 2

SUBJECTIVE TYPE QUESTIONS

Very Short Answer Type Questions (1 mark each) Q. 1. If a, b and c are the sides of a triangle and s = semiperimeter, then calculate the area of triangle. R

| s( s − a )( s − b )( s − c ) | unit2 Sol.

1

Q. 2. Write the name of formula for finding the area of a triangle when sides are given. R [Board Term I, 2014] Sol. Heron’s Formula.

1

Q. 3. Two sides of a triangle are 13 cm and 14 cm and its semi-perimeter is 18 cm, then what will be the third side of the triangle ? R

Sol.

or,

s =

a+b+c 13 + 14 + c ⇒ 18 = 2 2

c = 36 – 27

= 9 cm. 1 Q. 4. The sides of ∆ABC are 8 cm, 7 cm and 5 cm respectively. Find out its semi-perimeter. R a+b+c 8+7+5 Sol. s = = 2 2 = 10 cm. 1 Q. 5. The sides of a triangle are 7 cm, 24 cm and 25 cm. What will be its area ? U





AREAS

Sol.

s =

7 + 24 + 25 2



= 28 cm \

Area = | 28( 28 − 7 )( 28 − 24 )( 28 − 25) |

= 28 × 21 × 4 × 3 = 84 cm2.

1

Q. 1. Find the area of a triangle, two sides of which are 8 cm and 11 cm and the perimeter is 32 cm. U [KVS 2019]

Let the side AB(c) = 8 cm Side AC(b) = 11 cm A

8c

m

11



= | 42( 42 − 26 )( 42 − 28 )( 42 − 30 ) |



= | 42 × 16 × 14 × 12 |



= | 14 × 3 × 16 × 14 × 4 × 3 | 1

= 336 cm2

Q. 3. Sides of a triangle are 70 cm, 80 cm and 90 cm. Find its area. (Use 5 = 2.23) U [Board Term I, 2014] Sol.



or,

s =

a+b+c 2

s =

70 + 80 + 90 2

= 120 cm \ Area = | s( s − a )( s − b )( s − c ) |

cm

C

Perimeter of DABC = 32 cm a + b + c = 32 cm

=

120(120 − 70 )(120 − 80 )(120 − 90 )

=

120 × 50 × 40 × 30

=

40 × 3 × 5 × 10 × 4 × 10 × 3 × 10

= 40 × 10 × 3 ×

a + 8 + 11 = 32 a = 32 – 19 = 13

\

a = 13 cm



s =



ar(DABC) =

s( s − a )( s − b )( s − c )



ar(DABC) =

16(16 − 13)(16 − 8 )(16 − 11)

= 2676 cm2 1 Q. 4. The semi perimeter of a triangle is 132 cm and the product of the differences of semi perimeter and its respective sides (in cm) is 13200. Find the area of triangle. U [Board Term I, 2016]

32 = 16 2

Sol. Here, s = 132 cm, (s – a)(s – b)(s – c) = 13200

=

16 × 3 × 8 × 5

=

4×4×2×2×2×3×5

1

Area of D = | s( s − a )( s − b )( s − c ) |

= 132 × 13200 = 132 × 10 = 1320 cm2 1 [CBSE Marking Scheme, 2016]

= 4 × 2 2 × 3 × 5 2



Q. 2. If the sides of a triangle are 26 cm, 28 cm and 30 cm. Find the area of triangle.

Q. 5. The longest side of a right angled triangle is 125 m and one of the remaining two sides is 100 m. Find its area using Heron’s formula. U [Board Term I, 2015] Sol.

Third side =

R [Board Term I, 2016]

Sol. Here, a = 26 cm, b = 28 cm, c = 30 cm a+b+c s = 2 26 + 28 + 30 84 = = 2 2 = 42 cm

½

5

= 1200 × 2.23



\ ar(DABC) = 8 30 cm 2

½



B



Area of D = | s( s − a )( s − b )( s − c ) |

[CBSE Marking Scheme, 2016]

Short Answer Type Questions-I (2 marks each)

Sol.

147

1

s =

(125)2 − (100 )2 = 75

300 2

1

 100 + 75 + 125    2  

= 150 cm Area of ∆ = 150 × (150 - 125)(150 − 100 )(150 − 75)

Area of ∆ =

150 × 25 × 50 × 75 = 3750 m2 [CBSE Marking Scheme, 2015] 1

148

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-IX

Short Answer Type Questions-II (3 marks each) Q. 1. An umbrella is made by stitching ten triangular pieces of cloth, each measuring 60 cm, 60 cm and 20 cm. Find the area of the cloth required for the umbrella. A [Board Term I, 2014] Sol. Area of cloth required = 10 × Area of cloth for one piece ½ Area of one piece of cloth having sides 60 cm, 60 cm and 20 cm. 60 + 60 + 20 s = = 70 cm ½ 2 \

70(70 − 60 )(70 − 60 )(70 − 20 ) | Area =|

=

70 × 10 × 10 × 50

=

7 × 10 × 10 × 10 × 5 × 10

51 + 37 + 20 108 = 2 2 = 54 m

Sol.

1

s =



Area = | s( s − a )( s − b )( s − c ) |

\



=

= =

54( 54 − 51)( 54 − 37 )( 54 − 20 ) 54 × 3 × 17 × 34

9 × 3 × 2 × 3 × 17 × 17 × 2 1 = 306 m2 No. of rose beds =

306 = 51 6

1

Q. 4. The triangular side walls of a flyover have been used for advertisements. The sides of the walls are 122 m, 22 m and 120 m. The advertisements yield an earning of Rs 5,000 per m2 per year. A company hired one of its walls for 3 months. How much rent did it pay?

= 10 × 10 × 35 2 = 100 35 cm Area of cloth required

1

122m 22m 120m

= 10 × 100 × 35 2 1 = 1000 35 cm Q. 2. Find the percentage increase in the area of a

triangle, if its each side is doubled. Sol. Let a, b, c be the given sides, Semi perimeter of original triangle s =



A

 Sol.

a+b+c 2

Area of original triangle :

=

= | s( s − a )( s − b )( s − c ) | 1



s' =



=

2 a + 2b + 2 c 2 2( a + b + c ) = 2s 2

\ New Area = | 2 s( 2 s − 2 a )( 2 s − 2b )( 2 s − 2 c ) |

=

2 × s × 2( s − a ) × 2( s − b ) × 2( s − c )

= 4 s( s − a )( s − b )( s − c ) = 4 × original area 1 ∴ Increase in area = 4 × original area – original area = 3 × original area 3 × original area×100 % Increase in area = original area = 300% 1 Q. 3. The sides of a triangular field are 51 m, 37 m and 20 m. Find the number of rose beds that can be prepared in the field if each rose bed occupies a space of 6 sq. m. A [Board Term I, 2012]

132(132 − 122)(132 − 22)(132 − 120)

=

Again, 2a, 2b, 2c be the new sides, then

Semi-perimeter of new triangle,

U [NCERT Ex. 12.1, Q. 2, Page 202] = a 122 = m, b 22 = m, c 120 m a + b + c 122 + 22 + 120 = s = 2 2 264 = = 132 m 2 areaof triangle = s( s − a )( s − b )( s − c )

132 × 10 × 110 × 12 = 1320 m 2



[2] Rent of 1 m2 area per year = ` 5,000 5000 Rent of 1 m2 area per month = 12 Rent of 1320 m2 area for 3 months

 5000  = ` × 3 × 1320 [1]  12   The company had to pay = ` 1,650,000 Q. 5. There is a slide in a park. One of its side walls has been painted in some colour with a message “KEEP THE PARK GREEN AND CLEAN”. If the sides of the wall are 15 m, 11 m and 6 m, find the area painted in colour. 6c

m

m

11 c

KEEP THE PARK GREEN AND CLEAN 15 cm



Ap

[NCERT Ex. 12.1, Q. 3, Page 203]



AREAS

Sol. a = 11 m, b = 6 m, c = 15 m 11 + 6 + 15 32 = s = = 16 m 2 2              [1] Area painted in colour = s( s − a )( s − b )( s − c ) = 16(16 − 11)(16 − 6)(16 − 15)) = 16 × 5 × 10 × 1 = 20 2 m [2] 2

Long Answer Type Questions (5 marks each) Q. 1. Two identical circles with same inside design as shown in the figure are to be made at the entrance. The identical triangular leaves are to be painted red and the remaining are to be painted green. Find the total area to be painted red. A

41

cm

15 cm

Sol. For one identical triangular leaf, let a = 28 cm, b = 15 cm and c = 41 cm s =



=



a + b + c 28 + 15 + 41 = 2 2 84 = 42 cm 2

1

28 cm

= =

Using Heron’s formula, Area of one triangular leaf = | s( s − a )( s − b )( s − c ) | =

42( 42 − 28 )( 42 − 15)( 42 − 41)

=

42 × 14 × 27 × 1

=

3 × 14 × 14 × 3 × 9 = 9 × 14

= 126 cm2 There are 6 leaves in a circle.

1

10(10 − 5)(10 − 7 )(10 − 8 ) 1 10 × 5 × 3 × 2

= 100 × 1.73 = ` 173 1½ [CBSE Marking Scheme, 2016] Q. 3. A triangular park has sides 60 m, 40 m and 26 m. Gardener has to put a fence all around its boundary and also plant grass inside. Find the area in which grass will be planted. Also calculate the cost of fencing it with barbed wire at the rate of ` 30 per meter, leaving a spare 2 m wide for a gate on one side. A [Board Term I, 2016] 60 + 40 + 26 2 126 = = 63 2

15 cm



A [Board Term I, 2016]

Sol. Here, a = 5 m, b = 7 m, c = 8 m a+b+c s = 2 5 + 7 + 8 20 = = 2 2 s = 10 cm Area of D = | s( s − a )( s − b )( s − c ) |

Sol.

