Comprehensive CBSE Question Bank in Mathematics IX (Term-II)

CBSE II Question Bank in Mathematics CLASS 9 Features Short Answer Type Questions Long Answer Type Questions Strictly Based on the Latest CBSE Term-wise Syllabus Case Study Based MCQs Chapter-wise Important Results and Formulae Very Short Answer Type Questions



Comprehensive CBSE Question Bank in Mathematics Term–II (FOR CLASS IX) (According to the Latest CBSE Examination Pattern) By Dr. V.K. SONI M.Sc. (Maths), M.Ed., Ph.D Senior Lecturer Dept. of Mathematics Lovely Professional University, Phagwara Formerly, Head Dept. of Mathematics Army School, Amritsar Cantt. Punjab   LAXMI PUBLICATIONS (P) LTD (An ISO 9001:2015 Company) BENGALURU • CHENNAI • GUWAHATI • HYDERABAD • JALANDHAR KOCHI • KOLKATA • LUCKNOW • MUMBAI • RANCHI NEW DELHI

Comprehensive CBSE QUESTION BANK IN MATHEMATICS–IX (TERM–II) Copyright © by Laxmi Publications Pvt., Ltd. All rights reserved including those of translation into other languages. In accordance with the Copyright (Amendment) Act, 2012, no part of this publication may be reproduced, stored in a retrieval system, translated into any other language or transmitted in any form or by any means, electronic, mechanical, photocopying, recording or otherwise. Any such act or scanning, uploading, and or electronic sharing of any part of this book without the permission of the publisher constitutes unlawful piracy and theft of the copyright holder’s intellectual property. If you would like to use material from the book (other than for review purposes), prior written permission must be obtained from the publishers. Printed and bound in India Typeset at : Goswami Associates, Delhi New Edition ISBN : 978-93-93738-03-5 Limits of Liability/Disclaimer of Warranty: The publisher and the author make no representation or warranties with respect to the accuracy or completeness of the contents of this work and specifically disclaim all warranties. The advice, strategies, and activities contained herein may not be suitable for every situation. In performing activities adult supervision must be sought. Likewise, common sense and care are essential to the conduct of any and all activities, whether described in this book or otherwise. Neither the publisher nor the author shall be liable or assumes any responsibility for any injuries or damages arising here from. The fact that an organization or Website if referred to in this work as a citation and/or a potential source of further information does not mean that the author or the publisher endorses the information the organization or Website may provide or recommendations it may make. Further, readers must be aware that the Internet Websites listed in this work may have changed or disappeared between when this work was written and when it is read. All trademarks, logos or any other mark such as Vibgyor, USP, Amanda, Golden Bells, Firewall Media, Mercury, Trinity, Laxmi appearing in this work are trademarks and intellectual property owned by or licensed to Laxmi Publications, its subsidiaries or affiliates. Notwithstanding this disclaimer, all other names and marks mentioned in this work are the trade names, trademarks or service marks of their respective owners. & Bengaluru 080-26 75 69 30 & Chennai 044-24 34 47 26 Branches & Guwahati 0361-254 36 69 & Hyderabad 040-27 55 53 83 & Jalandhar 0181-222 12 72 & Kochi 0484-405 13 03 & Kolkata 033-40 04 77 79 & Lucknow 0522-430 36 13 Published in India by & Ranchi 0651-224 24 64 Laxmi Publications (P) Ltd. C—00000/021/12 Printed at : Ajit Printing Press, Delhi. (An ISO 9001:2015 Company) 113, GOLDEN HOUSE, GURUDWARA ROAD, DARYAGANJ, NEW DELHI - 110002, INDIA Telephone : 91-11-4353 2500, 4353 2501 www.laxmipublications.com [email protected]

Contents Unit 1. Polynomials ............................................................................................................................ 1 Unit 2. Quadrilaterals ....................................................................................................................... 21 Unit 3. Circles .................................................................................................................................... 43 Unit 4. Constructions ....................................................................................................................... 64 Unit 5. Surface Areas and Volumes ............................................................................................... 76 Unit 6. Probability ........................................................................................................................... 128

Syllabus Term – II (Class IX) No. Unit Name Marks I ALGEBRA(Cont.) 12 II GEOMETRY(Cont.) 15 III MENSURATION(Cont.) 9 IV STATISTICS & PROBABILITY(Cont) 4 40 Total 10 Internal Assessment 50 Total UNIT – ALGEBRA 1. Polynomials 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. 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. (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) Verification of identities and their use in factorization of polynomials. UNIT – GEOMETRY 2. Quadrilaterals 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. 3. Circles Through examples, arrive at definition of circle and related concepts-radius, circumference, diameter, chord, arc, secant, sector, segment, subtended angle.

1. (Prove) Equal chords of a circle subtend equal angles at the centre and (motivate) its converse. 2. (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. 3. (Motivate) Equal chords of a circle (or of congruent circles) are equidistant from the centre (or their respective centres) and conversely. 4. (Motivate) 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. 5. (Motivate) Angles in the same segment of a circle are equal. 6. (Motivate) The sum of either of the pair of the opposite angles of a cyclic quadrilateral is 180° and its converse. 4. Constructions 1. Construction of bisectors of line segments and angles of measure 60°, 90°, 45° etc., equilateral triangles. 2. Construction of a triangle given its base, sum/difference of the other two sides and one base angle. UNIT – MENSURATION 5. Surface Areas and Volumes Surface areas and volumes of cubes, cuboids, spheres (including hemispheres) and right circular cylinders/cones. UNIT – STATISTICS & PROBABILITY 6. Probability History, Repeated experiments and observed frequency approach to probability. Focus is on empirical probability. (A large amount of time to be devoted to group and to individual activities to motivate the concept; the experiments to be drawn from real - life situations, and from examples used in the chapter on statistics). Internal Assessment Marks Total Marks 10 marks for the term Periodic Tests 3 Multiple Assessments 2 Portfolio 2 Student Enrichment Activities-practical work 3



Unit 1. Polynomials  IMPORTANT RESULTS AND FORMULAE  In this chapter, we will study the following points to remember: 1. Polynomial: A polynomial f (x) in one variable x is an algebraic expression in x of the form f(x) = an xn + an – 1 xn – 1 + ...... + a2x2 + a1x + a0, an  0. where ‘n’ is a non-negative integer/whole number, the degree of the polynomial ‘x’ is the variable of the polynomial f (x). a0, a1, a2, ......, xa0n, are constants, called the co-efficients of the polynomial f (x), are the co-efficients of x, x2, ......, xn, respectively. Each of anxn, an – 1 xn – 1, ......, a1x, a0x0, an  0 is called a term of the polynomial f (x). 2. Degree of a Polynomial: It is the highest power (or exponent) of the variable, in any term of the polynomial. The phrase “degree of f (x)” is denoted by deg ( f (x)) or simply deg f (x) e.g., In, f (x) = an xn + an – 1 xn – 1 + ......... + a0, deg f (x) = n In, p(x) = 4x6 – 5x3 + 3x2 + 3x – 2, deg p(x) = 6 (i) Constant Polynomial: A zero degree polynomial is called a constant polynomial. (ii) Linear Polynomial: A polynomial of degree one is called a linear polynomial. e.g., f (x) = x – 1; h(x) = ax + b, (a  0), etc. (iii) Quadratic Polynomial: A polynomial of degree two is called a quadratic polynomial. e.g., f (x) = ax2 + bx + c, (a  0, a, b, c  R) h(x) = 3 – x – x2, etc. (iv) Cubic Polynomial: A polynomial of degree three is called a cubic polynomial. e.g., g(x) = 4x3 – 3x2 + 5x + 12 h(x) = ax3 + bx2 + cx + d (a  0), etc. 3. Types of a Polynomial: There are the following important types of polynomial: (i) Monomial: A polynomial is said to be a monomial, if it contains only one term. (ii) Binomial: A polynomial is said to be a binomial, if it contains two terms. (iii) Trinomial: A polynomial is said to be a trinomial, if it contains three terms. (iv) Quadratic Polynomial: A polynomial of degree two is called a quadratic polynomial. (v) Linear Polynomial: A polynomial of degree one is called a linear polynomial. (vi) Cubic Polynomial: A polynomial of degree three is called a cubic polynomial. (vii) Bi-quadratic (or quartic): A polynomial of degree four is called a bi-quadratic (or quartic) polynomial. 1

2 Mathematics—IX 4. Value of a Polynomial: The value of a polynomial f (x), at x = , (a real number) is obtained on replacing ‘x’, in the polynomial by ‘’. It is denoted by f (). 5. Zeros of a Polynomial: Any value of x for which f (x), becomes zero (i.e., vanishes) is called a zero of the polynomial f (x). 6. Number of Zeros of a Polynomial: Every polynomial of the nth degree can have at most ‘n’ real zeros. 7. Uniqueness of Zeros of a Polynomial: Every linear polynomial in one variable has a unique zero, a non-zero constant polynomial has no zero, and every real number is a zero of the zero polynomial. 8. Factors and Multiples: Let f (x), g(x)  0 be two polynomials. Then, non-zero polynomial, g(x), is called a factor (or divisor) of f (x), if there exists a polynomial q(x) such that f(x) = g(x) q(x) q(x) Multiple Factors g(x) f (x) We write it as g(x)|f (x).  f (x) g(x) is called the factor of f (x), 0 f (x) is called the multiple of g(x) q(x) is called the quotient of f (x) by g(x) and is written as q(x) = f (x) . g(x) Note: q(x) is also a factor of f (x). Remainder Theorem: If a polynomial f (x) of degree  1 (i.e., greater than or equal to 1) is divided by a linear binomial (x – ), then the remainder is f(), where  is any real number. 9. New Identities: (i) Trinomial-Square Identity: (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca (ii) Binomial-Cube Identities: (a) (a + b)3 = a3 + b3 + 3ab (a + b) (b) (a – b)3 = a3 – b3 – 3ab (a – b)

Polynomials 3 PART – A  VERY SHORT ANSWER TYPE QUESTIONS  Illustrative Examples EXAMPLE 1. (i) Is it correct to say that 1 1 +1 is a polynomial? Justify your answer. 5 x2 3 (ii) Is it correct to say that 6 x  x 2 x is a polynomial x  0? Justify your answer. SOLUTION. (i) No! because the exponent of the variable is 1 , which is not a whole-number. 2 (ii) Yes! x 3 6  x2  x 6 x, which is a polynomial. EXAMPLE 2. Which of the following expressions are polynomials in one variable and which are not? State reasons for your answer. (i) 4x3 – 3x + 7 (ii) y2  2 (iii) 3 t  t 2 (iv) y  2 . [NCERT] y SOLUTION. (i) Here, the given expression is: f(x) = 4x3 – 3x + 7  f (x) is a polynomial in x since there occurs only one variable ‘x’ and the exponent of x in every term is a non-negative integer. (ii) Here, the given expression is: f(y) = y2  2  f (y) is a polynomial in y since there occurs only one variable ‘y’ and the exponent of y is a non-negative integer. (iii) Here, the given expression is: f(t) = 3 t  t 2  f(t) is not a polynomial in t since the exponent of t in the first term is 1 , which 2 is not a non-negative integer. (iv) Here, the given expression is: f(y) = y 2 = y + 2 . y–1 y  f(y) is not a polynomial in y since the exponent of y in the second term is –1, which is not a non-negative integer.

