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A right circular cone is inscribed in a hemisphere so that [#permalink]
09 Jul 2012, 03:41

Expert's post

3

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

15% (low)

Question Stats:

68% (01:48) correct
32% (01:01) wrong based on 323 sessions

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) \(2:1\)

Diagnostic Test Question: 20 Page: 23 Difficulty: 600

Re: A right circular cone is inscribed in a hemisphere so that [#permalink]
09 Jul 2012, 03:42

1

This post received KUDOS

Expert's post

SOLUTION

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) \(2:1\)

Note that a hemisphere is just a half of a sphere.

Now, since the cone is a right circular one, then the vertex of the cone must touch the surface of the hemisphere directly above the center of the base (as shown in the diagram below), which makes the height of the cone also the radius of the hemisphere.

Re: A right circular cone is inscribed in a hemisphere so that [#permalink]
09 Jul 2012, 05:28

1

This post received KUDOS

Bunuel wrote:

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?

Re: A right circular cone is inscribed in a hemisphere so that [#permalink]
13 Jul 2012, 02:11

Expert's post

SOLUTION

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) \(2:1\)

Note that a hemisphere is just a half of a sphere.

Now, since the cone is a right circular one, then the vertex of the cone must touch the surface of the hemisphere directly above the center of the base (as shown in the diagram below), which makes the height of the cone also the radius of the hemisphere.

Re: A right circular cone is inscribed in a hemisphere so that [#permalink]
25 Sep 2013, 15:09

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Re: A right circular cone is inscribed in a hemisphere so that [#permalink]
06 May 2014, 11:57

Bunuel wrote:

SOLUTION

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) \(2:1\)

Note that a hemisphere is just a half of a sphere.

Now, since the cone is a right circular one, then the vertex of the cone must touch the surface of the hemisphere directly above the center of the base (as shown in the diagram below), which makes the height of the cone also the radius of the hemisphere.

Attachment:

Cone.png

Answer: B.

Hi Bunuel, how did you conclude the underlined in above statement? Thanks!

Re: A right circular cone is inscribed in a hemisphere so that [#permalink]
07 May 2014, 03:25

Expert's post

Dienekes wrote:

Bunuel wrote:

SOLUTION

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) \(2:1\)

Note that a hemisphere is just a half of a sphere.

Now, since the cone is a right circular one, then the vertex of the cone must touch the surface of the hemisphere directly above the center of the base (as shown in the diagram below), which makes the height of the cone also the radius of the hemisphere.

Attachment:

The attachment Cone.png is no longer available

Answer: B.

Hi Bunuel, how did you conclude the underlined in above statement? Thanks!

Consider the cross-section. We'd have an isosceles triangle inscribed in a semi-circle:

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