If the triangle $ \mathrm{ABC} $ in the Question 7 above is revolved about the side $ 5 \mathrm{~cm} $, then find the volume of the solid so obtained. Find also the ratio of the volumes of the two solids obtained in Questions 7 and 8 .

Given:

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

To do:

We have to find the ratio of the volumes of the two solids 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 $BC$, we get a cone as shown in the below figure.

Therefore,

Height of the cone $h=5\ cm$

Radius of the cone $r=12\ cm$

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 $=\frac{1}{3} \times \pi (12)^2 \times 5$

$=240\pi\ cm^3$

The volume of the cone so formed is $240\pi\ cm^3$.

 The ratio of the volumes of the two solids so obtained $=100\pi\ cm^3:240\pi\ cm^3$

$=100:240$

$=5:12$

 The ratio of the volumes of the two solids so obtained is $5:12$.