Let ABC be the right angled triangle such that

AB = 15 cm and AC = 20 cm

Using Pythagoreas theorem in ∆ABC we have

BC² = AB² + AC²

BC² = ² + ²

BC² = + =

BC = √( ) = cm

Let OA = x and OB = y.

In triangles ABO and ABC, we have ∠BOA = ∠BAC and ∠ABO = ∠ABC

So, by angle-angle-criterion of similarity, we have ∆BOA ∼ ∆BAC

⇒ y = and x = .

Thus, we have

OA = cm and OB = cm

Also OC = BC - OB = 25 - 9 = 16 cm

When the ABC is revolved about the hypotenuse.we get a double cone as shown in figure

Volume of the double cone = Volume of the cone CAA' + Volume of the cone BAA'

= ( ) cm³

Surface area of the doubled cone = (Curved surface area of Cone CAA') + (Curved surface area of cone BAA')

= (π × OA × AC) + (π × OA × AB)

= (π × 12 × 20) + (π × 12 × 15) cm²

= ( )π cm²

= cm²