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

Area of$$\begin{gathered} =\frac{1}{3}\times \frac{22}{7}\times \frac{12}{5}\times \frac{12}{5}\times 5 \\ =\frac{1056}{35} \\ =30\frac{6}{35} \end{gathered}$$

$$\begin{gathered} \frac{1}{2}\times 3\times 4=\frac{1}{2}\times 5\times OB \\ OB=\frac{12}{5} \end{gathered}$$

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

$$\begin{gathered} =\frac{1}{3}\times \frac{22}{7}\times \frac{12}{5}\times \frac{12}{5}\times 5 \\ =\frac{1056}{35} \\ =30\frac{6}{35}{\mathrm{cm}}^2 \end{gathered}$$