**Question:**

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?

**Solution:**

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.