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A right circular cone is inscribed in a hemisphere so that

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A right circular cone is inscribed in a hemisphere so that

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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.

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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.

Answer: B

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Scott Woodbury-Stewart Founder and CEO

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