41 cm



3 = 1·73)

2 ½ = 10 3 m \ Cost of levelling = Rate × Area = 10 × 10 3 1 = ` 100 3

28 cm

Also,

So, total number of leaves in 2 circles = 2 × 6 = 12 ∴ Area of 12 leaves = (12 × 126) cm2 = 1512 cm2 Hence, total area to be painted red = 1512 cm2 1½ Q. 2. The sides of a triangular park are 5 m, 7 m and 8 m respectively. Find the cost of levelling the park at the rate of `10 per m2. (Use

= 4×5× 2

149

1

s =

Area of D = | s( s − a )( s − b )( s − c ) |

= 63 × 3 × 23 × 37 = 401 m2 (Approx.) Length of wire needed for fencing

1

= 60 + 40 + 26 – 2

1

= 124 m

1

\ Cost of fencing = 124 × 30 = ` 3720

1½

1



1

[CBSE Marking Scheme, 2016]

150

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-IX

OBJECTIVE TYPE QUESTIONS A Multiple Choice Questions Q. 1. An isosceles right triangle has area 8 cm2. The length of its hypotenuse is (A) 16 cm 32 cm (B) (C) 48 cm (D) 24 cm Ans. Option (A) is correct. Explanation:

∵ Area of isosceles ∆ =

A

1 × Base × height 2

=

Q. 6. The sides of a triangle are 35 cm, 54 cm and 61 cm, respectively. The length of its longest altitude is (A) 16 5 cm (B) 10 5 cm

( H= )2 ( 4 )2 + ( 4 )2 H = 32 cm

Q. 2. The perimeter of an equilateral triangle is 60 m. The area is (A) 10 3 m 2 (B) 15 3 m 2 (D) 100 3 m 2

(C) 24 5 cm  Ans. Option (C) is correct. Explanation:

3 2 a = 4

35 + 54 + 61 = 75 cm 2 Area = s( s − a )( s − b )( s − c ) s=

3 × 20 × 20 = 100 3 m 2 4

Q. 3. The area of an equilateral triangle with side 2 3 cm is (A) 5.196 cm (B) 0.866 cm (C) 3.496 cm2 (D) 1.732 cm2 Ans. Option (A) is correct. Explanation: Area of equilateral triangle 2

(

)

75( 75 − 35)( 75 − 54)( 75 − 61) 75 × 40 × 21 × 14

Area = 420 5

A

2 3 × 2 3 4 3 = × 4 × 3= 3 3 4 = 3 × 1.732 = 5.196 cm 2

= =

= 420 5 cm 2

2

=

(D) 28 cm A [NCERT Exemp.]

A

Ans. Option (D) is correct. Explanation: Perimeter of equilateral triangle = 60 m 3a = 60 m a = 20 m Its Area =

84 × 28 × 24 × 32

⇒ 1344 cm 2

By Pythagoras theorem ( B )2 + ( P )2 = ( H )2

(C) 20 3 m 2

Q. 4. If the area of an equilateral triangle is 16 3 cm2, then the perimeter of the triangle is (A) 48 cm (B) 24 cm A (C) 12 cm (D) 36 cm Ans. Option (B) is correct. Explanation: Area of equilateral ∆ = 16 3 cm 2 16 3 × 4 = a 2 = 64 3 = a = 64 8 cm P = 3 × a = 3 × 8 = 24 cm Q. 5. The sides of a triangle are 56 cm, 60 cm, and 52 cm long. Then the area of the triangle is (A) 1322 cm2 (B) 1311 cm2 2 (C) 1344 cm (D) 1392 cm2  A [NCERT Exemp.] Ans. Option (C) is correct. Explanation: 56 + 60 + 52 168 = s = = 84 cm 2 2 Area = 84(84 − 56)(84 − 60)(84 − 52)

1 8 × ( B )2 = 2 1 × ( B )2 = 8 2 B = 16 = 4 cm

 

(1 mark each)

1 × 35 × h = 420 5 2 h = 24 5 cm Q. 7. The area of an isosceles triangle having base 2 cm and the length of one of the equal sides 4 cm, is 15 cm2 (A) 15 cm 2 (B) 2



AREAS

(C) 2 15 cm 2

(D) 4 15 cm 2

 Ans. Option (A) is correct. Explanation:

A [NCERT Exemp.]

A

=

1 cm

D

1 cm

C

cm, b 4= cm, c 2 cm = a 4= a+b+c 4+4+2 = s = = 5 2 2 Area = s( s − a )( s − b )( s − c ) = =

5(5 − 4)(5 − 4)(5 − 2) 5 ×1×1× 3

a + b + c 6 + 8 + 10 24 = s = = = 12 cm 2 2 2 s( s − a )( s − b )( s − c ) 12(12 − 6)(12 − 8)(12 − 10)

=

12 × 6 × 4 × 2

=

= 24 cm 2 Cost of painting 24 cm2 = 0.09 × 24 = ` 2.16.

B Assertion & Reason

Directions: In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Mark the correct choice as: (A) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A). (B) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A). (C) Assertion (A) is true but reason (R) is false. (D) Assertion (A) is false but reason (R) is true. Q. 1. Assertion (A): The height of a triangle is 18 cm and its area is 72 cm2. Its base is 8 cm.

Reason (R): Area of a triangle =

1 × Base × Height 2

Ans. Option (A) is correct. Explanation: In case of assertion (A):

Area of D =

1 × Base × Height 2

Hence, Both A and R are true and R is the correct Explanation of A. Q. 2. Assertion (A): If the area of an equilateral triangle is 81 3 cm2, then semi perimeter of triangle is 20 cm. Reason (R): Semi-perimeter of a triangle is a+b+c s= where a, b, c are sides of triangle. 2

= 15 cm 2 Q. 8. The edges of a triangular board are 6 cm, 8 cm, and 10 cm. The cost of painting it at the rate of 9 paise per cm2 is (A) ` 2.00 (B) ` 2.16 (C) ` 2.48 (D) ` 3.00  A [NCERT Exemp.] Ans. Option (B) is correct. Explanation: = a 6= cm, b 8= cm, c 10 cm

Area =

1 × 8 cm × 18 cm 2

= 72 cm2 \ Assertion is correct. In case of reason (R): Reason is correct. Area of a triangle =

B

151

1 × Base × Height 2

Ans. Option (D) is correct. Explanation: In case of assertion (A): Area of equilateral triangle =

3 2 a 4

81 3 =

3 2 a 4

Þ Now

81 × 4 = a2 a = 18 cm 18 + 18 + 18 s = 2

= 27 cm \ Assertion is false. In case of reason (R): a+b+c as semi-perimeter = where a, b, c are sides 2 of triangle. Reason is correct Q. 3. Assertion (A): Area of a rhombus whose side is 20 cm and one diagonal is 24 cm is 384 cm2. Reason (R): All sides of a rhombus are equal. Ans. Option (A) is correct. Explanation: In case of assertion (A):

Semi perimeter of DABD a+b+c s = 2 =

20 + 20 + 24 2

=

64 2

= 32 cm

152

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-IX



Area of DABD = =

=

s( s − a )( s − b )( s − c ) 32( 32 − 24 )( 32 − 20 )( 32 − 20 ) 32 × 8 × 12 × 12

= 8 × 2 × 12 = 192 cm2 \ Area of rhombus ABCD = 2 × Area of DABD = 2 × 192 = 384 cm2 In case of reason (R):

Reason is true.

16 + 16 + 16 2

s =

All sides of a rhombus are equal. Hence, Both A and R are true and R is the correct Explanation of A.

=

equilateral triangle with side 16 cm is 64 3 cm2. Reason (R): Heron's formula =



where, s =

s( s − a )( s − b )( s − c )

Area =

Q. 4. Assertion (A): Using Heron's formula, area of an

48 = 24 cm 2

=

\ AB = BC = CD = AD = 20 cm

=

s( s − a )( s − b )( s − c )

24( 24 − 16 )( 24 − 16 )( 24 − 16 ) 24 × 8 × 8 × 8

2 = 64 3 cm

a+b+c and a, b, c are sides a triangle. 2

\ Assertion is true. Reason is correct. Area of triangle is calculated using Heron's formula

Ans. Option (A) is correct. Explanation: In case of assertion (A): Side = 16 cm

= s( s − a )( s − b )( s − c ) Hence, Both A and R are true and R is the correct Explanation of A.

COMPETENCY BASED QUESTIONS A Case based MCQs

Read the following passage and answer any four questions of the following : I. Isosceles triangles were used to construct a bridge in which the base (unequal side) of an isosceles triangle is 4 cm and its perimeter is 20 cm.

(4 marks each)



2x = 20 – 4



2x = 16



x = 16 = 8 cm 2

½

Q. 2. What is the Heron's formula for the area of? (A) | s( s + a )( s − b )( s − c ) |



(B) | s( s + a )( s + b )( s + c ) |

(C) | s( s − a )( s − b )( s − c ) |

Q. 1. What is the length of equal sides ? (A) 2 cm (B) 3 cm (C) 8 cm (D) 10 cm Ans. Option (C) is correct. Explanation: Let x cm be the length of equal sides of the isosceles triangle. So, x + x + 4 = 20 ½

2x + 4 = 20

(D) | s( s . a )( s . b )( s . c ) | Ans. Option (C) is correct. Q. 3. What is the semi perimeter of the highlighted triangle ? (A) 30 cm (B) 40 cm (C) 10 cm (D) 50 cm Ans. Option (C) is correct. Explanation: Perimeter 20 Required semi perimeter = = 2 2 = 10 m Q. 4. What is the area of highlighted triangle ? (A) 4 15 cm 2

(B) 4 cm2

2 (C) 15 cm

(D) 20 cm2



AREAS

Ans. Option (A) is correct. Explanation: Since, semi perimeter, s = 10 cm Thus, area of the triangle

s( s − a )( s − b )( s − c ) =

10(10 − 8 )(10 − 8 )(10 − 4 ) ½ 

= 10( 2 )( 2 )( 6 ) = 4 15 cm2

½

Q. 5. If the sides of a triangle are in the ratio 3 : 5 : 7 and its perimeter is 300 m. Find its area.  2 (A) (B) 500 2 m2 100 2 m (C) 1500 3 m2 (D) 200 3 m2 Ans. Option (C) is correct. Explanation: Let the sides of a triangle are a = 3x, b = 5x, c = 7x then

a + b + c = 300



3x + 5x + 7x = 300



15x = 300



x = 20

So, a = 60, b = 100, c = 140

s = a + b + c

2

= 300 2 = 150 km Area of triangle =

½

s( s − a )( s − b )( s − c )

= 150(150 − 60 )(150 − 100 )(150 − 140 ) = 150 × 90 × 50 × 10 2 ½ = 1500 3 m II. Shakshi prepared a Rangoli in triangular shape on Diwali. She makes a small triangle under a big triangle as shown in figure.

153

Ans. Option (B) is correct. Explanation: P = sum of all 3 sides = (25 + 26 + 28) cm = 79 cm  1 Now, semi-perimeter = ×P 2 = 1 × 79 2 = 39.5 cm Q. 2.

1

1 of AB = 2 

(A) QR (B) RP (C) QP (D) QC Ans. Option (C) is correct. Explanation: 1 of AB = QP 1 2 Q. 3. What is the length of RQ ? (A) 16 cm (B) 15 cm (C) 13 cm (D) 14 cm Ans. Option (D) is correct. Explanation: Length of RQ = 1 of BC 2 = 1 × 28 2 = 14 cm 1 Q. 4. If colourful rope is to be placed along the sides of small DPQR. What is the length of the rope ? (A) 34.5 cm (B) 39.5 cm (C) 32.5 cm (D) 31.5 cm Ans. Option (B) is correct. Explanation: Length of rope = Perimeter of DPQR = (12.5 + 13 + 14) cm = 39.5 cm 1 Q. 5. Area of DPQR =  (A) | s( s − 12.5)( s − 13)( s − 14 ) | cm2 2 (B) | s( s − 25)( s − 26 )( s − 28 ) | cm (C) | s( s + 12.5)( s + 13)( s + 14 ) | cm2 (D) | s( s + 25)( s + 26 )( s + 28 ) | cm2 where s is the semi-perimeter of DPQR. Ans. Option (A) is correct. Explanation: Area of DPQR =

s( s − 12.5)( s − 13)( s − 14 ) cm 2 1 

B Cased Based Subjective

Sides of big triangle are 25 cm, 26 cm and 28 cm. Also, DPQR is formed by joining mid points of sides of DABC.

