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The interior of a building is in the form of cylinder of diameter $4.3 m$ and $3.8 m$ height, surmounted by a cone whose vertical angle is a right angle. What is the area of the surface?

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71 m 2

70.73 m 2

72 m 2

71.83 m 2

answer is D.

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Detailed Solution

Given that cylinder of diameter

4.3 m

and its height is

3.8 m

.
We have the following formulae,
Surface area of cylinder

= 2 π r h

Surface area of cone

= π r l

sin θ = h e i g h t h y p o t e n u s e

In △VOA,

sin 45 o

O A V A

12 = 2.15 V A

V A = 2 × 2.15 m = 3.04 m

We see that

Δ V O A

is an isosceles triangle, then,

V O = O A = 2.15 m

We take

r 1, h 1

as the radius and the height of the cylinder.
The height and radius of the cylinder will be,

h 1 = 3.8 r 1 = 2.15

Height of the cone will be,

h 2 = V O h 2 = 2.15 m

Slant height of the cone will be,

l 2 = V A l 2 = 3.04 m

Surface area of the building = Surface area of the cylinder + Surface area of cone

= 2 π r 1 h 1 + π r 2 l 2 m 2

= 2 π r 1 h 1 + π r 1 l 2 m 2

= π r 1 2 h 1 + l 2 m 2

= 3.14 × 2.15 × (2 × 3.8 + 3.04) m 2

= 3.14 × 2.15 × 10.64 m 2

= 71.83 m 2

Thus, the surface area of the building is

71.83 m 2

.
Therefore the correct option is 4.

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