Question

A road roller has a diameter 0.7 M and width is 1.2 M Find the least number of revolution that the roller must take in the order to level a playground of size 120m X 44 m

Answer 1

area of playground =120*44 m^2
in cylinder,
height =1.2 m
diameter =0.7 m
radius= 0.35
c. s. a =2πrh
=2*22/7*0.35*1.2
=2.64
no of revolution =area of playground/c.s.a
=120*44/2.64
=2000

Answer 2

Answer:

2000 m^2

Step-by-step explanation:

area of playground = 120× 44

height = 1.2 m

diameter = 0.7 m

radius = 0.35

c.s.a = 2πrh

        =2×22/7×0.35× 12/10

        =2.64

no. of revolutions area = area of playground/c.s.a

=120×44/2.64

=2000 m^2