The diameter of a roller is 84 cm and its length is 120 cm.
Question. 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 $\mathrm{m}^{2}$ ? $\left[\right.$ Assume $\left.\pi=\frac{22}{7}\right]$

Solution:

It can be observed that a roller is cylindrical.

Height $(h)$ of cylindrical roller $=$ Length of roller $=120 \mathrm{~cm}$

Radius $(r)$ of the circular end of roller $=\left(\frac{84}{2}\right) \mathrm{cm}=42 \mathrm{~cm}$

CSA of roller $=2 \pi r h$

$=\left(2 \times \frac{22}{7} \times 42 \times 120\right) \mathrm{cm}^{2}$

$=31680 \mathrm{~cm}^{2}$

Area of field $=500 \times$ CSA of roller

$=(500 \times 31680) \mathrm{cm}^{2}$

$=15840000 \mathrm{~cm}^{2}$

$=1584 \mathrm{~m}^{2}$
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