4 Mathematics—IX EXAMPLE 3. Write the co-efficients of x2 and x in each of the following: (i) 2 + x2 + x (ii) 2 – x2 + x3 (iii) x2 x (iv) 2 x2  1. [NCERT] 2 | Writing in the standard form SOLUTION. (i) Here, the given polynomial is: f (x) = 2 + x2 + x = x2 + x + 2 Co-efficient of x2 = 1; co-efficient of x = 1. (ii) Here, the given polynomial is: f(x) = 2 – x2 + x3 = x3 – x2 + 0 . x + 2 | Writing in the standard form Co-efficient of x2 = –1; co-efficient of x = 0 (iii) Here, the given polynomial is: f(x) = 1 x2  x | Which is in the standard form 2 Co-efficient of x2 = 1 ; co-efficient of x = 1 2 (iv) Here, the given polynomial is f(x) = 2x2  1 | Which is in the standard form Co-efficient of x2 = 2 ; co-efficient of x = 0. EXAMPLE 4. Classify the following as linear, quadratic and cubic polynomials: (i) x2 + x (ii) x – x3 (iii) x + x2 + 4 (iv) 1 + x (v) 3t [NCERT] SOLUTION. (i) Here, the given polynomial is: f (x) = x2 + x degree of f (x) = 2  f(x) is a quadratic polynomial. (ii) Here, the given polynomial is: f (x) = x – x3 = –x3 + x | Writing in the standard form degree of f (x) = 3  f(x) is a cubic polynomial. (iii) Here, the given polynomial is: f (x) = x + x2 + 4 = x2 + x + 4 | Writing in the standard form degree of f (x) = 2  f(x) is a quadratic polynomial. (iv) Here, the given polynomial is: f(x) = 1 + x degree of f(x) = 1  f(x) is a linear polynomial. (v) Here, the given polynomial is: f(t) = 3t degree of f(t) = 1  f(t) is a linear polynomial.

Polynomials 5 EXAMPLE 5. Find the value of the polynomial 5x – 4x2 – 3 at: (i) x = 0 (ii) x = 2. [NCERT] SOLUTION. Here, let f (x) = 5x – 4x2 – 3 = – 4x2 + 5x – 3 (i) The value of f (x) at x = 0 is given by f (0) = –4(0)2 + 5(0) – 3 = –0 + 0 – 3 = –3 (ii) The value of f (x) at x = 2 is given by f (2) = – 4(2)2 + 5(2) – 3 = –4 . 4 + 10 – 3 = –16 + 10 – 3 = –9. EXAMPLE 6. Find the value of the polynomial 5x – 4x2 + 3 when x = –1. [CBSE 2014] SOLUTION. Here, let f(x) = 5x – 4x2 + 3 = – 4x2 + 5x + 3 The value of f (x) at x = –1 is given by f (–1) = – 4(–1)2 + 5(–1) + 3 = – 4 . 1 – 5 + 3 = – 6 EXAMPLE 7. Find the zero of the polynomial in each of the following cases: (i) p(x) = x + 5 (ii) p(x) = x – 5 (iii) p(x) = 2x + 5 (iv) p(x) = 3x – 2 (v) p(x) = 3x (vi) p(x) = ax, a  0 [NCERT] (vii) p(x) = cx + d, c  0, c, d are real numbers. SOLUTION. (i) Here, we have the polynomial: p(x) = x + 5 Putting p(x) = 0, we get x + 5 = 0 Solving for x, we get x = –5 Hence x = – 5 is a zero of the given polynomial. (ii) Here, we have the polynomial: p(x) = x – 5 Putting p(x) = 0, we get x – 5 = 0 Solving for x, we get x = 5. Hence, x = 5 is a zero of the given polynomial. (iii) Here, we have the polynomial: p(x) = 2x + 5 Putting p(x) = 0, we get 2x + 5 = 0 Solving for x, we get 2x = –5  x =  5 2 Hence, x=  5 is the required zero of the given polynomial. 2 (iv) Here, we have the polynomial: p(x) = 3x – 2 Putting p(x) = 0, we get 3x – 2 = 0

6 Mathematics—IX Solving for x, we get 3x = 2  x= 2 3 Hence, x= 2 is the required zero of the given polynomial. 3 (v) Here, p(x) = 3x Putting p(x) = 0, we get 3x = 0  x = 0  x=0 3 Hence, x = 0 is the required zero of the given polynomial. (vi) Here, p(x) = ax, a  0 Putting p(x) = 0, we get ax = 0  x = 0 |a0 Hence, x = 0, is the required zero of the given polynomial. (vii) Here, we have the polynomial: p(x) = cx + d, c  0 Putting p(x) = 0, we get cx + d = 0, c  0 Solving for x, we get cx + d = 0  cx = – d  x =  d |c0 c [NCERT] Hence, x=  d is the required zero of the given polynomial. c EXAMPLE 8. Write the following cubes in expanded form (2a – 3b)3. SOLUTION. Given expression = (2a – 3b)3 y = 3b = (x – y)3, where x = 2a, = x3 – y3 – 3xy(x – y) | By using ‘Binomial-Cube’ Identity: (x – y)3 = x3 – y3 – 3xy(x – y) = (2a)3 – (3b)3 – 3(2a)(3b)(2a – 3b) | Replacing x by 2a and y by 3b = 8a3 – 27b3 – 18ab(2a – 3b) = 8a3 – 27b3 – 36a2b + 54ab2 = 8a3 – 36a2b + 54ab2 – 27b3 | Writing the terms in the descending powers of a EXAMPLE 9. If (2x + 5y)3 = 8x3 + 125y3 + ax2y + bxy2, then find the value of b – a. [CBSE 2014] SOLUTION. Here, (2x + 5y)3 = 8x3 + 125y3 + ax2y + bxy2 ...(1) | Given ...(2) LHS = (2x + 5y)3 = (2x)3 + (5y)3 + 3  (2x)  (5y) (2x + 5y) = 8x3 + 125y3 + 30xy(2x + 5y) = 8x3 + 125y3 + 60x2y + 150xy2

Polynomials 7 Comparing (1) and (2), we get a = 60 and b = 150  b – a = 150 – 60 = 90 Hence, the required value of (b – a) is 90. EXAMPLE 10. Factorise: (i) x2 + 9x + 18 [NCERT (EP)] (ii) 20x2 – 9x + 1 [CBSE 2016] SOLUTION. (i) Given expression = x2 + 9x + 18 = x2 + 6x + 3x + 18 | Splitting the middle term: 6 + 3 = 9 and 6 × 3 = 18 = (x2 + 6x) + (3x + 18) = x(x + 6) + 3(x + 6) Hence, = (x + 3) (x + 6) x2 + 9x + 18 = (x + 3) (x + 6) (ii) Given expression = 20x2 – 9x + 1 = 20x2 – 5x – 4x + 1 Hence, | Splitting the middle term: (–5) + (–4) = –9 and (–5) × (–4) = 20 × 1 = 20 = (20x2 – 5x) – (4x – 1) = 5x (4x – 1) – (4x – 1) = (5x – 1) (4x – 1) 20x2 – 9x + 1 = (5x – 1) (4x – 1)  PRACTICE EXERCISE  1. Which of the following expressions are polynomials? Justify your answer: (i) 8 (ii) 3 x2  2x (iii) 1  5 x (iv) 1  5x 7 (v) (x  2) (x  4) (vi) 1 5x2 x x1 (vii) 1 a3  2 a2  4a  7 (viii) 1 . [NCERT (EP)] 7 3 2x 2. Indicate whether the following are monomials, binomials or trinomials and give the degree of each. Justify your answer. (ii) x2 + x + 1 (iii) 5x3 + 8x3 (i) 10 3. Find the value of each of the following polynomials at the indicated value of variables: (i) p(x) = 5x2 – 3x + 7 at x = 1 (ii) q(y) = 3y3 – 4y + 11 at y = 2 (iii) p(t) = 4t4 + 5t3 – t2 + 6 at t = a [NCERT] 4. Find the value of the polynomial 3x3 – 4x2 + 7x – 5, when x = 3 and also x = –3. [NCERT (EP)] 5. Find p(1), p(– 1), p(– 2) for the following polynomials: (i) p(x) = 10x – 4x2 – 3 (ii) p(y) = (y + 2) (y – 2) [NCERT (EP)] 6. Find f ( 2) , if f (x) = x2  2x  1. [CBSE 2014] 7. If p(x) = x2 + 3x + 2, find the value of p(1) + p(2) + p(0). [CBSE 2014]

8 Mathematics—IX 8. Find the zeros of the polynomials in each of the following: (i) q(x) = 2x – 7 (ii) h(y) = 2y [NCERT (EP)] 9. Write each of the following in expanded form, using a suitable algebraic identity: (i) (3x + 4y + 5z)2 [NCERT ] (ii) (4a – 3b – 3c)2 [CBSE 2012] 10. Write each of the following in expanded form, using a suitable algebraic identity: (i) (4a – b + 2c)2 [NCERT (EP)] (ii) (3a – 5b – c)2 [NCERT (EP)] 11. Write each of the following in expanded form, using a suitable algebraic identity: (i) (–x + 2y – 3z)2 [NCERT (EP)] (ii) (3a – 7b – c)2 12. Factorise the following: (i) 84 – 2r – 2r2 [NCERT (EP)] (ii) 6x2 + 7x – 3 [NCERT (EP)] (iii) 9x2 – 12x + 3 [NCERT (EP)] (iv) 9x2 – 12x + 4 [NCERT (EP)] PART – B  VERY SHORT ANSWER TYPE QUESTIONS  Illustrative Examples EXAMPLE 1. Verify whether the indicated numbers are zeros of the polynomial corresponding to them in the following cases: (i) p(x) = 3x + 1, x=  1 (ii) p(x) = 5x – 4, x= 4 3 5 (iii) p(x) = x2 – 1, x = 1, –1 (iv) p(x) = (x + 1) (x – 2), x = –1, 2 (v) p(x) = x2, x = 0 (vi) p(x) = lx + m, x=  m . [NCERT] l SOLUTION. (i) Here, we have the polynomial: p(x) = 3x + 1 Putting x =  1 , we get 3 p   1  = 3   1   1 = –1 + 1 = 0  3   3    1 is a zero of the given polynomial. 3 (ii) Here, we have the polynomial: p(x) = 5x – 4 Putting x = 4 , we get 5 p  4  = 5  4   4 =4–4=0 5   5 

