Question

A cone of base radius 10 cm is cut down from the mid-point of its height.find the ratio of the two cones.

Answer 1

Radius of larger cone(R) = 10 cm
height = H
volume(V) = $\frac{1}{3} \pi R^2H=\frac{100}{3} \pi H$

after cutting from mid-point of its height, 
height of smaller cone(h) = H/2
radius (r) = R/2 = 10/2 = 5 cm
volume(v) = $\frac{1}{3} \pi r^2h= \frac{1}{3} \pi *5^2*( \frac{H}{2} )= \frac{25}{6} \pi H$

$Ratio= \frac{V}{v}= \frac{ \frac{100}{3} \pi H}{ \frac{25}{6} \pi H} = \frac{100}{3} \times \frac{6}{25}=8:1\\ \\Ratio\ of\ volume\ of\ larger\ cone\ to\ smaller\ cone\ is\ \boxed{8:1}$