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 cm and length is 1 m.

Answer 1

In one revolution, the roller will cover an area equal to its lateral surface area. 
Thus, in 1 revolution, area of the road covered = \(2\pi rh\)
\(=2\times \frac{22}{7}\times42\) cm \(\times1\)m

\(=2\times \frac{22}{7}\times\frac{42}{100}\) m x1m

\(=\frac{264}{100}\) m²

In 750 revolutions, area of the road covered
\(=(\frac{264}{100}×750)\)m²
\(= 1980\) m²

Answer 2

Given,
\(D= 84\) cm 
\(⇒ r = 42\) cm
\(l= 1\) m
Area of road covered in 1 revolution,
\(= 2\pi rl\)
\(=2\times \frac {22}{7} \times 42 \times 1\)
\(=\frac {264}{100} \ sq \ m\)

Area of road covered in 75 revolution,
\(=750 \times \frac {264}{100}\)
\(=1980 \ sq\  m\)

So, the answer is 1980 sq m.