The height of a cone is $ 10 \mathrm{~cm} $. The cone is divided into two parts using a plane parallel to its base at the middle of its height. Find the ratio of the volumes of two parts.

Given:

The height of a cone is \( 10 \mathrm{~cm} \). The cone is divided into two parts using a plane parallel to its base at the middle of its height.

To do:

We have to find the ratio of the volumes of two parts.

Solution:

Radius of the base of the 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 of the volumes of two parts is $1:7$. 

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Updated on: 10-Oct-2022

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