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

`=sqrt3^2 +4^2`

`=5 cm.`

Area of

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

\[ = \frac{1056}{35}\]

\[ = 30\frac{6}{35}\]

`Delta ABC =1/2 xx AB xx AC`

`1/2 xx AC xx OB`

`=1/2 xx 4 xx 3`

`=1/2 xx 5 xx OB`

`=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

`=1/3pir^2h_1 + 1/3 pir^2 h_2`

`=1/3pir^2 (h_1 +h_2)`

`=1/3pix^2 (OA + OC)`

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

\[ = \frac{1056}{35}\]

\[ = 30\frac{6}{35} {cm}^2\]