Question

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Answer 1

Given that the diameter of the road roller d = 28cm.

Then the radius of the road roller r = d/2

                                                           = 28/2

                                                           = 14cm.


Given that the length of the road roller = 150cm.

Distance covered by the road roller in 1 revolution = 2pirh

                                                                                     = 2 * 22/7 * 14 * 150

                                                                                     = 13200cm^2


Then, the distance covered by the road roller in 500 revolutions = 

= 13200 * 500

= 6600000cm^2.

= 660m^2.


Therefore the area of the land = 660m^2.


Given that the cost of leveling per sq.m = 100.


Therefore the cost of leveling the ground = 660 * 100

                                                                        = 66000.

Answer 2

ANSWER
......................








diameter of the road ROLLER ( d) = 28cm.

radius of the road roller (R) = d
____
2
= 28/2

= 14cm.






## length of the road roller = 150cm.




## Distance covered by the road roller in 1 revolution =
$2\pi \: r$




$2 \times \frac{22}{7} \times 14 \times 150 \\ \\ = 13200 \: cm \: {}^{2}$


the distance covered by the road roller in 500 revolutions =



$13200 \times 500 \\ \\ \\ \frac{6600000}{100} \\ \\ = \\ \\ 660 \: m \: {}^{2}$


the area of the land = 660m^2.



cost of leveling per sq.m = 100.


the cost of leveling the ground = 660 * 100

= 66000