The inner diameter of a circular well is 3.5 m . It is 10 my deep . Find
1 it's inner curved surface area
2 the cost of plastering this curved surface at the rate of rupee 40 m square
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curved surface area of well=2πrh
that employs area=2×3.14×3.5\2×10
area =110m^2
as cost of plastering 1 m^2=40 RS
THEREFORE COST OF 11Om^2=110×40
This becomes=4400RS
that employs area=2×3.14×3.5\2×10
area =110m^2
as cost of plastering 1 m^2=40 RS
THEREFORE COST OF 11Om^2=110×40
This becomes=4400RS
Answered by
2
Answer:
Step-by-step explanation:
Inner curved surface area=2πrh
Radius=35/20=7/4m
Put values
2π*7/4*10=110m is the CSA
Cost of plastering at 1msquare =40
Cost of plastering at 110m square= 110*40=4400rupees
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