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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