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# RD Sharma Solution for Class 9 Mathematics Chapter 13 - Quadrilaterals

Our RD Sharma Textbook Solutions are considered extremely helpful for solving the tough questions which are asked in the CBSE Class 9 exam. TopperLearning Textbook Solutions are compiled by subject experts. Herein, you can find all the answers to the textbook questions for Chapter 13 - Quadrilaterals.

All our solutions are created in accordance with the latest CBSE syllabus, and they are amended from time to time to provide the most relevant answers. Our free RD Sharma Solutions for CBSE Class 9 Mathematics will strengthen your fundamentals in Mathematics and will help you in your attempts to score more marks in the final examination. CBSE Class 9 students can refer to our solutions any time — while doing their homework and while preparing for the exam.

Exercise/Page

## RD Sharma Solution for Class 9 Mathematics Chapter 13 - Quadrilaterals Page/Excercise 13.1

Solution 1

Solution 2

Solution 3

Let the common ratio between the angles is x. So, the angles will be 3x, 5x, 9x and 13x respectively.
Since the sum of all interior angles of a quadrilateral is 360o.
3x + 5x + 9x + 13x = 360o
30x = 360o
x = 12o
Hence, the angles are
3x = 3  12 = 36o
5x = 5  12 = 60o
9x = 9  12 = 108o
13x = 13  12 = 156o

Solution 4

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

i. F

ii. T

iii. F

iv. F

v. T

vi. F

vii. F

viii. T

Solution 9

Solution 10

## RD Sharma Solution for Class 9 Mathematics Chapter 13 - Quadrilaterals Page/Excercise 13.3

Solution 1

C and D are cosecutive interior angles on the same side of the transversal CD. Therefore,

C + D = 180o

Solution 2

Solution 3

Since, diagonals of a square bisect each other at right angle. Therefore, AOB = 90o

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

## RD Sharma Solution for Class 9 Mathematics Chapter 13 - Quadrilaterals Page/Excercise 13.4

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Let ABCD is a quadrilateral in which P, Q, R and S are mid-points of sides AB, BC, CD and DA respectively.
Join PQ, QR, RS, SP and BD.
In ABD, S and P are mid points of AD and AB respectively.
So, By using mid-point theorem, we can say that
SP || BD and SP =   BD             ... (1)
Similarly in BCD
QR || BD and QR = BD               ... (2)
From equations (1) and (2), we have
SP || QR and SP = QR
As in quadrilateral SPQR one pair of opposite sides are equal and parallel to
each other.
So, SPQR is a parallelogram. Since, diagonals of a parallelogram bisect each other.
Hence, PR and QS bisect each other.

Solution 13

(i) isosceles

(ii) right triangle

(iii) parallelogram

Solution 14

Solution 15

Solution 16

Solution 17

Solution 18

Solution 19

Solution 20

## RD Sharma Solution for Class 9 Mathematics Chapter 13 - Quadrilaterals Page/Excercise 13.70

Solution 1

The opposite sides AB and DC, AD and BC have no common point.

Hence, correct option is (a).

Solution 2

Consecutive sides of a Quadrilateral ABCD are

AB and BC,

BC and CD,

which have only one point in common

i.e the joint point of their ends.

Hence, correct option is (b).

## RD Sharma Solution for Class 9 Mathematics Chapter 13 - Quadrilaterals Page/Excercise 13.71

Solution 3

Solution 4

For a rhombus, the angle between the diagonals is 90° and not 60°.

Hence, correct option is (d).

Solution 5

Diagonals necessarily bisect opposite angles in a square.

Hence, correct option is (d).

Solution 6

The two diagonals are equal in a rectangle (property).

Hence, correct option is (c).

Solution 7

Solution 8

Solution 9

AR, BR, CP, DP are the bisectors of angles of parallelogram.

Because two bisectors of adjacent angles make 90° between them So  PQRS is a Rectangle

Because DP and BR are acute angle bisectors so the distance between them PQ < PS (The distance between other two bisectors)

So PQ ≠ PS (So PQRS  is not a square, but only a rectangle)

Hence, correct option is (c).

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

Solution 15

Solution 16

Solution 17

## RD Sharma Solution for Class 9 Mathematics Chapter 13 - Quadrilaterals Page/Excercise 13.72

Solution 18

Solution 19

Solution 20

Sum of all angles of a Quadrilateral = 360°

4x + 7x + 9x + 10x = 360°

30x = 360°

x = 12°

So, sum of smallest and largest angle,

i.e. 4x + 10x = 14x = 14 × 12 = 168°

Hence, correct option is (c).

Solution 21

Solution 22

Solution 23

Solution 24

Solution 25

Solution 26

Solution 30

Solution 27

Solution 28

Solution 29

Solution 31

Solution 32

## Browse Study Material

TopperLearning provides step-by-step solutions for each question in each chapter in the RD Sharma textbook for class 9. Access the CBSE Class 9 Mathematics Chapter 13 - Quadrilaterals for free. The textbook questions have been solved by our subject matter experts to help you understand how to answer them. Our RD Sharma Textbook Solutions will help you to study and revise, and you can easily clear your fundamentals of Chapter 13 - Quadrilaterals.

# Text Book Solutions

CBSE IX - Mathematics

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