Question

Please help me to solve 15th question

Answer 1

Answer :

Given diameter of cylinderical tank = 21 m

Height of cylinderical tank = 4 m

Radius of cylinderical tank = Diameter ÷ 2

= 21 ÷ 2 => 10.5 m

Now, Given that, the cylinderical tank is closed

So the area of sheet required = Total Surface Area of the tank.

As the tank is in a cylinderical shape.

Therefore, Total Surface Area of cylinder =>

2πr(r + h)

where, r is the radius and h is the height of Cylinder. The value of π is constant. It can either be 3.14 or 22/7

Here, I've taken π = 22/7

So Total Surface Area =
$2 \times \frac{22}{7} \times 10.5 (10.5 + 4) \\ \\ 2 \times \frac{22}{7} \times \frac{21}{2} \times 14.5\\ \\ 2 \times \frac{22}{7} \times \frac{21}{2} \times \frac{29}{2} \\ \\ 22 \times 3 \times \frac{29}{2} \\ \\ 11 \times 3 \times 29 \\ \\ 957 \: {m}^{2} \\ \\ so \:total \: surface \: area \: of \: tank \: = {957 \: m}^{2} \\ \\ therefore \: \\ area \: of \: sheet \: required \: = 957 \: {m}^{2}$

HOPE IT WOULD HELP YOU