# Find the lateral curved surface area of a cylindrical petrol storage tank that is $4.2\ m$ in diameter and $4.5\ m$ high. How much steel was actually used, if $\frac{1}{12}$ of steel actually used was wasted in making the closed tank?

Given:

A cylindrical petrol storage tank is $4.2\ m$ in diameter and $4.5\ m$ high.

To do:

We have to find the amount of steel used if $\frac{1}{12}$ of steel actually used was wasted in making the closed tank.

Solution:

Diameter of the cylindrical tank $= 4.2\ m$

This implies,

Radius $(r)=\frac{4.2}{2}$

$=2.1 \mathrm{~m}$

Height $(h)=4.5 \mathrm{~m}$

Therefore,

Lateral surface area $=2 \pi r h$

$=2 \times \frac{22}{7} \times 2.1 \times 4.5$

$=59.4 \mathrm{~m}^{2}$

Total surface area $=2 \pi r(h+r)$

$=2 \times \frac{22}{7} \times 2.1(4.5+2.1)$

$=13.2 \times 6.6$

$=87.12 \mathrm{~m}^{2}$

Let the total area of the sheet be $x \mathrm{~m}^{2}$

Wastage $=\frac{1}{12} x$

Remaining area of sheet $=x-\frac{1}{12} x$

$=\frac{11}{12} x$

This implies,

$\frac{11}{12} x=87.12$

$x=\frac{87.12 \times 12}{11}$

$x=95.04$

Hence, $95.04\ m^2$ of steel was actually used.

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