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## Download PDF of RD Sharma Solutions for Class 9 Maths Chapter 20 Surface Area and Volume of A Right Circular Cone Exercise 20.2

### Access Answers to Maths RD Sharma Class 9 Chapter 20 Surface Area and Volume of A Right Circular Cone Exercise 20.2 Page number 20.20

**Question 1: Find the volume of the right circular cone with:**

**(i) Radius 6cm, height 7cm**

**(ii)Radius 3.5cm, height 12cm**

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

**Solution: **

**(i)** Radius of cone(r)=6cm

Height of cone(h)=7cm

We know, Volume of a right circular cone = 1/3 Ï€r^{2}h

By substituting the values, we get

= 1/3 x 3.14 x 6^{2} x 7

= 264

Volume of a right circular cone is 264 cm^{3}

**(ii)** Radius of cone(r)=3.5 cm

Height of cone(h)=12cm

Volume of a right circular cone = 1/3 Ï€r^{2}h

By substituting the values, we get

= 1/3 x 3.14 x 3.5^{2} x 12

=154

Volume of a right circular cone is 154 cm^{3}

**(iii)** Height of cone(h)=21 cm

Slant height of cone(l) = 28 cm

Find the measure of r:

We know, l^{2 }= r^{2 }+ h^{2}

28^{2} = r^{2 }+ 21^{2}

or r = 7âˆš7

Now,

Volume of a right circular cone = 1/3 Ï€r^{2}h

By substituting the values, we get

= 1/3 x 3.14 x (7âˆš7)^{2} x 21

=7546

Volume of a right circular cone is 7546 cm^{3}

**Question 2: Find the capacity in litres 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(r) =7 cm

Slant height of the cone (l) =25 cm

As we know that, l^{2 }= r^{2 }+ h^{2}

25^{2} = 7^{2 }+ h^{2}

or h = 24

Now, Volume of a right circular cone = = 1/3 Ï€r^{2}h

By substituting the values, we get

= 1/3 x 3.14 x (7)^{2} x 24

= 1232

Volume of a right circular cone is 1232 cm^{3} or 1.232 litres

^{3}= 0.01 liter]

**(ii)** Height of cone(h)=12 cm

Slant height of cone(l)=13 cm

As we know that, l^{2 }= r^{2 }+ h^{2}

13^{2} = r^{2 }+ 12^{2}

or r = 5

Now, Volume of a right circular cone = 1/3 Ï€r^{2}h

By substituting the values, we get

= 1/3 x 3.14 x (5)^{2} x 12

= 314.28

Volume of a right circular cone is 314.28 cm^{3} or 0.314 litres.

^{3 }= 0.01 liters]

**Question 3: Two cones have their heights in the ratio 1:3 and the radii of their bases in the ratio 3:1. Find the ratio of their volumes.**

**Solution:**

Let the heights of the cones be h and 3h and radii of their bases be 3r and r respectively. Then, their volumes are

Volume of first cone (V1) = 1/3 Ï€(3r)^{2}h

Volume of second cone (V2) = 1/3 Ï€r^{2}(3h)

Now, V1/V2 = 3/1

Ratio of two volumes is 3:1.

**Question 4: The radius and the height of a right circular cone are in the ratio 5:12. If its volume is 314 cubic meter, find the slant height and the radius. (Use Ï€=3.14).**

**Solution:**

Let us assume the ratio of radius and the height of a right circular cone to be x.

Then, radius be 5x and height be 12x

We know, l^{2} = r^{2} + h^{2}

= (5x)^{ 2} + (12x)^{2}

= 25 x^{2} + 144 x^{2}

or l = 13x

Therefore, slant height is 13 m.

Now it is given that volume of cone = 314 m^{3}

â‡’ 1/3Ï€r^{2}h = 314

â‡’ 1/3 x 3.14 x (25x^{2} ) x (12x) = 314

â‡’ x^{3}=1

or x = 1

So, radius = 5x 1 = 5 m

Therefore ,

Answer: Slant height = 13m

Radius = 5m

**Question 5: The radius and height of a right circular cone are in the ratio 5 : 12 and its volume is 2512 cubic cm. Find the slant height and radius of the cone. (Use Ï€=3.14).**

**Solution:**

Let the ratio of radius and height of a right circular cone be y.

Radius of cone(r) = 5y

Height of cone (h) =12y

Now we know, l^{2} = r^{2} + h^{2}

= (5y)^{ 2} + (12y)^{2}

= 25 y^{2} + 144 y^{2}

or l = 13y

Now, volume of the cone is given 2512cm^{3}

â‡’ 1/3Ï€r^{2}h=2512

â‡’ 1/3 x 3.14 x (5y)^{2} x 12y = 2512

â‡’ y^{3} = (2512 x 3)/(3.14 x 25 x 12) = 8

or y = 2

Therefore,

Radius of cone = 5y = 5×2 = 10cm

Slant height (l) =13y = 13×2 = 26cm

**Question 6: The ratio of volumes of two cones is 4 : 5 and the ratio of the radii of their bases is 2 : 3. Find the ratio of their vertical heights.**

**Solution:**

Let the ratio of the radius be x and ratio of the volume be y.

Then, Radius of 1st cone (r_{1}) =2x

Radius of 2nd cone (r_{2}) =3x

Volume of 1st cone (V_{1})= 4y

Volume of 2nd cone (V_{2})= 5y

We know formula for volume of a cone = 1/3Ï€r^{2}h

Let h_{1} and h_{2} be the heights of respective cones.

Therefore, heights are in the ratio of 9 : 5.

**Question 7: A cylinder and a cone have equal radii of their bases and equal heights. Show that their volumes are in the ratio 3:1.**

**Solution:**

We are given, a cylinder and a cone are having equal radii of their bases and heights.

Let, radius of the cone = radius of the cylinder = r and

Height of the cone = height of the cylinder = h

Now,

Therefore, ratio of their volumes is 3:1.

## Access other exercise solutions of Class 9 Maths Chapter 20 Surface Area and Volume of A Right Circular Cone

## RD Sharma Solutions for Class 9 Maths Chapter 20 Exercise 20.2

Class 9 Maths Chapter 20 Surface Area and Volume of A Right Circular Cone Exercise 20.2 is based on topic – Volume of a Right Circular Cone. RD Sharma class 9 solutions is one of the best tools to practice Maths concepts and get ready for exams.