# RD Sharma Chapter 20 Class 9 Maths Exercise 20.2 Solutions

RD Sharma Chapter 20 Class 9 Maths Exercise 20.2 Solutions is based on the Volume of a Right Circular Cone, which is concerned under Surface Area And Volume of A Right Circular Cone. The solutions provided here can be useful for the students in preparing their own comfort zone strategically. Along with providing the answers, it can be useful in terms of a better understanding of the concepts.

We have attached the PDF with this article, which helps students practice different types of questions related to the exercise. The PDF is free to access, prepared by our experts with the help of Previous Year’s Question Paper, Text Book, and RD Sharma of Class 9.

Learn about RD Sharma Class 9 Chapter 20 Surface Area And Volume of A Right Circular Cone

## Download RD Sharma Chapter 20 Class 9 Maths Exercise 20.2 Solutions PDF

Solutions for Class 9 Maths Chapter 20 Surface Area and Volume of A Right Circular Cone Exercise 20.2

## Important Definitions RD Sharma Chapter 20 Class 9 Maths Exercise 20.2 Solutions

In this article, we will discuss the Volume of a Right Circular Cone with proof and examples. Go down for more information on the RD Sharma Chapter 20 Class 9 Maths Exercise 20.2 Solutions.

### The volume of a Right Circular Cone

The volume of a cone is one-third (⅓) of the outcome of the area of the base and the height of the cone. The volume is measured in the names of cubic units.

The volume of a right circular cone can be determined (calculated) by the following formula-

The volume of a right circular cone = ⅓ (Base area × Height)

Where,

Base Area = π r2

Hence, Volume = ⅓ π r2h

### Examples of Volume of a Circular Cone of RD Sharma Chapter 20 Class 9 Maths Exercise 20.2 Solutions

Ques- Determine the volume of the right round cone with:

(iii) Height is 21cm and slant height 28cm

Solution-

Height of cone(b)=7cm

We understand, Volume of a right circular cone = 1/3 πa2b

By changing the values, we get

= 1/3 x 3.14 x 62 x 7

= 264

The volume of the right circular cone is 264 cm3

Height of cone(b)=12cm

The volume of the right circular cone = 1/3 πa2b

By changing the values, we get

= 1/3 x 3.14 x 3.52 x 12

= 154

The volume of the right circular cone is 154 cm3

(iii) Height of cone(a)=21 cm

Slant height of cone(b) = 28 cm

Find the measure of c:

We know, b2 = c2 + a2

282 = c2 + 212

or c = 7√7

Now,

The volume of the right circular cone = 1/3 πc2a

By changing the values, we get

= 1/3 x 3.14 x (7√7)2 x 21

=7546

The volume of the right circular cone is 7546 cm3

Ques- Determine the capacity in liters of a conical vessel with:

(i) radius 7 cm, slant height 25 cm

(ii) height 12 cm, slant height 13 cm.

Solution-

(i) Radius of the cone(a) =7 cm

Slant height of the cone (b) =25 cm

As we know that, b2 = a2 + c2

252 = 72 + c2

or c = 24

Now, Volume of the right circular cone = = 1/3 πa2c

By changing the values, we get

= 1/3 x 3.14 x (7)2 x 24

= 1232

The volume of a right circular cone is 1232 cm3 or 1.232 liters.

[1 cm3 = 0.01 liter]

(ii) Height of cone(a)=12 cm

Slant height of cone(b)=13 cm

As we know that, b2 = c2 + a2

132 = c2 + 122

or c = 5

Now, Volume of the right circular cone = 1/3 πc2a

By changing the values, we get

= 1/3 x 3.14 x (5)2 x 12

= 314.28

The volume of a right circular cone is 314.28 cm3 or 0.314 liters.

[1 cm3 = 0.01 liters]

Ques- Two cones have their heights in the ratio 1:3 and the radii of their bases in the ratio 3:1. Determine the ratio of their volumes.

Sol-

Let the heights of the cones are a and 3a and, the radii of their bases are 3b and b, respectively. Then, their volumes are

The volume of the first cone (V1) = 1/3 π(3b)2a

Volume of second cone (V2) = 1/3 πb2(3a)

Now, V1/V2 = 3/1

The ratio of the two volumes is 3:1.