Question

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 $m^2$?
$[Assume\pi =\frac{22}{7}]$

Answer 1

A roller is cylindrical shaped.
Height (h) of cylindrical roller = Length of roller = 120 cm
Radius (r) of the circular end of roller
$=\left(\frac{84}{2}\right)cm=42cm$
$\left[0.5marks\right]$

CSA of roller
$=2\pi rh$
$=2\pi \times 42\times 120$
$=31680c{m}^2$
$\left[0.5marks\right]$

Area of field
$=Numberofrevolutions\times CSAofroller$$=500\times CSAofroller$
$=(500\times 31680)c{m}^2=15840000c{m}^2=1584m^2$
$\left[1marks\right]$