A right circular cone of base radius r and height h is placed inside a hollow hemisphere of radius R such that the cone's base coincides with the hemisphere's base circle and its apex touches the hemisphere's inner surface. If r = frac R 2 , what is the ratio frac h R ?

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EEJ MAIN Aptitude QUESTION #10987

Question 1

A right circular cone of base radius $r$ and height $h$ is placed inside a hollow hemisphere of radius $R$ such that the cone's base coincides with the hemisphere's base circle and its apex touches the hemisphere's inner surface. If $r = \frac{R}{2}$, what is the ratio $\frac{h}{R}$?

$\frac{\sqrt{3}}{2}$✔️
$\frac{\sqrt{2}}{2}$
$\frac{\sqrt{3}}{4}$
$\frac{1}{2}$

Correct Answer Explanation

The hemisphere's center is at the base center. The apex is at distance $R$ from center. Using Pythagoras: $h^2 + r^2 = R^2$. With $r = R/2$, we get $h^2 = R^2 - R^2/4 = 3R^2/4$, so $h = \frac{\sqrt{3}R}{2}$. Thus $\frac{h}{R} = \frac{\sqrt{3}}{2}$.

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