Question

A road roller takes \[750\] to complete revolutions to move once over to level a road. Find the area of the road if the diameter of a road roller is \[84cm\] and length is \[1m\]?

Answer 1

Hint: To find the total length of the road to be rolled we need to find the patch of road that is to be rolled when the roller takes one revolution, to find the area of road after one revolution of roller, we need to find the circumference (in centimeters or meters) and multiply it with the length of the roller the product will give us the area of the patch of road rolled after a single revolution and then multiply that area with the total number of complete revolutions. To find the formula for the total area of the road we use:
Area of the road after \[750\] revolutions \[\text{= Revolutio}{{\text{n}}_{\text{Total}}}\times \text{Are}{{\text{a}}_{roller}}\]
where \[\text{Revolutio}{{\text{n}}_{\text{Total}}}\] is the total number of revolutions the roller rolled and \[\text{Are}{{\text{a}}_{roller}}\] is the area of the roller rolled after single revolution.

Complete step-by-step answer:
To find the total area of the road we use the formula:
Area of the road after \[750\] revolutions \[\text{= Revolutio}{{\text{n}}_{\text{Total}}}\times \text{Are}{{\text{a}}_{roller}}\].
Now placing the values in the formula i.e. \[\text{Revolutio}{{\text{n}}_{\text{Total}}}=750\] and \[\text{Are}{{\text{a}}_{roller}}=\pi d\times h\] but before that let us find the area of the roller after single revolution.
The circumference of the roller is \[\pi d\] where \[d=84cm,(0.84m)\] is the diameter (in centimeter or meter):
\[=\pi d\]
\[=\pi \times 0.84m\]
\[=2.64m\]
Hence, area of a single patch of rolled road after single revolution is
\[\text{Are}{{\text{a}}_{roller}}=2.64m\times l\]
\[l\] is the length of the roller sideways as \[l=1m\].
\[=2.64m\times 1m\]
\[=2.64{{m}^{2}}\]
Now calculating the area of the road rolled after \[750\] revolutions:
\[\text{= Revolutio}{{\text{n}}_{\text{Total}}}\times \text{Are}{{\text{a}}_{roller}}\]
\[\text{= 750}\times 2.64m\]
\[\text{= }1979.5{{m}^{2}}\approx 1980{{m}^2}\]

\[\therefore \] The area of the road rolled after \[750\] revolutions is \[1980{m^2}\].

Note: Another method to solve the question is by considering the roller as cylinder and taking the lateral surface area of the cylinder and multiplying it with the total number of revolutions which is :
Lateral Area of the cylinder \[=2\pi rh\]
\[=2\times 3.1415\times \dfrac{0.84}{2}\times 1\]
Here radius is \[\dfrac{0.84}{2}m\] and all the dimensions are in meters.
Now multiplying the total number of revolutions by the lateral surface area, we get the total area of the road as:
\[=750\times 2\times 3.1415\times \dfrac{0.84}{2}\times 1\]
\[=1980{{m}^{2}}\]