# 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? - Mathematics

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?

#### Solution

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

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#### APPEARS IN

RD Sharma Class 8 Maths
Chapter 22 Mensuration - III (Surface Area and Volume of a Right Circular Cylinder)
Exercise 22.1 | Q 14 | Page 11