An exhibition tent is in the form of a cylinder surmounted by a cone. The height of the tent above the ground is 85m and height of the cylinderical part is 50m. If the diameter of the base is 168m, find the quantity of canvas required to make the tent. Allow 20% extra for fold and for stitching. Give your answer to the nearest m^2
Answers
Answer:
total canvas required will be equal to the sum of the lateral surface area of cone and cylinder.
Given, Diameter of base of cylinder and cone 168m⇒r=84m
Total height of tent = 85m
Height of cylinder = 50m
Height of cone 35m
slant height (l) of cone =
35
2
+84
2
=
h
2
+r
2
l=
8281
=91m
Lateral surface area of cylinder 2πrh
=2×π×84×50
Lateral surface area of cone = πrl
=π×84×91
It is given that add 20m
2
extra for fold and for a stitching
∴ Total canvas required =Lateral surface area of (cylinder+cone) + 20m
2
Total canvas required = 2×π×84×+π×84×91+ 20 % extra
=26389.378+24014.334+20 % extra
=
7
22
×84×(2×50+91)+20 % extra
=264×191+
100
120
=60508.8m
2
=60509m
2
Hence,the total area of canvas needed = 60509 sq. meters
Step-by-step explanation:
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