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Re: Cube and Squares of the Cube [#permalink]
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]
option 4 i.e sqrt 3 should be the answer
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Re: Cube and Squares of the Cube [#permalink]
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Re: Cube and Squares of the Cube [#permalink]
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