Use the above data to help her in resolving below doubts. Q. 1. What is the semi-perimeter of DABC ? (A) 39 cm

(B) 39.5 cm

(C) 40 cm

(D) 40.5 cm

Questions



Read the following passage and answers the following questions:

I. The School Principal gave a contract to a company to plant grass in the park at the rate of ` 4500 per hectare. The sides of a triangular park of a school are 120m, 100m and 110m.

154

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-IX

6 5 m Inv m en tor y

12 B

3m

C

100 m

[CBSE SAS] Q. 1. What is the perimeter of the park? Ans. Perimeter = AB + BC + CA

Ans.

= 120 + 100 + 110 = 330 m



Ans. Area of the park =

s =



Area =

=

Wa re

ho u 5 m se Canteen 4m

Q. 1. Find the area of Inventory?

1

Q. 2. What is the area of the park?

5m

2

Area =



s( s - a )( s - b )( s - c )

=

=

a + b + c 330 = = 165 m 2 2

s( s - a )( s - b )( s - c ) 8( 8 - 6 )( 8 - 5)( 8 - 5) 8´2´3´3

= 12 m2 Q. 2. Find area allotted for canteen?

165(165 - 120 )(165 - 100 )(165 - 110 )

Ans.

165 ´ 45 ´ 65 ´ 55

6( 6 - 5)( 6 - 4 )( 6 - 3)

= 5152.12 m2 (approx)



Q. 3. How much does the school have to pay to the company? Give you answer to the nearest ` 100. 1

= 6 m2

Area = =

6 ´1´ 2 ´ 3

Q. 3. Find cost of whole land at the rate of ` 500 per m2? 2

Ans. Cost of planting = Area × Rate Area = 5152.12 m2

Ans. Area of land = Area of canteen + Area of inventory + Area of warehouse

= 0.5152 hectare ( 1 hect = 10,000 m2)



1

5 + 4 + 3 12 s= =6m = 2 2





1

6 + 5 + 5 16 s= = =8m 2 2

= 825 39



8m

m 110

0m

A

Area of warehouse = s =

Cost = 4500 × 0.5152

= 2318



= ` 2300 (to nearest hundred) II. Ajay bought some land for carrying out his wholesale business as shown in the figure below. He plans to divide this land into 3 parts for warehouse, inventory, and canteen. Now using the given information, answer the following questions.

=

=

8 + 5 + 5 18 = =9 2 2

9( 9 - 8 )( 9 - 5)( 9 - 5) 9 ´1´ 4 ´ 4

= 12 m2 Total Area = 12 + 6 + 12 = 30 m2

Cost = 30 × 500

= ` 15000







Study Time: Max. Time: 2.30 Hrs Max. Questions: 39

CHAPTER

11

Syllabus

SURFACE AREAS AND VOLUMES

Surface areas and volumes of spheres (including hemispheres) and right cones.



Topic-1

Surface Area and Volume of Sphere (Including Hemisphere)

List of Topics Topic-1: Surface Area and Volume of Sphere (Including Hemisphere) Page No. 155 Topic-2 : Surface Area and Volume of Right circular Cone. Page No. 160

Revision Notes  Sphere: A sphere is a perfectly round geometrical object in three-dimensional space, such as the shape of a round ball.

 Hemisphere: A hemisphere is half of a sphere

 The total surface area of any object will be equal or greater than its curved surface area.  Volume is the capacity or the space occupied by a body.  The unit of measurement of both volume and capacity is cubic unit such as cubic feet, cubic cm and cubic m, etc.  When an object of certain volume is recast into a new shape, the volume of the new shape, formed will always be equal to the volume of the original object.  The solids having the same curved surface do not necessarily occupy the same volume.  When an object is dropped into a liquid, the volume of the displaced liquid is equal to the volume of the object that is dipped. Important Formulae  Sphere : Surface area = 4pr2 4 Volume = pr3 3

Scan to know more about this topic

Surface area of sphere and hemisphere

156

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-IX



SURFACE AREAS AND VOLUMES

 Hemisphere :

157

Curved surface area = 2pr2 Total surface Area = 3pr2 2 Volume = pr3 3



SUBJECTIVE TYPE QUESTIONS Very Short Answer Type Questions (1 mark each) Q. 1. Compute the curved surface area of a hemisphere whose diameter is 14 cm. R [Board Term II, 2015, NCERT] Sol. Given diameter of hemisphere = 14 cm \ Curved surface area 1 = 2pr2 22 × 7 × 7 = 308 cm2 = 2 × 7 Q. 2. If the number of square centimetres in the surface area of a sphere is equal to the number of cubic cm in its volume. Find the diameter of the sphere ? A [Board Term II, 2014] Sol. Given, Area of sphere = Volume of Sphere 4 4pr2 = pr3 3 where r is the radius of sphere or, r = 3 cm [on solving] \ Diameter = 2r = 6 cm 1 Q. 3. Find the amount of water displaced by a solid spherical ball of diameter 4.2 cm, when it is completely immersed in water. U [Board Term II, KVS 2016] Sol. Amount of water displaced

If r denotes radius of a football and r’ that of a cricket ball, then we have 2r = 5 × (2r’) 2r =5 2r ' r or = 5 r ' Now, ratio of surface areas



2

=

\ Volume of solid spherical ball =



\ Volume of solid spherical ball =

Q. 1. Find the radius of a sphere whose surface area is 616 cm2. R [Board Term II, 2016] Sol. Surface area of sphere = 4pr2 \ 4pr2 = 616 or, pr2 = 154 1 22   154 × 7 or, r2 = ∵ π =  7 22  or, r2 = 49 \ r = 7 Hence, the radius of sphere is 7 cm. 1 Q. 2. Find the volume of a sphere whose surface area is 154 cm2. R [Board Term II, 2015, NCERT] Sol.

= 38.808 ml



Volume =

4 3 pr 3 1

But surface area = 4pr2 = 154



r2 =



r =

\

Volume =

154 × 7 154 = 4 × 22 4p

22   ∵ p = 7   

7 2 539 or 179.67 cm3 (Approx) 1 3

[CBSE Marking Scheme, 2015] 1

( 1 cm3 = 1 ml)

Q. 4. The diameter of a football is five times the diameter of a cricket ball. Ratio of surface areas of football and cricket ball is .................. . A [Board Term II, 2013] Sol. Given, diameter of football =





4 π(2.1)3 3

38808 = cm 3 1000 \ Amount of water displaced = 38.808 cm3



(given)

4 22 × ( 2.1)3 cm3 = × 3 7

½

Short Answer Type Questions-I (2 marks each)

4 3 πr 3

4.2 r = = 2.1 cm 2

4 πr 2 25 r =  = 1 4 π( r ')2  r ' 

= 25 : 1

= Volume of solid spherical ball

½

5 × diameter of cricket ball

Q. 3. The total surface area of a solid hemisphere is 5940 cm2. Find the diameter of the hemisphere.  U [Board Term II, 2012] Sol. Let the radius of hemisphere be r ∴ 3pr2 = 5940 1 5940 × 7 or, r2 = = 630 3 × 22

or,

r =

630 or 3 70 cm.

158

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-IX

So,

d = 2r = 6 70 cm.

Hence, diameter of the hemisphere is 6 70 cm. Q. 4. A solid shotput is a metallic sphere of radius 4.9 cm. Find the volume of the shotput. U [Board Term II, 2012] OR A solid shotput is a metallic sphere of radius 4.9 cm. Find the mass of shotput, if density is 7.8 gm/cm3. [Board Term II, 2012, NCERT] Sol. Given: r = 4.9 cm 4 Volume V = πr3 3 4 22 × 4.9 × 4.9 × 4.9 = × 3 7 = 493 cm3 (approx.) Mass of the shot-put (V × d) = 3845.44 gm = 3.85 kg (Approx.)

1

or, r2 = 39.69 or, r = 6.3 m \ Volume of air inside it 2 = πr 3 3 =

1

2 22 × × 6.3 × 6.3 × 6.3 3 7

= 523.90 m3. 1 [CBSE Marking Scheme, 2015] Q. 3. The internal and external diameters of a hollow hemispherical vessel are 24 cm and 25 cm respectively. If the cost of painting 1 cm2 of the surface area is ` 0.05, find the total cost of painting the vessel all over. A [Board Term II, 2013] Sol. Internal radius (r) = 12 cm External radius (R) = 12.5 cm S.A. = 2πr2 + 2πR2 + π(R2 – r2)

1

Short Answer Type Questions-II (3 marks each) Q. 1. How many litres of milk can a hemispherical bowl of diameter 10.5 cm hold ? (Use p = 3.14) U [Board Term II, KVS, 2016]

Sol. Given, diameter of hemispherical bowl = 10.5 cm \ Radius of hemispherical bowl 10.5 = 5.25 cm 1 r = 2 Volume of hemispherical bowl =



1

2 3 pr 3

22 7

= (600.50 + 12.25) ×

1

= 1925.79 cm2

Cost of painting 1925.79 cm2 at the rate of ` 0.05/cm2 = 1925.79 × 0.05 1

= ` 96.29

Q. 4. A hemispherical bowl made of brass has inner diameter 0·105 m. Find the cost of tin-plating it on the inside at the rate of ` 16 per 100 cm2. U [Board Term II, 2017]

Sol. Diameter of hemispherical bowl = 0·105 m \

Its radius = 0·0525 m = 5·25 cm

CSA of hemisphere = 2pr2

= 2×





= 173·25 cm2





1

22 525 525 × × cm2 7 100 100



1

16 Total cost = 173·25 × = ` 27·72 100 1

Q. 5. Metallic spheres of radii 6 m, 8 m and 10 m, respectively are melted to form a single solid sphere. Find the radius of the resulting sphere.

