=1/3×22/7×12(9² 4² 9×4)
= 22/21×12(81 16 36)
= =88/7×133
=418cm³
Written on Dec. 21, 2018, 8:28 p.m.
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Answer of this question
If the areas of circular bases of a frustum of a cone are 4cm2 and 9cm
Shweta MauryaVolume of frustum =1/3×pie×h(R² r² Rr)
=1/3×22/7×12(9² 4² 9×4)
= 22/21×12(81 16 36)
= =88/7×133
=418cm³
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