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Right circular cone question?

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Right circular cone question?

by phoenixhazard » Sun Oct 24, 2010 1:17 pm
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?


So the answer states its 1:1, which only makes sense if you know that the height of the one is exactly the height of the hemisphere. How does this have to be true? It sounds like the cone is only inside the hemisphere but doesn't tell you how high?
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by vijaynaik » Sun Oct 24, 2010 1:51 pm
When it says 'inscribed' means all the vertices of the cone should be on the 'hemisphere' (semi cercle).

Refer to this u'll get an idea.

https://www.google.com/images?client=fir ... 60&bih=591
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by GMATGuruNY » Mon Oct 25, 2010 3:04 am
phoenixhazard 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?


So the answer states its 1:1, which only makes sense if you know that the height of the one is exactly the height of the hemisphere. How does this have to be true? It sounds like the cone is only inside the hemisphere but doesn't tell you how high?
When no figure is provided, draw your own. The attached drawing shows why the radius of the hemisphere = the height of the cone, giving us a 1:1 ratio.
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by phoenixhazard » Mon Oct 25, 2010 10:49 am
GMATGuruNY wrote:
phoenixhazard 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?


So the answer states its 1:1, which only makes sense if you know that the height of the one is exactly the height of the hemisphere. How does this have to be true? It sounds like the cone is only inside the hemisphere but doesn't tell you how high?
When no figure is provided, draw your own. The attached drawing shows why the radius of the hemisphere = the height of the cone, giving us a 1:1 ratio.
Thanks, I just didn't know the definition of inscribed. I assumed it just meant that the cone was in the area of the hemisphere not that the vertices had to touch.
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by jeetu_vishnoi » Mon Oct 25, 2010 2:42 pm
Since cone is inscribed in a hemisphere conciding with the base of hemisphere then the height of the cone must be touching the side of hemisphere hence the height will be equal to the radius of the hemisphere and ratio is 1:1.
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by frank1 » Wed Oct 27, 2010 5:12 am
phoenixhazard wrote:
Thanks, I just didn't know the definition of inscribed. I assumed it just meant that the cone was in the area of the hemisphere not that the vertices had to touch.
There is another term called circumcribed as well.(sometime used.So will be better to know the defination of it as well)

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by goyalsau » Wed Oct 27, 2010 6:45 am
jeetu_vishnoi wrote:Since cone is inscribed in a hemisphere conciding with the base of hemisphere then the height of the cone must be touching the side of hemisphere hence the height will be equal to the radius of the hemisphere and ratio is 1:1.
I really doubt why the ratio is 1:1 ,
Just imagine that the base of the cone and the base of hemisphere is coinciding but what if the height of the cone is just half what it is right now, ( half of what in picture of Gmat Guru )
Even in that case the base with coincide but still ratio will not be same,

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