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

Go back to  'Ex.13.2'

## Question

Find

1. 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.
2. How much steel was actually used, if $$1/12$$  of the steel actually used was wasted in making the tank?

Video Solution
Surface-Areas-And-Volumes
Ex exercise-13-2 | Question 9

## 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} \end{align}

Curved surface area
\begin{align}&=\,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:

Total surface area
\begin{align} &=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}

Learn from the best math teachers and top your exams

• Live one on one classroom and doubt clearing
• Practice worksheets in and after class for conceptual clarity
• Personalized curriculum to keep up with school