## RD Sharma Solutions Class 9 Maths Chapter 20 – Free PDF Download

**RD Sharma Solutions Class 9 Chapter 20** comes with questions and answers related to cone and its dimensions. The subject experts at BYJU’S outline the concepts in a clear and precise manner based on the IQ level of students. The solutions to all questions in RD Sharma Solutions are given here in a detailed and step by step way to help the students understand more effectively. This helps students to get a good score in the examinations, while also providing extensive knowledge about the subject. Check the detailed RD Sharma Solutions for Class 9 for the chapter “Surface areas and volumes of a right circular cone” given below.

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

### Access Answers to Maths RD Sharma Solutions for Class 9 Chapter 20 Surface Area and Volume of A Right Circular Cone

### Exercise 20.1 Page No: 20.7

**Question 1: **Find the curved surface area of a cone, if its slant height is 60 cm and the radius of its base is 21 cm.

Solution:

Slant height of cone (l) = 60 cm

Radius of the base of the cone (r) = 21 cm

Now,

Curved surface area of the right circular cone = πrl = 22/7 x 21 x 60 = 3960 cm^{2}

Therefore the curved surface area of the right circular cone is 3960 cm^{2}

**Question 2: The radius of a cone is 5cm and vertical height is 12cm. Find the area of the curved surface.**

**Solution:**

Radius of cone (r) = 5 cm

Height of cone (h) = 12 cm

Find Slant Height of cone (l):

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

l^{2 }= 5^{2 }+12^{2}

l^{2 }= 25 + 144 = 169

Or l = 13 cm

Now,

C.S.A = πrl =3.14 x 5 x 13 = 204.28

Therefore, the curved surface area of the cone is 204.28 cm^{2}

**Question 3 : The radius of a cone is 7 cm and area of curved surface is 176 cm ^{2} .Find the slant height.**

**Solution:**

Radius of cone(r) = 7 cm

Curved surface area(C.S.A)= 176cm^{2}

We know, C.S.A. = πrl

⇒ πrl = 176

⇒ 22/7 x 7 x l = 176

or l = 8

Therefore, slant height of the cone is 8 cm.

**Question 4: The height of a cone 21 cm. Find the area of the base if the slant height is 28 cm.**

**Solution:**

Height of cone(h) = 21 cm

Slant height of cone (l) = 28 cm

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

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

r^{2}=28^{2}−21^{2}

or r= 7√7 cm

Now,

Area of the circular base = πr^{2}

= 22/7 x (7√7 )^{2}

=1078

Therefore, area of the base is 1078 cm^{2}.

**Question 5: Find the total surface area of a right circular cone with radius 6 cm and height 8 cm.**

**Solution:**

Radius of cone (r) = 6 cm

Height of cone (h) = 8 cm

Total Surface area of the cone (T.S.A)=?

Find slant height of cone:

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

=6^{2}+8^{2}

= 36 + 64

= 100

or l = 10 cm

Now,

Total Surface area of the cone (T.S.A) = Curved surface area of cone + Area of circular base

= πrl + πr^{2}

= (22/7 x 6 x 10) + (22/7 x 6 x 6)

= 1320/7 + 792/7

= 301.71

Therefore, area of the base is 301.71cm^{2}.

**Question 6: Find the curved surface area of a cone with base radius 5.25 cm and slant height 10 cm.**

**Solution:**

Base radius of the cone(r) = 5.25 cm

Slant height of the cone(l) = 10 cm

Curved surface area (C.S.A) = πrl

=22/7 x 5.25 x 10

= 165

Therefore, curved surface area of the cone is 165cm^{2}.

**Question 7: Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.**

**Solution:**

Diameter of the cone(d)=24 m

So, radius of the cone(r)= diameter/ 2 = 24/2 m = 12m

Slant height of the cone(l) = 21 m

T.S.A = Curved surface area of cone + Area of circular base

= πrl+ πr^{2}

= (22/7 x 12 x 21) + (22/7 x 12 x 12)

= 1244.57

Therefore, total surface area of the cone is 1244.57 m^{2}.

**Question 8: The area of the curved surface of a cone is 60 π cm ^{2}. If the slant height of the cone be 8 cm, find the radius of the base.**

**Solution:**

Curved surface area(C.S.A)= 60 π cm^{2}

Slant height of the cone(l) = 8 cm

We know, Curved surface area(C.S.A )=πrl

⇒ πrl = 60 π

⇒ r x 8 = 60

or r = 60/8 = 7.5

Therefore, radius of the base of the cone is 7.5 cm.

