VeritasPrepKarishma wrote:

gmatzac wrote:

Why do we have to go along the diagonal of the cube?

I am getting E by going straight from the corner towards the sphere. The way I'm doing it would be the same if it were a circle/square rather than sphere/cube.

Why is this wrong? I can't figure it out

That's not correct. Draw a cube and see how you will inscribe a sphere in it. Note that the sphere will not touch any edges/corners of the cube. It will touch only the 6 faces of the cube at one point each. This point will lie in the center of the face of the cube.

If you go the usual two dimensional way, you are assuming that the sphere is lying flat on the face of the cube which is not correct. The sphere only touches the face of the cube on one point i.e. the point where the diagonals of the square face intersect. Hence, actually the distance of this diagonal to the sphere will be half the length of the diagonal. On the other hand, the diagonal of the cube (from one vertex to the opposite vertex across the cube will go right through the center of the sphere. It will stick a little bit out on both sides close to the vertex but will predominantly lie within the sphere on its diameter. So we find the length of the cube diagonal, subtract the sphere diameter out of it and divide the rest of the diagonal by 2 to get length of each little piece.

Think of a globe and its inclined axis. Imagine making a cube around it such that the globe touches each face of the cube. The shortest distance between a vertex of the cube and the globe will be the part of the inclined axis sticking out of the globe touching a vertex of the cube.

Responding to a pm:

**Quote:**

CAN YOU please explain why is the diagonal root-square 10^2*10^2*10^2 and not just 10^2*10^2 (applying P.theor.)?

Diagonal of a square will be \(\sqrt{(10^2 + 10^2)}\) (shown by 'd' in the diagram)

Diagonal of a cube will be 3 dimensional (shown by 'D' in the figure - the green line). We will need to use pythagorean theorem again on it. It will be the hypotenuse when the legs are height of the cube (a) and the diagonal of the square face (d).

Attachment:

cube111.PNG [ 3.19 KiB | Viewed 4300 times ]
\(D = \sqrt{d^2 + a^2} = \sqrt{{sqrt(10^2 + 10^2)}^2 + 10^2}} = \sqrt{10^2 + 10^2 + 10^2}\)

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Karishma

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