The slant height of a frustum of a cone is 7 m and the perimeter of circular ends are 24 m and 20 m . The cost of painting it's curved surface area at Rs. 50 sq. m *
Answers
Given :- The slant height of a frustum of a cone is 7 m and the perimeter of circular ends are 24 m and 20 m . The cost of painting it's curved surface area at Rs. 50 sq. m ?
Solution :-
let radius of both circular ends is r1 and r2 .
so,
→ 2πr1 = 24 m
→ πr1 = 12 m ---- Eqn.(1)
and,
→ 2πr2 = 20 m
→ πr2 = 10 m ---- Eqn.(2)
adding Eqn.(1) and Eqn.(2),
→ πr1 + πr2 = 12 + 10
→ π(r1 + r2) = 22
then,
→ CSA of frustum of cone = π(r1 + r2) * slant height
→ CSA of frustum of cone = 22 * 7
→ CSA of frustum of cone = 154 m² .
therefore,
→ Cost of painting the CSA = Rate / m² * CSA = 50 * 154 = Rs.7700 (Ans.)
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Given : The slant height of a frustum of a cone is 7 m and the perimeter of circular ends are 24 m and 20 m .
To Find : The cost of painting it's curved surface area at Rs. 50 sq. m
Solution:
curved surface area of frustum of a cone = π(R + r) l
R = Radius of Base
r = Radius of top
l = slant height
perimeter of circular ends are 24 m and 20 m
=> 2πR = 24 and 2πr = 20
=> 2πR + 2πr = 24 + 20
=> 2π(R + r) = 44
=> π(R + r) = 22
l = 7
curved surface area of frustum of a cone = π(R + r) l
= 22 * 7
= 154 m²
cost of painting Rs. 50 sq. m
Hence total cost = 154 * 50 = 7700 Rs
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