# Q: 4    The diameter of a roller is $\small 84\hspace{1mm}cm$ and its length is $\small 120\hspace{1mm}cm$. It takes $\small 500$ complete revolutions to move once over to level a playground. Find the area of the playground in $\small m^2$.

H Harsh Kankaria

Given,

The diameter of the cylindrical roller = $\small d = 84\hspace{1mm}cm$

Length of the cylindrical roller = $\small h = 120\hspace{1mm}cm$

The curved surface area of the roller = $2\pi r h = \pi dh$

$\\ = \frac{22}{7}\times84\times120 \\ \\ = 31680\ cm^2$

$\therefore$ Area of the playground = $Area\ covered\ in\ 1\ rotation \times500$

$\\ = 31680 \times500 \\ = 15840000 cm^2 \\ = 1584\ m^2$

Therefore, the required area of the playground = $1584\ m^2$

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