due to sudden floods some welfare association jointly requested the government to get 100 tents fixed immediately and offered to contribute 50percent of the cost. If the lower part of each of each tent is of the form of cylinder of diameter 4.2 m and height 4 m with the conical upper part of same diameter but height 2.8 m and the canvas to be used cost 100 rupees per sq m find the amount the association will have to pay
Answers
Association has to pay 379500 Rs
•cylinder :
height = 4m
diameter = 4.2 m
radius = 2.1 m
•CSA of cylinder = 2πrh
= 2×22×2.1×4/7
= 176 × 0.3
= 52.8 m²
•Cone :
radius = 2.1 m
height = 2.8 m
l = √(r²+h²)
l = √ (4.41+7.84)
l = √ (12.25)
l = 3.5m
•LSA of cone = πrl
= 22×2.1×3.5/7
= 23.1 m²
•surface area of tent = CSA of
cylinder + CSA of cone
•surface area of tent = 52.8 + 23.1
= 75.9 m²
•surface area of 100 tents = 7590
m²
•cost of 1m² cloth = 100 Rs
•cost of 7590 m² = 759000 Rs
•Association has to pay 50% i.e. half of the total value = 759000/2
= 379500 Rs
Answer:
Step-by-step explanation:
Given: height of cylinder = 4m
Height of conical = 2.8m
Diameter = 4.2m
Radius = 2.1 m
Surface area of cylinder part :2×22/7×r × h
= 2× 22/7 ×2.1×4
=52.80
Surface area of conical part:
= 22/7 × r (r^2 +h^2)^1/2
= 23.10
Surface area of tent = surface area of cylinder + surface area of conical
=52.80 + 23.10 = 75.90
Surface area of 100 tent = =100× 75.90
= 7590
Cost of 100 tent = 100 × 7590
= 759000
Association has to pay 50 % =
= 759000×0.5
=379500