# 2. A right triangle, whose sides are 3 cm and 4 cm (other than hypotenuse) is made to revolve about its hypotenuse. Find the volume and surface area of the double cone so formed. (Choose value of $\pi$ as found appropriate.)

The volume of the double cone will be   =    Volume of cone 1  +  Volume of cone 2.

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

$=\ \frac{1}{3}\ \pi \times 2.4^2\times 5$        (Note that sum of heights of both the cone is 5 cm  -  hypotenuse).

$=\ 30.14\ cm^3$

Now the surface area of a double cone is :

$=\ \pi rl_1\ +\ \pi rl_2$

$=\ \pi \times 2.4 \left ( 4\ +\ 3 \right )$

$=\ 52.8\ cm^2$

Exams
Articles
Questions