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The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to move once over to level a playground. Find the area of the playground in $${m}^2$$.

Given :
Height (h) of cylindrical roller = Length of roller = 120 cm
And radius (r) of the circular end roller = $$\frac{84}{2}$$ = 42 cm

It can be observed that a roller is cylindrical.

Now, we know that,
CSA of rollar
= $$2{\pi}rh$$
= $$2 × \frac{22}{7} × 42 × 120$$ $${cm}^2$$
= $$2 × 22 × 6 × 120$$
= 31680 $${cm}^2$$

Therefore, Area of field
= 500 x CSA of roller
= $$31680 × 500$$ $${cm}^2$$
= 1584 $${m}^2$$
$$\therefore$$ the area of the playground in $${m}^2$$ is 1584.