Polynomials 9 ...(1)  x= 4 is a zero of the given polynomial. 5 [NCERT] (iii) Here, we have the polynomial: p(x) = x2 – 1 Putting x = 1, we get p(1) = (1)2 – 1 = 1 – 1 = 0 Putting x = –1, we get p(–1) = (–1)2 – 1 = 1 – 1 = 0  x = 1, –1 are both zeros of the given polynomial. (iv) Here, we have the polynomial: p(x) = (x + 1) (x – 2) = x2 – x – 2 Putting x = –1 in (1), we get p(–1) = (–1)2 – (–1) – 2 = 1+1–2=2–2=0  x = –1 is a zero of the given polynomial. Putting x = 2 in (1), we get p(2) = (2)2 – 2 – 2 = 4 – 2 – 2 = 4 – 4 = 0  x = 2 is a zero of the given polynomial. (v) Here, the polynomial is: p(x) = x2 Putting x = 0, we get p(0) = (0)2 = 0  x = 0 is a zero of the given polynomial. (vi) Here, the polynomial is: p(x) = lx + m Putting x =  m , we get l p   m  = l   m   m = –m +m=0  l   l   x=  m is a zero of the given polynomial. l EXAMPLE 2. Evaluate the following using suitable identities  y 2  3  y2  3  .  2   2  SOLUTION. Here, we have  y 2  3   y 2  3 =  x  3  x  3 where x = y2  2   2   2   2  = x2 + Sx + P,

10 Mathematics—IX where S =  3    3  = 0, P =  3     3  =  9  2   2   2   2  4  x2 + Sx + P = x2  9 4   y2  3   y2  3  = (y2)2  9 | Putting back x = y2 2 2 4 = y4  9 . 4 EXAMPLE 3. Expand each of the following using suitable identities: (i) (–2x + 5y – 3z)2 [CBSE 2014] (ii)  1 a  1 b  12 `[NCERT]  4 2 SOLUTION. (i) Given expression = (–2x + 5y – 3z)2 = [(–2x) + 5y + (–3z)]2 = (a + b + c)2, where a = –2x, b = 5y, c = –3z = a2 + b2 + c2 + 2ab + 2bc + 2ca | Using ‘Trinomial Square’ Identity = (–2x)2 + (5y)2 + (–3z)2 + 2 . (–2x) . (5y) + 2 . (5y) . (–3z) + 2 . (–3z) . (–2x) | Putting back the values of a, b and c = 4x2 + 25y2 + 9z2 – 20xy – 30yz + 12zx. (ii) Given expression =  1 a  1 b  1 2 = a   b   12  4 2   4  2  = (x + y + z)2, where x = a , y= b , z=1 4 2 = x2 + y2 + z2 + 2xy + 2yz + 2zx | Using ‘Trinomial-Square’ Identity =  a 2    b 2  (1)2  2  a     b   2   b   (1)  2(1)   a   4   2   4   2   2   4  | Putting back the values of x, y, z = a2  b2  1  ab  b  a 16 4 4 2 EXAMPLE 4. Factorise each of the following expressions: (i) 27y3 + 125z3 [NCERT] (ii) 64m3 – 343n3. [NCERT] (iii) 3x5 + 1536x2y3. [CBSE 2015]

Polynomials 11 SOLUTION. (i) Given expression = 27y3 + 125z3 = (3y)3 + (5z)3 = a3 + b3, where a = 3y, b = 5z = (a + b) [a2 + b2 – ab] = (3y + 5z) [(3y)2 + (5z)2 – (3y)(5z)] = (3y + 5z)(9y2 + 25z2 – 15yz) (ii) Given expression = 64m3 – 343n3 = (4m)3 – (7n)3 = a3 – b3, where a = 4m, b = 7n = (a – b) [a2 + b2 + ab] = (4m – 7n) [(4m)2 + (7n)2 + (4m)(7n)] = (4m – 7n) (16m2 + 49n2 + 28mn) (iii) Given expression = 3x5 + 1536x2y3 = 3x2 [x3 + 512y3] = 3x2 [(x)3 + (8y)3] = 3x2(x + 8y)(x2 + 64y2 – 8xy)  PRACTICE EXERCISE  1. Verify whether 2 and 0 are zeros of the polynomial p(x) = x2 – 2x. [NCERT] 2. Check whether –2 and 2 are zeros of the polynomial f(x) = x + 2. [NCERT] 3. Evaluate the following using suitable identities: (i) (x2 + 4) (x2 + 9) (ii) (z3 + 14) (z3 + 1) (iii) (p2 + 16)  p2  1  4  (iv)  y2  5  y2  14  (v) (2x – 3) (2x + 5) (vi) (3x – 7) (3x + 5)  7   5  4. Evaluate the following, without directly multiplying: (i) 101 × 102 [NCERT (EP)] (ii) 83 × 79 5. (i) If x  1 = 3, find the value of x3  1 . [CBSE 2012] x x3 [CBSE 2011] (ii) If x  1 = 3, find the value of x3  1 . x x3 (iii) If x  1 = 7, then find the value of x3  1 . [CBSE 2011] x x3 6. Factorise the following polynomials: (i) 1 + 64x3 [NCERT (EP)] (ii) a3  2 2b3 [NCERT (EP)] (iii) 64x3 + 125y3 [CBSE 2013] 7. Factorise the following: [NCERT] (i) 4x2 + y2 + z2 – 4xy – 2yz + 4xz. [NCERT (EP)] (ii) 9x2 + 4y2 + 16z2 + 12xy – 16yz – 24xz. [NCERT (EP)] (iii) 25x2 + 16y2 + 4z2 – 40xy + 16yz – 20xz. (iv) 16x2 + 4y2 + 9z2 – 16xy + 12yz – 24xz. [CBSE 2014]

12 Mathematics—IX  SHORT ANSWER TYPE QUESTIONS  Illustrative Examples EXAMPLE 1. Show that 1, 2 and 3 are zeros of f(x) = x3 – 6x2 + 11x – 6. SOLUTION. Here, we have f(x) = x3 – 6x2 + 11x – 6  f(1) = (1)3 – 6(1)2 + 11(1) – 6 = 1 – 6 + 11 – 6 = 12 – 12 = 0 f(2) = (2)3 – 6(2)2 + 11(2) – 6 = 8 – 6(4) + 11(2) – 6 = 8 – 24 + 22 – 6 = 30 – 30 = 0 f(3) = (3)3 – 6(3)2 + 11(3) – 6 = 27 – 6(9) + 11(3) – 6 = 27 – 54 + 33 – 6 = 60 – 60 = 0 f(1) = 0, f(2) = 0 and f(3) = 0  1, 2 and 3 are the zeros of the polynomial f(x). EXAMPLE 2. If x =  1 is a zero of a polynomial p(x) = 27x3 – ax2 – x + 3, then find the value 3 of a. [CBSE 2012, 14] SOLUTION. Here, p(x) = 27x3 – ax2 – x + 3  p   1  = 27   1 3  a   1 2    1   3  3   3   3   3   0 =  27  a  1 3    1  is a zero of p(x) 27 9 3  3   1  a  1  3 = 0   a  1  2 = 0 9 3 9 3   a  3  18 =0  – a + 3 + 18 = 0 9  a = 21 EXAMPLE 3. Find the value of x2 + y2, if (i) x + y = 4, xy = 3, (ii) x – y = 2, xy = 15. [CBSE 2012] SOLUTION. (i) Here, we have ...(1) xy = 3 ...(2) x+y = 4 Squaring both sides of (1), we get (x + y)2 = (4)2  x2 + y2 + 2xy = 16  By using ‘Binomial Square’ Identity (a  b)2  a2  b2  2ab Putting the value of xy from (2), we get x2 + y2 + 6 = 16 x2 + y2 + 2(3) = 16  x2 + y2 = 10  x2 + y2 = 16 – 6 

Polynomials 13 (ii) Here, we have x – y = 2 ...(1) xy = 15 ...(2) Squaring both sides of (1), we get  By using ‘Binomial Square’ Identity (x – y)2 = (2)2 (a  b)2  a2  b2  2ab  x2 + y2 – 2xy = 4 Putting the value of xy from (2), we get x2 + y2 – 2(15) = 4  x2 + y2 – 30 = 4 x2 + y2 = 34.  x2 + y2 = 30 + 4  EXAMPLE 4. Simplify:  x  2 3   x + 2 y 3 [CBSE 2012]    3   3y  SOLUTION. Given expression =  x  2 y 3   x  2 y 3  3   3  =  x 3   2 y 3  3  (x)  2 y   x  2 y  –  x3   2 y 3  3  (x)   2 y   x  2 y   3   3   3    3   3   3    | Expanding by ‘Binomial Cube’ Identities =  x3  8 y3  2x2y  4 xy2  – x3  8 y3  2x2y  4 xy 2   27 3  27 3  =  16 y 3  4x2y . 27 EXAMPLE 5. (i) If 2x – 5y = 2 and xy = 12, find the value of 8x3 – 125y3. [CBSE 2014] (ii) If x + 2y = 10 and xy = 15, find the value of x3 + 8y3. SOLUTION. (i) Here, we have 2x – 5y = 2 ...(1) and xy = 12 ...(2) Cubing both sides of (1), we get (2x)3 – (5y)3 – 3 (2x) (5y) (2x – 5y) = 8 (2x – 5y)3 = (2)3  | Using (1) and (2)  8x3 – 125y3 – 30 (12) (2) = 8  8x3 – 125y3 = 728.  8x3 – 125y3 = 720 + 8 (ii) Here, we have x + 2y = 10 ...(1) and xy = 15 ...(2) Cubing both sides of (1), we get (x + 2y)3 = (10)3  x3 + (2y)3 + 3 · (x)(2y)(x + 2y) = 10 × 10 × 10 | Using ‘Binomial-Cube’ Identity (a + b)3 = a3 + b3 + 3ab (a + b)

