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a cube has edges . say 4 vertices of one face are A,B,C,D

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Current Student
Joined: 11 May 2008
Posts: 553
a cube has edges . say 4 vertices of one face are A,B,C,D [#permalink]

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31 Aug 2008, 08:27
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a cube has edges . say 4 vertices of one face are A,B,C,D and on the opp face, the vertices above A is E, above B is F . 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
* $$\sqrt{2}$$
* $$\sqrt{3}$$
* $$2\sqrt{3}$$
Current Student
Joined: 11 May 2008
Posts: 553

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31 Aug 2008, 08:52
for those who want to understand better , here is the diagram.
the blue lines so constructed , give away the ans...
so if u want to solve it on your own, dont see this...
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cube.doc [19.5 KiB]

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Joined: 07 Nov 2007
Posts: 1789
Location: New York

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31 Aug 2008, 08:59
arjtryarjtry wrote:
a cube has edges . say 4 vertices of one face are A,B,C,D and on the opp face, the vertices above A is E, above B is F . 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
* $$\sqrt{2}$$
* $$\sqrt{3}$$
* $$2\sqrt{3}$$

Distance between Mid point of AB (say X) and E. = $$sqrt{{sqrt(2)}^2 +{sqrt{1/2}}^2}$$ = sqrt{ 2+1/2}= sqrt{5/2}

Distance beween midpoint of AB and midpoint of EH = $$sqrt{sqrt(5/2)^2+sqrt(1/2)^2)$$$$= sqrt(3)$$
D
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Re: gmat club   [#permalink] 31 Aug 2008, 08:59
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a cube has edges . say 4 vertices of one face are A,B,C,D

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