Question

A solid cone of radius 20 cm is cut from its middle, parallel to its base, into two parts as shown. Find:

 The ratio between the volumes of the two parts obtained.

Answer 1

Let the radius of the base of the cone be R = 20 cm and its height be H cm.

The cone is cut at the middle of its height by a plane parallel to the base. So the cut is at a height \[\frac{H}{2}\] from the apex.

The upper part is a small cone similar to the whole cone. Since the height is halved, by similar triangles its radius is also halved.

∴ Radius of small cone \[(r) = \frac{R}{2} = 10\ \text{cm}\] and height \[= \frac{H}{2}.\]

Volume of small cone (upper part) :

\[V_1 = \frac{1}{3}\, \pi r^2 \left(\frac{H}{2}\right) = \frac{1}{3}\, \pi (10)^2 \times \frac{H}{2} = \frac{1}{3}\, \pi \times 50H.\]

Volume of the whole cone :

\[V = \frac{1}{3}\, \pi R^2 H = \frac{1}{3}\, \pi (20)^2 H = \frac{1}{3}\, \pi \times 400H.\]

Volume of lower part (frustum) = Volume of whole conce − Volume of small cone

\[V_2 = V - V_1 = \frac{1}{3}\, \pi \times 400H - \frac{1}{3}\, \pi \times 50H = \frac{1}{3}\, \pi \times 350H.\]

∴ Ratio of the two parts:

\[V_1 : V_2 = \frac{1}{3}\, \pi \times 50H : \frac{1}{3}\, \pi \times 350H = 50 : 350 = 1 : 7.\]

Hence, the ratio between the volumes of the two parts (smaller cone : frustum) is \[1 : 7.\]