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Question
The slant height of a frustum of a cone is 4 m and the perimeter of circular ends is 18 m and 16 m. Find the cost of painting its curved surface area at ₹ 100 per sq. m
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Solution
Slant height of a frustum (l) = 4 m
Perimeter of the top part = 18 m
2πR = 18
`2 xx 22/7 xx "R"` = 18
R = `(18 xx 7)/(2 xx 22) xx 63/22`
Perimeter of the bottom = 16 m
2πr = 16
`2 xx 22/7 xx "r"` = 16
r = `(16 xx 7)/(2 xx 22) = 28/11`
C.S.A of the frustum = π l (R + r) sq.units
= `22/7 xx 4 (63/22 + 28/11)`
= `22/7 xx 4 ((63 + 56)/22)`
= `22/7 xx 4 xx 119/22`
= 4 × 17
= 68 sq.units
Cost of painting
= ₹ 100 × 68
= ₹ 6800
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