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 ${m^2}$.

Answer 1

Hint: Area of the playground will be the area covered by the roller revolving $500$ times. In other words, it will be $500$ times the curved surface area of the roller.

Complete step-by-step answer:
We know that the shape of the roller is cylindrical and the length of the roller in this case is its height.
So, according to the question, the height of the cylindrical roller is $120cm$ i.e. $h = 120cm$.
And its diameter is $84cm$. Thus, its radius will be:
$
   \Rightarrow r = \dfrac{{84}}{2}cm, \\
   \Rightarrow r = 42cm \\
$
We know that the curved surface area of a cylinder is $2\pi rh$. Surface area of roller is:
$
   \Rightarrow A = 2 \times \pi \times 42 \times 120, \\
   \Rightarrow A = 2 \times \dfrac{{22}}{7} \times 42 \times 120, \\
   \Rightarrow A = 31680c{m^2} \\
$
Thus total area covered by the roller in $500$ revolutions will be:
$
   \Rightarrow Area = 500 \times 31600c{m^2}, \\
   \Rightarrow Area = 15840000c{m^2}, \\
   \Rightarrow Area = \dfrac{{15840000}}{{10000}}{m^2}, \\
   \Rightarrow Area = 1584{m^2} \\
$
Therefore, the area of the playground is $1584{m^2}$.

Note: Here we are considering only the curved surface area of the roller because the cross-sectional portion will not come in contact with the ground.