In this problem use π=3.14.
This question was asked in 2012 cbse board exam. Question from RS Aggarwal book chapter volume and surface area of solid, page number 823, exercise 17C, problem number 13
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In this problem use π=3.14.
This question was asked in 2012 cbse board exam. Question from RS Aggarwal book chapter volume and surface area of solid, page number 823, exercise 17C, problem number 13
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Solution
given data
Greater radius = R = 33 cm
Smaller radius = r = 27 cm
Slant height = l = 10 cm
Surface area of the frustum = π(R+r)l +πR² +πr²
=π(R²+r²+l[R+r])
= [33²+27²+10[33+27]]π
= [1089+729+1060]π
= 2418×3.14
= 7592.52 cm²
∴ total surface area of frustum is 7592.52 cm²