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 `1/(12)` of steel actually used was wasted in making the closed tank?

Explanation

(i) Radius of the cylinder, `r = (4.2)/(2)m = 2.1 m`
Height of the cylinder,`h = 4.5 m`
Curved surface area of the cylinderical tank
`= (2pi rh) m^(2) = (2 xx (22)/(7) xx 2.1 xx 4.5) m^(2)`
`= (2 xx (22)/(7) xx (21)/(10) xx (45)/(10)) m^(2) = (297)/(5) m^(2) = 59.4 m^(2)`
(ii) Total surface area of the tank
`= (2 pi rh + 2pi r^(2)) m^(2) = 2pi r(h +r) m^(2)`
`= {2 xx (22)/(7) xx 2.1 xx (4.5 + 2.1)} m^(2) = ((44)/(7) xx (21)/(10) xx (66)/(10)) m^(2)`
`= (8712)/(100) m^(2) = 87.12 m^(2)`
Let the actual area of steel used be `x` `m^(2)`
Then, wasted steel `= (x)/(12) m^(2)`
Area of steel used in the tank `= (x - (x)/(12)) m^(2) = (11x)/(12) m^(2)`
`:. (11x)/(12) = 87.12 rArr x = ((87.12 xx 12)/(11)) m^(2) = (7.92 xx 12) m^(2)`
`= 95.04 m^(2)`
Hence, `95.04 m^(2)` of steel was actually used in the tank