Question

A road roller takes 700 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 2 m.

Answer 1

Given :

• Length of road roller, h = 2 m

• Diameter of road roller, d = 84 cm

• Total revolution taken by road roller to move once over to level a road = 700

To Find :

• Area of road = ?

Solution :

We have diameter of road roller, d = 84 cm

We know that :

$\underline{\boxed{\sf radius, \: r = \dfrac{d}{2}}}$

$\sf : \implies r = \dfrac{84}{2}$

$\sf : \implies r = \cancel{\dfrac{84}{2}}$

$\sf : \implies r = 42 \: cm$

$\sf : \implies r = \dfrac{42}{100} \: m$

$\sf : \implies r = 0.42 \: m$

$\underline{\boxed{\sf r = 0.42 \: m}}$

Now, we are given that road roller takes 700 complete revolutions to level the road.

$\sf : \implies Area \: of \: road =700 \times Area \: covered \: in \: 1 \: round$

$\sf : \implies Area \: of \: road =700 \times Lateral \: surface \: area \: of \: road \: roller$

$\sf : \implies Area \: of \: road =700 \times 2 \pi r h$

By substituting values :

$\sf : \implies Area \: of \: road =700 \times 2 \times \dfrac{22}{7} \times 0.42 \times 2$

$\sf : \implies Area \: of \: road =7\cancel{00} \times 2 \times \dfrac{22}{\cancel{7}} \times \dfrac{\cancel{42}}{1\cancel{00}} \times 2$

$\sf : \implies Area \: of \: road =7 \times 44 \times 6 \times 2$

$\sf : \implies Area \: of \: road = 3696$

$\underline{\boxed{\sf Area \: of \: road = 3696 \: m^{2}}}$

Hence, Area of road is 3696 m².