U [Board Term II, 2013, 2015]

Sol. Cost of white washing hemispherical dome = ` 997.92 Cost of white washing per square meter = 400 paise = ` 4 \ CSA = 997.92 ÷ 4 = 249.48 m2 2pr2 = 249.48 1 22 2 2× × r = 249.48 7

2

= 2π (144 + 156.25) + π (12.5 + 12)( 12.5 – 12)

2 = × 3.14 × (5.25)3 3 = 302.91 = 303 cm3 1 \ Amount of milk that the hemispherical bowl can hold = 0.303 litres 1 Q. 2. A dome of a building is in the form of a hemisphere From inside, it was white washed at the cost of ` 997.92. If the cost of white washing is 400 paisa per square meter, find the volume of air inside the 22 dome.  Take π =   7

1

= 2π(r + R ) + π(R – r)(R + r) 2



U [Board Term II, 2017]

Sol. Let r1 = 6 m, r2 = 8 m, r3 = 10 m Let the radius of resulting sphere be R. Then volume of resulting sphere = sum of volume of small spheres i.e.,

4 3 4 4 4 pR = πr 3 + πr 3 + πr 3 1 3 3 1 3 2 3 3 4 3 3 3 = π( r1 + r2 + r3 ) 3



SURFACE AREAS AND VOLUMES



4 = π( 6 3 + 8 3 + 10 3 ) 1 3 4 = π( 216 + 512 + 1000 ) 3



4 = π(1728) 3



or,

=

1

Long Answer Type Questions (5 marks each) Q. 1. The water for a industry is stored in a hemispherical tank of internal diameter 14 m. The tank contains 40 kilolitres of water. Water is pumped into the tank to fill it to full capacity. Calculate the volume of water pumped into the tank.  U [Board Term II, 2016] Sol. Volume of hemispherical tank 2 3 = πr 3 2 22 × × 7 × 7 × 7 m3 3 7 2156 3 = m 3

1 = 64

or,

Volume of moon =

or,

Q. 3. A hemispherical dome, open at base is made from sheet of fibre. If the diameter of hemispherical 13 of sheet actually used was dome is 80 cm and 170 wasted in making the dome, then find the cost of dome at the rate of ` 35/100 cm2 A [Board Term II, 2014]

Q. 2. The diameter of the moon is approximately onefourth the diameter of earth. What fraction of volume of earth is the volume of moon ? A [KVS 2014, NCERT]

Sol. Let the diameter of earth be d. ∴ The radius of the earth will be, r1 =



Diameter of moon will be



d 2

1

d and radius of moon 4

(r2) =

d 8



1

r = 40 cm

= 2 ×



since,

C.S.A. =

22 × 40 × 40 7

70400 cm2 7

1

13 of sheet was wasted, 170

Area of sheet wasted

13 70400 × = 170 7 = ∴

Total area =

915200 cm2 1190

1

70400 915200 + 7 1190

= 10826.21 cm2

1

Cost of sheet per square metre = ` 3

4 d = π  3 8 1 4 × πd 3 × = 512 3

∴

C.S.A. of the dome = 2πr2



4 3 Volume of moon = πr2 3

4 Volume of earth = πr13 3

1

Sol. Given, Diameter = 80 cm

= 718.67 m3 = 718.67 kl. (Approx) 2½ Volume of water already present = 40 kl \ Volume to be pumped = 718.67 – 40 = 678.67 m3 (Approx) 2½ [CBSE Marking Scheme, 2016]



1 (Volume of earth) 64

Volume of earth 64 = Volume of moon 1

=



1

1 4 × πd 3 × Volume of moon 512 3 = 1 4 3 Volume of earth × πd 8 3

3

R = 1728 = 12 m

3

1 4 = × πd 3 8 3

R3 = 1728

4 d π  3 2

159

1

35 100

1

∴ Total cost of sheet

35 = × 10826.21 100 = ` 3789.17.

1

160

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-IX

Topic-2 Surface Area and Volume of Right Circular Cone Revision Notes Scan to know more about this topic

 Cone: Cone is a pyramid with a circular base.

Surface Area of Cone

 Right circular cone :

Slant height (l) =

Scan to know more about this topic

h2 + r 2

Area of curved surface = prl = pr h 2 + r 2

Total surface area = Area of curved surface + Area of base = prl + pr2 = pr(l + r) 1 Volume = pr2h 3

Volume of Cone

Example 1 There are two cones. The ratio of their radii are 4 : 1 Also, the slant height of the second cone is twice that of the former. Find the relationship between their curved surface area. Solution: Step I : First consider the unknown variables. Let r1 and l1 be the radius and slant height of first cone. Let r2 and l2 be the radius and slant height of second cone. Step II : Write the formula for curved surface area for both the cones. Curved surface area of first cone (CSA1) = pr1l1 and curved surface area of second cone (CSA2) = pr2l2 Step III : Use the given condition and simplify it. According to the question, r1 : r2 = 4 : 1

or,





\

r1 r2

= 4 and 1 l 1 l2 = 2l1 or, 1 = l2 2

r l  πr l CSA 1 = 1 1 =  1  1  π r l  r2   l2  CSA 2 2 2

41 =    12 2 = 1 \ CSA1 = 2CSA2 i.e., curved surface area of first cone is twice of the second cone.

SUBJECTIVE TYPE QUESTIONS Very Short Answer Type Questions (1 mark each) Q. 1. Find the volume of a right circular cone with radius 6 cm and height 7 cm. R [Board Term II, 2012, NCERT] 1 Sol. Volume of right circular cone = πr 2 h 3

=

1 22 × × ( 6 )2 × 7 3 7

1 22 × 36 × 7 = × 3 7 = 264 cm3 1 Q. 2. If the height and the radius of cone is tripled, then find ratio of volume of new cone and that of original. A [Board Term II, 2016]



Sol. Let h and r be the height and radius of original cone and let h' and r' be the height and radius of new cone. Given, h' = 3h and r' = 3r 1 Volume of original cone, V = πr 2 h 3

1

π( r ')2 h '

Volume of new cone 3 = 1 2 Volume of original cone πr h 3

( r ')2 h ' ( 3r )2 3 h = = 2 r 2h r h 27 = 1 

1

Hence, the ratio of new cone to the original cone is 27 : 1.

Short Answer Type Questions-I (2 marks each) Q. 1. The radius and slant height of a cone are in the ratio 4 : 7. If its curved surface area is 792 cm2, find its radius. U [Board Term II, 2017] Sol. Let the radius of a cone r = 4x and slant height l = 7x ∵ CSA = 792 cm2 1 ∵ prl = 792 22 or, × 4 x × 7 x = 792 7

x2 =

792 × 7 =9 22 × 4 × 7

or, x = 3 cm \ radius = 4 × 3 = 12 cm 1 Q. 2. How much ice-cream can be put into a cone with base radius 3·5 cm and height 12 cm ? U [Board Term II, 2017] Sol. r = 3·5 cm, h = 12 cm 1 \ Amount of ice-cream = πr 2 h 1 3 =

1 22 × × 3·5 × 3·5 × 12 3 7

= 154 cm3



\

l =

r 2 + h2

= (12 )2 + (16 )2 = 144 + 256 = 400 = 20 cm \ T.S.A. = pr(l + r) = 3.14 × 12 (20 + 12)

1

2 Volume of new cone, V' = 3 π( r ') h '



161

SURFACE AREAS AND VOLUMES

1

Short Answer Type Questions-II (3 marks each) Q. 1. The height of a cone is 16 cm and its base radius is 12 cm. Find the total surface area of the cone. (Use p = 3.14) U [NCERT] [Board Term II, KVS, 2016] Sol. Total Surface area of cone = pr(l + r) Given, r = 12 cm and h = 16 cm

1

= 3.14 × 12 × 32 = 1205.76 cm2. 2 Q. 2. The radius and height of a right circular cone are in the ratio 4 : 3 and its volume is 2156 cm3. Find the curved surface area of the cone. U [Board Term II, 2013] Sol. Let the radius of the cone = 4x and the height of the cone = 3x Volume of the cone = 2156 cm3 1 2 or, πr h = 2156 3 1 22 or, × × 4x × 4x × 3x = 2156 3 7

or,



or,



∴



22 × 16x3 = 2156 1 7 x3 =

7 ×7 ×7 7  =  2×2×2 2

3

7 = 3.5 cm 2 ∴ Radius of the cone r = 4x = 4 × 3.5 = 14 cm Height of the cone h = 3x = 3 × 3.5 = 10.5 cm Slant height of the cone x =

l=

h 2 + r 2 = 10.5 2 + 14 2



l = 17.5 cm C.S.A of the cone = πrl 1 22 = × 14 × 17.5 cm2 7 = 44 × 17.5 cm2 \ C.S.A of the cone = 770 cm2. 1 Q. 3. A joker's cap is in the form of right circular cone of base radius 7 cm and slant height 25 cm. Find the area of sheet required for 10 such caps. U [Board Term II, KVS 2014] OR A joker's cap is in the form of right circular cone of base radius 7 cm and height 24 cm. Find the area of the sheet required to make 10 such caps. [NCERT] Sol. Given, l = 25 cm, r = 7 cm l2 = h2 + r2 or, h2 = l2 – r2 or, h2 = (25)2 – (7)2 or, h = 24 cm Area required = C.S.A of cone 1 = prl 22 = × 7 × 25 7 = 550 cm2 1 Area required to make 10 such caps = 10 × 550 = 5500 cm2 1 = 0.55 m2.

162

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-IX

Q. 4. Bhavya has a piece of canvas whose area is 552 m2. She uses it to make a conical tent with a base radius of 7 m. Assuming that all the stitching margins and the wastage incurred while cutting amounts to approximately 2 m2. Find the volume of the tent that can be made with it. 22 A [Board Term II, 2014, 2012] (Take π = ) 7 Sol. Curved surface area of the tent = 552 – 2 = 550 m2 ½ Radius (r) = 7 m ∴ π × 7 × l = 550 ½ or, l = 25 m



h =

l2 − r2



∴

h =

252 − 7 2

Q. 2. A right triangle ABC with sides 5 cm, 12 cm and 13 cm is revolved about the side 5 cm. Find the volume of the solid so obtained. If, it is revolved about the side 12 cm, what would be the ratio of volumes of two solids obtained in two cases ? 

A [Board Term II, 2014, NCERT]

Sol. Case I : When revolved about the side 5 cm.

Here,



r = 12 cm, h = 5 cm 1 1 1 Volume = πr 2 h = π × (12 )2 × 5 . 3 3

13

5

= 24 m ½ 1 22 × 7 × 7 × 24 ½ Volume of the tent = × 3 7 = 1232 m3.

