Here, r = 4. 2/ 2 m = 2.1 m, h = 4.5 m

(i) Curved surface area of the storage tank = 2πrh

= 2 × 22 7 × 2.1 × 4.5 m^{2} = 59.4 m^{2} Ans.

(ii) Total surface area of the tank = 2πr (h + r)

= 2 × 22/ 7 × 2.1 (4.5 + 2.1) m^{2 }

= 44 × 0.3 × 6.6 m2 = 87.12 m^{2 }

Let the actual area of steel used be x m^{2}.

Area of steel wasted = 1/ 12 of x m^{2} = x/ 12 m^{2}. ... (i)

∴ area of the steel used in the tank =( x − x /120 )m^{2} = 11/ 12 x m^{2}

⇒ 87.12 = 11/ 12 x

⇒ x = 87. 12x 12/ 11 = 95.04 m^{2 }

Hence, 95.04 m^{2} of steel was actually used** Ans.**