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

Question 26 Mensuration Exercise 17.2

The height of a cone is 30 cm. A small cone is cut off at the top by a plane parallel to its base. If its volume be 1/27 of the volume of the given cone, at what height above the base is the section cut?

Solution:

Assume that a cone has a height of "h" and a circular base with radius "r." This cone will have a volume that is one-third of the product of the base's area and its height.

Given height of the cone, H = 30 cm

Let R be the radius of the given cone and r be radius of small cone.

Let h be the height of small cone.

Volume of the given cone = (1/3)π R2H

Volume of the small cone = 1/27 th of the volume of the given cone.

(1/3)π r2h = (1/27)× (1/3)π R2H

Substitute H = 30

(1/3)π r2h = (1/27)× (1/3)π R2×30

r2h/R2 = 30/27

r2h/R2 = 10/9 ….(i)

From figure, r/R = h/H

r/R = h/30 ….(ii)

Substitute (ii) in (i)

(h/30)2×h = 10/9

h3/900 = 10/9

h3 = 900×10/9 = 1000

Taking cube root on both sides.

h = 10 cm

H-h = 30-10 = 20

The small cone is cut at a height of 20 cm above the base.

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