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.