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A sphere is inscribed in a cube

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by rsarashi » Sun May 07, 2017 4:14 am
A sphere is inscribed in a cube with an edge of 10. What is the shortest possible distance from one of the vertices of the cube to the surface of the sphere?

A) 10(underroot 3 - 1)

B) 5

C) 10(underroot 2 - 1)

D) 5(underroot 3 - 1)

E) 5(underroot 2 - 1)

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by Jay@ManhattanReview » Sun May 07, 2017 5:08 am
rsarashi wrote:A sphere is inscribed in a cube with an edge of 10. What is the shortest possible distance from one of the vertices of the cube to the surface of the sphere?

A) 10(underroot 3 - 1)

B) 5

C) 10(underroot 2 - 1)

D) 5(underroot 3 - 1)

E) 5(underroot 2 - 1)

OAD
Since the length of the edge = 10, diameter of the inscribed sphere = 10

The shortest possible distance from one of the vertices of the cube to the surface of the sphere = 1/2 of [Distance from one vertex to the diagonally opposite vertex - Diameter of the sphere] = 1/2[10√3 - 10] = [spoiler]5[√3 - 1][/spoiler]

The correct answer: D

Hope this helps!

Relevant book: Manhattan Review GMAT Geometry Guide

-Jay
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by rsarashi » Mon May 08, 2017 8:55 am

The shortest possible distance from one of the vertices of the cube to the surface of the sphere = 1/2 of [Distance from one vertex to the diagonally opposite vertex - Diameter of the sphere] = 1/2[10√3 - 10] = [spoiler]5[√3 - 1][/spoiler]
Hi Jay ,

Thank you so much for your reply.

I really don't understand the above part.

Can you please explain?

Thanks.
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by GMATGuruNY » Mon May 08, 2017 8:58 am
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