A right circular cone has a height that is double its radius, and its volume is measured at 2,250π cm³. If the radius is reduced to one-third and the height to half of their original sizes, what would be the new volume of the cone?
Let the radius of right circular cone be 'R' cm. So, height of cone = '2R' cm Volume of a right circular cone = (1/3) X π X (radius) 2 X height So, (1/3) X π X R2 X 2R = 2,250π Or, R3 = 3,375 Or, 'R' = 15 cm Height of cone = 2 X 15 = 30 cm New radius of cone = (15/3) = 5 cm New height of cone = (30/2) = 15 cm So, required volume = (1/3) X π X 52 X 15 = 125π cm3
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