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

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03 Apr 2012, 16:49

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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. \(2 : 1\)

I've never heard of a "hemisphere". I know that the cone forms a 30-60-90 right triangle and therefore the height of the cone is /3 but because I don't know what to do with the hemisphere, I'm stuck.

A right cone does not necessarily form a 30-60-90 triangle. In this case, the cone's is formed from an a isosceles right triangle rotated about the center - so the radius of the cone's base (also the hemisphere's base) = the height of the cone.

1:1 ratio - B

I noticed some "gimme" geometry problem on my exams - learn this stuff well & you can shave minutes off the time spent on geometry questions.

A right cone does not necessarily form a 30-60-90 triangle. In this case, the cone's is formed from an a isosceles right triangle rotated about the center - so the radius of the cone's base (also the hemisphere's base) = the height of the cone.

1:1 ratio - B

I noticed some "gimme" geometry problem on my exams - learn this stuff well & you can shave minutes off the time spent on geometry questions.

I take back what I said regarding the 30-60-90 triangle, that was a silly assumption on my part.

Thanks for the explanation regarding the hemisphere, it's so simple!

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

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04 Apr 2012, 00:48

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NvrEvrGvUp 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?

A. \(\sqrt{3} : 1\)

B. \(1 : 1\)

C. \(\frac{1}{2} : 1\)

D. \(\sqrt{2} : 1\)

E. \(2 : 1\)

I've never heard of a "hemisphere". I know that the cone forms a 30-60-90 right triangle and therefore the height of the cone is /3 but because I don't know what to do with the hemisphere, I'm stuck.

Thanks, Rich

As mentioned above hemisphere is just a half of a sphere. Now, since the cone is a right circular cone, the vertex of the cone must touch the surface of the hemisphere directly above the center of the base, 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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23 Nov 2012, 02:10

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Not sure if anyone else was scratching their heads wondering why the height couldn't be less than the radius, but just in case... it has to do with the word "inscribed": to draw within a figure so as to touch in as many places as possible <a regular polygon inscribed in a circle>

A right cone does not necessarily form a 30-60-90 triangle. In this case, the cone's is formed from an a isosceles right triangle rotated about the center - so the radius of the cone's base (also the hemisphere's base) = the height of the cone.

1:1 ratio - B

I noticed some "gimme" geometry problem on my exams - learn this stuff well & you can shave minutes off the time spent on geometry questions.

"gimme" geometry problems? might be because I'm new on the forum, but can you tell me what are you referring to?

A right cone does not necessarily form a 30-60-90 triangle. In this case, the cone's is formed from an a isosceles right triangle rotated about the center - so the radius of the cone's base (also the hemisphere's base) = the height of the cone.

1:1 ratio - B

I noticed some "gimme" geometry problem on my exams - learn this stuff well & you can shave minutes off the time spent on geometry questions.

"gimme" geometry problems? might be because I'm new on the forum, but can you tell me what are you referring to?

"Gimme" is slang for quick/easy. If you know geometry well, you can avoid doing any calculation on some problems.

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

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08 Jun 2013, 21:12

I know there are solutions to this question posted, however I have a follow-up clarifying question.

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?

Answer is 1:1

My question is, if you have a cone with a radius of 5 and a height of 1, how could the ratio be 1:1? You could have a very wide and short hemisphere. Am I missing something with regards to the cone being designated as a "right circular cone"? Can someone please explain?

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

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08 Jun 2013, 22:36

It's the combination of "right circular cone" and "inscribed". The cone shares a base with the hemisphere and it's tip touches the top of the hemisphere. If you had a right circular cone of a height to radius ratio of anything other than 1:1 you would not be able to inscribe it in a hemisphere. Essentially, what you've described is impossible for this particular problem.

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

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08 Jun 2013, 22:39

Sorry, forgot to add that you cannot have a hemisphere of the type you've described. It wouldn't be a hemisphere. A sphere's radius is the same throughout. A hemisphere is half of a sphere (so it's height must equal it's radius.

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

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01 Aug 2014, 04:01

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

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12 Oct 2015, 11:19

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).

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