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Answer :
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