14 Mathematics—IX  x3 + 8y3 + 6xy (x + 2y) = 1000  x3 + 8y3 + 6 . (15) . (10) = 1000 | By using (1) and (2)  x3 + 8y3 + 900 = 1000  x3 + 8y3 = 1000 – 900  x3 + 8y3 = 100. EXAMPLE 6. If x and y are two positive real numbers such that 8x3 + 27y3 = 730 and 2x2y + 3xy2 = 15, then evaluate (2x + 3y). [CBSE 2012] SOLUTION. Here, 8x3 + 27y3 = 730 ...(1) and 2x2y + 3xy2 = 15 ...(2) Then, (2x + 3y)3 = (2x)3 + (3y)3 + 3 · (2x) · (3y) (2x + 3y) | Using ‘Binomial-Cube’ Identity = 8x3 + 27y3 + 18 (2x2y + 3xy2) = 730 + 18 (15) | Using (1) and (2) = 730 + 270 = 1000 = (10)3  (2x + 3y)3 = (10)3 |  a3 = b3  a = b  2x + 3y = 10. EXAMPLE 7. Find the value of x3 – 8y3 – 36xy – 216 when x = 2y + 6. [CBSE 2011, 12, 14] SOLUTION. Here, x = 2y + 6 ...(1)  (x)3 = (2y + 6)3 | Cubing both sides  x3 = (2y)3 + (6)3 + 3 · (2y) (6) (2y + 6) = 8y3 + 216 + 36y (x) | Using (1) = 8y3 + 36xy + 216  x3 – 8y3 – 36xy – 216 = 0. EXAMPLE 8. Factorise : 27a3 – 8b3 – 6a + 4b. [CBSE 2014] SOLUTION. Here, given expression = 27a3 – 8b3 – 6a + 4b = (3a)3 – (2b)3 – 2 (3a – 2b) = (3a – 2b) {(3a)2 + (2b)2 + (3a) · (2b)} – 2 (3a – 2b) = (3a – 2b) (9a2 + 4b2 + 6ab – 2) | Taking (3a – 2b) outside throughout Hence, the required factorisation of the given expression is (3a – 2b) (9a2 + 4b2 + 6ab – 2). EXAMPLE 9. Factorise: 9x2 + 4y2 + z2 – 12xy + 4yz – 6zx [CBSE 2012] Hence, find value when x = 1, y = 2 and z = –1. SOLUTION. (i) Given expression = 9x2 + 4y2 + z2 – 12xy + 4yz – 6zx

Polynomials 15 = (–3x)2 + (2y)2 + (z)2 + 2 · (–3x) · (2y) + 2 (2y) (z) + 2 (z) (–3x) = (–3x + 2y + z)2 ...(1) (ii) Putting x = 1, y = 2 and z = –1 in (1), we get Given expression = [– 3(1) + 2(2) + (–1)]2 = (– 3 + 4 – 1)2 = (0)2 = 0. EXAMPLE 10. Factorise each of the following expressions (x2 – 4x) (x2 – 4x – 1) – 20. [CBSE 2012] SOLUTION. Given expression = (x2 – 4x)(x2 – 4x – 1) – 20 = y(y – 1) – 20, putting x2 – 4x = y = y2 – y – 20 = y2 – 5y + 4y – 20 | Splitting the middle term: (–5) + (+4) = –1 and (–5) × (+4) = –20 = (y2 – 5y) + (4y – 20) = y(y – 5) + 4(y – 5) = (y – 5) (y + 4) = (x2 – 4x – 5) (x2 – 4x + 4) |  y = x2 – 4x = [x2 – 5x + x – 5](x – 2)2 = [x(x – 5) + (x – 5)](x – 2)2 = (x – 5)(x + 1)(x – 2)2 Hence, (x2 – 4x)(x2 – 4x – 1) – 20 = (x – 5)(x + 1)(x – 2)2. EXAMPLE 11. Give possible expressions for the length and breadth of the following rectangles, in which their area is given by: Area : 25a2  35a  12 [CBSE 2013] SOLUTION. Here, we have A = 25a2 – 35a + 12 Coefficient of the middle term = –35 = (–20) + (–15) Product of extreme coefficients = 25 × 12 = 300 = (– 20) × (–15)  A = 25a2 + (–20 – 15)a + 12 = 25a2 – 20a – 15a + 12 = (25a2 – 20a) – (15a – 12) = 5a(5a – 4) – 3(5a – 4) = (5a – 4)(5a – 3) ...(1)  Area of a rectangle, A = Length × Breadth = l×b ...(2) Comparing (1) and (2), we get l = (5a – 4) units, b = (5a – 3) units Thus, possible expressions for the dimensions of the given rectangle are (5a – 4) and (5a – 3).  PRACTICE EXERCISE  1. Find the zeros of the polynomial p(x) = (x – 2)2 – (x + 2)2. [NCERT (EP)] 2. If 2 is a zero of polynomial 4y2 – 6y – k, find the value of k. Also, find the other zero. [CBSE 2011]

16 Mathematics—IX 3. Show that 4 and 1 are the zeros of the polynomial 12x3 – 55x2 + 29x – 4. Also, find 4 the third zero of the polynomial. [CBSE 2014] 4. If x = –2 is the root of the equation 2 (x  p) = 0 and is also the zero of the polynomial px2 + kx + 2 2 , then find the value of k. [CBSE 2011] 5. Find the value of x2  1 , if (i) x  1  6, (ii) x  1  3. [CBSE 2012] x2 x x [CBSE 2012] 6. Simplify (p + q + r)2 + (p – q – r)2 [CBSE 2012] 7. Find the value of x3 + 1 , if x3 (i) x2  1 =4 [CBSE 2012] (ii) x2 + 1 =7 x2 x2 (iii) x2 + 1 = 23 [CBSE 2012] x2 8. Simplify æçèçç x + y ÷÷÷öø3 - çççæè x - y ÷÷÷øö3 [CBSE 2012] 3 5 3 5 9. Find the value of p3 – q3, if p–q= 10 and pq = 5 . [CBSE 2012] 9 3 [CBSE 2015] 10. If a – b = 7, a2 + b2 = 85, find the value of a3 – b3. [CBSE 2012] 11. (i) If 2x + y = – 5, prove that 8x3 + y3 – 30xy + 125 = 0. [CBSE 2012] [CBSE 2011] (ii) If x + y = – 4, then, find the value of x3 + y3 – 12xy + 64. [CBSE 2016] 12. Factorise 9(2a – b)2 – 4(2a – b) – 13 13. If x2 + y2 = 221 and x – y = 1, then find the value of x3 – y3. [CBSE 2012] 14. If p(x) = x2 – 4x + 3, evaluate: p(2) – p(–1) + p  1  .  2   LONG ANSWER TYPE QUESTIONS  Illustrative Examples EXAMPLE 1. If a + b + c = 20, a2 + b2 + c2 = 90, find the value of ab + bc + ca. [CBSE 2012] SOLUTION. Since a + b + c = 20 | Given Squaring both sides, we get (a + b + c)2 = (20)2

Polynomials 17  a2 + b2 + c2 + 2ab + 2bc + 2ca = 400 | Expanding LHS, by using ‘Trinomial-Square’ Identity  (a2 + b2 + c2) + 2(ab + bc + ca) = 400  90 + 2(ab + bc + ca) = 400  2(ab + bc + ca) = 400 – 90  ab + bc + ca = 310  155 2  ab + bc + ca = 155. [CBSE 2014] EXAMPLE 2. Factorise: 27a3 – 64b3 – 108a2b + 144 ab2 + 9a – 12b. SOLUTION. Given expression: = 27a3 – 64b3 – 108a2b + 144ab2 + 9a – 12b = {(3a)3 – (4b)3 – 3(3a) (4b) (3a – 4b)} + 3(3a – 4b) = {(3a – 4b)3} + 3(3a – 4b) = (3a – 4b)3 + 3(3a – 4b) = (3a – 4b) [(3a – 4b)2 + 3] Hence, the required factorisation of the given expression is (3a – 4b) [(3a – 4b)2 + 3]. EXAMPLE 3. Factorise the following: [CBSE 2014] 12(x2 + 7x)2 – 8(x2 + 7x)(2x – 1) – 15(2x – 1)2. SOLUTION. Let x2 + 7x = p and 2x – 1 = q  Given expression = 12p2 – 8pq – 15q2 = 12p2 – 18pq + 10pq – 15q2 = 6p(2p – 3q) + 5q(2p – 3q) = (6p + 5q)(2p – 3q) = [6(x2 + 7x) + 5(2x – 1)] [2(x2 + 7x) – 3(2x – 1)] = (6x2 + 42x + 10x – 5)(2x2 + 14x – 6x + 3) = (6x2 + 52x – 5)(2x2 + 8x + 3) Hence, the required factors of the given expression are (6x2 + 52x – 5) and (2x2 + 8x + 3).  PRACTICE EXERCISE  [NCERT (EP)] [CBSE 2016] 1. If a + b + c = 9, and ab + bc + ca = 26, find a2 + b2 + c2. 2. Factorise x6 + 6x3 + 8 [CBSE 2013, 2014] 3. Factorise (p + q)2 – 20 (p + q) – 125 [CBSE 2014] 4. Factorise (a2 – 2a)2 – 23(a2 – 2a) + 120 [CBSE 2011] 5. Factorise x4 + x3 – 7x2 – x + 6

18 Mathematics—IX  CASE STUDIES  1. Let p(x) be a polynomial given by p(x) = x3 – kx – 16, where k is a constant real number. Based on the above information, answer the following questions: (i) p(0) = (b) – 16 (c) 16 (d) None of these (a) 0 (ii) If x + 2 is a factor of p(x), then k = (a) 0 (b) 4 (c) 8 (d) 12 (iii) If p(x) is multiple of x – 3, then k = (a) 0 (b) 11 (c) 11/3 (d) None of these (iv) If p(x), when divided by x + 1, leaves remainder –10, then k = (a) 27 (b) 7 (c) – 7 (d) – 17 (v) Which of the following is not a polynomial ? (a) 3x2  5 (b) (x  2)2 (c) x x  3 (d) None of these 2. Given that a + b + 3 = 0 and ab = – 4. Based on the above information, answer the following questions: (i) a2 + b2 = (b) 13 (c) 9 (d) 17 (a) 1 (ii) a3 + b3 = (b) – 63 (c) 27 (d) 63 (a) – 27 (iii) | a – b | = (a) 5 (b) 2 (c) 1 (d) 9 3. Class IX of Kendriya Vidhyala has x students. On a 2nd saturday, 1 th times the square of the 12 total number of students planned to visit historical monuments. 7 th times the number of 12 students planned to visit old age homes while 10 students decided to teach poor children. Based on the above information, answer the following questions: (i) The total number of students in a polynomial in terms of x is: (a) x2 + 7x + 22 (b) 1 x2  7 x  12 12 12 (c) 1 x2  7 x  10 (d) None of these 12 12 (ii) Degree of the polynomial is: (a) 1 (b) 2 (c) 3 (d) 4

