A closed cylindrical tank, made of thin iron-sheet, has diameter = 8.4 m and height 5.4 m. How much metal sheet, to the nearest m², is used in making this tank, if $\dfrac{1}{15}$ of the sheet actually used was wasted in making the tank ?

Radius of cylindrical tank = $\dfrac{8.4}{2}$ = 4.2 m.

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

= 2 × $\dfrac{22}{7}$ × 4.2 × (5.4 + 4.2)

= 2 × 22 × 0.6 × 9.6

= 253.44 m².

Given,

$\dfrac{1}{15}$ of the sheet actually used was wasted in making the tank.

∴ Fraction of sheet used in making the tank = $1 - \dfrac{1}{15}$

$= \dfrac{15 - 1}{15} \\[1em] = \dfrac{14}{15}$

Let metal sheet used be $x$ m²

$\dfrac{14}{15}$ of $x$ = TSA of Cylindrical tank

$\Rightarrow \dfrac{14}{15} \times x = 253.44 \\[1em] \Rightarrow x = \dfrac{253.44}{\dfrac{14}{15}} \\[1em] \Rightarrow x = 253.44 \times \dfrac{15}{14} \\[1em] \Rightarrow x \approx 271.54 \text{m}^2$

Area of metal sheet in nearest m² = 272 m²

Hence, total metal (iron) sheet used in nearest m² is 272 m².