# 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 .

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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 $5:12$.

Updated on 10-Oct-2022 13:46:38