RD Sharma Class 9 Solutions Chapter 20 Exercise 20.2 (Updated for 2021-22)

RD Sharma Class 9 Solutions Chapter 20 Exercise 20.2 solutions are provided here. These solutions are prepared by our experts and provide accurate answers to all the practice questions that students can refer to and prepare to solve during the exam. The chapter of RD Sharma Class 9 Solutions Chapter 20 exercise 20.2 shows students many problems related to the surface area of ​​straight cones.

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RD Sharma Class 9 Solutions Chapter 20 Exercise 20.2

Access RD Sharma Class 9 Solutions Chapter 20 Exercise 20.2

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 πr²h

By substituting the values, we get

= 1/3 x 3.14 x 6² x 7

= 264

Volume of a right circular cone is 264 cm³

(ii) Radius of cone(r)=3.5 cm

Height of cone(h)=12cm

Volume of a right circular cone = 1/3 πr²h

By substituting the values, we get

= 1/3 x 3.14 x 3.5² x 12

=154

Volume of a right circular cone is 154 cm³

(iii) Height of cone(h)=21 cm

Slant height of cone(l) = 28 cm

Find the measure of r:

We know, l² = r² + h²

28² = r² + 21²

or r = 7√7

Now,

Volume of a right circular cone = 1/3 πr²h

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 cm³

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, l² = r² + h²

25² = 7² + h²

or h = 24

Now, Volume of a right circular cone = = 1/3 πr²h

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 cm³ or 1.232 litres

[1 cm³ = 0.01 liter]

(ii) Height of cone(h)=12 cm

Slant height of cone(l)=13 cm

As we know that, l² = r² + h²

13² = r² + 12²

or r = 5

Now, Volume of a right circular cone = 1/3 πr²h

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 cm³ or 0.314 litres.

[1 cm³ = 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)²h

Volume of second cone (V2) = 1/3 πr²(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, l² = r² + h²

= (5x) ² + (12x)²

= 25 x² + 144 x²

or l = 13x

Therefore, slant height is 13 m.

Now it is given that volume of cone = 314 m³

⇒ 1/3πr²h = 314

⇒ 1/3 x 3.14 x (25x² ) x (12x) = 314

⇒ x³=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, l² = r² + h²

= (5y) ² + (12y)²

= 25 y² + 144 y²

or l = 13y

Now, volume of the cone is given 2512cm³

⇒ 1/3πr²h=2512

⇒ 1/3 x 3.14 x (5y)² x 12y = 2512

⇒ y³ = (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 (r₁) =2x

Radius of 2nd cone (r₂) =3x

Volume of 1st cone (V₁)= 4y

Volume of 2nd cone (V₂)= 5y

We know formula for volume of a cone = 1/3πr²h

Let h₁ and h₂ 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.

We have included all the information regarding CBSE RD Sharma Class 9 Solutions Chapter 20 Exercise 20.2. If you have any query feel free to ask in the comment section. 

FAQ: RD Sharma Class 9 Solutions Chapter 20 Exercise 20.2