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
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A sphere is inscribed in a cube
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Jay@ManhattanReview
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Since the length of the edge = 10, diameter of the inscribed sphere = 10rsarashi 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
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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rsarashi
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Hi Jay ,
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]
Thank you so much for your reply.
I really don't understand the above part.
Can you please explain?
Thanks.
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GMATGuruNY
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I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
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