Question

A solid cone of base radius $10cm$ is cut into two parts through the mid-point of its height, by a plane parallel to its base. Find the ratio in the volumes of two parts of the cone.

Answer 1

$△OAB\sim △OCD$ (By AA similarity)

$\therefore \frac{OA}{OB}=\frac{AB}{CD}\Rightarrow \frac{h/2}{h}=\frac{r}{10}\Rightarrow r=\frac{10}{2}=5cm$

Given that cone cut into two parts through the mid point of its height $H=\frac{h}{2}$

Volume of frustum of cone $=\frac{\pi H}{3}[R^2+r^2+Rr]$

$=\frac{\pi h}{3\times 2}[{10}^2+5^2+10\times 5]=\frac{175\pi h}{6}$

Volume of cone $OAB=\frac{1}{3}\pi r^2\frac{h}{2}=\frac{1}{3}\times \pi \times 5^2\times \frac{h}{2}=\frac{25\pi h}{6}$

Required ratio $=\frac{\frac{25\pi h}{6}}{\frac{175\pi h}{6}}=\frac{1}{7}$