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Sphere is inscribed in a cube of side length 10 cms. What is

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Sphere is inscribed in a cube of side length 10 cms. What is [#permalink]

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New post 03 Aug 2008, 23:08
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Sphere is inscribed in a cube of side length 10 cms. What is the shortest distance from any of the vertices of the cube to the sphere surface?

a. 5
b. 10
c. 5-sqrt(3)
d. 10-sqrt(3)
e. 10-s*sqrt(3)
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Re: Shortest Length [#permalink]

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New post 03 Aug 2008, 23:23
GMAT TIGER wrote:
Sphere is inscribed in a cube of side length 10 cms. What is the shortest distance from any of the vertices of the cube to the sphere surface?

a. 5
b. 10
c. 5-sqrt(3)
d. 10-sqrt(3)
e. 10-s*sqrt(3)


diagonal of the cube = 10sqrt(3)
half of it = 5sqrt(3)
radius of the sphere = 5
shortest distance = 5sqrt(3) - 5
... this is not an option. :? ... could you check the question...

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Manager
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Re: Shortest Length [#permalink]

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New post 03 Aug 2008, 23:44
I AGREE..

10SQRT3 - 10/2
WHICH GIVES 5 SQRT3 - 5

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Senior Manager
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Re: Shortest Length [#permalink]

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New post 04 Aug 2008, 00:58
Agree,
\(5*sqrt{3} - 5\)

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VP
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Re: Shortest Length [#permalink]

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New post 05 Aug 2008, 19:28
GMAT TIGER wrote:
Sphere is inscribed in a cube of side length 10 cms. What is the shortest distance from any of the vertices of the cube to the sphere surface?

a. 5
b. 10
c. 5-sqrt(3)
d. 10-sqrt(3)
e. 10-s*sqrt(3)


radius of the largest sphere in a cube of length 5 is 5
dist between vertex of cube to center of sphere is 5(3^1/2)
hence shortest dis between vertex of cube to surface = 5*sqrt(3)-5

OA =???
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Re: Shortest Length   [#permalink] 05 Aug 2008, 19:28
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