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.)

Let us consider the ∆ ABC

Here,

AB = 3 cm, AC = 4 cm

So, Hypotenuse BC = 5 cm

We have got 2 cones on the same base AA’ where the radius = DA or DA’

Now, \( \frac{AD}{CA} = \frac{AB}{CB} \)

By putting the value of CA, AB and CB we get,

AD = \( \frac{12}{5} \) cm

We also know,

\( \frac{DB}{AB} = \frac{AB}{CB} \)

So, DB = \( \frac{9}{5} \) cm

As, CD = BC - DB,

CD = \( \frac{16}{5} \) cm

Now, volume of double cone will be

\( [ \frac{1}{3} \pi \ × \ ( \frac{12}{5})^2 \frac{9}{5} \ + \ \frac{1}{3} \pi \ × \ ( \frac{12}{5})^2 \ × \ \frac{16}{5} ] \) cm³

Solving this we get,

V = 30.14 cm³

The surface area of the double cone wil be

\( ( \pi \ × \ \frac{12}{5} \ × \ 3) \ + \ ( \pi \ × \ \frac{12}{5} \ × \ 4) \) \( = \ \pi \ × \ \frac{12}{5}[3 \ + \ 4] \)

\( = \ 52.75 \) cm³