A sphere is inscribed in a cube with an edge of 10. What is the shortest possible distance from one of the vertices of

[VJesus12]

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) \(10(\sqrt3−1)\)
(B) \(5\)
(C) \(10(\sqrt2−1)\)
(D) \(5(\sqrt3−1)\)
(E) \(5(\sqrt2−1)\)

[spoiler]OA=D[/spoiler]

Source: Manhattan GMAT

[Scott@TargetTestPrep]

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) \(10(\sqrt3−1)\)
(B) \(5\)
(C) \(10(\sqrt2−1)\)
(D) \(5(\sqrt3−1)\)
(E) \(5(\sqrt2−1)\)

[spoiler]OA=D[/spoiler]

Solution:

When a sphere is inscribed in a cube, the edge of the cube is also the diameter of the sphere. Therefore, the diameter of the sphere is 10. The shortest possible distance from one of the vertices (corners) of the cube to the surface of the sphere is half the difference between the space diagonal of the cube and the diameter of the sphere. Since the space diagonal of the cube is √[(10^2 + 10^2 + 10^2)] = √300 = 10√3, the shortest distance is therefore ½ x (10√3 - 10) = ½ x 10(√3 - 1) = 5(√3 - 1).

Answer: D