Polynomials 19 (iii) If the total number of students is 96, then how many students planned to visit the historical monuments ? (a) 768 (b) 762 (c) 760 (d) 764 (iv) Number of students planning to visit old age homes is: (a) 54 (b) 52 (c) 56 (d) 60 (v) Total number of students planned to visit historical monuments and old age homes is: (a) 804 (b) 824 (c) 816 (d) 814 4. Rakesh sells x kg of apples at the rate of ` 80 per kg, some oranges at the rate of ` 48 per kg, whose quantity is equal to the square of the apple quantity. Along that he sells 10 kg pomegranate at the rate of ` 120 per kg. Based on the above information, answer the following questions: (i) Equation of the total cost of the quantity is: (a) C(x) = 48x2 + 80x + 1200 (b) C(x) = 20x2 + 40x + 600 (c) C(x) = 40x2 + 40x + 800 (d) C(x) = 40x2 + 80x + 400 (ii) Equation of the total quantity is: (b) q(x) = x2 + x + 10 (a) q(x) = x2 + x + 5 (d) q(x) = x2 + 3x + 15 (c) q(x) = x2 + 2x + 5 (iii) The mathematical concept used in this case is: (a) number system (b) polynomial concept (c) linear equation (d) real number concept (iv) Degree of an equation of total cost of quantity is: (a) 1 (b) 3 (c) 2 (d) 0 (v) If Rakesh sells 5 kg of apples, then how much money does he earn ? (a) ` 2400 (b) ` 2200 (c) ` 2600 (d) ` 2800 Answers Part – A Very Short Answer Type Questions Practice Exercise 1. (i), (ii), (iii), (iv), (vii) because the exponent of the variable, after simplification, in each of these is a whole number. 2. (i) Monomial; 0 (ii) Trinomial; 2 (iii) Monomial; 3 3. (i) 9 (ii) 16  11 (iii) 4a4 + 5a3 – a2 + 6 4. 61, –143 5. (i) 3, –17, –39 (ii) –3, –3, 0. 6. 5 7. 20. 8. (i) 7 (ii) 0 2 9. (i) 9x2 + 16y2 + 25z2 + 24xy + 40yz + 30zx 10. (i) 16a2 + b2 + 4c2 – 8ab – 4bc + 16ac (ii) 16a2 + 9b2 + 9c2 – 24ab + 18bc – 24ca (ii) 9a2 + 25b2 + c2 – 30ab + 10bc – 6ac

20 Mathematics—IX 11. (i) x2 + 4y2 + 9z2 – 4xy – 12yz + 6xz (ii) 9a2 + 49b2 + c2 – 42ab + 14bc – 6ca 12. (i) 2(7 + r) (6 – r) (ii) (3x – 1) (2x + 3) (iv) (3x – 2) (3x – 2) (iii) 3(x – 1) (3x – 1) Part – B Very Short Answer Type Questions Practice Exercise 1. Yes, 2 and 0 are both zeros 2. –2 is a zero, but 2 is not. 3. (i) x4 + 13x2 + 36 (ii) z6 + 15z3 + 14 (iii) p4  63 p2  4 4 (iv) y4  73 y 2  2 (v) 4x2 + 4x – 15 (vi) 9x2 – 6x – 35 35 4. (i) 10302 (ii) 6557 5. (i) 18 (ii) 36 (iii) 322 6. (i) (1 + 4x)(1 – 4x + 16x2)    (ii) a  2b a2  2ab  2b2 (iii) (4x + 5y)(16x2 + 25y2 – 20xy) 7. (i) (2x – y + z) (2x – y + z) (ii) (3x + 2y – 4z)(3x + 2y – 4z) (iii) (–5x + 4y + 2z)(–5x + 4y + 2z) (iv) (4x – 2y – 3z)(4x – 2y – 3z) Short Answer Type Questions Practice Exercise 1. 0 2. k = 4;  1 3. x=  1 4. 4  2 5. (i) 34 (ii) 5 2 3 6. 2p2 + 2q2 + 2r2 + 4qr 7. (i) 52 (ii) 18 (iii) 110. 8. 2y3 + 2x2y 11. 0 125 15 13. 331 9. 5050 10. 721 729 12. (18a – 9b – 13) (2a – b + 1) 14. – 31 4 Long Answer Type Questions Practice Exercise 1. 29 2. (x3 + 4) (x3 + 2) 3. (p + q – 25) (p + q + 5) 4. (a + 2) (a – 4) (a – 5) (a + 3) 5. (x – 1) (x + 1) (x – 2) (x + 3) 1. (i) (b) (ii) (d) (iii) (c) Case Studies 2. (i) (d) (ii) (b) (iii) (a) (iv) (b) (v) (c) 3. (i) (c) (ii) (b) (iii) (a) 4. (i) (a) (ii) (b) (iii) (b) (iv) (c) (v) (b) (iv) (c) (v) (d)

Unit 2. Quadrilaterals  IMPORTANT RESULTS AND FORMULAE  In this chapter, we will study the following points to remember: 1. Quadrilateral: It is a plane, closed, geometric figure with four sides, bounded by four line-segments. It is formed by joining four points in an order. D C Vertex Diagonals AB 2. Angle-Sum-Property of a Quadrilateral: “The sum of angles of a quadrilateral is 360°”. 3. Parallelogram Theorem: A diagonal of a parallelogram divides it into two congruent triangles. 4. Properties of a Parallelogram: (i) Opposite sides are equal. (ii) Opposite angles are equal. (iii) Diagonals bisect each other. (iv) Each diagonal bisects the parallelogram 5. Criterion For a Quadrilateral to be a gm: A is a gm, if its (i) Opposite sides are equal. (ii) Opposite angles are equal. (iii) Diagonals bisect each other. (iv) A pair of opposite sides is equal and parallel. 6. Diagonal Property of Rectangle, Rhombus and Square: (i) Diagonals of a rectangle bisect each other and are equal and vice versa. (ii) Diagonals of a rhombus bisect each other at right angles and vice versa. (iii) Diagonals of a square bisect each other at right angles and are equal, and vice versa. 7. The Mid-point Theorem: The line segment joining the mid-points of two sides of a triangle is parallel to the third side. 8. The Converse of Mid-point Theorem: The line drawn through the mid-point of one side of a triangle, parallel to another side bisects the third side. 9. Equal Intercept Theorem: If there are three or more parallel lines and the intercepts made by them on one transversal are equal, then the corresponding intercepts on any other transversal are also equal. 21

22 Mathematics—IX PART – A  VERY SHORT ANSWER TYPE QUESTIONS  Illustrative Examples EXAMPLE 1. In parallelogram ABCD of Fig. 2.1,  DAB = 60° and  DBC = 80°. Find,  ABD. [CBSE 2014] SOLUTION. In parallelogram ABCD, DC AD  BC | Opposite sides of a parallelogram 60° 80° and BD is a transversal to them A B   ADB = alt.  DBC = 80° Fig. 2.1 ...(1) In  DAB by Angle-Sum-Property of a triangle, we have | Using (1)  ABD = 180° – [ A +  ADB] = 180° – (60° + 80°) = 180° – 140° = 40°. Hence, the required  ABD = 40°. EXAMPLE 2. ABCD is a parallelogram in which  ADC = 75° and D C side AB is produced to point E as shown in Fig. 2.2. Find 75° x (x + y). [CBSE 2015] SOLUTION. In parallelogram ABCD, y A BE   C and  D are consecutive angles.   C +  D = 180° Fig. 2.2  x + 75° = 180°  x = 180° – 75° = 105° ...(1) Also, DC  AB and BC is a transversal to them | alternate angles  y=x ...(2)  y = 105° | From (1) and (2)  x + y = 105° + 105°  x + y = 210°. EXAMPLE 3. Three angles of a quadrilateral are 60°, 110° and 86°. Find the fourth angle of the quadrilateral. [CBSE 2012] SOLUTION. Let the measure of the fourth angle be x. Then, by the Angle-Sum Property of a quadrilateral, we have 60° + 110° + 86° + x = 360°  256° + x = 360°  x = 360° – 256° = 104° Thus, the required 4th angle is 104°.

Quadrilaterals 23 EXAMPLE 4. The angles of a quadrilateral are in the ratio 3 : 5 : 9 : 13. Find all the angles of the quadrilateral. [NCERT; CBSE 2012] SOLUTION. Let the angles  A,  B,  C and  D of the quadrilateral ABCD be 3x, 5x, 9x and 13x. Using ‘Angles-Sum-Property’ of a quadrilateral, we have  A +  B +  C +  D = 360° C D  3x + 5x + 9x + 13x = 360° or 30x = 360°  x = 12° Hence,  A = (3x) = (3 × 12)° = 36° A B  B = (5x) = (5 × 12)° = 60°  C = (9x) = (9 × 12)° = 108° Fig. 2.3 and  D = (13x) = (13 × 12)° = 156° Hence, the required four angles of the quadrilateral are 36°, 60°, 108º and 156°. EXAMPLE 5. In  ABC, D and E are the mid-points of A D sides AB and AC respectively as shown in Fig. 2.4. Find the length DE. [CBSE 2016] 4.6 cm 6.4 cm SOLUTION. In  ABC, D and E are the mid-points of E AB and AC respectively  By Mid-point Theorem, B C DE  BC and DE = 1 BC 8.2 cm 2 Fig. 2.4  DE = 1 × 8.2 = 4.1 cm. 2  PRACTICE EXERCISE  1. The angles A, B, C and D of a quadrilateral have measures in the ratio 2 : 4 : 5 : 7. Find the measurements of these angles. What type of quadrilateral is it ? Give reasons. [CBSE 2012] 2. In a parallelogram PQRS, if  QRS = 2x,  PQR = 4x and  PSQ = 4x, find the angles of the parallellogram. [CBSE 2012] 3. ABCD is a rhombus with  ABC = 50°. Determine  ACD. [CBSE 2012] 4. ABCD is a parallelogram in which  ADC = 75° and D C side AB is produced to point E as shown in x Fig. 2.5. Find (x + y). [CBSE 2015] 75° 5. Two consecutive angles of a parallelogram are (x + 60)° and (2x + 30)°. What special name can you give to this parallelogram? [CBSE 2016] y 6. One angle of a quadrilateral is 108° and the remaining A B E three angles are equal. Find each of the three equal Fig. 2.5 angles. [NCERT (EP)]

