A right triangle whose sides are 15 cm and 20 cm is made to revolve about its hypotenuse. Find the volume and surface area of the double cone so formed. (π=3.14)
Answers
Answer: volume = 3768 cm cube and
surface area = 1318.8 cm square
Step-by-step explanation:
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