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The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to move once over to level a playground. Find the area of the playground in m^{2}?

`["Assume "pi=22/7]`

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#### Solution

It can be observed that a roller is cylindrical.

Height (*h*) of cylindrical roller = Length of roller = 120 cm

Radius (*r*) of the circular end of roller = (84/2)cm = 42cm

CSA of roller = 2π*rh*

`=(2xx22/7xx42xx120)cm^2`

= 31680 cm^{2}

Area of field = 500 × CSA of roller

= (500 × 31680) cm^{2}

= 15840000 cm^{2}

`=15840000/(100xx100) m^2`

= 1584 m^{2}

Concept: Surface Area of Cylinder

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