**Question 9: The curved surface area of a cone is 4070 cm ^{2} and diameter is 70 cm .What is its slant height? (Use π =22/7)**

**Solution:**

Diameter of the cone(d) = 70 cm

So, radius of the cone(r)= diameter/2 = 70/2 cm = 35 cm

Curved surface area = 4070 cm^{2}

Now,

We know, Curved surface area = πrl

So, πrl = 4070

By substituting the values, we get

22/7 x 35 x l = 4070

or l = 37

Therefore, slant height of cone is 37 cm.

**Question 10: The radius and slant height of a cone are in the ratio 4:7. If its curved surface area is 792 cm ^{2}, find its radius. (Use π =22/7)**

**Solution:**

Curved surface area = 792 cm^{2}

The radius and slant height of a cone are in the ratio 4:7 (Given)

Let 4x be the radius and 7x be the height of cone.

Now,

Curved surface area (C.S.A.) = πrl

So, 22/7 x (4x) x (7x) = 792

or x^{2 }= 9

or x = 3

Therefore, Radius = 4x = 4(3) cm = 12 cm

### Exercise 20.2 Page No: 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 πr^{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 πr^{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 = 7√7

Now,

Volume of a right circular cone = 1/3 πr^{2}h

By substituting the values, we get

= 1/3 x 3.14 x (7√7)^{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 πr^{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 πr^{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 π(3r)^{2}h

Volume of second cone (V2) = 1/3 πr^{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 π=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}

⇒1/3πr^{2}h = 314

⇒1/3 x 3.14 x (25x^{2} ) x (12x) = 314

⇒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 π=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}

⇒1/3πr^{2}h=2512

⇒1/3 x 3.14 x (5y)^{2} x 12y = 2512

⇒ 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 = 1/3πr^{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.

### Exercise VSAQs Page No: 20.23

**Question 1: The height of a cone is 15 cm. If its volume is 500π cm ^{3}, then find the radius of its base.**

**Solution: **

Height of a cone = 15 cm

Volume of cone = 500 π cm^{3}

We know, Volume of cone = 1/3 πr^{2}h

So, 500π = 1/3 π r^{2} x 15

r^{2} = 100

or r = 10

Radius of base is 10 cm.

**Question 2: If the volume of a right circular cone of height 9 cm is 48π cm ^{3}, find the diameter of its base.**

**Solution: **

Height of a cone = 9 cm

Volume of cone = 48 π cm^{3}

We know, Volume of cone = 1/3 πr^{2}h

So, 48π = 1/3 π r^{2} x 9

r^{2} = 16

or r = 4

Radius of base r = 4 cm

Therefore, Diameter = 2 Radius = 2 x 4 cm = 8 cm.

**Question 3: If the height and slant height of a cone are 21 cm and 28 cm respectively. Find its volume.**

**Solution: **

Height of cone (h) = 21 cm

Slant height of cone (l) = 28 cm

Find radius of cone:

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

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

or r = 7√7 cm

Now,

We know, Volume of cone = 1/3 πr^{2}h

= 1/3 x π x (7√7 )^{2} x 21

= 2401 π

Therefore, Volume of cone is 2401 π cm^{3}.

**Question 4: The height of a conical vessel is 3.5 cm. If its capacity is 3.3 litres of milk. Find the diameter of its base.**

**Solution: **

Height of a conical vessel = 3.5 cm and

Capacity of conical vessel is 3.3 litres or 3300 cm^{3}

Now,

We know, Volume of cone = 1/3 πr^{2}h

3300 = 1/3 x 22/7 x r^{2} x 3.5

or r2 = 900

or r = 30

So, radius of cone is 30 cm

Hence, diameter of its base = 2 Radius = 2×30 cm = 60 cm

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

In the 20th chapter of Class 9 RD Sharma Solutions students will study important concepts listed below:

- Right Circular Cone Introduction
- Some important terms definition – Vertex, Base, Axis, Radius and Slant Height.
- Surface Area of a Right Circular Cone
- Volume of a Right Circular Cone

Let us discuss some of the important terms used in this chapter.

**Total surface area**: The total area occupied by the surface including the curved part and the base(s).

**Curved surface area:** The area occupied by the surface excluding the base(s) is known as curved surface area.

**Volume:** The space occupied by an object, which is measured in cubic units.

A right circular cone is a geometric figure having a circular base. Consider a right circular cone having ‘r’ as the base area, ‘h’ as the height of a cone and ‘l’ as a slant height. Then,

- Total surface area = πr(r + l),
- Curved surface area = πrl and
- Volume = 1/3πr
^{2}h

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### What are the topics covered under RD Sharma Solutions for Class 9 Maths Chapter 20?

1. Right Circular Cone Introduction

2. Surface Area of a Right Circular Cone