A right triangle having sides 15 cm and 20 cm is made to revolve about its hypotenuse. Find the Volume and Surface Area of the double cone so formed.

(Use π=3.14).

When a right-angled triangle is revolved around its hypotenuse, a double cone is formed with same radius but with different heights.

Given,

AB = 15 cm

AC = 20 cm

Let, OB = x and OA = y

Observe from the figure,

In right-angled triangle ABC, By Pythagoras theorem [Hypotenuse² = Base² + Perpendicular²]

BC² = AC² + AB²

⇒ BC² = 20² + 15²

⇒ BC² = 400 + 225

⇒ BC² = 625

⇒ BC = 25 cm

In ΔOAB

AB² = OA² + OB²

⇒ 15² = x² + y² ……[1]

In ΔAOC

AC² = OA² + OC²

⇒ 20² = y² + (BC – OB)²

⇒ 400 = y² + (25 – x)²

⇒ 400 = y² + 625 – 50x + x²

⇒ 400 = 15² + 625 – 50x

⇒ 400 = 225 + 625 – 50x

⇒ 50x = 450

⇒ x = 9 cm

Putting in [1], we get

15² = 9² + y²

⇒ y² = 225 – 81

⇒ y² = 144

⇒ y = 12 cm

Also, OC = 25 – x = 25 – 13 = 12 cm²

Now, Volume of cone

Where r denotes base radius and h denotes height of cone

Hence, volume of double cone

Curved surface area of cone = πrl

Where r denotes base radius and l denotes slant height

Also,

Surface area of double cone = CSA of left cone + CSA of right cone

Putting values,

Surface area = π(OA)(AB) + π(OA)(AC)

= 3.14 × 12 × (15 + 20)

= 1318.8 cm²