Find the lateral surface area of a petrol storage tank is $4.2 \mathrm{~m}$ in diameter and $4.5 \mathrm{~m}$ high. How much steel was actually used, if $1 / 12^{\text {th }}$ of the steel actually used was wasted in making the closed tank?
It is given that
Diameter of cylinder = 4.2 m
Radius of cylinder = 4.22 m
= 2.1 m
Height of cylinder = 4.5 m
Therefore,
Lateral or Curved surface area = 2πrh
$=2 * 3.14 * 2.1 * 4.5=59.4 \mathrm{~m}^{2}$
Total surface area of tank $=2^{\star} ?^{\star} r(r+h)$
$=2^{\star}\left(22 / 7^{*} 2.1(2.1+4.5) \mathrm{m}^{2}\right.$
$=87.12 \mathrm{~m}^{2}$
Let, $\mathrm{A} \mathrm{m}^{2}$ steel be actually used in making the tank
Area of iron present in cylinder $=\left(\mathrm{y}-\frac{\mathrm{A}}{12} \mathrm{~m}^{2}\right)$
$=11 / 12 \mathrm{Am}^{2}$
Hence,
11/12A = Total surface area of cylinder
⟹ A = 12/11*Total surface area
$\Rightarrow \mathrm{A}=(12 / 11 * 87.12) \mathrm{m}^{2}$
$=95.04 \mathrm{~m}^{2}$
Thus, $m^{2}$ steel was actually wasted while constructing a tank.