The diameter of a roller is 72 cm and its length is 120 cm. It takes 200 complete revolutions to move over to level playground. Find the area of the playground.

Given:

Diameter of the roller$=72\ cm$

Length of the roller$=120\ cm$.
Number of revolutions taken to level the playground$=200$.
To do: 

We have to find the area of the playground.

Solution:

Radius of the roller$=\frac{72}{2}\ cm=36\ cm$.

We know that,

Curved surface area of a cylinder of radius r and height h is $2 \pi rh$.

Therefore,

Area covered in 1 revolution$=$Curved surface area of the roller

Area of the playground$=$Number of revolutions$\times$Curved surface area of the roller

$=200\times2\times3.14\times36\times120\ cm^2$

$=5425920\ cm^2$

$=\frac{5425920}{10000}\ m^2$

$=542.592\ m^2$

The area of the playground is $542.592\ m^2$.

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Updated on: 10-Oct-2022

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