Q. 333.7( 7 Votes )

Answer :

__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 m^{2}→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 m^{2}→eqn4

Outer surface area of building = eqn3 + eqn4

⇒ Outer surface area = 7.35π + 16.8π

= 24.15π

∴ Outer surface area =75.9 m^{2}

__The outer surface area of building is 75.9 m ^{2}.__

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