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

## 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 \(\begin{align}\rm{{m^2}} \end{align}\).

## 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 \(=1584 \rm{m^{2}}\)