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Cube and Squares of the Cube

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VP
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Cube and Squares of the Cube [#permalink] New post 09 May 2009, 15:30
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Square ABCD is the base of the cube while square EFGH is the cube's top facet such that point E is above point A, point F is above point B etc. What is the distance between the midpoint of edge AB and the midpoint of edge EH if the area of square ABCD is 2?

* \(\frac{1}{\sqrt{2}}\)
* 1
* \(\sqrt{2}\)
* \(\sqrt{3}\)
* \(2\sqrt{3}\)
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Re: Cube and Squares of the Cube [#permalink] New post 09 May 2009, 23:57
icandy wrote:
Square ABCD is the base of the cube while square EFGH is the cube's top facet such that point E is above point A, point F is above point B etc. What is the distance between the midpoint of edge AB and the midpoint of edge EH if the area of square ABCD is 2?

* \(\frac{1}{\sqrt{2}}\)
* 1
* \(\sqrt{2}\)
* \(\sqrt{3}\)
* \(2\sqrt{3}\)


mathmatical approach:

say midpoint of AB = X
mid point of AD =Y
mid point of EF= Z
distance between X AND y = sqrt ((1/sqrt(2) )^2 +(1/sqrt(2) )^2) =1
distance between Z ABD X = sqrt((1)^2 +(sqrt(2))^2) =sqrt(3)
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Manager
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Affiliations: CFA Level 2 Candidate
Joined: 29 Jun 2009
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Re: Cube and Squares of the Cube [#permalink] New post 15 Oct 2009, 08:20
I'm going to say sqrt 3.

The distance between midpoint AB and midpoint AD is a pythagorum theorum.

Since Area of ABCD is 2 we know each side is sqrt 2.
Therefore midpoint AB = .5 \sqrt{2} and AD = .5 \sqrt{2}

Therefore distance between midpoints AB and AD is sqrt 2
Midpoint between EH and AD is also sqrt 2 (straight line) Add them together and we get 2

We now know that since it is being cut through the square the distance is less 2 but definately more than sqrt 2.

Only sqrt 3 give this answer.
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Re: Cube and Squares of the Cube [#permalink] New post 15 Oct 2009, 10:21
option 4 i.e sqrt 3 should be the answer
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Re: Cube and Squares of the Cube [#permalink] New post 18 Oct 2009, 16:44
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YZ= 1
XY= \sqrt{2}
xz=\sqrt{3}
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Re: Cube and Squares of the Cube   [#permalink] 18 Oct 2009, 16:44
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