A road roller has to make 1200 revolutions to level a stretch of a road. If the length of the roller is 1.2metres and the diameter of its circular face is 98cm. Find the area of the road leveled.
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Answer:
The area of road levelled = 4435.2 m^2
Step-by-step explanation:
Given:
length( or height) = 1.2m
diameter = 98 cm
=> radius = 49 cm = 49/100 m
In one revolution, it covers an area on the road equal to the shaded part in my diagram.
It covers an area on the road equal to its own circumference into the height of the road roller.
Therefore,
The area road levelled in one revolution = 2πrh
= 2 x 22/7 x 49/100 x 1.2
= 3.696 m^2
Now, the area of road levelled in 1200 revolutions
= 3.696 x 1200
= 4435.2 m^2
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