Question

The diameter of a roller is $84cm$ and its length is $120cm$. It takes $500$ complete revolutions to move once over to level a playground. Find the area of the playground in $m^2$?(Assume $\mathrm{\pi }=\frac{22}{7}$)

Answer 1

Step 1: Given Information

Diameter of roller $=84cm$

Therefore, radius $\left(r\right)=\frac{84}{2}=42cm$

Length of the roller $\left(h\right)=120cm$

Step 2. Finding curved surface area, $CSA$ of roller.

$CSA$ of roller $=CSA$ of cylinder

$$\begin{gathered} =2\mathrm{\pi rh} \\ =2\times \frac{22}{7}\times 42\times 120 \\ =44\times 6\times 120 \\ =44\times 720 \\ =31680{\mathrm{cm}}^2 \end{gathered}$$

That means area covered by roller in one revolution $=31680c{m}^2$

Therefore area covered by roller in $500$ revolution$=\frac{500\times 31680}{10000}\left[\because 1c{m}^2=\frac{1}{10000}{m}^2\right]$

$=1584m^2$

Hence the area of the playground is $1584m^2$.