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ABC is a right angled triangle in which ∠A = 90°, AB = 5 cm and AC = 12 cm. What is the approximate volume (in cm3) of the double cone formed by rotating the triangle about its hypotenuse?
- A145
- B408
- C435
- D580
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CCommunity
Applying Pythagoras theorem in triangle ABC
BC2 = AC2 + AB2
⇒ BC = √(122 + 52) = √169
⇒ BC = 13 cm
Therefore the radius is BC/2 = 13/2 = 6.5 cm
Now drop a perpendicular on CB from A, and name it as D on side CB
Using the properties of triangle = AC/BC = AD/AB
⇒ 12/13 = AD/5
⇒ AD = 4.61 cm
AD is equal to the height of the cone
Now the volume of cone can be given as (1/3)πr2h
There are 2 cones so, their volume is = 2 × (1/3)πr2h
⇒ Volume = (2/3) π (6.5)2 × 4.61 = 408 cm3 (approximately)
∴ The volume of the double cone formed is approximately 408 cm3
BC2 = AC2 + AB2
⇒ BC = √(122 + 52) = √169
⇒ BC = 13 cm
Therefore the radius is BC/2 = 13/2 = 6.5 cm
Now drop a perpendicular on CB from A, and name it as D on side CB
Using the properties of triangle = AC/BC = AD/AB
⇒ 12/13 = AD/5
⇒ AD = 4.61 cm
AD is equal to the height of the cone
Now the volume of cone can be given as (1/3)πr2h
There are 2 cones so, their volume is = 2 × (1/3)πr2h
⇒ Volume = (2/3) π (6.5)2 × 4.61 = 408 cm3 (approximately)
∴ The volume of the double cone formed is approximately 408 cm3