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A sphere is inscribed in a cube with an edge of 10. What is

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Director
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A sphere is inscribed in a cube with an edge of 10. What is [#permalink]

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New post 30 Aug 2008, 15:56
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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(sqrt3 – 1)

5

10(sqrt2 – 1)

5(sqrt3– 1)

5(sqrt2– 1)

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Intern
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Re: another mgmat [#permalink]

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New post 30 Aug 2008, 16:19
bigtreezl 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(sqrt3 – 1)

5

10(sqrt2 – 1)

5(sqrt3– 1)

5(sqrt2– 1)


Ans: D

Good question....this is a bit tricky....easy to fall trap for E

Diameter of sphere is 10 (because sphere is inscribed in the cube....and for the shortest distance you would need the largest sphere which could fit in the cube)

Think in 3 dimensional. You wanna get the distance between a vertice of cube and the surface of sphere.

Each side of cube is 10
Hence, imaginary line connecting vertices on the same face of cube (2 dimensional)
= 10 sqrt 2

Also imaginary line going diagonally through the sphere and connecting opposite vertices (3 dimensional)
= sqrt [ (10^2) + (10sqrt2)^2] (one side of cube and other hypotenuse on 2-dimensional side of cube)
= 10 sqrt 3

shortest possible distance from one of the vertices of the cube to the surface of the sphere
= [(10 sqrt 3)/2] - 5
= 5 ( sqrt 3 - 1 )

I hope this helps

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Director
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Re: another mgmat [#permalink]

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New post 30 Aug 2008, 16:22
bigtreezl 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(sqrt3 – 1)

5

10(sqrt2 – 1)

5(sqrt3– 1)

5(sqrt2– 1)



D

the radius of the sphere is 5, diameter is 10

the diagonal of a square side is 10sqrt(2)
the diagonal of the cube is 10sqrt(3), use a^2 + b^2 = c^2 ---> (10^2) + (10sqrt(2))^2 = c^2
c^2 = 300
c = 10sqrt(3)

10sqrt(3) - 10 = 10(sqrt(3) - 1)
this is the difference between the diagonal of the cube and the diameter of the sphere
divide by 2 to get the distance of the vertex to the surface of the sphere.

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Re: another mgmat [#permalink]

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New post 31 Aug 2008, 09:13
bigtreezl 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(sqrt3 – 1)

5

10(sqrt2 – 1)

5(sqrt3– 1)

5(sqrt2– 1)



Shortest distance between to opposite vertices = 10 sqrt(3)

Mid point of distance between to oppsite vertices= 1/2 * 10 sqrt(3) = 5 sqrt(3)

shortest possible distance from one of the vertices of the cube to the surface of the sphere
= 5 sqrt(3)-5

D
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Re: another mgmat   [#permalink] 31 Aug 2008, 09:13
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A sphere is inscribed in a cube with an edge of 10. What is

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