Question

Find

(i) the lateral or curved surface area of a closed cylindrical petrol storage tank that is$4.2m$ in diameter and $4.5m$high.

(ii) How much steel was actually used if $\frac{1}{12}$ of the steel actually used was wasted in making the tank. (Assume $\pi =\frac{22}{7}$)

Answer 1

Step 1: Find the curved surface area of the cylinder.

We are given,

The diameter of the storage tank $=4.2m$

$\Rightarrow$radius $r=2.1m$

The height of the storage tank $=4.5m$

The curved surface area of a cylinder $\mathbf{=}\mathbf{2}\mathit{\pi }\mathit{r}\mathit{h}$

where,

$r=$radius of cylinder

$h=$height of cylinder

So, the lateral surface area of the cylinder

$\Rightarrow 2\times \left(\frac{22}{7}\right)\times 2.1\times 4.5=59.4m^2$

Step 2: Find the amount of steel used in making the tank.

Area of steel required to make the tank = Total surface area of a cylindrical tank

The total surface area of the cylinder $\mathbf{=}\mathbf{2}\mathit{\pi }\mathit{r}\mathbf{(}\mathbf{r}\mathbf{+}\mathbf{h}\mathbf{)}$

$$\begin{gathered} =2\pi r(r+h) \\ =2\times \frac{22}{7}\times 2.1\times (2.1+4.5) \\ =87.12m^2....\left(1\right) \end{gathered}$$

Let actually used steel be $x{m}^2$.

$\frac{1}{12}$ of actually used steel was wasted.

Hence, the tank is made up of$1-\frac{1}{12}=\frac{11}{12}$ part of the steel.$...\left(2\right)$

Hence, from$\left(1\right)$ and $\left(2\right)$, we get,

$$\begin{gathered} x\left(\frac{11}{12}\right)=87.12m^2 \\ \Rightarrow x=\frac{87.12\times 12}{11}=95.04m^2 \end{gathered}$$

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

Hence, the required answer is

1. The curved surface area of the tank is $59.4m^2$
2. $95.04m^2$ steel was used in actual while making the cylindrical tank.