24 Mathematics—IX PART – B  VERY SHORT ANSWER TYPE QUESTIONS  Illustrative Examples DX C EXAMPLE 1. In Fig. 2.6, ABCD is a parallelogram in which X and Y are the mid-points of the sides DC and AB respectively. Prove that AXCY is a parallelogram. [CBSE 2012] SOLUTION. In parallelogram ABCD, AY B AB = DC | Opposite sides of a gm Fig. 2.6  1 AB = 1 DC 2 2  AY = CX ...(1) |  X and Y are the midpoints of DC and AB respectively Also, AY  CX ...(2) |  AB  DC (1) and (2) implies AXCY is a parallelogram. |  A pair of opposite sides equal and parallel EXAMPLE 2. In Fig. 2.7, AY and CX are respectively the bisectors D Y C of the opposite angles A and C of a gm ABCD. Show that AX  CY. [CBSE 2012] SOLUTION.  ABCD is a gm in which  A and  C are a A X B pair of opposite angles. Fig. 2.7  A = C  1   A = 1 C   YAX =  YCX ...(1) Also, 2 2   AYC +  YCX = 180° ...(2) |  YA  CX From (1) and (2), we get  AYC +  YAX = 180°  AX  CY ...(3) |  Interior angles on the same side of the transversal are supplimentary EXAMPLE 3. In quadrilateral ABCD of Fig. 2.8, X and Y are points DC on diagonal AC such that AX = CY and BXDY is a parallelogram. Show that ABCD is a parallelogram. Y O [CBSE 2015, 16] X SOLUTION.  BXDY is a parallelogram B  XO = YO ...(1) A Fig. 2.8 and ...(2) But, |  Diagonals bisect each other ...(3) | Given BO = D O AX = CY

Quadrilaterals 25 Adding (1) and (3), we get XO + AX = YO + CY  AO = C O ...(4) From (2) and (4), ABCD is a parallelogram. |  Diagonals bisect each other EXAMPLE 4. Lines are drawn through vertices P, Q and R of a  PQR C P B which are parallel to the sides QR, PR and PQ and form a  ABC as shown in Fig. 2.9. Show that the perimeter of  PQR is equal to half the perimeter of  ABC. [CBSE 2016] Q R SOLUTION. In  ABC,  PC  QR and QC  PR | Given A  PCQR is a gm.  | Fig. 2.9 PC = QR Opposite sides of a gm are equal Similarly, PB = QR  P is the mid-point of BC Similarly, Q is the mid-point of AC  By Mid-point Theorem in  PQR, PQ  AB and PQ = 1 AB ...(1) 2 Similarly, QR  BC and QR = 1 BC ...(2) 2 and PR  AC and PR = 1 AC ...(3)  Perimeter of 2  PQR = PQ + QR + PR = 1 AB + 1 BC + 1 AC | By using (1), (2) and (3) 2 2 2 = 1 (AB + BC + AC) = 1 × (perimeter of  ABC) 2 2 Hence proved.  PRACTICE EXERCISE  1. Show that each angle of a rectangle is right angle. [NCERT, CBSE 2012] 2. Show that the diagonals of a rhombus are perpendicular AD to each other. [NCERT, CBSE 2013] 75° 3. In the quadrilateral ABCD in Fig. 2.10, AB  DC and AB = BC DC. If  B = 75°, find the measurement of other angles. Fig. 2.10 [CBSE 2015]

26 Mathematics—IX 4. ABCD is a parallelogram. The angle bisectors of  A and  D intersect at O. Find the measure of  AOD. [CBSE 2012] DC O AB Fig. 2.11  SHORT ANSWER TYPE QUESTIONS  Illustrative Examples D C EXAMPLE 1. Diagonal AC of a parallelogram ABCD bisects  A (see Fig. 2.12) Show that: (i) it bisects  C also, [CBSE] (ii) ABCD is a rhombus. [CBSE 2012] A B SOLUTION. Given. A gm ABCD in which diagonal AC bisects Fig. 2.12 A ...(1) i.e.,  1 =  2 ...(2) | Given To prove. The diagonal AC bisects  C i.e.,  3 =  4 Proof. (i) Since AB  CD and AC is a transversal to them  By Alternate Angles Theorem  1 = alt.  4  2 = alt.  3 But  1 =  2 From (1) and (2), 3 = 4 Hence, the diagonal AC bisects  C. (ii) Now, in gm ABCD, AC is a diagonal  AC divides gm ABCD into two congruent triangles. DC   ADC   ABC 4 3  By CPCT  1 = 4 and  2 =  3 2 But 1 = 2 and  3 =  4 2 = 4 and  1 =  3 1 B  CD = AD and BC = AB A Fig. 2.13 But AB = CD and BC = AD |  Opposite sides of a gm are equal

Quadrilaterals 27 Thus, in gm ABCD, we have Q.E.D. AB = BC = CD = DA.  gm ABCD is a rhombus. EXAMPLE 2. Show that the diagonals of a square are equal and bisect each other at right angles. [NCERT, CBSE 2012] SOLUTION. Given. A square ABCD in which DC (i) AC and BD are the diagonals (ii) AC and BD intersect each other at the point O. To prove. O (i) AC = BD AB (ii)  AOB =  AOD = 90° Proof. In  ABC and  BAD, we have Fig. 2.14 S  AB = BA | Common A   ABC =  BAD | Each = 90° S  BC = AD Sides of a square are equal |  By SAS Congruency Theorem  ABC   BAD  By CPCT AC = BD Now, in  AOB and  AOD, we have S  OA = OA | Common S  AB = AD |  Sides of a square are equal S  OB = OD |  The diagonals of a square bisect each other  By SSS Congruency Rule,  AOB   AOD  By CPCT  AOB =  AOD By Linear-Pair-Axiom, we have  AOB +  AOD = 180°   AOB =  AOD = 90° Hence, the diagonals of a square are equal and bisect each other at right angles. EXAMPLE 3. If the diagonals of a parallelogram are equal then, prove that it is a rectangle. [NCERT, CBSE 2012] SOLUTION. Given. A gm ABCD in which DC (i) AC and BD are the diagonals such that AC = BD (ii) AC and BD intersect each other at the point O. O To prove. gm ABCD is a rectangle AB Proof. In  ABC and  DCB, we have Fig. 2.15 |  Opposite sides of the a gm S  AB = DC S  BC = CB | Common S  AC = DB | Given

28 Mathematics—IX By SSS Congruency Rule,  ABC   DCB.  By CPCT,  ABC =  DCB ...(1) But, DC  AB and BC is a transversal to them forming  ABC and  DCB a pair of interior angles  By Interior Angles Axiom,  ABC +  DCB = 180° ...(2) From (1) and (2), we have  ABC =  DCB = 90°  ABCD is a gm whose one angle is 90°. Hence, ABCD is a rectangle. Q.E.D. EXAMPLE 4. ABCD is a parallelogram and AP and CQ are D C perpendiculars from vertices A and C on diagonal BD P respectively (see Fig. 2.16). Show that B (i)  APB   CQD. C (ii) AP = CQ. [CBSE 2012] Q SOLUTION. Given. A gm. ABCD in which AP and CQ are A Fig. 2.16 perpendicular from A and C on the diagonal BD respectively. To prove. (i)  APB   CQD (ii) AP = CQ Proof. Since AB  DC and a transversal BD intersects D them, forming  1 and  2 as a pair of alt.  s 2P 3  By Alternate Angles Theorem 4 Q1 1 = 2 ...(1) A B Fig. 2.17 Now, in  APB and  CQD, we have A   1 =  2 | By (1) A   3 =  4 |  Each = 90° (Given) S  AB = CD |  Opposite sides of a gm are equal  By AAS Congruence Criterion  APB   CQD  By CPCT AP = C Q EXAMPLE 5. Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus. [NCERT, CBSE] SOLUTION. Given. A ABCD in which (i) the diagonals AC and BD bisect each other at O i.e., OA = OC, OB = OD

Quadrilaterals 29 (ii) AC  BD i.e.,  AOB =  BOC =  COD =  DOA = 90° To prove. ABCD is a rhombus D C i.e., AB = BC = CD = DA. Proof. In  AOB and  BOC, we have S  OA = OC | Given O A   AOB =  BOC | Given S  OB = OB | Common  By SAS Congruence Rule, AB   AOB   COB Fig. 2.18 ...(1)  By CPCT AB = BC Similarly,  BOC   DOC  BC = CD ...(2) and  COD   AOD  CD = DA ...(3) From (1), (2) and (3), we get AB = BC = CD = DA  ABCD is a rhombus. D FC EXAMPLE 6. In a parallelogram ABCD, E and F are the mid-points of P sides AB and CD respectively (see Fig. 2.19), show that the line segments AF and EC trisect the diagonal BD. [NCERT, CBSE 2013] Q SOLUTION. Let the diagonal BD be intersected by CE and AF at Q and P respectively. AEB  AB  DC | Given Fig. 2.19 and AB = D C |  Opposite sides of a gm are equal  AE  FC and AE = 1 AB, FC = 1 CD 2 2  AE  FC and AE = FC |  AB = CD  AECF is a gm.  AF  CE and EQ  AP and FP  CQ. Now, in  BAP, E is the mid-point of AB and EQ  AP.  By the Converse of Mid-point Theorem Q is the mid-point of BP.  BQ = PQ ...(1) Now, in  DQC, F is the mid-point of DC and FP  CQ, so P is the mid-point of DQ  DP = PQ ...(2) From (1) and (2), we get BQ = QP = DP Hence, CE and AF trisect the diagonal BD of gm ABCD.

