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

RD Sharma Class 9 Solutions Chapter 20 VSAQS: This exercise solution will help students better understand this chapter. After practicing the RD Sharma solution class 9 Chapter 20, the student will be able to easily solve complex problems based on the right cone. These solutions will also help students to understand the concepts in depth and lay a solid foundation for further study. Students can obtain a copy of their practice VSAQ by clicking the link below.

• RD Sharma Class 9 Maths Solutions

Access RD Sharma Class 9 Solutions Maths Chapter 20 VSAQS

Question 1: The height of a cone is 15 cm. If its volume is 500π cm³, then find the radius of its base.

Solution:

Height of a cone = 15 cm

Volume of cone = 500 π cm³

We know, Volume of cone = 1/3 πr²h

So, 500π = 1/3 π r² x 15

r² = 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 cm³, find the diameter of its base.

Solution:

Height of a cone = 9 cm

Volume of cone = 48 π cm³

We know, Volume of cone = 1/3 πr²h

So, 48π = 1/3 π r² x 9

r² = 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, l² = r² + h²

28² = r² + 21²

or r = 7√7 cm

Now,

We know, Volume of cone = 1/3 πr²h

= 1/3 x π x (7√7 )² x 21

= 2401 π

Therefore, Volume of cone is 2401 π cm³.

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 cm³

Now,

We know, Volume of cone = 1/3 πr²h

3300 = 1/3 x 22/7 x r² 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

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