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ML Aggarwal Solutions Class 10 Mathematics Solutions for Mensuration Exercise 17.4 in Chapter 17 - Mensuration
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An exhibition tent is in the form of a cylinder surmounted by a cone. The height of the tent above the ground is 85 m and the height of the cylindrical part is 50 m. If the diameter of the base is 168 m, find the quantity of canvas required to make the tent. Allow 20% extra for folds and stitching. Give your answer to the nearest m².
Solution:
As the tent is the combination of cone and cylinder, so its surface area will be equal to sum of curved surface area of both cone and cylinder.
Given height of the tent above the ground = 85 m
Height of the cylindrical part, H = 50 m
height of the cone, h = 85-50
h = 35 m
Diameter of the base, d = 168 m
Radius of the base of cylindrical part, r = d/2 = 168/2 = 84 m
Radius of cone, r = 84 m
Slant height of the cone, l = √(h2+r2)
l = √(352+842)
l = √(1225+7056)
l = √(8281)
l = 91 m
Surface area of tent = curved surface area of cylinder + curved surface area of cone
= 2π rH+ π rl
= π r(2H+l)
= (22/7)×84×(2×50+91)
= (22/7)×84×(100+91)
= ((22×84)/7)×191
= (1848/7)×191
= 264×191
= 50424 m2
Adding 20% extra for folds and stitches,
Area of canvas = 50424+ 20% of 50424
= 50424+ (20/100) ×50424
= 50424+ 0.2×50424
= 50424+ 10084.8
= 60508.8 m2
= 60509 m2
Hence the the quantity of canvas required to make the tent is 60509 m2.
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