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

Given :
Height (h) of the cylindrical tank = 4.5 m
radius(r) of the circular end of cylindrical tank = $$\frac{4.2}{2}$$ m = 2.1 m

i) lateral or curved surface area of tank
= $$2{\pi}rh$$
= $$2 × \frac{22}{7} × 2.1 × 4.5$$ $${m}^2$$
= (44 × 0.3 × 4.5) $${m}^2$$
= 59.4 $${m}^2$$

Therefore , CSA of tank is 59.4 $${m}^2$$.

ii) Now, so as to find steel used,

total surface area
= $$2{\pi}r(r + h) {m}^2$$
= $$2 × \frac{22}{7} × 2.1(2.1 + 4.5) {m}^2$$
= $$4.4 × 0.3 × 6.6 {m}^2$$
= 87.12 $${m}^2$$

Now, Let actual area of x $${m}^2$$ sheet be used in making the tank.

Since, $$\frac{1}{12}th$$ of the actual steel wasted, the area of steel which has gone into the tank = $$\frac{11}{12}th$$ of x.

This means that the actual area used
= $$\frac{12}{11}th × 87.12$$ $${m}^2$$ = 95.04 $${m}^2$$.

Therefore, 95.04 $${m}^2$$ steel was used in actual while making such a tank.