1

Long Answer Type Questions (5 marks each) Q. 1. A right angled DABC with sides 3 cm, 4 cm and 5 cm is revolved about the fixed side of 4 cm. Find the volume of the solid generated. Also, find the total surface area of the solid. A [Board Term II, 2015]

1

= 240π cm3

12



Case II : When revolved about 12 cm,



∴

r = 5 cm, h = 12 cm Volume =

1

1 2 πr h 3

12 13

Sol. r





5

1 = × 5 × 5 × 12 π 3 = 100 π cm3

rcone = 3 cm hcone = 4 cm lcone = 5 cm

1

4 cm 5 cm

r 3 cm



1

∴ Ratio of the two volume = 240π : 100π

= 12 : 5 1 Q. 3. What length of tarpaulin 3 m wide will be required to make conical tent of height 8 m and base radius 6 m ? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately 20 cm (Use p = 3.14). Sol. Conical tent : height = 8 m base radius = 6 m

1 Above given cone is formed with radius 3 cm, height 4 cm and slant height 5 cm when revolved about the fixed side of 4 cm. 1 V = πr 2 h 3

1 22 = . .( 3)( 3)( 4 ) 3 7 = 37.71 cm3 Total surface area = prl + pr2 = pr(l + r) 22 = × 3( 5 + 3) 7

1

= 75.43 cm2. 2 [CBSE Marking Scheme, 2015]



l2 = r2 + h2 l2 = 82 + 62 l =

64 + 36 = 10 m

C.S.A of tent = prl unit2 C.S.A. of Tent = 3.14 × 6 × 10 m2 = 188.4 m2



SURFACE AREAS AND VOLUMES

Area of Tarpaulin = C.S.A of tent width × length of tarpaulin = 188.4 m2 3 × length of tarpaulin = 188.4 m2 188.4 length of tarpaulin = = 62.8 m 3

Extra length required for stitching and wastage of cutting = 20 cm = 0.20 m \ Total length of tarpaulin = 62.8 + 0.2 = 63 m  5

OBJECTIVE TYPE QUESTIONS

A Multiple Choice Questions Q. 1. The radius of a hemispherical balloon increases from 6 cm to 12 cm as air is being pumped into it. The ratios of the surface areas of the balloon in the two cases is (A) 1 : 4 (B) 1:3 (C) 2 : 3 (D) 2:1  A [NCERT Exemp.] Ans. Option (A) is correct. Explanation :  Radius r = 6cm Surface area = 3pr2 = 3p62 = 3p × 6 × 6     Radius R = 12 cm Surface area = 3pR2 = 3p 122 = 3p × 12 × 12    Ratio = 3p × 6 × 6 : 3p × 12 × 12 = 1 : 4 Q. 2. The radius of a sphere is 2r, and then its volume will be 4 pr 3 (A) 3 (C) 

4 3 pr 3

Q. 3. The surface areas of two spheres are in the ratio 16 : 9. The ratio of their volumes is (A) 4 : 3 (B) 64 : 27 (C) 16 : 9 (D) 163 : 93 Ans. Option (B) is correct. Explanation : Surface Aarea of sphere = 4pr2



4 3 pr1 4 3 64 = 3 = Ratio = 3 4 3 3 27 pr 3 2

Ans. Option (C) is correct. Explanation : Volume of hemisphere = So,

2 3 pr 3

2 3 pr = 19404 3

19404 ´ 7 ´ 3 r3 = 2 ´ 22 r3 = 441 × 3 × 7 r = 21 Total surface Area of hemisphere = 3pr2 22 = 3 ´ ´ 21 ´ 21 7

= 4158 cm2

But radius r = 2r Therefore, volume of the sphere 4 32 3 = p(2r)3 = pr . 3 3

\

4 3 pr 3

= 64 : 27 Q. 4. The volume of a hemisphere is 19404 cm3. The total surface area of the hemisphere is (A) 16632 cm2 (B) 3696 cm2 2 (C) 4158 cm (D) 8316 cm2  A [NCERT Exemp.]

8 pr 3 32 3 (D) pr 3 3 A [NCERT Exemp.]

Volume of the sphere =

(1 mark each)

Volume of sphere =

\

(B) 4pr3

Ans. Option (D) is correct. Explanation :

163

Ratio = r1 = r2

4 pr12 4 pr22

=

16 9

16 4 = 9 3

Q. 5. The total surface area of a cone whose radius is 2r and slant height 2l is (A) 4pr (l + r) (B) pr (l + 4r) (C) pr (l + r) (D) 2prl  A [NCERT Exemp.] Sol. Option (A) is Correct : Explanation :     Radius = 2r Slant height = 2l     TSA = pr(l + r)         = p(2r) (2l + 2r)         = p(2r) × 2(l + r)         = 4pr(l + r) Q. 6. A cone is 8.4 cm high and the radius of its base is 2.1 cm. It is melted and recast into a sphere. The radius of the sphere is (A) 4.2 cm (B) 2.1 cm (C) 2.4 cm (D) 1.6 cm  Ans. Option (B) is Correct :

A [NCERT Exemp.]

164

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-IX

Explanation. Height h = 8.4 cm Radius r = 2.1 cm Volume of the sphere = Volume of cone 4 1 pR3 = pr2h 3 3   4 22 1 22 × × R3 = × × 2.1 × 2.1 × 8.4 3 7 7 3 4 1 × R3 = 2.1 × 2.1 × 8.4 × 3 3     R3 = (2.1)3   R = 2.1 cm. Q. 7. In a right circular cone, the cross section made by a plane parallel to the base is a (A) Sphere (B) Hemisphere (C) Circle (D) Semicircle Ans. Option (C) is correct. Q. 8. A triangle having sides equal to 7 cm, 24 cm and 25 cm forms a cone when revolved about 24 cm side. What is the volume of a cone formed? (A) 1225 cm3 (B) 1232 cm3 3 (C) 4000 cm (D) 3696 cm3  A [NCERT Exemp.] Ans. Option (B) is correct. Explanation : As shown when the cone is formed, its base radius is 14 cm and height is 24 cm.



Reason (R): The total surface area of a hemisphere is 3pr2. Ans. Option (A) is correct. Explanation : In case of Assertion (A): Total surface Area = 3pr2 22 = 3 ´ ´ 7 ´ 7 = 462 cm2 7 Cost of painting = `(465 × 5) = `2310 \ Assertion is true and Reason is also true as well correct explanation of A. Q. 2. Assertion (A): If the volumes of two spheres are in the ratio 27 : 8 then their surface areas are in the ratio 3 : 2. 4 Reason (R): Volume of sphere = pr 3 3 Surface area of a sphere = 4pr2 Ans. Option (D) is correct. Explanation : 4 Volume of sphere = pr 3 3 4 3 pr1 27 = Ratio of Volume = 3 4 3 8 pr 3 2



25

cm

= 24 cm

\

7 cm

= 1232 cm3

B Assertion & Reason

Directions: In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Mark the correct choice as: (A) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A). (B) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A). (C) Assertion (A) is true but reason (R) is false. (D) Assertion (A) is false but reason (R) is true. Q. 1. Assertion (A): A hemisphere of radius 7 cm is to be painted outside on the surface. The total cost of painting at `5 per cm2 is `2310.

r23

=

27 8

r1 3 = = 3 : 2 r2 2

Ratio of surface Area = 1 Volume = pr 2 h 3 1 22 = ´ ´ 7 ´ 7 ´ 24 3 7

r13

4 pr12 4 pr22

=

32 22

=

9 4

= 9 : 4 \ Assertion is false but Reason is true as volume of 4 sphere = pr 3 and Surface area = 4pr2 3 Q. 3. Assertion (A): If the height of cone is 24 cm and diameter of base is 14 cm, then the slant height of cone is 25 cm. Reason (R): If r be radius and h be the slant height of cone then the slant height = æç h 2 + r 2 ö÷ è ø Ans. Option (A) is correct. Explanation : In case of Assertion (A):

In cone h = 24 cm, r =

\

l2 =

h2 + r 2

=

24 2 + 7 2

=

625

\ Assertion is true.

l = 25 cm

14 = 7 cm 2



SURFACE AREAS AND VOLUMES

In case of Reason (R): h2 + r 2 It is true and correct explanation of Assertion. Hence, A and R are true and R is correct explanation of A.

l2 =

Q. 4. Assertion (A): The curved surface area of a cone of base radius 3 cm and height 4 cm is (15p) cm2. Reason (R): Volume of a cone = pr2h Ans. Option (C) is correct. Explanation : In case of Assertion (A): Curved surface area = prl

l2 =

h2 + r 2

=

4 2 + 32

=

25



165

l = 5 cm \ CSA = p × 3 × 5 = (15p) cm2 \ Assertion is true. In case of Reason (R) : 1 Volume = pr 2 h 3 \ Reason is False.

COMPETENCY BASED QUESTIONS A Case based MCQs

Read the following passage and answer any four questions of the following : I. Nikita has to make her project on 'Monument in India'. She decided to make her project on Gol Gumbaz monument. She already knows following things about it :  It is located in a small town in Northern Karnataka.  It reaches up to 51 meters in height while the giant dome has an external diameter of 44 meters, making it one of the largest domes ever built.  At each of the four corners of the cube is a dome shaped octagonal tower seven stories high with a staircase inside.



Help her in making project by answering the following questions :

Q. 1. What is the curved surface area of hemispherical dome ? (A) 908p m2 (B) 968p m2 2 (C) 340p m (D) 780p m2 Ans. Option (B) is correct. Explanation : Diameter = 44 m Radius = 22 m Curved surface area of hemispherical dome = 2pr2½ = 2p(22)2 = 968p m2½ Q. 2. What is the circumference of the base of the dome ? (A) 34p (B) 22p (C) 44p (D) 55p

Ans. Option (C) is correct. Explanation : Circumference of the base of the dome = 2pr = 2p(22) = 44p 1 Q. 3. Find the cost of painting the dome, given the cost of painting is ` 100 per cm2. (A) ` 980p (B) ` 9690p (C) ` 968000p (D) ` 9700p Ans. Option (C) is correct. Explanation  : Curved surface area of hemispherical dome = 968p m2 Cost of painting 100 cm2 = ` 10 C ost of painting 1 m2 = ` 1000 ½ Thus, cost of painting the dome = ` 1000 × 968p cm2 = ` 968000p½ II. A school organised an educational trip to Taj Mahal. Mathematics teacher of the school took her 9th standard students to it. The teacher had interest in history as well. She narrated the facts to Taj Mahal to students. Then the teacher said in this monument one can find combination of solid figures. There are 4 pillars which are cylindrical in shape. The Taj Mahal has a larger white dome surrounded by four smaller domes.

166

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-IX

Q. 1. How much cloth material will be required to cover 4 small domes each of radius 2 m? (A) 99 m2 (B) 100.57 m2 2 (C) 98.23 m (D) 75 m2 Ans. Option (B) is correct. Explanation: Required cloth material = 4 × Curved surface area of hemispherical domes ½ = 4 × 2pr2 22 = 4 × 2 × × 22 7



= 100.57 m2 ½ Q. 2. How much is the volume of a hemisphere if the radius of the base is 3.5 m? (A) 87.53 m3 (B) 90.22 m3 3 (C) 75.34 m (D) 89.83 m3 Ans. Option (D) is correct. Explanation: Volume of the hemisphere 2 = pr 3 3

=

2 22 × × ( 3.5)3 3 7

= 89.83 m3

1

Case based Subjective

following questions: I. Mr. Kumar a mathematics teacher brings some green clay in the classroom to teach the topic mensuration. First he forms a cone of radius 6 cm and height 3 cm with the clay. Then he mould that clay into a sphere similarly, he mould the sphere into hemisphere. Q. 1. When clay changes into on shape to other what remains same. Ans. Whenever a shape changes to other 'volume' remains the same. 1 Q. 2. What is the volume of hemisphere?

Q. 3. What is the radius of sphere. Ans.

Volume of cone = Volume of sphere



1 6´6´3´3 = r3 ´ 3 4



3 × 3 × 3 = r3



r = 3 cm 2

II. A farmer Rajesh grows a corn cob in his farm. Corn cob contains valuable Vitamin B, antioxidants, Carotenoids, lutein and Zeaxanthin which are useful for body growth.

20 cm

2.1 cm A corn cob (above figure). Shaped some what likes a cone, has the radius of its broadest end as 2.1 cm and length as 20 cm. Then, answer the following questions. Q. 1. Write the formula to find the curved surface area



Ans. Curved surface of cone is prl.

