Question

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).

Answer 1

When a right-angled triangle is revolved around its hypotenuse, a double cone is formed with same radius but with different heights.
it is given that, 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
from equation (1),
15²  = 9²  + y²
⇒ y² = 225 – 81
⇒ y² = 144
⇒ y = 12 cm
Also, OC = 25 – x = 25 – 13 = 12 cm²
Now, Volume of cone , V = 1/3 πr²h

Hence, volume of double cone 
= 1/3 π(OA)² × BO + 1/3 π(OA)² × OC
= 1/3 π(12)² × (OB + OC)
= 1/3 × 3.14 × 144 × 25
= 3768 cm³

Curved surface area of cone = πrl

Surface area of double cone = CSA of left cone + CSA of right cone
= π(OA)(AB) + π(OA)(AC)
= 3.14 × 12 × (15 + 20)
= 1318.8 cm²

Answer 2

hope it helps u a lot ⤴⤴⤴⤴