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A solid cone of base radius $ 10 \mathrm{~cm} $ 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.
Given:
A solid cone of base radius \( 10 \mathrm{~cm} \) is cut into two parts through the mid-point of its height, by a plane parallel to its base.
To do:
We have to find the ratio in the volumes of two parts of the cone.
Solution:
Radius of the base of the solid cone $r = 10\ cm$
Let the total height of the cone be $h$.
In $\triangle AOB$,
$C$ is the mid point of $AO$ and $CD\ \parallel\ OB$
Therefore,
$\frac{\mathrm{OB}}{\mathrm{CD}}=\frac{\mathrm{AO}}{\mathrm{AC}}$
$\Rightarrow \frac{10}{\mathrm{CD}}=\frac{h}{\frac{h}{2}}$
$\Rightarrow \frac{10}{\mathrm{CD}}=\frac{2}{1}$
$\Rightarrow \mathrm{CD}=\frac{10}{2}=5 \mathrm{~cm}$
This implies,
$r_{2}=5 \mathrm{~cm}$
Volume of the smaller cone $=\frac{1}{3} \pi r_{2}^{2} \frac{h}{2}$
$=\frac{1}{3} \pi \times 5^2 \times \frac{h}{2}$
$=\frac{25}{6} \pi h$
Volume of the frustum $=\frac{1}{3} \pi \frac{h}{2}(r_{1}^{2}+r_{1} r_{2}+r_{2}^{2})$
$=\frac{h \pi}{6}(10^{2}+10 \times 5+5^{2})$
$=\frac{\pi h}{6}(100+50+25)$
$=\frac{175}{6} \pi h$
Ratio between the volumes of the upper part and lower part $=\frac{25}{6} \pi h: \frac{175}{6} \pi h$
$= 1: 7$
The ratio in the volumes of two parts of the cone is $1:7$.