Read the following passage and answer the

2 Ans. Volume of hemisphere = pr 2 3



1 22 4 22 ´ ´ 6 ´ 6 ´ 3 = ´ ´ r3 3 7 3 7

of cone?

B Questions



\ Radius of sphere = 3 cm

22 = 8× ×4 7 22 = 32 × 7



1 2 4 pr h = pr 3 3 3

1

Q. 2. Slant height of the conical corn cob will be? h2 + r 2

Ans.

Slant height (l) =





l =

( 2.1)2 + ( 20 )2





=

4.41 + 400





=

404.41



l = 20.11 cm

1

Q. 3. Curved surface area of the corn cob in cm2 is? Ans.



CSA = prl 22 ´ 2.1 ´ 20.11 = 7

= 132.726 1

= 132.73 cm2

2



SURFACE AREAS AND VOLUMES

Artificial Intelligence DESCRIPTION

PARAMETERS Chapter Covered

Chapter 13: Surface Areas and Volumes

Name of the book

Mathematics, Class 9 NCERT

AI CONCEPTS INTEGRATED

Subject and Artificial Understanding the concept of surface area of solids: cube, cuboid, Intelligence Integrated right circular cylinder and right circular cone using AI. Objectives

● To understand the concept of surface area. ● To derive formulas for calculating surface area of given solids using their nets. ● To calculate the surface area of cube and cuboid using their formula. ● To calculate the curved surface area of cylinder and cone using their formula. ● To calculate the total surface area of cylinder and cone using their formula. ● To estimate the surface area of different prisms.

Time Required

4 sessions of 40 minutes each

Classroom Arrangement

Flexible

Material Required

Pen, Paper, White Board, Markers, Laptop, Internet Connection.

Pre-Preparation Activity

Students will be asked to recall all 2D shapes using Autodraw and their area and perimeter. Making of 3D shapes with the nets.

Previous knowledge

Questioning will be used to check students’ previous knowledge in the form of a quiz.

Methodology

Activity-1: Identifying Solids from Their Nets and Finding Their Surface Area: Students will be divided into groups of 4 or 5 and each student will make cube, cuboid, cylinder and cone using paper folding and cutting. They will then open and see the nets of each and find the area of the all 2D shapes obtained. Adding the area of all shapes of a given net of solid, they will arrive at the formula of surface area of that solid under the guidance of their teacher. Exploring Nets of Solids:

Autodraw

Cube Cone Cylinder Activity-2: Students will be working in pairs with their partners in computer lab and individually at their home to understand the formulae of surface area of solids by using Geogebra tool wherein they can change measurements of the dimensions and explore the corresponding change in their SA. This will help them to understand change in SA in problems related to increase and decrease of dimensions. Activity-3: Model making: Students will be divided into groups of 4 or 5 and they have to create a model which includes all solids discussed USING 3D MODELING APP. It can be a classroom scene, factory model, colony/society model, temple, galaxy model....etc. and will have to explain the need and use of the solids used to create it.

167

168

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-IX

Activity-4: Students will be using the formula derived for the SA of solids to find the surface area of objects in the real life problems of NCERT exercises and the assignments given. Activity-5: Research on shapes of beaker, test Tube, conical flask & gas cylinders: Students will do research in the use of various laboratory equipment and their shapes. They will identify the combination of solids used and will research on why beakers and test tubes are generally cylindrical but flasks are conical at the bottom and has cylindrical neck? Also, students will also do research on why gas cylinders and boilers are cylindrical in shape? Discussion on the Text

Open discussion on all new terms related to Surface Area: Lateral/ Curved Surface Area, Total Surface Area, Cuboid, Cube, right circular cylinder, Right circular Cone, Right circular Cone, Slant Height.

Learning Outcomes

Students will be able to ● Calculate the surface area of given solids. ● Solve real life word problems involving finding surface area of the solids done. ● Estimate the change in surface area due to change in their dimensions.

Self-Evaluation and Follow Teacher will observe students work and give individual feedback. up Also, models made and research done will be assessed. Peer assessment: Asking questions to each other in pairs and peer tutoring wherever required. Flip teaching Worksheets/Assignments







SELF ASSESSMENT PAPER - 05 Time: 1 hour

MM: 30

UNIT-V I. Multiple Choice Questions 

[1 × 6 = 6]

1. Volume of cone is given by the formula 4 1 2 (A) pr2h (B) πr 2 h (C) (D) 2prh πr h 3 3 2. Complete the curved surface area of a hemisphere whose diameter is 42 cm. (A) 308 cm2 (B) 1232 cm2 (C) 716 cm2 (D) 2772 cm2 3. Find the volume of a cone whose height is 3 cm and radius is 4 cm. (A) 52.6 cm3 (B) 50.3 cm3 (C) 51.2 cm2 (D) 50 cm3 4. Find the volume of a shot put where mass is 3.85 kg and density is 7.8 g/cm2. (A) 408 cm2 (B) 490 cm3 (C) 493 cm3 (D) 500 cm3 II. Assertion and Reason Based MCQs Directions: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as. (A) Both A and R are true and R is the correct explanation of A. (B) Both A and R are true but R is NOT the correct explanation of A. (C) A is true but R is false. (D) A is false and R is True. 1. Assertion (A): The height of a cone is 15 cm. If its volume is 500p cm3, then the radius of its base is 10 cm. 2 Reason (R): Volume of hemisphere is pr 3 3 2. Assertion (A): Radius 6 cm, slant height 7 cm then the curved surface of right circular cone is 132 cm2 Reason (R): Curved surface of cone is prl. III. Very Short Answer Type questions

[1 × 5 = 5]

1. If area of sphere is equal to volume of the sphere. Then find the diameter of the sphere. 2. Find the amount of water displaced by a solid spherical ball of radius 1.4 cm. 3. Find the volume of right circular cone with radius 7 cm and height 6 cm. 4. How many feed balls each of radius 1 cm can be made from a sphere where radius is 8 cm. 5. The volume of a sphere is 113

1 cm3. Find its diameter. 7

IV. Short Answer Type questions-I

[2 × 2 = 4]

1. A spherical ball has 21 cm as diameter, calculate its surface area. 2. Find the area of curved surface of the cone whose slant height is 8 cm and base diameter is 12 cm. V. Short Answer Type questions -II

[3 × 2 = 6]

1. Find the height of a cone whose base radius is 5 cm and volume 50 pcm . 3

2. The volume of a sphere is 4851 cm3. find its surface Area. VI. Long Answer Type questions

[5 × 1 = 5]

1. The radius of a spherical balloon increase from 8 cm to 16 cm as air is pumped into it find the ratio of surface areas of the balloon in two cases. 2. The cost of polishing the dome at the cost of `270 cm2 of radius is 4.2 cm is.

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Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-IX

VI. Case Study Based Questions

[1 × 4 = 4]

A school took for an education trip to see Museum. The model of Gol gumbaz movement was displayed. The leader told that the movement is a large dome and the dome is the shape of half of a sphere. 1. How much cloth will be required to cover 2 small domes of radius 4.2 m each? (A) 52.08 cm2

(B) 52.8 m2

(C) 52 m2

(D) None of these

2. What is the formula of volume of hemispherical dome? (A)

2 3 πr 3

(B) 2pr2h

(C)

4 3 πr 3

(D) pr2h

3. What is the volume of hemispherical dome of radius 7 m? (A) 718.66 cm3

(B) 152.8 m3

(C) 718.66 m3

(D) 56 cm3

4. What is the surface area of hemispherical dome of radius 14 m? (A) 1234 m2

(B) 1312 m2

(C) 1232 m2

(D) 1223 m2

qq

UNIT-VI

Statistics & Probability

CHAPTER

12

Syllabus

Study Time: Maximum time: 2:30 Hrs Maximum questions: 21

STATISTICS

Bar graphs, histograms (with varying base lengths), frequency polygons.

Revision Notes  Graphical Representation of Data can be represented graphically in following ways : (a) Bar Graph (b) Histogram (c) Frequency polygon.  Bar Graph: A bar graph is a pictorial representation of data in which rectangular bars of uniform width are drawn with equal spacing between them on one axis, usually the x-axis. The value of the variable is shown on the other axis that is the y-axis. Following Bar graph depicts number of books sold per month. Bar Graph

500

466

Scan to know more about this topic

415

402 400

371 302

293

300

305

305

310

June

July

Aug

Introduction to Statistics

200

200

100

0

Jan

Feb

Mar

Apr

May

Sep

Oct

Month vs Number of books sold

 Bar charts are used for comparing two or more values.  Histogram: A histogram is one of the most commonly used graphs. A histogram is a vertical bar-graph with no spacing between the bars. The histogram is constructed by the following steps : 1. The values of the observations are taken on the x-axis with the class-limits clearly marked. 2. The frequencies are taken along the y-axis.

172

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-IX



STATISTICS

173

 A histogram is a set of adjacent rectangles whose areas are proportional to the frequencies of a given continuous frequency distribution. The height of rectangles corresponds to the numerical value of the data and base corresponds to a particular class.  The histogram is drawn only for exclusive/continuous frequency distributions.  If classes are not of equal width, then the height of the rectangle is calculated by the ratio of the frequency of that class, to the width of that class.  A histogram is different from a bar chart, as in the former case it is the area of the bar that denotes the value, not the height.  When the scale on the x-axis starts at a higher value and not from the origin, a kink is indicated near the origin to signify that the graph is drawn to a scale beginning at a higher value and not at the origin.

 ‘Kinks’ are a tool used to express areas in a graph. In this case, the kink tells us that there is no observation which takes the value less than 200.  Frequency Polygon: The frequency polygon of a frequency distribution is a line-graph drawn by plotting the class marks on the x-axis against the frequencies on the y-axis.  In case of grouped data, where the classes are of equal width, the frequency polygon is obtained by joining the mid-points of the top edges of the rectangles in the histogram. Two extra lines are drawn by introducing two extra classes (or values).  One class is introduced before the first class and the other is introduced after the last class. These classes have zero frequencies.  Frequency polygons are used for understanding the shape of distributions.  If both a histogram and a frequency polygon are to be drawn on the same graph, then first draw the histogram and then join the mid-points of the tops of the adjacent rectangles in the histogram with line-segments to get the frequency polygon.  The cumulative frequency of a class-interval is the sum of frequencies of that class and the classes which preceed (come before) it. Range  Class size = Number of classes

174

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-IX



Class size = Upper limit – Lower Limit D 20 18 16 14 12 C 10 E 8 B 6

F

4 G 2 A 0

H

135 145 155 165 175 185 195 205 215

Example 1 The following data on the number of girls (to the nearest ten) per thousand boys in different sections of Indian Society is given below : Number of girls per Sections of Society thousand boys Schedule Caste (SC) 940

Rural 930 Urban 910 (i) Represent the information above by a bar graph. (ii) In the classroom discuss what conclusions can be arrived at from the graph?

