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 Post subject: GMAT 3D Geometry
PostPosted: Tue Apr 13, 2010 2:04 pm 
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A cubic box with an edge length of 4 inches holds a ball with a radius of 2 inches, such that the ball touches the box on all six sides (it is tangent to all six sides). What is the distance, in inches, between the center of the ball and a corner of the box?
A. 6√3
B. 4√3
C. 2√3
D. 2√2
E. √2

(C) The distance between the center of the ball and any of the 8 corners of the cube is simply one-half the distance between two of the opposite corners of the cube. To imagine opposite corners in a cube, pick any corner, and think about drawing a line through the center of the cube to the furthest other corner. The distance between any two opposite corners in a 3 dimensional rectangular solid (whether a perfect cube or not) can be determined very quickly using the 3 Dimensional Pythagorean theorem:
D² = a² + b² + c², where a, b, and c represent the dimensions of the solid. Since all of the edge lengths of a cube are the same, we can simplify the equation to: D² = 3a², where a represents the edge length of the cube. So the distance between two opposite corners in the cube is:
D² = 3 × 4²
D² = 48.
D = √48
D = 43
Now as we said before, to get the distance between the center of the ball and any corner of the cube, we divide this distance by 2:
4√3 / 2 = 2√3. The correct answer is choice (C).
If you chose√choice (B), you probably did not divide by 2 at the end.
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Hello. What is wrong with the following ? The radius of the cube is 2. The vertical distance between the point where the sphere touches the cube and the top of the cube is 2. so applying the Pythagoras theorem, don't we get the right answer as 2√2. ? What is wrong here ?


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 Post subject: Re: GMAT 3D Geometry
PostPosted: Thu Apr 15, 2010 1:08 pm 
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Joined: Fri Apr 09, 2010 2:11 pm
Posts: 459
First of all, please, be careful with terms. There is no such concept as "the radius of the cube", but I assume "radius of the inscribed sphere" is meant instead.
In this case let's take a look at the following sketch to visualize the problem:
Image
We need to find the length of OB.

OB²=OC² + CB²
in its turn
CB² = CA² + AB²
so
OB²=OC² + CA² + AB²
That is what solution tells us about.

Lets take a look at proposed reasoning, considering that by "vertical distance between the point where the sphere touches the cube and the top of the cube is 2" it is meant distance CA.
The radius of inscribed sphere is OC.

And when we calculate √(OC² + CA²) = 2√2 we find OA, but not OB.

It is important to clearly visualize the promblem when we solve it.
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