Question

A cone of radius 4 cm is divided into two parts by drawing a plane through the mid point of its axis and parallel to its base . Compare the volumes of two parts. 

Answer 1

Radius of the cone, R = 4 cm
Let the height be H. 
Since the plane divides the cone into two parts through the mid point so, a small cone and a frustum will be formed.

\[OC = CA = \frac{H}{2}\]

Let the radius of the smaller cone be r cm.

In ∆OCD and ∆OAB,

∠OCD = ∠OAB  (90°)

∠COD = ∠AOB  (Common)

∴∆OCD ∼ ∆OAB  (AA Similarly criterion)

\[\Rightarrow \frac{OA}{OC} = \frac{AB}{CD} = \frac{OB}{OD} \left( \text { corresponding sides are proportional } \right)\]

\[ \Rightarrow \frac{H}{\frac{H}{2}} = \frac{4}{r}\]

\[ \Rightarrow r = 2 cm\]

Now volume of the smaller cone =  \[\frac{1}{3}\pi \left( CD \right)^2 \times OC = \frac{1}{3}\pi \left( 2 \right)^2 \times \frac{H}{2} = \frac{2\pi H}{3} cm\]

Height of the frustum of the cone = \[\frac{H}{2}\]

Volume of frustum of cone = \[\frac{1}{3}\pi h\left[ r_1^2 + r_1 r_2 + r_2^2 \right]\]

\[= \frac{1}{3}\pi\left( \frac{H}{2} \right)\left[ \left( 4 \right)^2 + \left( 2 \right)^2 + 4 \times 2 \right]\]

\[ = \frac{14\pi H}{3} {cm}^3\]

\[\frac{\text { Volume of the smaller cone }}{\text { Volume of the frustum of the cone }} = \frac{\frac{2\pi H}{3}}{\frac{14\pi H}{3}}\]

\[ \Rightarrow \frac{\text { Volume of the smaller cone }}{\text { Volume of the frustum of the cone }} = \frac{1}{7}\]

Hence, the ratio of the volumes of the two parts will be 1 : 7.