Discussion

Question:

A right circular cone of height h is cut by a plane parallel to the base and at a distance h/3 from the base, then the volumes of the resulting cone and the frustum are in the ratio

Explanation:

Let the radius and height of original cone be ‘r’ and ‘h’ respectively.
∴ The volume of the original cone (V) = $\frac{π r^{2} h}{3}$ .

The height and radius of the smaller cone are $\frac{2 h}{3}$ and $\frac{2 r}{3}$ respectively.

So its volume = $\frac{π}{3} \times {(\frac{2 r}{3})}^{2} \times \frac{2 h}{3} = \frac{8 V}{27}$ .

∴ Volume of frustum = $(V - \frac{8 V}{27}) = \frac{19 V}{27} .$

∴ Ratio of the volumes = 8 : 19.

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