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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

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]

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09 Jul 2012, 06:28

1

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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]

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25 Sep 2013, 16:09

1

This post received KUDOS

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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]

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06 May 2014, 12:57

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!

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:

Re: A right circular cone is inscribed in a hemisphere so that [#permalink]

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22 Jul 2015, 20:12

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Re: A right circular cone is inscribed in a hemisphere so that [#permalink]

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11 May 2016, 13:32

Hemisphere is just half of a sphere if you can visualize the figure , you can easily deduce that height of cone will be equal to radius of hemisphere So the ratio would be 1:1 correct answer - B

Re: A right circular cone is inscribed in a hemisphere so that [#permalink]

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20 Apr 2017, 19:04

Bunuel wrote:

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:

Attachment:

The attachment Untitled.png is no longer available

Why cone can't be like the image below?

Attachment:

File comment: cone in hemisphere

cone in hemisphere.png [ 36.85 KiB | Viewed 406 times ]

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