A right triangle, whose sides are $ 3 \mathrm{~cm} $ and $ 4 \mathrm{~cm} $ (other than hypotenuse) is made to revolve about its hypotenuse. Find the volume and surface area of the double cone so formed. (Choose value of $ \pi $ as found appropriate.)

Given:

A right triangle, whose sides are \( 3 \mathrm{~cm} \) and \( 4 \mathrm{~cm} \) (other than hypotenuse) is made to revolve about its hypotenuse. 

To do:

We have to find the volume and surface area of the double cone so formed.

Solution:

Let $AB=3\ cm$ and $BC=4\ cm$ be the two sides of the right triangle $ABC$.

This implies, using Pythagoras theorem,

$AC^2=AB^2+BC^2$

$AC^2=3^2+4^2$

$=9+16$

$=25$

$\Rightarrow AC=\sqrt{25}$

$=5\ cm$

When the triangle $ABC$ is revolved about the hypotenuse $AC$, the following double cone is formed.

In $\triangle \mathrm{ABC}$ and $\triangle \mathrm{BDC}$,

$\angle \mathrm{ABC}=\angle \mathrm{CDB}=90^{\circ}$           ($\mathrm{BD} \perp \mathrm{AC})$)

$\angle \mathrm{BCA}=\angle \mathrm{BCD}$           (Common)

Therefore, by AA similarity,

$\triangle \mathrm{ABC} \sim \triangle \mathrm{BDC}$

This implies,

$\frac{\mathrm{AB}}{\mathrm{BD}}=\frac{\mathrm{AC}}{\mathrm{BC}}$                          (Corresponding sides of similar triangles are in

proportion)

$\mathrm{BD}=\frac{\mathrm{AB} \times \mathrm{BC}}{\mathrm{AC}}$

$=\frac{3 \mathrm{~cm} \times 4 \mathrm{~cm}}{5 \mathrm{~cm}}$

$=\frac{12}{5} \mathrm{~cm}$

$=2.4 \mathrm{~cm}$

We know that,

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

The volume of the double cone in the figure $=$ Volume of cone ABB' $+$ Volume of cone BCB'

$=\frac{1}{3} \times \pi(\mathrm{BD})^{2} \times \mathrm{AD}+\frac{1}{3} \pi(\mathrm{BD})^{2} \times \mathrm{DC}$

$=\frac{1}{3}\times \pi(\mathrm{BD})^{2}(\mathrm{AD}+\mathrm{DC})$

$=\frac{1}{3} \times \pi(\mathrm{BD})^{2} \times \mathrm{AC})$

$=\frac{1}{3}\times 3.14 \times (2.4)^2 \mathrm{~cm} \times 5 \mathrm{~cm}$

$=30.144 \mathrm{~cm}^{3}$

The curved surface area of a cone $=\pi rl$

The curved surface area of the double cone $=$ curved surface area of the cone ABB' $+$ curved surface area of the cone BCB'

$=\pi \times \mathrm{BD} \times \mathrm{AB}+\pi \times \mathrm{BD} \times \mathrm{BC}$

$=\pi \times \mathrm{BD}(\mathrm{AB}+\mathrm{BC})$

$=3.14 \times 2.4 \mathrm{~cm} \times(3 \mathrm{~cm}+4 \mathrm{~cm})$

$=3.14 \times 2.4 \mathrm{~cm} \times 7 \mathrm{~cm}$

$=52.752 \mathrm{~cm}{ }^{2}$

$=52.75 \mathrm{~cm}{ }^{2}$

The volume and surface area of the double cone so formed are $30.144 \mathrm{~cm}^{3}$ and $52.75 \mathrm{~cm}{ }^{2}$ respectively.