Question

A road roller 750 complete revolutions to move once over to level a road. find the area of the road if the diameter of a road too is 84cm and length is 1m.
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Answer 1

Given:

• Diameter of the road roller = 84cm = 84/100 = 0.8m
• Length (Height) of the road roller = 1m
• No. of revolutions taken by road roller to move once over to load a road = 750

To find:

• Area of the road

$\huge \mathbb{ \underline{SOLUTION:-}}$

The shape of the road roller is like a cylinder.

Area of the road roller = C.S.A of cylinder

$\large \implies{2 \pi \: rh} \\ \large \implies{2 \times \frac{22}{7} \times \frac{0.84}{2} \times 1} \\ \large \implies{ \bold{2.64m}}$

Area of the road = Area of the road roller × No. of revolutions taken by it.

$\large \implies{2.64 \times 750} \\ \large \implies{1980.00} \\ \large \implies{ \bold{1980 {m}^{2} }}$

Hence,area of the road is $\large { \bold{1980 {m}^{2} }}$

Answer 2

Solution:-

Diameter of roller = 84 then radius = 84/2 = 42 or 0.42 m and height = 1 m

In one revolution, the road roller will cover an area equal to its lateral surface area.

So, in one revolution the area of the road covered = 2πrh

= 2*22/7*0.42*1

Area of the road in one revolution = 2.64 sq m

Therefore, in 750 revolutions, the area of the road = 2.64*750

= 1980 sq m