Question

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The diameter of a roller is 84 cm and its length is 120 cm.It takes 500 complete revolutions to move once to level a playground.Find the area of the playground in m^2.


This answer is 1584 m^2

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Answer 1

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Radius of the roller (r) = 84/ 2 cm = 42 cm 

Length of the roller (h) = 120 cm 

Curved surface area of the roller = 2πrh 

= 2 × 22/ 7 × 42 × 120 cm2 = 44 × 720 cm2 = 31680 cm2 

∴ area covered by the roller in 1 revolution = 31680 cm2 

∴ area covered by the roller in 500 revolutions = 31680 × 500 cm2 = 15840000 cm

Hence, area of the playground = 15840000/ 100× 100 m2 
= 1584 m2 Ans.

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Answer 2

diameter of a roller =84cm
length of a roller is 120 cm
radius = 42cm
CSA of a roller =2pie r h
2*22/7*42*120
=31680 cm
area of ground level in one rotation is 1368 m square therefore area of ground level in 500 rotations is equals to 500*31680=15840000cm2
$1cm2 = 10000cm2$
=1584m2
cost of levelling the ground per metre square is equals to 2 therefore total cost of levelling the ground is equals to 1584 into 2 =3168