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# A right circular cone is inscribed in a hemisphere so that

tagged by: AAPL

00:00

A

B

C

D

E

## Global Stats

Difficult

Official Guide

A right circular cone is inscribed in a hemisphere so that the base of the cone coincides with the base of the hemisphere. What is the ratio of the height of the cone to the radius of the hemisphere?

$$A.\ \sqrt{3}:1$$
$$B.\ 1:1$$
$$C.\ \frac{1}{2}:1$$
$$D.\ \sqrt{2}:1$$
$$E.\ 2:1$$

OA B.

### GMAT/MBA Expert

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AAPL wrote:
Official Guide

A right circular cone is inscribed in a hemisphere so that the base of the cone coincides with the base of the hemisphere. What is the ratio of the height of the cone to the radius of the hemisphere?

$$A.\ \sqrt{3}:1$$
$$B.\ 1:1$$
$$C.\ \frac{1}{2}:1$$
$$D.\ \sqrt{2}:1$$
$$E.\ 2:1$$
If a right circular cone is inscribed in a hemisphere so that the base of the cone coincides with the base of the hemisphere, then the height of the cone is exactly the radius of the hemisphere. So the ratio is 1:1.

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Scott Woodbury-Stewart Founder and CEO

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