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Sphere inscribed in a cube

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by GmatKiss » Sun Oct 30, 2011 10:44 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?
10( root(3)- 1)

5

10( root(2)- 1)

5( root(3)- 1)

5( root(2)- 1)

Explain pls!
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by rijul007 » Sun Oct 30, 2011 10:58 am
The sphere that would completely fit the cube with an edge of 10 will have a diameter of 10
and the figure can be represented like this




Image



Acc to the figure, the shortest possible distance would be
[AB - (diameter of the sphere)]/2
[10*root(3) - 10]/2
5[root(3)-1]

Correct ans- 4th option
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by smackmartine » Sun Oct 30, 2011 11:14 am
IMO D
I would say this is more of a visualization problem. Because the diameter of the sphere is a part of the diagonal of the cube,in words , calculate the diagonal of the cube, subtract the diameter of the sphere and divide it by 2.

mathematically,

calculate the diagonal of the cube --> [(10)^2 + (10sqrt(2))^2]^1/2 = 10sqrt(3)
subtract the diameter of the sphere-> 10sqrt(3)-10
divide the result by 2 --> 5( root(3)- 1)
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by GMATGuruNY » Mon Oct 31, 2011 7:42 am
GmatKiss 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?
10( root(3)- 1)

5

10( root(2)- 1)

5( root(3)- 1)

5( root(2)- 1)

Explain pls!
I received a PM asking me to clarify the solution.

Image

The formula for the diagonal of a cube = √(3e²).

In the figure above, x = the distance between the cube and the surface of the sphere.
The diagonal of the cube = 2x + the diameter of the sphere.
Thus, x = (diagonal of the cube - diameter of the sphere)/2.

The diagonal of the cube = √(3e²) = √(3*10²) = 10√3.
The diameter of the sphere = the edge of the cube = 10.
Thus, x = (10√3 - 10)/2 = 5√3 - 5 = 5(√3 - 1).

The correct answer is D.
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