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Question
A tent consists of a frustum of a cone capped by a cone. If the radii of the ends of the frustum be 13 m and 7 m , the height of the frustum be 8 m and the slant height of the conical cap be 12 m, find the canvas required for the tent. (Take : π = 22/7)
Solution
The height of the frustum cone is h= 8 m. The radii of the end circles of the frustum are r1 = 13m and r2 =7m.
The slant height of the frustum cone is
`l=sqrt((r_1-r_2)^2+h^2`
`=sqrt((13-7)^2+8^2`
`=sqrt(100)`
= 10 meter
The curved surface area of the frustum is
`S_1=pi(r_1+r_2)xxl`
= π x (13+7) x 10
= π 20 x 10
= 200π m2
The base-radius of the upper cap cone is 7m and the slant height is 12m. Therefore, the curved surface area of the upper cap cone is
S2 = π x 7 x 12
`=22/7xx7xx12`
= 22 x 12
= 264 m2
Hence, the total canvas required for the tent is
S1 + S2 = 200π + 264
= 892.57 m2
Hence total canvas is 892.57 m2
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