# Ex.13.2 Q4 Surface Areas and Volumes - NCERT Maths Class 9

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## Question

The diameter of a roller is $$84\,\rm{ cm}$$ and its length is $$120\,\rm{ cm}.$$. It takes $$500$$ complete revolutions to move once over to level a playground. Find the area of the playground in \begin{align}\rm{{m^2}} \end{align}.

Video Solution
Surface Areas And Volumes
Ex 13.2 | Question 4

## Text Solution

Reasoning:

The roller is cylindrical in shape and hence it is considered as a right circular cylinder.

The curved surface area of a right circular cylinder of base radius $$r$$ and height  $$h$$  is

\begin{align}2\pi rh \end{align}

What is the known?

Radius and height of the cylinder.

What is the unknown?

Area of the playground in\begin{align}\rm{{m^2}}. \end{align}

Steps:

Curved surface area of the cylinder = \begin{align}2\pi rh \end{align}

\begin{align}{\rm{diameter}} ={ 2r} &= {84\,\,\rm{cm}}\\r &= {42\,\,\rm{cm}}\\h &={ 120\,\,\,\rm{cm}}\\{} \end{align}

Area of the playground leveled in taking  $$1$$  complete revolution = The curved surface area of the cylinder having $$r =$$ $$42\,\rm{ cm}$$ and $$h =$$ $$120\,\rm{ cm}.$$

\begin{align} &= 2\pi rh\\ &= 2 \times \frac{{22}}{7} \times 42 \times 120\\ &= 31680\,\,\rm{c{m^2}} \end{align}

Area of the playground = Area levelled by the cylinder in 500 revolutions.

\begin{align} &= 500 \times 31680\\ &= 15840000\,\rm{c{m^2}}\\\ \rm\end{align}

Since,

\begin{align}&100\,\rm{cm} = 1\,m\\&{100^2}\rm{c{m^2}} = 1\,\rm{{m^2}}\end{align}

\begin{align} &= \frac{{15840000}}{{100 \times 100}}\\ &= 1584\,\rm{{m^2}} \end{align}

Area of the play ground $$=1584 \rm{m^{2}}$$

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