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# 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