The interior of a building is in the form of cylinder of diameter $4.3 m$ and height $3.8 m$ surrounded by a cone whose vertical angle is a right angle. Find the area of the surface and the volume of the building (Take $π = 3.14$ )

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Solution

We have
Radius of the base of the cylinder $r_{1} = \frac{4.3}{2} m = 2.15 m$

Radius of base of the cone $= r_{1} = 2.15 m$

Height of the cylinder $h_{1} = 3.8 m$

In $△ V O A$ we have

$sin 45^{o} = \frac{O A}{V A}$

$\Rightarrow \frac{1}{\sqrt{2}} = \frac{2.15}{V A}$

$\Rightarrow V A = (\sqrt{2} \times 2.15) m = 3.04 m$

Clearly $△ V O A$ is an isosceles triangle

Therefore, $V O = O A = 2.15 m$

Thus, we have

height of the cone $= h_{2} = V O = 2.15 m$

Slant height of the ocne $l_{2} = V A = 3.04 m$

Surface area of the building = Surface area of the cylinder + Surface area of cone
$= (2 π r_{1} h_{1} + π r_{2} l_{2}) m^{2} = (2 π r_{1} h_{1} + π r_{1} l_{2}) m^{2}$
$= π r_{1} (2 h_{1} + l_{2}) m^{2} = 3.14 \times 2.15 \times (2 \times 3.8 + 3.04) m^{2} = 3.14 \times 2.15 \times 10.64 m^{2} = 71.83 m^{2}$

Volume of the building = volume of the cylinder + volume of the cone

$= (π r_{1}^{2} h_{1} + \frac{1}{3} π r_{2}^{2} h_{2}) m^{3} = (π r_{1}^{2} h_{1} + \frac{1}{3} π r_{1}^{2} h_{2}) m^{3} [∵ r_{2} = r_{1}]$

$= π r_{1}^{2} (h_{1} + \frac{1}{3} h_{2}) m^{3} = 3.14 \times 2.15 \times 2.15 \times (3.8 + \frac{2.15}{3}) m^{3} = 65.55 m^{3}$

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The interior of a building is in the form of cylinder of diameter 4.3m and height 3.8m surrounded by a cone whose vertical angle is a right angle. Find the area of the surface and the volume of the building (Take π=3.14)

The interior of a building is in the form of cylinder of diameter $4.3 m$ and height $3.8 m$ surrounded by a cone whose vertical angle is a right angle. Find the area of the surface and the volume of the building (Take $π = 3.14$ )