The interior of a building is in the form of a right circular of diameter 4.2 m and height 4 m surmounted by a cone of same diameter. The height of the cone is 2.8 m. Find the outer surface area of the building.
Explanation: Here in order to find th outer surface area of building we need to simply add the curved surface areas of cone and cylinder.
Given height of cylinder = h = 4 m
Height of cone = h’ = 2.8 m
Diameter of cylinder = diameter of cone = d = 4.2 m
⇒ Radius of cone = Radius of cylinder = d/2 =4.2m/2 = 2.1 m
Outer surface area of building = C.S.A of cylinder + C.S.A of cone
Now, C.S.A of cylinder = 2πrh →eqn1
Where r = radius of base of cylinder, h = height of cylinder
And C.S.A of cone = πrℓ →eqn2
Where r = radius of base of cone, ℓ = Slant height of cone
We know in a cone
(Slant height)2 = (height)2 + (radius)2 (put the given values)
(ℓ)2 = (2.8)2 + (2.1)2
⇒ (ℓ)2 = 7.84 + 4.41
⇒ (ℓ)2 = 12.25
⇒ ℓ = √(12.25)
∴ ℓ = 3.5 m
Now, C.S.A of cone = π × 2.1 × 3.5 (putting the values in eqn2)
⇒ C.S.A of cone = 7.35π m2→eqn3
C.S.A of cylinder = 2 × π × 2.1 × 4 (putting the values in eqn1)
⇒ C.S.A of cylinder = 2 × π × 4.41 × 4
∴ C.S.A of cylinder = 16.8π m2→eqn4
Outer surface area of building = eqn3 + eqn4
⇒ Outer surface area = 7.35π + 16.8π
= 24.15π
∴ Outer surface area =75.9 m2
The outer surface area of building is 75.9 m2.