(iii) What step should be taken to improve the situation ? Solution: Step I : Choose the appropriate data for horizontal axis (i.e., x-axis) and vertical axis (i.e., y-axis). Here, we represent the sections on horizontal axis choosing any scale, since width of bar is not important but for clarity, we take equal widths for all bars and maintain equal gap between them. Let one section be represented by one unit. We represent the number of girls per thousand boys on vertical axis. Here, we can choose the scale as 1 unit = 10. Step II : Draw the graph as per given information. (i) Now, the graph is as shown below according to the given data.

Step III : Draw the conclusion for part (ii). (ii) From the graph, we observe that in Scheduled Tribe (ST), there is maximum number of girls per thousand boys among different sections of Indian Society, i.e., 970 whereas there are minimum

number of girls per thousand boys in urban area. Step IV : Suggest one positive step to improve the situation. (iii) Prenatal sex determination should be strictly banned in urban areas.

Schedule Tribe (ST)

970

Non–SC/ST

920

Backward districts

950

Non-backward districts

920



175

STATISTICS

SUBJECTIVE TYPE QUESTIONS Very Short Answer Type Questions (1 mark each) Q. 1. In a histogram the class intervals or the groups are taken along which axis? Sol. Class intervals are taken along x-axis.



Construct a grouped frequency distribution table with classes 84-88, 88-92 etc. R [Board Term II, 2012] Sol. Class interval

Tally Marks

Frequency

||

2 4

1

84 - 88

Q. 2. What are the graphical representation of statistical data.

88 - 92

||||

Sol. Histogram, Bar graph and Frequency Polygons are all graphical representation of statistical data. 1

92 - 96

|||| |||| |||

13

96 - 100

|||| ||||

11

Q. 3. What are the facts or information collected with a definite purpose called? R [Board Term II, 2012] Sol. Data

|



½×4=2

Short Answer Type Questions-II (3 marks each)

1

Q. 4. Two consecutive class marks of a distribution are 52 and 57, Find the class size. Q. 1. A company manufactures car tyres of a particular R [Board Term II, 2012] type. The lives (in years) of 40 such tyres are as Sol. 57 – 52 = 5

1

Short Answer Type Questions-I (2 marks each) Q. 1. Read the bar graph. Find the percentage of excess expenditure on wheat than pulses and ghee taken together. R y

Percentage Expenditure



follows : 2.6, 3.0, 3.7, 3.2, 2.2, 4.1, 3.5, 4.5, 3.5, 2.3, 3.2, 3.4, 3.8, 3.2, 4.6, 3.7, 2.5, 4.4, 3.4, 3.3, 2.9, 3.0, 4.3, 2.8, 3.5, 3.2, 3.9, 3.2, 3.2, 3.1, 3.7, 3.4, 4.6, 3.8, 3.2, 2.6, 2.5, 4.2, 2.9, 3.6. Construct a continuous grouped frequency distribution for the above data of equal class size and with first class interval as 2-2.5, (2.5 is not included)

50

R [Board Term II, 2016]

Sol.

40

Class Intervals 2.0 - 2.5

35%

30 20%

20

15%

20%

10%

10 0

||

Number of Tyres 2

2.5 - 3.0

|||| ||

7

3.0 - 3.5

|||| |||| ||||

14

3.5 - 4.0

|||| ||||

10

4.0 - 4.5 4.5 - 5.0 Total

|||| |||

Tally Marks

x Rice

Wheat Pulses

Ghee

Others

Food Items

Sol. Expenditure on pulses and ghee = 10% + 20% = 30% 1 Expenditure on wheat = 35% ½ \ Excess expenditure on wheat = 35% – 30% = 5% ½ Q. 2. The class marks of a distribution are 37, 42, 47, 52, 57. Determine the class size and the class limits of one last class mark. U [Board Term II, 2012] Sol. Class size = 42 – 37 = 5 1 5 Lower limit of last class mark = 57 − = 54.5 ½ 2 5 Upper limit of last class mark = 57 + = 59.5 ½ 2 Q. 3. The relative humidity (in %) of a certain city of month of 30 days was as follows : 98

98

99

90

86

95

92

96

94

95

89

92

97

93

92

95

97

93

95

97

96

92

84

90

95

98

97

96

92

89



4 3 40 [CBSE Marking Scheme, 2016] 3

Q. 2. The blood groups of 30 students of class IX are recorded as follows : A, B, O, O, AB, O, A, O, B, A, O, B, A, O, O, A, AB, O, A, A, O, O, AB, B, A, O, B, A, B, O. (i) Represent this data in the form of a frequency distribution table. (ii) Which is the most common and which is the rarest blood group among these students ?

A [NCERT][Board Term II, 2012]

176

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-IX

Sol. (i) Frequency Distribution Table : Blood group

Tally Marks

No. of students (frequency)

A

|||| ||||

9

B

|||| |

6

O |||| |||| || AB

Q. 5. Two coins were tossed 20 times simultaneously. Each time the number of "Heads" occurring was noted down as follows : 0, 1, 1, 2, 0, 1, 2, 0, 0, 1, 2, 2, 0, 2, 1, 0, 1, 1, 0, 2. Prepare a frequency distribution table for the data. U [Board Term II, 2017] Sol.

12 3

|||

Total

2

30

(ii) Blood group 'O' is most common as it has highest frequency i.e., 12. Blood group AB is rarest. ½+½



Q. 3. Represent the following frequency distribution by means of a histogram. R Marks

10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 60 - 70

Number of Students

7

11

9

13

16

Observation (Number of heads)

Tally marks

Frequency

0

|||| ||

7

1

|||| ||

7

2

|||| |

6

Total

20 3

4

Long Answer Type Questions (5 marks each)

Sol.

Q. 1. The following table gives the life time of 400 neon lamps : Life Time (in hours)

300 - 400 14 400 - 500 56 500 - 600 60 600 - 700 86 700 - 800 74 800 - 900 62 900 - 1000 48 (i) Represent the given information with help of histogram. (ii) How many lamps have life time of more than 700 U [NCERT][Board Term II, KVS, 2016] hours ? Sol. (i)

3 Q. 4. A family with a monthly income of ` 20,000 had planned the following expenditure per month under various heads. R [Board Term II, 2012] Expenditure (in thousands rupees)

Heads Grocery Rent Education Medicine Fuel Entertainment Miscellaneous

04 05 05 02 02 01 01

Draw a bar graph for the data above.



Miscellaneous

Entertainment

Fuel

Medicine

Education

Expenditure (in thousand rupees)

7 6 5 4 3 2 1 0

Rent

y

Sol.

Grocery



Number of Lamps

x Heads

3



3½ (ii) Lamps having life time for more than 700 hours are = 74 + 62 + 48 = 184 1½



STATISTICS

177

Q. 2. Draw a histogram to represent the following grouped frequency. Age 5-9 (in yrs) No. of persons

10-14

15-19

20-24

25-29

3034

3539

28

32

48

50

35

12

10

Also draw frequency polygon. U [Board Term-II, 2016]

Sol. Age (in years) 5-9 10-14 15-19 20-24 25-29 30-34 35-39

Number of Persons

Continuous Age (in years)

Class Marks

10 28 32 48 50 35 12

4.5-9.5 9.5-14.5 14.5-19.5 19.5-24.5 24.5-29.5 29.5-34.5 34.5-39.5

7 12 17 22 27 32 37

3 [CBSE Marking Scheme, 2016]

2

Q. 3. Draw a histogram to represent the following frequency distribution.

[KVS 2019]

Marks

0 – 20

20 – 30

30 – 40

40 – 50

50 – 60

60 – 70

70 – 100

Number of students

7

10

10

20

20

15

8

Sol.



Frequency density

40 - 50

20

20 × 10 = 20 10

7

7 × 10 = 3.5 20

50 - 60

20

20 × 10 = 20 10

20 - 30

10

10 × 10 = 10 10

60 - 70

15

15 × 10 = 15 10

30 - 40

10

10 × 10 = 10 10

70 - 100

8

8 × 10 = 2.66 = 2.6 30

Marks

Number of Students

0 - 20

 Marks →

5

178

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-IX

OBJECTIVE TYPE QUESTION A Multiple Choice Question

Q. 1. For drawing a frequency polygon of a continuous frequency distribution, we plot the points whose ordinates are the frequencies of the respective classes and abscissae are

(A) Upper limits of the classes (B) Continuous classes (C) Discontinuous classes (D) Maximum frequency Ans. Option (B) is correct.

B Assertion & Reason

(A) upper limits of the classes. (B) lower limits of the classes.



(C) class marks of the classes. (D) upper limits of preceding classes. 

[NCERT Exemp.]

Ans. Option (C) is correct. Explanation: We have to take only class marks while drawing the polygons. Q. 2. In a bar graph, the width of bars (A) Are proportional to the corresponding frequencies (B) Have no significance (C) Are proportional to the space between two consecutive bars. (D) Are proportional to the corresponding heights. Ans. Option (B) is correct. Explanation: Width of bar's are equal only from clarity point of view otherwise they have no significance. Q. 3. In a histogram, which of the following is proportional to the frequency of the corresponding class? (A) Area of the rectangle (B) Length of the rectangle

(1 mark each)

Directions: In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Mark the correct choice as:

(A) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A). (B) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A). (C) Assertion (A) is true but reason (R) is false. (D) Assertion (A) is false but reason (R) is true. Q. 1. Assertion (A): According to statistics more female children are born each year than male children in India. Reason (R): In India the death rate of a male child is higher than that of the female child. Ans. Option (C) is correct. Q. 2. Assertion (A): The class interval needs to be continuous while drawing a Histogram. Reason (R): Histogram is a rectangular diagram using frequency distributions which are joined to one another. Ans. Option (B) is correct.

(C) Width of the rectangle

Explanation : In case of Assertion (A):

(D) Perimeter of the rectangle

Histograms depicts continuous data which is taken in the form of class interval.

Ans. Option (A) is correct. Q. 4. Histogram graphically represent the grouped frequency distribution with

\ Assertion is true, but Reason is not correct explanation of Assertion.

COMPETENCY BASED QUESTIONS

(4 marks each)

A Case based MCQs

Read the following passage and answer any four questions of the following : I. Child labour refers to any work or activity that deprives children of their childhood. It is a violation of children's rights. This can harm them mentally or physically. It also exposes them to hazardous situations or stop them from going to school. Naman got data on number of child labours (in million) in different country that is given below.



STATISTICS

Q. 1. Which country has the highest child labour ? (A) Peru (B) United States

1200,0

Q. 2. Which country has the lowest child labour ? (A) Brazil (B) Kenya (C) United States (D) India Ans. Option (C) is correct.

10300

1

1

Q. 3. Name the countries having more than 4 million child labour. (A) India and Bangladesh (B) India and Peru (C) Bangladesh and Brazil (D) Bangladesh and Kenya Ans. Option (A) is correct. 1 Q. 4. Which country has more than 3 million but less than 6 million child labour ? (A) India (B) Peru (C) Egypt (D) Kenya Ans. Option (B) is correct. 1 Q. 5. Name the country which has 1 million child labour. (A) Mexico (B) Kenya (C) Peru (D) India Ans. Option (B) is correct. 1 II. Ladli Scheme was launched by the Delhi Government in the year 2008. This scheme helps to make women strong and will empower a girl child. This scheme was started in 2008.

