Question

The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions moving once over to level a playground. What is the area of the playground?

Answer 1

\[\text{ Given: }  \]
\[\text{ Diameter of the roller = 84 cm} \]
\[ \therefore \text{ Radius } , r = \frac{\text{ Diameter}}{2} = 42 cm\]
\[\text{ In 1 revolution, it covers the distance of its lateral surface area } . \]
\[\text{ Roller is a cylinder of height, h = 120 cm } \]
\[ \text{ Radius = 42 cm} \]
\[\text{ Lateral surface area of the cylinder }  = 2\pi rh\]
\[ = 2 \times \frac{22}{7} \times 42 \times 120\]
\[ = 31680 {\text{ cm } }^2 \]
\[\text{ It takes 500 complete revolutions to level a playground } . \]
\[ \therefore \text{ Area of the field } = 31680 \times 500 = 15840000 {\text{ cm } }^2 \left( {1 \text{ cm } }^2 = \frac{1}{10000} m^2 \right)\]
\[ \therefore {15840000\text{ cm } }^2 =1584 \text{ m } ^2 . \]
\[\text{ Thus, the area of the field in }  m^2 \text{ is 1584 m} ^2 .\]