# ML Aggarwal Solutions Class 10 Mathematics Solutions for Mensuration Exercise 17.2 in Chapter 17 - Mensuration

Question 19 Mensuration Exercise 17.2

The height and the radius of the base of a right circular cone is half the corresponding height and radius of another bigger cone.

Find:the ratio of their volumes.

Solution:

Consider a cone with a height of "h" and a circular base with radius "r." This cone's volume will be equal to one-third of the base's area multiplied by its height.

Let r be the radius of bigger cone. Then the radius of smaller cone is r/2.

Let h be the height of bigger cone. Then the height of smaller cone is h/2.

Volume of bigger cone, V1 = (1/3)π r2h

Volume of smaller cone, V2 = (1/3)π (r/2)2 (h/2) = (1/3)π r2h/8

Ratio of volume of smaller cone to bigger cone, V2/V1 = ( 1/3)π r2 h/8÷ (1/3)π r2h

= (1/24) r2 h ×(3/r2 h)

= 1/8

Hence the ratio of their volumes is 1:8.

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