Read the above bar graph and answer the following questions :

1000,0

Rs. (in Million)

(C) Egypt (D) India Ans. Option (D) is correct.

179

9060

9160

800,0 600,0 400,0 200,0 5, 4 0

2007-08

2008-09 2009-10 Estimated Budget

2010-11

Q. 1. Which statement is incorrect about bar graph? (A) A bar graph is a pictorial representation of data. (B) A bar graph can be vertical or horizontal. (C) Heights of the bars depend on the values of the variable. (D) There are no gaps in between consecutive rectangles. Ans. Option (D) is correct. Explanation: There are no gaps in between consecutive rectangles. Given statement is incorrect as there is gap in between consecutive rectangles in a bar graph. 1 Q. 2. In which year the budget was minimum? (A) 2008 - 09 (B) 2007 - 08 (C) 2010 - 11 (D) 2009 - 10 Ans. Option (B) is correct. 1 Q. 3. In which year the budget was maximum? (A) 2007 - 08 (B) 2008 - 09 (C) 2009 - 10 (D) 2010 - 11 Ans. Option (D) is correct.

1

Q. 4. What was the budget in 2010 - 11? (A) 9060 Million (B) 10300 Million (C) 9160 Million (D) 54 Million Ans. Option (B) is correct.

1

180

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-IX

Q. 5. What was the difference in budget in the year 2008 - 09 and 2009 - 10 ? (A) 100 Million (B) 1000 Million (C) 50 Million (D) 150 Million Ans. Option (A) is correct. 1

Case based Subjective

B Questions



Read the following passage and answer the following questions: I. India's national animal Tiger (which has also been under the radar of the government as its population declined in the country) has witnessed an increase in its population. A survey was done by the Ministry of Environment, Forests and Climate change, according to which tiger population is on rise at the rate of 6 per cent every year from 2006 and 2018. Year

1900

1960

2002

2007

No. of Tigers

40,000

18,000

3642

1411

Pranav has found a report on status of tigers in India in last 100 years. He has some queries as follow : Q. 1. In which year, number of tigers were minimum? Ans. According to graph in 2007 number of tigers were minimum. 1

Q. 2. What does this bar graph show ? Ans. This bar graph shows decline in number of tigers. 1 Q. 3. How much decrease in number of tigers was there between 1900 and 1960 ? Ans. Number of tigers in 1900 = 40,000 Number of tigers in 1960 = 18000 Decrease in number of tigers = 22000 2 II. Electricity energy consumption is the form of energy consumption that uses electric energy. Global electricity consumption continues to increase faster than world world population, leading to an increase in the average amount of electricity consumed per person (per capita electricity consumption). A survey is conducted for 56 families of a colony A. The following bar graph gives the weekly consumption of electricity of these families.

Q. 1. How many families weekly consumption is 50-60 units? Ans. 0 families have weekly consumption 50-60 (units). 1 Q. 2. What is the weekly consumption of maximum number of families? Ans. From bar graph 18 families are maximum number of families having 20-30 (units) of weekly consumption. 1 Q. 3. What is the difference in the number of families which consumes (30-40) and (40-50) units? Also tell whether families are increasing or decreasing? Ans. As we can see from graph No. of families consuming (30-40) units = 6 No. of families consuming (40-50) units = 4 \ Difference = 6 – 4 = 2 families As we can see number of families are decreasing.

Artificial Intelligence PARAMETERS

DESCRIPTION

Chapter Covered

Chapter 14: Statistics - Understanding Frequency Table

Name of the book

Mathematics, Class 9 NCERT

Subject and Artificial Analyzing the frequency tables using AI Tools. (Statistical Data). Intelligence Integrated

AI CONCEPTS INTEGRATED



STATISTICS

Learning Objectives

Students will able to ● Prepare and use the frequency tables. ● Analyze the frequency tables.

Time Required

2 periods of 40 minutes each

Classroom Arrangement

Seating arrangement -In pairs for both the sessions.

Material Required

Pen, paper, Laptops/ desktops/ Tabs and Internet connection.

Pre-Preparation Activity

● Students will be asked to look at some frequency tables in the handouts. ● Students will be asked a few questions based on the frequency tables. Like: What is the info given in the table? How many….. What is the Highest/Lowest?

Previous knowledge

Preparing a frequency table and reading it.

Methodology

After a preliminary round of pre -knowledge testing, students will be guided to play the Markov Data Game. Students, Play Rock, Paper, Scissors with the evil Dr. Markov to save the dog. A video will be shown for the guidelines.

181

Introduction to Markov - Part I Students will play the game:

codap.concord While playing the games the students will be asked to make a list of their moves and the Markov’s moves. After round 1, they will be asked to prepare the frequency tables for their moves and try to analyze the same. Video 2 will be shown:

Introduction to Markov - Part II, Advanced Gameplay Now the students will be in a better position to win the game and save the dog. Learning Outcomes

Students prepared the tables with the data collected while playing the game. They analyzed the data in tables and won the game.

Follow up Activities

The following worksheet will be shared. Markov Student Worksheet ● Students may be asked to search and share more such games either online or offline. ● Same tables can be later used to draw bar graphs. ● We can collect data of win or loss from all pairs and it can be used to: - Create discrete or grouped tables. - Calculate the measures of central tendency: mean/median or mode.

Reflections

● Markov uses a different strategy on each new level, but his strategy remains consistent throughout each game played on a single level. ● There is increasing variability in Markov’s moves as the student progresses through the levels, which makes it more challenging to win. ● Discussion of the students’ responses in the worksheet.

qqq

182

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-IX

PARAMETERS

DESCRIPTION

AI CONCEPTS INTEGRATED

Chapter Covered

Chapter 14: Statistics - Understanding the Concept of Data Handling

Name of the book Subject and Artificial Intelligence Integrated Learning Objectives

Mathematics, Class 9 NCERT Understanding the concept of data handling using AI Experiential applications To understand the concept of representing data in the form of various graphs. To understand the process of Data Handling: ● Sources of data ● Data Acquisition ● Exploration To understand the process of Data Handling in real life situations using AI project cycle process of Data Acquisition. 2 periods of 40 minutes each Flexible Pen, paper, black board, chalk, Laptop/Desktop, and internet connections. The students are divided into three groups and asked to collect information on AQI. (pollution level) The students should know the terms like average, range, median, and mode. Divide the class into two teams Activity 1- Different types of graphs Ask the students to collect data in different types of graphs. Activity 2- Ask the students to collect and compare data on Fine Particulate Matter PM2.5 across SAFAR cities Ask the students to observe and write the activities that pollute the environment. Data Acquisition Activity 3- Ask the students to collect the data on temperature and humidity and deduce the correlation between them. Activity 4- To make a chart on the weather forecast using http://safar.tropmet.res.in/ The students would be able to understand that using AI and past data, the weather forecasts are done. Ask the students to draw and compare AQI of various cities of the world. Activity 5: To check which graph is suited for such type of data https://datavizcatalogue.com/ Data Exploration The students will ● Understand the importance of data collection in real life ● Appreciate the importance of data analysis and forecasting using AI. ● Develop skills of factual representation of data using AI ● Be able to interpret various graphs Ask students to make a presentation Stating facts and using pictorial representation of Data collected and analyzed. Discussion with Students: How do you like the site http://safar.tropmet.res.in/? Try to analyze the data of past 10 years of a place. Observe the climate change. How AI and machine learning are transforming weather forecasting. Do you know of any other tool/app that can help you to access data?

Time Required Classroom Arrangement Material Required Pre-Preparation Activities Previous Knowledge Methodology

Learning Outcomes

Follow up Activities Reflections







SELF ASSESSMENT PAPER - 06 Time: 1 hour

MM: 30

UNIT-VI I. Multiple Choice Questions

[1 × 6 = 6]

1. The height of the rectangle in a histogram is (A) width of the class

(B) upper limit of the class

(C) lower limit of the class

(D) frequency of the class

2. Graphical representation of numerical data by a number of base of uniform width with equal spacing between then is (A) Bar graph

(B) Histogram

(C) Frequency polygon

(D) pie-chart

3. To draw a histogram to represent frequency for the class 24-45. The adjusted frequency is ................... if frequency of class 25-45 is 8. Class internal

5-10

10-15

15-25

25-45

Frequency

6

12

10

8

(A) 6

(B) 5

(C) 3

(D) 2

4. A vertical graph with no spacing between then is (A) Bar graph

(B) frequency polygon (C) Histogram

(D) None of these

II. Assertion and Reason Based MCQs

[1×6 = 6]

Directions: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as. (A) Both A and R are true and R is the correct explanation of A. (B) Both A and R are true but R is NOT the correct explanation of A. (C) A is true but R is false. (D) A is false and R is True. 1. Assertion (A): A histogram is a two dimensional graph. Reason (R): Histogram is a two dimensional graph of frequency distribution that emphasises both lengths and widths of the rectangle. 2. Assertion (A): A frequency polygon can not be drawn with the help of a Histogram. Reason (R): A frequency histogram is a type of bar graph that shows the frequency or number of times an outcome occurs. III. Very Short Answer Type questions [1 × 2 = 2] 1. Is it correct to say that in a histogram, the area of each rectangle is proportional to the class size of the corresponding class interval? If not, correct the statement. 2. The class marks of a continuous distribution are : 1.04, 1.14, 1.24, 1.34, 1.44, 1.54 and 1.64. Is it correct to say that the last interval will be 1.55–1.73? Justify your answer. IV. Short Answer Type questions–I

[2 × 3 = 6]

1. Find the given frequency table construct the bar graph: Marks

0-20

20-40

40-60

60-100

No. of Students

10

15

20

25

Represent the information above by a bar graph. 2. In the given Figure, there is a histogram depicting daily wages of workers in a factory. Construct the frequency distribution table.

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-IX

No of Workers

184

50 40 30 20 10



50 100 150 200 250 300 350 400 Wage in `

3. Draw a frequency polygon graph for the given data : Weight (kg)

60

61

62

63

64

65

No. of person

15

18

14

16

15

12

V. Long Answer Type questions

[5 × 1 = 5]

1. Draw a histogram for the following data : Age (in years)

0-8

8-16

16-24

24-32

32-40

40-48

9

13

15

20

10

5

Number of person VI. Case Study Based Questions

[1 × 4 = 4]

Locate the bars as shown given below :

see the above bar graph and given the answer of following questions : 1. How many students use bus as the mode of transports ? (A) 25

(B) 27

(C) 20

(D) 44

2. How many total students use car and scooter as the mode of transport ? (A) 15

(B) 10

(C) 20

(D) 30

3. How many students use rickshaw as the mode of transport ? (A) 10

(B) 12

(C) 8

(D) 15

4. How many students use Bicycle as made of transport ? (A) 15

(B) 20

(C) 30

(D) 25

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