30 Mathematics—IX EXAMPLE 7. In Fig. 2.20, D, E and F are, respectively the mid-points of the sides BC, CA and AB of an equilateral triangle ABC. Show that  DEF is also an equilateral triangle. [CBSE 2012] SOLUTION. In  ABC,  D and E are the mid-points of BC and CA respectively  By Mid-point Theorem, DE = 1 AB ...(1) A 2 Similarly, FE = 1 BC ...(2) 2 ...(3) | Given FE and DF = 1 CA. 2 B D Also,  ABC is an equilateral triangle ...(4) Fig. 2.20 C  AB = BC = CA  1 AB = 1 BC = 1 CA 2 2 2 From (1), (2), (3) and (4), we have  DE = FE = DF Thus, in  DEF, all the three sides are equal.   DEF is an equilateral triangle.  PRACTICE EXERCISE  1. Prove that the quadrilateral formed by the bisectors of the P angles of a parallelogram is a rectangle. [NCERT (EP), CBSE] A D 2. ABC is an isosceles triangle in which AB = AC. AD bisects  PAC and CD  AB (see Fig. 2.21) show that: (i)  DAC =  BCA. BC (ii) ABCD is a parallelogram. [CBSE 2012] Fig. 2.21 3. In a parallelogram, show that the angle bisector of two adjacent angles intersect at right angles. [CBSE 2012] 4. If a diagonal of a gm bisects one of the angles of the gm, it also bisects the second angle and then prove that the two diagonals are perpendicular to each other. 5. Prove that the diagonals of a rectangle are equal. [CBSE 2012] [CBSE 2011] 6. The diagonals of rectangle ABCD intersect at O. If  OAD = 68°, then find  BOC. [CBSE 2016] 7. In  ABC, D, E and F are respectively the mid-points of the sides AB, BC and CA, prove that  ABC is divided into four congruent triangles by joining D, E and F. (see Fig. 2.22) [CBSE 2012] Fig. 2.22

Quadrilaterals 31 8. In  ABC, AD is a median through A, E is the mid-point of AD and BE produced meets AC in X in Fig. 2.23. Prove that AX = 1 AC. [CBSE 2012, 14] 3 X Fig. 2.23 Fig. 2.24 9. In Fig. 2.24, points A, B are on the same side of a line l. AD  l and BE  l, meet l in D and E respectively, C is the mid-point of AB. Prove that CD = CE. [CBSE 2012]  LONG ANSWER TYPE QUESTIONS  Illustrative Examples EXAMPLE 1. ABCD is a rhombus. Show that diagonal AC bisects  A as well as  C and diagonal BD bisects  B as well as  D. [NCERT, CBSE 2012, 15] SOLUTION. Given. A rhombus ABCD in which AC and BD are the diagonals. To prove. (i) AC bisects  A and  C (ii) BD bisects  B and  D Proof. (i) Since AB  DC and AC is a transversal to them  By Alternate Angles Theorem D C 2  1 = alt.  2 ...(1) 4 Similarly,  3 = alt.  4 Now, in rhombus ABCD, AB and BC are two consecutive sides.  AB = BC 3 1  By Isosceles Triangle Property in  ABC, A  4 = 1 ...(2) B Fig. 2.25 (1) and (2)   1 =  3 and  2 =  4 Hence, the diagonal AC bisects  A and  C (ii) Similarly, the diagonal BD bisects  B and  D. EXAMPLE 2. Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square. [NCERT, CBSE 2012] SOLUTION. Given. A PQRS in which PR and QS are the diagonals S R intersecting each other at O such that. (i) PR = QS O (ii) PR and QS bisect each other at O. i.e., PO = RO and OQ = OS PQ Fig. 2.26

32 Mathematics—IX (iii) PR  QS i.e.,  POQ =  POS = 90º To prove. PQRS is a square Proof. In  POQ and  POS, we have S  OQ = OS | Given A   PO Q =  POS | Each = 90° (Given) S  PO = PO | Common By SAS Congruency Rule,  POQ   POS  By CPCT PQ = PS Similarly, QR = RS Now, in  POQ and  ROS, we have S   PO = RO | Given A  POQ =  ROS | Vertically opposite angles are equal S  O Q = OS | Given By SAS, Congruency Axiom,  POQ   ROS  By CPCT PQ = RS Similarly, QR = PS ...(1) Thus, in PQRS, opposite sides are equal.  PQRS is a gm ...(2) Now, in a gm, the sum of consecutive angles is 180°.   QPS +  PQR = 180° ...(3) Now, in  PQS and  QPR, we have S  PQ = QP | Common S  QS = PR | Given S  PS = QR | By (1)  By SSS Congruency Rule,  PQS   QPR. By CPCT,  QPS =  PQR ...(4) From (3) and (4), we get  QPS =  PQR = 90° | Each (2) and (3)  gm PQRS is a square. EXAMPLE 3. In a parallelogram ABCD, two points P and Q AD are taken on diagonal BD such that DP = BQ as shown in P Fig. 2.27. Q Show that: (ii) AP = CQ B C (i)  APD   CQB (iv) AQ = CP [NCERT, CBSE 2012] Fig. 2.27 (iii)  AQB   CPD (v) APCQ is a parallelogram.

Quadrilaterals 33 SOLUTION. Given. A gm ABCD in which (ii) AP = CQ (i) two points P and Q lie on BD (iv) AQ = CP (ii) DP = BQ To prove. (i)  APD   CQB (iii)  AQB   CPD (v) APCQ is a gm . Proof. (i) In s APD and CQB, we have S   DP = BQ | Given A  ADP =  CBQ | Alternate angles,  BC  AD S  AD = CB Opposite sides of a gm are equal |  By SAS Congruency Rule,  APD   CQB (ii) Since  APD   CQB  By CPCT AP = CQ (iii) Now, in  AQB and  CPD, we have S  AB = CD |  Opposite sides of a  gm are equal A   ABQ =  CDP | Alternate angles since AB  CD | Given S  BQ = DP By SAS Congruency Rule,  AQB   CPD (iv) Since,  AQB   CPD  By CPCT AQ = CP (v) Now, in APCQ, we have | Proved AQ = CP | Proved and AP = CQ  APCQ is a gm. EXAMPLE 4. ABCD is a trapezium in which AB  CD and AD = BC [See Fig. 2.28(i)]. Show that: (i)  A =  B (ii)  C =  D (iii)  ABC   BAD. [CBSE 2012] (iv) Diagonal AC = Diagonal BD. [NCERT] SOLUTION. Given. A trapezium  ABCD in which AB (i) AB  DC and (ii) AD = BC To prove. (i)  A =  B DC (ii)  C =  D (iii)  ABC   BAD Fig. 2.28 (i) (iv) Diagonal AC = Diagonal BD.

34 Mathematics—IX Construction. (i) Through C, draw CX  DA (ii) Also, produce AB and let it intersect CX at a point E. Proof. (i) In trapezium ABCD, since AB  CD | Given and AD  EC | By Const.  AECD is a gm  AD = CE |  Opposite sides of a gm are equal But AD = BC | Given  BC = CE X  By Isosceles Triangle Property in  CBE,  1 = 2 A B Since, BC stands on AE forming  2 and  3 a 4 32 1 E linear pair of angles  By Linear Pair Axiom  2 +  3 = 180° ...(1) Also, since AD  EC | By Const. D C and AE is a transversal to them forming  1 and  4 Fig. 2.28 (ii) a pair of interior angles ...(2)  By Interior Angles Axiom |   1 =  2  1 +  4 = 180° From (1) and (2), we get 2 + 3 = 1 + 4  3 = 4   ABC =  DAB   B =  A (ii) Again, since AB  CD and AD is a transversal to them forming  A and  D a pair of interior angles  By Interior Angles Axiom  A +  D = 180°   B +  D = 180° ...(1) |   A =  B, proved in part (i) Also, AB CD and BC is a transversal to them forming  B and  BCD a pair of interior angles.  By Interior Angles Axiom  B +  BCD = 180° ...(2) From (1) and (2), we get  B +  BCD =  B +  D or  BCD =  D   C =  D (iii) Now, in  ABC and  BAD, we have S  AB = AB | Common A   ABC =  BAD | Proved S  BC = AD | Given  By SAS Congruence Axiom  ABC   BAD

Quadrilaterals 35 (iv) Since,  ABC   BAD | Proved in (iii) Q.E.D.  By CPCT AC = BD Hence, diagonal AC = diagonal BD. EXAMPLE 5. In a quadrilateral ABCD, the line segments bisecting  C and  D meet at E. Prove that  A +  B = 2 CED. [CBSE 2012] SOLUTION. Let CE and DE be the bisectors of  C and  D D C respectively. 2 1 Then, 1 = 1  C and 2 = 1  D E 2 2 In  DEC, by Angle Sum Property of a Triangle, we have A B  1 +  2 +  CED = 180° Fig. 2.29   CED = 180° – ( 1 +  2) ...(1) Now, by Angle Sum Property of a Quadrilateral, we have  A +  B +  C +  D = 360°  1 ( A +  B) + 1  C + 1  D = 360 = 180° | On dividing both the sides by 2 2 2 2 2  1 ( A +  B) + 1 + 2 = 180° 2  1 ( A +  B) = 180° – ( 1 +  2) ...(2) 2 From (1) and (2), we get 1 ( A +  B) =  CED   A +  B = 2 CED 2 Hence proved. EXAMPLE 6. ABCD is a quadrilateral in which P, Q, R and S are the DR mid-points of the sides AB, BC, CD and DA respectively (see Fig. 2.30) AC is a diagonal. Show that: C Q S (i) SR  AC and SR = 1 AC (ii) PQ = SR 2 (iii) PQRS is a parallelogram. [NCERT, CBSE 2012] A P B SOLUTION. In  ABC,  P and Q are the mid-points of sides AB Fig. 2.30 and CB respectively.  By Mid-point Theorem PQ  AC and PQ = 1 AC ...(1) 2 Again, in  ADC, S and R are the mid-points of sides AD and DC respectively.

