Manhattan Practice GMAT

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Manhattan Practice GMAT

by [email protected] » Sat Apr 21, 2012 6:58 am
Sorry, i couldnt figure out how else to post this problem than to attach the image
why is the answer not E ?[/img]
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by aneesh.kg » Sat Apr 21, 2012 7:11 am
It can be derived from Pythagoras' Theorem that the length of the diagonal in a cuboid of edge lengths a,b and c is (a^2 + b^2 + c^2)^0.5.
For a cube, since all the edges are equal, the length of a diagonal for edge length 'a' is a(3)^0.5.
So, for a edge length 10, the length of the diagonal is 10(3)^0.5.

If you draw a cube with a sphere completely inscribed in it, you will see that the diameter of the sphere = edge of the cube and
The Shortest possible distance from one vertex to surface of sphere = Half of (length of diagonal - diameter of spehere)
= 1/2*(10(3)^0.5 - 10)
= 5(3)^0.5 - 5

(D) is the answer
Aneesh Bangia
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by [email protected] » Sat Apr 21, 2012 7:20 am
Thanks Aneesh,
I now realized my mistake,
i was trying to visualize a two dimensional figure and was calculating the distance of the vertex to the surface of the "circle" and not the sphere

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by [email protected] » Sun Apr 22, 2012 5:18 am
Yes I have understood now as to what mistakes I was committing:

The longest diagonal in a Cube is l x root 3.

So the correct answer as D is the perfect answer...
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