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 π as found appropriate)
Let us consider the ABA
AS = 3 cm, AC = 4 cm
Hypotenuse BC = 5 cm
We have got 2 cones on the same base AA’ where the radius = DA or DA’
AD/CA = AB/CB
By putting the value of CA, AB and CB we get,
AD = 2/5 cm
We also know,
DB/AB = AB/CB
So, DB = 9/5 cm
As, CD = BC-DB,
CD = 16/5 cm
Now, volume of double cone will be
= \(\big[ \frac{1}{3} \pi \times (\frac{12}{5})^2 \frac{9}{5} + \frac{1}{3} \pi \times (\frac{12}{5})^2 \times \frac{16}{5} \big ] cm^3\)
Solving this we get,
V = 30.14 cm3
The surface area of the double cone will be
= \((\pi \times \frac{12}{5} \times 3) + (\pi \times \frac{12}{5} \times 4) cm^2\)
= \(\pi \times \frac{12}{5}[3 + 4] cm^2\)
= 52.75 cm2