Question

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.
how much steel was actually used, if 1/12
of the steel actually used was wasted in making the tank​

Answer 1

Given:-

• (I)The lateral or curved surface area of a closed cylindrical petrol storage tank's diameter= 4.2 m.
• High = 4.5m
• (ii) 1/12 of the steel actually used was wasted in making the tank.

To Find:-

• How much steel was actually used ?

Solution:-

(I) Diameter of a cylindrical (closed) petrol tank = 4.2m

Radius = 4.2/2 = 2.1 m

Height = 4.5 m

${\boxed{\pmb{\sf{\red{curved \: surface \: area = 2\pi \: rh}}}}}$

Now,

$\sf \: c.s.a = 2 \times \frac{22}{7} \times 2.1 \times 4.5 \\ \\ \sf \: c.s.a = 59.4 {m}^{2}$

(ii)

${\boxed{\pmb{\sf{\green{total \: surface \: area = 2\pi \: r(r + h)}}}}}$

Now,

$\sf \: t.s.a = 2 \times \frac{22}{7} \times 2.1 \times (2.1 + 4.5) \\ \\ \sf \: t.s.a = 2 \times \frac{22}{7} \times 2.1 \times 6.6 \\ \\ \sf \: t.s.a = 87.12 {m}^{2}$

Let the amount of steel used in making the tank = x.

steel used= x/12

Given:

Amount of steel required-Amount of steel wasted = T.S.A of the tank

$\sf \implies \: x - \frac{x}{12} = 87.12 \\ \\ \sf \implies \: \frac{12x - x}{12} = 87.12 \\ \\ \sf \implies \: \frac{11x}{12} = 87.12 \\ \\ \sf \implies \: x = \frac{12}{11} \times 87.12 = 95.04 {m}^{2}$

Therefore,95.04 m^2 steel was used in actual while making such a tank.