# Ex.13.2 Q9 Surface Areas and Volumes - NCERT Maths Class 9

## Question

**Q9.** Find

- The lateral or curved surface area of a closed cylindrical petrol storage tank that is \(4.2 \,\rm{m}\) in diameter and \(4.5 \,\rm{m}\) high.
- How much steel was actually used, if \(1/12\) of the steel actually used was wasted in making the tank?

## Text Solution

**Reasoning:**

The curved surface area of a right circular cylinder of base radius \(r\) and height \(h\) is \(\begin{align}2\pi rh \end{align}\)

and its total surface area is\(\begin{align}2\pi r(r + h) \end{align}\).

**What is the known?**

The diameter and height of the storage tank.

i. The lateral or curved surface area

**Steps:**

\(\begin{align} \rm Diameter\text{ }=2r &=4.2\,\rm{m} \\ r&=\frac{4.2}{2} \\ \,\,\,\,&=2.1\,\,\rm{m} \\ {\rm height}=h&=4.5\,\,\rm{m} \\ \text {Curved surface area}&=\,2\pi rh \\ & =2\times \frac{22}{7}\times 2.1\times 4.5 \\ & =59.4\,\,\rm{{{m}^{2}}} \\ \end{align}\)

ii. Steel actually used

**Steps:**

\(\begin{align} \text{Total surface area }&=2\pi r(r+h) \\ & =2\times \frac{22}{7}\times 2.1(4.5+2.1) \\ &=2\times \frac{22}{7}\times 2.1\times 6.6 \\ &=87.12\,\,\rm{{{m}^{2}}} \\ \end{align}\)

This is the steel used\(\begin{align} = 87.12\rm{{m^2}} \end{align}\), \(\begin{align}\frac{1}{{12}} \end{align}\) of this is wasted.

The area of the steel which has gone into the task\(\begin{align} = 1 - \frac{1}{{12}} = \frac{{11}}{{12}} \end{align}\) of the steel used to make the tank \(= 87.12\)

This means actual area of steel used \(\begin{align} = \frac{{12}}{{11}} \times 87.12 = 95.04\,\,\rm{{m^2}} \end{align}\)

Curve surface area \(\begin{align} = 59.4\,\rm{{m^2}} \end{align}\)

Steel actually used \(\begin{align} = 95.04\,\rm{{m^2}} \end{align}\)