Cube and Squares of the Cube

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Cube and Squares of the Cube [#permalink]

09 May 2009, 16:30

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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]

10 May 2009, 00:57

icandy wrote:

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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Re: Cube and Squares of the Cube [#permalink]

15 Oct 2009, 09:20

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]

15 Oct 2009, 11:21

option 4 i.e sqrt 3 should be the answer

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Re: Cube and Squares of the Cube [#permalink]

18 Oct 2009, 17:44

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D

YZ= 1
XY= \sqrt{2}
xz=\sqrt{3}

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Re: Cube and Squares of the Cube [#permalink] 18 Oct 2009, 17:44

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