Question

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

Answer 1

$△AOB\sim △BOC$ [By AA similarity]

$\therefore \frac{r}{4}=\frac{h_2}{3}\Rightarrow h_2=\frac{3}{4}r$

Similarly, $△AOD\sim △COD$

$\therefore \frac{h_1}{4}=\frac{r}{3}\Rightarrow h_1=\frac{4r}{3}$

$h_1+h_2=\sqrt{4^2+3^2}=\sqrt{25}=5$

$\Rightarrow \frac{4r}{3}+\frac{3}{4}r=5\Rightarrow \frac{25r}{12}=5\Rightarrow r=\frac{12}{5}cm$

Volume of double cone$=\frac{1}{3}\pi r^2[h_1+h_2]$

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

$=\frac{1056}{35}=30\frac{6}{35}c{m}^3$