A right triangle $ \mathrm{ABC} $ with sides $ 5 \mathrm{~cm}, 12 \mathrm{~cm} $ and $ 13 \mathrm{~cm} $ is revolved about the side $ 12 \mathrm{~cm} $. Find the volume of the solid so obtained.

Given:

A right triangle \( \mathrm{ABC} \) with sides \( 5 \mathrm{~cm}, 12 \mathrm{~cm} \) and \( 13 \mathrm{~cm} \) is revolved about the side \( 12 \mathrm{~cm} \).

To do:

We have to find the volume of the solid so obtained.

Solution:

Let in a triangle $ABC$,

$AB=13\ cm$

$BC=5\ cm$

$CA=12\ cm$

On revolving the right angled triangle $ABC$ about the side $CA$, we get a cone as shown in the below figure.

Therefore,

Height of the cone $h=12\ cm$

Radius of the cone $r=5\ cm$

We know that,

Volume of a cone of radius $r$ and height $h$ is $\frac{1}{3} \pi r^2h$

This implies,

Volume of the cone $=\frac{1}{3} \times \pi \times 5^2 \times 12$

$=100\pi\ cm^3$

The volume of the cone so formed is $100\pi\ cm^3$.