tricky question

[DavoodBeater]

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)

OA: D

[spoiler]half of the length of the diagonal of the cube minus the radius of the sphere.[/spoiler]

[logitech]

Diameter of Cube is same as length of the cube = 10

The farthest two point in a cube is C which is LENGHTxSQRT3

https://en.wikipedia.org/wiki/Cube

so the distance is (C-D)/2 = (10sqrt3-10)/2

Choose D

[DavoodBeater]

can you use wikipedia during test?
but the point is exactly what you said.
it is not length * sqrt2, it is length * sqrt3.
but you can count this:
rectangular with lengths of 10 and 10 * sqrt2 which gives us 10 * sqrt3, is it clear?

[logitech]

can you use wikipedia during test?
but the point is exactly what you said.
it is not length * sqrt2, it is length * sqrt3.
but you can count this:
rectangular with lengths of 10 and 10 * sqrt2 which gives us 10 * sqrt3, is it clear?

what do you mean it is clear ? I put that link as a reference to people who does now know how to derive the equation.

[vivek.kapoor83]

so the distance is (C-D)/2 = (10sqrt3-10)/2

Choose D

why u hv divided it by 2 ?

[DavoodBeater]

come one man, is it clear refers to my last solution not to wiki.

you must divide it by 2 since you have the same distance from the other side .

[welcome]

It will be = Lnght of vertice from the center - redius of sphere
=> Diagonal of cube/2 - length of side of cube/2
=>10Sqrt3/2 - 10/2
=>5Sqrt3-5
=>5(Sqrt3-1)

Answer: D