36 Mathematics—IX  By Mid-point Theorem SR  AC and SR = 1 AC ...(2) 2 From (1) and (2), we get Thus, in PQ = SR and PQ  SR PQRS one pair of opposite sides is equal and parallel.  PQRS is a gm. Hence, (i) SR  AC and SR = 1 AC 2 (ii) PQ = SR (iii) PQRS is a gm. EXAMPLE 7. Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other. [CBSE 2012] SOLUTION. Let ABCD be a quadrilateral in which P, Q, R and S are RC the mid-points of AB, BC, CD and DA. Join B and D. D Also, join P, Q, R and S, in pairs, to form PQRS. S Q Now, in  ABD, P is the mid-point of AB and S is the mid-point of AD  By Mid-point Theorem, A PB 1 SP  BD and SP = 2 BD ...(1) Fig. 2.31 Again, in  BCD, Q is the mid-point of BC and R is the mid-point of CD.  By Mid-point Theorem, QR  BD and QR = 1 BD ...(2) 2 From (1) and (2), we have SP  QR and SP = QR Thus, in PQRS one pair of its opposite sides is equal and parallel.  PQRS is a gm in which PR and SQ are its diagonals.  PR and SQ bisect each other. |  The diagonals of a gm bisect each other. EXAMPLE 8. l, m and n are three parallel lines intersected by transversals p and q such that l, m, and n cut off equal intercepts AB and BC on p and q (As shown in the figure). Show that l, m and n cut off equal intercepts DE and EF on q also. [NCERT, CBSE 2012] SOLUTION. Given. Three lines l, m and n which are such that l  m  n. l, m and n are intersected by a transversal p at A, B and C respectively such that AB = BC.

Quadrilaterals 37 A transversal q intersects l, m and n at D, E and F respectively. To prove. DE = EF. Construction. Join A to F so that AF intersects m at G. Proof. Trapezium ACFD is divided into two triangles namely,  ACF and  AFD In  ACF,  B is the mid-point of AC | Given And BG  CF |  m  n  By, the Converse of Mid-point Theorem, G is the mid-point of AF Similarly, in  AFD, l  G is the mid-point of AF, and GE  AD By the Converse of Mid-point Theorem, E is the mid-point of DF,  DE = EF Hence, l, m and n cut off equal intercepts on q. Remark: Fig. 2.32 We notice that Example 8.41 leads us to an important result, called Equal Intercept Theorem, which is stated below: If there are three or more parallel lines and the intercepts made by them on one transversal are equal, then the corresponding intercepts on any other transversal are also equal. EXAMPLE 9. ABCD is a trapezium in which AB  DC. BD is a D C diagonal and E is the mid-point of AD. A line is drawn through E F E, parallel to AB, intersecting BC at F (see Fig. 2.33). Show that A B F is the mid-point of BC and EF = 1 (AB + DC). [CBSE 2012] 2 SOLUTION.  ABCD is a trapezium in which AB  DC Fig. 2.33 and E is mid-point of AD Let EF intersects BD at P. Then, in triangle DAB, through E, EP  AB intersects BD at P  By the Converse of Mid-point Theorem, P is the mid-point of BD. D C Now, in triangle BCD, P is the mid-point of BD. ...(1) Through P, PF  DC intersects BC at F. P By the Converse of Mid-point Theorem, EF F is the mid-point of BC. ...(2) A B ...(3) From (1) and (2) EF = EP + PF Fig. 2.34 where EP = 1 AB and PF = 1 CD 2 2  From (3) EF = 1 AB  1 CD = 1 (AB + CD) 2 2 2

38 Mathematics—IX Aliter:  ABCD is a trapezium in which AB  CD and AB  EF, | Given  AB  EF  CD. Also, AD and BC are transversal to AB  EF  CD. |  E is the mid-point of AD Now, AE = ED  By Equal Intercept Theorem, FC = FB  F is the mid-point of BC Hence, proved. EXAMPLE 10. P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively of a quadrilateral ABCD in which AC = BD. Prove that PQRS is a rhombus. [CBSE 2015] SOLUTION. Given. A  ABCD in which (i) P, Q, R and S are the mid-points of sides AB, BC, CD and DA respectively (ii) P, Q, R and S are joined, in pairs, to form  PQRS (iii) AC = BD To prove.  PQRS is a rhombus. Proof. In  ABC, | Given  P and Q are the mid-points of AB and BC respectively  PQ  AC and PQ = 1 AC ...(1) | By Mid-point theorem 2 In  ADC,  S and R are the mid-points of AD and DC respectively | Given  SR  AC and SR = 1 AC ...(2) | By Mid-point theorem From (1) and (2), we have 2 PQ  SR |  Two lines parallel to the same line are parallel to each other and PQ = SR   PQRS is a parallelogram ...(3) |  One pair of opposite sides is equal and parallel In  ABD,  P and S are the mid-points of AB and AD respectively | Given D R C  PS  BD and PS = 1 BD ...(4) 2 | By Mid-point theorem S O Q But, AC = BD | Given  1 AC = 1 BD  PQ = PS ...(5) A P B 2 2 | From (1) and (4) Fig. 2.35

Quadrilaterals 39 Thus,  PQRS is a parallelogram with PQ = PS | From (3) and (5)   PQRS is a rhombus |  A parallelogram with a pair of adjacent sides equal is a rhombus Hence, proved.  PRACTICE EXERCISE  1. In Fig. 2.36, PQRS is a parallelogram, in which PQ is produced to T such that PQ = QT. (a) Prove that TS bisects RQ. (b) PS = 10 cm, find OQ. [CBSE 2012] p SR Pl O Q S P QT m R Fig. 2.36 Fig. 2.37 2. In Fig. 2.37, l  m. Show that PQRS formed by the bisectors PQ, PS, RQ and RS of the interior angles is a rectangle. [CBSE 2012, 2013] 3. E and F are the mid-points of the non-parallel sides of the trapezium ABCD with AB  DC. Prove that EF  AB and EF = 1 (AB + CD). [NCERT (EP)] 2 4. In Fig. 2.38, X and Y are respectively the mid-points of the opposite sides AD and BC of a parallelogram ABCD. Also, BX and DY intersect AC at P and Q, respectively. Show that AP = PQ = QC. [NCERT (EP)] 5. Prove that the line segment joining the mid-points of the diagonals of a trapezium is parallel to the parallel sides. [CBSE 2012] A P B DC Q D QC XY P B A R Fig. 2.38 Fig. 2.39 6. P is the mid-point of side AB of a gm ABCD. A line through B parallel to PD meets DC at Q and AD produced at R (see Fig. 2.39) Prove that: (i) AR = 2BC (ii) BR = 2BQ [CBSE 2012]

40 Mathematics—IX 7. In Fig. 2.40, 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. [CBSE 2012] DC E A BF Fig. 2.40 Fig. 2.41 8. In  ABC, BE  AC, AD is any line drawn from A to BC intersecting BE in M as shown in Fig. 2.41. X, Y and Z are respectively the mid-points of AM, AB and BC. Prove that  XYZ = 90°. [CBSE 2016] 9. In  ABC and  DEF, AB = DE, AB  DE, BC = EF and A D BC  EF. Vertices A, B and C are joined to vertices D, E F and F respectively (see Fig. 2.42). Show that: C (i) Quadrilateral ABED is a parallelogram (ii) Quadrilateral BEFC is a parallelogram B E (iii) AD  CF and AD = CF Fig. 2.42 (iv) Quadrilateral ACFD is a parallelogram (v) AC = DF (vi)  ABC   DEF [NCERT, CBSE 2012] 10. ABCD is a rectangle and P, Q, R and S are the mid-points of sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus. [NCERT, CBSE 2012] 11. 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. (iii) CM = MA = 1 AB. [NCERT, CBSE 2012] 2 DX C 12. In the trapezium ABCD of Fig. 2.43, the line-segment joining the mid-points X and Y of sides DC and AB respectively is AYB perpendicular to AB and AB  CD. Prove that non parallel sides of the trapezium are equal. [CBSE 2014] Fig. 2.43 13. Prove that the line-segment joining the mid-points of any two sides of a triangle bisects the line-segment drawn from the common vertex of the two sides to its opposite side. [CBSE 2015]

Quadrilaterals 41  CASE STUDIES  1. Rehaan and Sheaan are celebrating Republic Day by flying their kites high in the sky. In Rehaan’s kite ABCD, the diagonal AC bisects BD perpendicular at O, while in Sheaan’s kite PQRS, both the diagonals PR and QS bisect each other at 90° as shown below in Fig. 2.44 (a) and (b) respectively: Based on the above information, answer the following questions: A L P B OD X M TS Q Y CR (a) (b) Fig. 2.44 (i) What could be the possible shape of Sheaan’s kite ? (a) Trapezium (b) Rectangle (c) Rhombus (d) None of these (ii) What could be the possible shape of Rehaan’s kite ? (a) Trapezium (b) Rectangle (c) Rhombus (d) None of these (iii) If PR = QS, then Sheaan’s kite will come in the shape of a (a) Square (b) Rectangle (c) Rhombus (d) None of these (iv) Which of the following is false about Rehaan’s kite ? (a) AB = AD (b) BC = DC (c) BO = DO (d) AO = CO (v) If midpoints of sides of Sheaan’s kite are joined, the quadrilateral obtained is a (a) Square (b) Rhombus (c) Rectangle (d) None of these 2. In Fig. 2.45, there is a door with a shedding. Door is a rectangle and the shedding is in shape of a trapezium. Beeding design on door consists of two entangled rhombus with a parallelogram inside. Also, corners of door are decorated using beeding to form shape of squares. Based on the above information, answer the following questions: (i) Which of the following is true ? Fig. 2.45 (a) a trapezium is a parallelogram (c) a rhombus is a rectangle (b) a rectangle is a rhombus (d) a parallelogram is a quadrilateral

42 Mathematics—IX (ii) Which of the following is not true for a rhombus ? (a) diagonals bisect each other (b) opposite sides are equal (c) opposite angles are equal (d) diagonals are equal (iii) Each angle of a quadrilateral is 90°, then which of the following is false about it? (a) Its diagonals bisect each other (b) Its opposite sides are equal (c) Its diagonals are equal (d) None of these (iv) Which of the following is not true ? (a) Area of a parallelogram = 1 × base × corresponding altitude 2 (b) Area of a rectangle = length × breadth (c) Area of a trapezium = 1 (sum of parallel sides) × distance between them 2 (d) None of these (v) If midpoints of sides of a rectangle are joined, the quadrilateral obtained is a (a) square (b) rhombus (c) rectangle (d) None of these Answers Part – A Very Short Answer Type Questions Practice Exercise 1. 40°, 80°, 100°, 140°, trapezium 2. 60°, 120°, 60° and 120° 3. 65° 4. 210° 5. Rectangle 6. 84° Part – B Very Short Answer Type Questions Practice Exercise 3.  A = 105°,  C = 105°,  D = 75° 4. 90° Short Answer Type Questions Practice Exercise 6. 44° Long Answer Type Questions Practice Exercise 1. (b) 5 cm Case Studies 1. (i) (c) (ii) (d) (iii) (a) (iv) (d) (v) (a) 2. (i) (a) (ii) (d) (iii) (d) (iv) (a) (v) (b)