AN EXBITION TENT IS IN THE FORM OF A CYLINDER SURMOUNTED BY A CONE .THE HEIGHT OF THE TENT ABOVE THE GROUND IS 85 m AND 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 FOR STICNG.GIVE YOUR ANSWER TO THE NEAREST sq cm .
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Given, the diameter of the base = 168 m.Then the radius, r = 168/2 = 84 m.Height of cylinder(H) = 50 m.
Height of the cone(h) = 85 – 50 = 35 m.
Using the Pythagoras theorem,
Slant height (l) = √(r^2 + h^2) = √(84^2 + 35^2) = √(7056 + 1225) = √8281 = 91 m.
Surface area of the tent = curved surface area of cylinder + curved surface area of cone
= 2πr(2H + l) = (22/7×84)( 2×50 + 91)
=( 22/7×84)(100 + 91) = 22×12×191 = 50424 m^2
It costs 20% extra for stitching and folding = 20% of 50424 = 20/100×50424 = 10084.8 m^2
Therefore, total canvas required = 50424 + 10084.8 = 60509 m^2.
Height of the cone(h) = 85 – 50 = 35 m.
Using the Pythagoras theorem,
Slant height (l) = √(r^2 + h^2) = √(84^2 + 35^2) = √(7056 + 1225) = √8281 = 91 m.
Surface area of the tent = curved surface area of cylinder + curved surface area of cone
= 2πr(2H + l) = (22/7×84)( 2×50 + 91)
=( 22/7×84)(100 + 91) = 22×12×191 = 50424 m^2
It costs 20% extra for stitching and folding = 20% of 50424 = 20/100×50424 = 10084.8 m^2
Therefore, total canvas required = 50424 + 10084.8 = 60509 m^2.
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