Question

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 $\pi =3.14$)

Answer 1

We have
Radius of the base of the cylinder $r_1=\frac{4.3}{2}m=2.15m$

Radius of base of the cone $=r_1=2.15m$

Height of the cylinder $h_1=3.8m$

In $△VOA$ we have

$\sin {45}^o=\frac{OA}{VA}$

$\Rightarrow \frac{1}{\sqrt{2}}=\frac{2.15}{VA}$

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

Clearly $△VOA$ is an isosceles triangle

Therefore, $VO=OA=2.15m$

Thus, we have

height of the cone $=h_2=VO=2.15m$

Slant height of the ocne $l_2=VA=3.04m$

Surface area of the building = Surface area of the cylinder + Surface area of cone
$=(2\pi r_1{h}_1+\pi r_2{l}_2)m^2=(2\pi r_1{h}_1+\pi r_1{l}_2)m^2$
$=\pi r_1(2h_1+l_2)m^2=3.14\times 2.15\times (2\times 3.8+3.04)m^2=3.14\times 2.15\times 10.64m^2=71.83m^2$

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

$=(\pi r_1^2{h}_1+\frac{1}{3}\pi r_2^2{h}_2)m^3=(\pi r_1^2{h}_1+\frac{1}{3}\pi r_1^2{h}_2)m^3[\because r_2=r_1]$

$=\pi r_1^2(h_1+\frac{1}{3}{h}_2)m^3=3.14\times 2.15\times 2.15\times (3.8+\frac{2.15}{3})m^3=65.55m^3$