Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 2.3 Junior Find the answer

A right circular cone has for its base a circle having the same radius as a given sphere.
The volume of the cone is one-half that of the sphere. The ratio of the altitude of the cone to the radius of its base is:
$\textbf{(A)}\ \frac{1}{1}$ $\textbf{(B)}\ \frac{1}{2}$ $\textbf{(C)}\ \frac{2}{3}$ $\textbf{(D)}\ \frac{2}{1}$ $\textbf{(E)}\ \sqrt{\frac{5}{4}}$

Your answer

Multiple choice: answer with the letter of the option you want.

Solution

Because the circle has the same radius as the sphere, the cylinder and sphere have the same radius. Then from the volume of cylinder and volume of a sphere formulas, we have $\frac{1}{3} \pi r^2 h= \frac{2}{3} \pi r^3 \implies h=2r\implies \frac{h}{r}=2$ $\boxed{(\textbf{D})}$

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.