Question

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 *​

Answer 1

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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Answer 2

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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