Question

A right angled triangle with sides 3 cm and 4 cm is revolved around its hypotenuse. Find the volume of the double cone thus generated.

Answer 1

The double cone so formed is as in figure.

Hypotenuse AC

${\displaystyle ={\sqrt{3}}^2+4^2}$

${\displaystyle =5cm.}$

Area of

$$=\frac{1}{3}\times \frac{22}{7}\times \frac{12}{5}\times \frac{12}{5}\times 5$$

$$=\frac{1056}{35}$$

$$=30\frac{6}{35}$$

${\displaystyle \Delta ABC=\frac{1}{2}\times AB\times AC}$

${\displaystyle \frac{1}{2}\times AC\times OB}$

${\displaystyle =\frac{1}{2}\times 4\times 3}$

${\displaystyle =\frac{1}{2}\times 5\times OB}$

${\displaystyle =6}$

$$\frac{1}{2}\times 3\times 4=\frac{1}{2}\times 5\times OB$$

$$OB=\frac{12}{5}$$

Volume of double cone = volume of cone 1 + cone 2

${\displaystyle =\frac{1}{3}\pi r^2{h}_1+\frac{1}{3}\pi r^2{h}_2}$

${\displaystyle =\frac{1}{3}\pi r^2(h_1+h_2)}$

${\displaystyle =\frac{1}{3}\pi x^2(OA+OC)}$

$$=\frac{1}{3}\times \frac{22}{7}\times \frac{12}{5}\times \frac{12}{5}\times 5$$

$$=\frac{1056}{35}$$

$$=30\frac{6}{35}{cm}^2$$