OG 12 Diagnostic Q. 20

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icrlp05: Newbie | Next Rank: 10 Posts; Posts: 2; Joined: Mon Nov 30, 2009 2:30 am

by icrlp05 » Sat Dec 12, 2009 3:36 am

20. 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) Root (3):1
(B) 1:1
(C) 0.5:1
(D) Root (2):1
(E) 2:1

Could anyone explain this question please? My understanding of inscribed is to etch onto the surface, but it wouldn't be possible to etch a right circular cone onto a hemisphere.

I also haven't found any hard and fast geometrical rule that relates the radial base of a right circular cone to its height, is there one? If, say, R=h for a right circular cone then of course this question is trivial.

Failing this, I don't see any information regarding the height of the cone!!

This question has really been perplexing me and I don't think the official answer is very good for it, so any help would be greatly appreciated!

Many thanks.[/spoiler]

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Source: Beat The GMAT — Problem Solving | Sat Dec 12, 2009 3:36 am

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Ian Stewart: GMAT Instructor; Posts: 2623; Joined: Mon Jun 02, 2008 3:17 amFollowed by: 355 members; Location: Montreal; Thanked: 1090 times; GMAT Score:780

by Ian Stewart » Sat Dec 12, 2009 5:15 am

When one figure is inscribed in another, this means that each vertex of the inscribed figure touches the surface of the outer figure. So the tip of the cone must touch the surface of the hemisphere. Since the cone is a right circular cone (it isn't slanted in some odd way), the tip of the cone must be directly above the center of the hemisphere, so the height of the cone is also a radius of the hemisphere; their lengths are equal.

It is not generally true that the radius and height of a right circular cone are equal; it only turns out to be true here because the cone is inscribed in the hemisphere.

For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

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gmatv09: Master | Next Rank: 500 Posts; Posts: 256; Joined: Mon Aug 10, 2009 6:31 pm; Thanked: 3 times

by gmatv09 » Sat Dec 12, 2009 6:37 am

IMO D....

what is the OA?

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thephoenix: Legendary Member; Posts: 1560; Joined: Tue Nov 17, 2009 2:38 amFollowed by: 5 members; Thanked: 137 times

by thephoenix » Sat Dec 12, 2009 7:45 am

icrlp05 wrote:20. 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) Root (3):1
(B) 1:1
(C) 0.5:1
(D) Root (2):1
(E) 2:1

Could anyone explain this question please? My understanding of inscribed is to etch onto the surface, but it wouldn't be possible to etch a right circular cone onto a hemisphere.

I also haven't found any hard and fast geometrical rule that relates the radial base of a right circular cone to its height, is there one? If, say, R=h for a right circular cone then of course this question is trivial.

Failing this, I don't see any information regarding the height of the cone!!

This question has really been perplexing me and I don't think the official answer is very good for it, so any help would be greatly appreciated!

Many thanks.[/spoiler]

it will be easy if one try to imagine or draw a 2-D figure , for this case it will be a trianglr drawn inside a semicircle with its dia as a base side of the triangle and the opposite vertex is on the circle

then its r=h so 1:1

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icrlp05: Newbie | Next Rank: 10 Posts; Posts: 2; Joined: Mon Nov 30, 2009 2:30 am

by icrlp05 » Sat Dec 12, 2009 8:28 am

Ian Stewart wrote:When one figure is inscribed in another, this means that each vertex of the inscribed figure touches the surface of the outer figure. So the tip of the cone must touch the surface of the hemisphere. Since the cone is a right circular cone (it isn't slanted in some odd way), the tip of the cone must be directly above the center of the hemisphere, so the height of the cone is also a radius of the hemisphere; their lengths are equal.

It is not generally true that the radius and height of a right circular cone are equal; it only turns out to be true here because the cone is inscribed in the hemisphere.

Brilliant - thanks very much

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