RD Sharma Solutions for Class 9 Maths Chapter 14 Quadrilaterals Exercise 14.2

RD Sharma Class 9 Mathematics Chapter 14 Exercise 14.2 Solutions for Quadrilaterals are provided here. In this exercise, students will learn about the various types of quadrilaterals like rectangles, squares, rhombuses and many more. RD Sharma Solutions given here include the solutions of all the questions enlisted under Exercise 14.2. Students are advised to click on the below link and download the PDF now.

RD Sharma Solutions for Class 9 Maths Chapter 14 Quadrilaterals Exercise 14.2

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Access Answers to Maths RD Sharma Solutions for Class 9 Chapter 14 Quadrilaterals Exercise 14.2 Page number 14.18

Question 1: Two opposite angles of a parallelogram are (3x – 2)⁰ and (50 – x) ⁰. Find the measure of each angle of the parallelogram.

Solution:

Given: Two opposite angles of a parallelogram are (3x – 2)⁰ and (50 – x) ⁰.

We know, opposite sides of a parallelogram are equal.

(3x – 2)⁰ = (50 – x) ⁰

3x + x = 50 + 2

4x = 52

x = 13

Angle x is 13⁰

Therefore,

(3x-2) ⁰ = (3(13) – 2) = 37⁰

(50-x) ⁰ = (50 – 13) = 37⁰

Adjacent angles of a parallelogram are supplementary.

x + 37 = 180⁰

x = 180⁰ − 37⁰ = 143⁰

Therefore, required angles are : 37⁰, 143⁰, 37⁰ and 143⁰.

Question 2: If an angle of a parallelogram is two-third of its adjacent angle, find the angles of the parallelogram.

Solution:

Let the measure of the angle be x. Therefore, measure of the adjacent angle is 2x/3.

We know, adjacent angle of a parallelogram is supplementary.

x + 2x/3 = 180⁰

3x + 2x = 540⁰

5x = 540⁰

or x = 108⁰

Measure of second angle is 2x/3 = 2(108⁰)/3 = 72⁰

Similarly measure of 3^rd and 4^th angles are 108⁰ and 72⁰

Hence, four angles are 108⁰, 72⁰, 108⁰, 72⁰

Question 3: Find the measure of all the angles of a parallelogram, if one angle is 24⁰ less than twice the smallest angle.

Solution:

Given: One angle of a parallelogram is 24⁰ less than twice the smallest angle.

Let x be the smallest angle, then

x + 2x – 24⁰ = 180⁰

3x – 24⁰ = 180⁰

3x = 108⁰ + 24⁰

3x = 204⁰

x = 204⁰/3 = 68⁰

So, x = 68⁰

Another angle = 2x – 24⁰ = 2(68⁰) – 24⁰ = 112⁰

Hence, four angles are 68⁰, 112⁰, 68⁰, 112⁰.

Question 4: The perimeter of a parallelogram is 22cm. If the longer side measures 6.5cm what is the measure of the shorter side?

Solution:

Let x be the shorter side of a parallelogram.

Perimeter = 22 cm

Longer side = 6.5 cm

Perimeter = Sum of all sides = x + 6.5 + 6.5 + x

22 = 2 ( x + 6.5 )

11 = x + 6.5

or x = 11 – 6.5 = 4.5

Therefore, shorter side of a parallelogram is 4.5 cm

RD Sharma Solutions for Class 9 Maths Chapter 14 Quadrilaterals Exercise 14.2

RD Sharma Solutions Class 9 Maths Chapter 14 Quadrilaterals Exercise 14.2 is based on the following topics and subtopics:

• Various Types of Quadrilaterals

– Trapezium

– Isosceles Trapezium

– Parallelogram

– Rhombus

– Rectangle

– Square

• Properties of Parallelogram
• Some important theorems on Quadrilaterals