# A right triangle whose sides are 15 cm and 20 cm (other than hypotenuse), is made to revolve about its hypotenuse.

Question:

A right triangle whose sides are 15 cm and 20 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 π">π as found appropriate)

Solution:

We have,

In $\Delta \mathrm{ABC}, \angle \mathrm{B}=90^{\circ}, \mathrm{AB}=l_{1}=15 \mathrm{~cm}$ and $\mathrm{BC}=l_{2}=20 \mathrm{~cm}$

Let $\mathrm{OD}=\mathrm{OB}=r, \mathrm{AO}=h_{1}$ and $\mathrm{CO}=h_{2}$

Using Pythagoras theorem,

$\mathrm{AC}=\sqrt{\mathrm{AB}^{2}+\mathrm{BC}^{2}}$

$=\sqrt{15^{2}+20^{2}}$

$=\sqrt{225+400}$

$=\sqrt{625}$

$\Rightarrow h=25 \mathrm{~cm}$

As, ar $(\Delta \mathrm{ABC})=\frac{1}{2} \times \mathrm{AC} \times \mathrm{BO}=\frac{1}{2} \times \mathrm{AB} \times \mathrm{BC}$

$\Rightarrow \mathrm{AC} \times \mathrm{BO}=\mathrm{AB} \times \mathrm{BC}$

$\Rightarrow 25 r=15 \times 20$

$\Rightarrow r=\frac{15 \times 20}{25}$

$\Rightarrow r=12 \mathrm{~cm}$

Now,

Volume of the double cone so formed $=$ Volume of cone $1+$ Volume of cone 2

$=\frac{1}{3} \pi r^{2} h_{1}+\frac{1}{3} \pi r^{2} h_{2}$

$=\frac{1}{3} \pi r^{2}\left(h_{1}+h_{2}\right)$

$=\frac{1}{3} \pi r^{2} h$

$=\frac{1}{3} \times 3.14 \times 12 \times 12 \times 25$

$=3768 \mathrm{~cm}^{3}$

Also,

Surace area of the solid so formed = CAS of cone $1+$ CSA of cone 2

$=\pi r l_{1}+\pi r l_{2}$

$=\pi \mathrm{r}\left(l_{1}+l_{2}\right)$

$=\frac{22}{7} \times 12 \times(15+20)$

$=\frac{22}{7} \times 12 \times 35$

$=1320 \mathrm{~cm}^{2}$