# Cones' radiuses ratio

Volume of the upper cone is equal to half volume of the whole cone. What is the ratio of the base radius of the whole cone to base radius of the upper cone?

I got: $\sqrt{2h \over H}$ where $h$ is the height of the upper cone and $H$ is the height of the whole cone. But it must be some constant according to the book from which this question was taken.

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## 1 Answer

The whole cone and the upper cone are similar. That means that one is just a scaled version of the other.

Let $\lambda$ be the ratio of the base radius of the whole cone to the base radius of the upper cone.

If the linear dimensions of an object are scaled by the factor $\lambda$, then areas are scaled by the factor $\lambda^2$, and volumes are scaled by the factor $\lambda^3$.

Thus the ratio of the volumes is $\lambda^3$. We know that $\lambda^3=2$, so $\lambda=2^{1/3}$.

Remark: We could let the base radius of the upper cone be $r$, and let its height be $h$. Suppose that the base radius of the whole cone is $\lambda r$. By using similar triangles you can then show that the height of the whole cone is $\lambda h$.

So the volume of the upper cone is $\frac{1}{3}\pi r^2 h$, and the volume of the whole cone is $\frac{1}{3}(\lambda r)^2(\lambda h)$. This is $\frac{\lambda^3}{3}r^2h$.

Find the ratio of the volumes. We get $\lambda^3$. Set this equal to $2$.

Same result, but a whole lot of unnecessary work!

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Thank you very much sir. Neat and simple indeed. How couldn't I think of it. –  crghp Feb 5 '13 at 22:50
In a remark, I added a sketch of the "ugly" solution. –  André Nicolas Feb 5 '13 at 22:54
Thanks for the brief explanation! I completely forgot of these similarity rules, and your "ugly" example cleared things out. Thank you very much again. I appreciate your support. Have a good day! –  crghp Feb 5 '13 at 23:02