# Ex.11.3 Q9 Mensuration Solution - NCERT Maths Class 8

## Question

A road roller takes \(750\) complete revolutions to move once over to level a road. Find the area of the road if the diameter of a road roller is \(84\,\rm{cm}\) and length is \(1\,\rm{ m.}\)

## Text Solution

**What is Known?**

Diameter of the road roller and length is given.

**What is unknown?**

Area of the road.

**Reasoning:**

In one revolution, the roller will cover an area equal to its lateral surface area.

**Steps:**

Radius of the road roller

\[\begin{align}r = \frac{{84}}{2}\,\rm{cm} = 42\,\rm{ cm} \end{align}\]

Length of the road roller, \(h = 1\,\rm{m} = 100\, \rm{cm}\)

In \(1\) revolution, area of the road covered

\[ \begin{align} &= 2\pi rh \\&= 2 \times \frac{{22}}{7} \times 42 \times 100\\&= 26400\,{\rm{c}}{{\rm{m}}^2} = 2.64\,{{\rm{m}}^2} \end{align}\]

In \(750\) revolutions area of the road covered

\[\begin{align} &= 750 \times 2.64\,\rm{m^2}\\&= 1980\,\rm{m^2} \end{align}\]

Thus, the area of the road is \(\begin{align}1980\,\rm{m^2}{.} \end{align}\)