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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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