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A cone and a cylinder are of the same height. If diameters of their bases are in the ratio 3 : 2, find the ratio of their volumes.

Mensuration

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Answer

Let height of cone and cylinder = h.

Diameter of base of cone = 3a, (r1) = 3a2\dfrac{3a}{2}.

Diameter of base of cylinder = 2a, (r2) = 2a2\dfrac{2a}{2}.

Volume of cone (V1) = 13πr12h\dfrac{1}{3}πr_1^2h.

Volume of cylinder (V2) = πr22hπr_2^2h.

∴V1V2=13πr12hπr22h=13×r12r22=13×(3a2)2(2a2)2=13×9a2×224a2×22=13×94=34=3:4.\therefore \dfrac{V1}{V2} = \dfrac{\dfrac{1}{3}πr1^2h}{πr2^2h} \\[1em] = \dfrac{1}{3} \times \dfrac{r1^2}{r2^2} \\[1em] = \dfrac{1}{3} \times \dfrac{\Big(\dfrac{3a}{2}\Big)^2}{\Big(\dfrac{2a}{2}\Big)^2} \\[1em] = \dfrac{1}{3} \times \dfrac{9a^2 \times 2^2}{4a^2 \times 2^2} \\[1em] = \dfrac{1}{3} \times \dfrac{9}{4} \\[1em] = \dfrac{3}{4} \\[1em] = 3 : 4.

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