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The diameter of a roller is $ 84 \mathrm{~cm} $ and its length is $ 120 \mathrm{~cm} $. It takes 500 complete revolutions to move once over to level a playground. Find the area of the playground in $ \mathrm{m}^{2} $.
Given:
Diameter of the roller$=84\ cm$
Length of the roller$=120\ cm$.
The number of revolutions taken to level the playground$=500$.
To do:
We have to find the area of the playground in $m^2$.
Solution:
Radius of the roller$=\frac{84}{2}\ cm=42\ cm$.
We know that,
The 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
$=500\times2\times(\frac{22}{7})\times42\times120\ cm^2$
$=15840000\ cm^2$
$=\frac{15840000}{10000}\ m^2$
$=1584\ m^2$
The area of the playground is $1584\ m^2$.
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