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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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