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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

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

Solving this we get,

V = 30.14 cm^{3}

The surface area of the double cone wil be

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

\( = \ 52.75 \) cm^{3}

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