## A right angled triangle whose sides are 15 cm and 20 cm , is made to revolve about its hypotenuse. What will be the volume of the cone so formed?

Dec 29, 2017

$V = 1200 \pi {\text{ cm}}^{3}$

#### Explanation: See Fig 1.
$A C = \sqrt{A {B}^{2} + B {C}^{2}} = \sqrt{{20}^{2} + {15}^{2}} = 25$,
Draw a line $B O$ perpendicular to $A C$,
Area of $\Delta A B C = \frac{1}{2} \cdot 20 \cdot 15 = \frac{1}{2} \cdot B O \cdot 25$
$\implies B O = \frac{20 \cdot 15}{25} = 12$
$\implies A O = \sqrt{{20}^{2} - {12}^{2}} = 16 , \implies O C = 25 - 16 = 9$
As shown in Fig 2,
after revolving about the hypotenuse $A C$, the sides $A B \mathmr{and} B C$ become two right circular cones, namely, cone 1 and cone 2, respectively.
volume of a cone is given by $V = \frac{1}{3} \pi {r}^{2} h$
where $r$ is the radius of the base and $h$ is the height of the cone.
Now you have 2 cones, cone 1 and cone 2.
cone 1 : ${h}_{1} = 16 , r = 12$,
cone 2 : ${h}_{2} = 9 , r = 12$,
let ${V}_{1} , {V}_{2}$ be the volume of cone 1 and cone 2, respectively.
total volume of the cones $V = {V}_{1} + {V}_{2}$
$\implies V = \frac{1}{3} \cdot \pi \cdot {12}^{2} \cdot 16 + \frac{1}{3} \cdot \pi \cdot {12}^{2} \cdot 9$
$= \frac{1}{3} \cdot \pi \cdot {12}^{2} \cdot \left(16 + 9\right) = 1200 \pi {\text{ cm}}^{3}$ 