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# M14-23

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Math Expert
Joined: 02 Sep 2009
Posts: 58464

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16 Sep 2014, 00:53
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Difficulty:

65% (hard)

Question Stats:

60% (03:06) correct 40% (03:17) wrong based on 50 sessions

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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?

A. $$\frac{1}{\sqrt{2}}$$
B. 1
C. $$\sqrt{2}$$
D. $$\sqrt{3}$$
E. $$2\sqrt{3}$$

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Joined: 02 Sep 2009
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16 Sep 2014, 00:53
2
Official Solution:

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?

A. $$\frac{1}{\sqrt{2}}$$
B. 1
C. $$\sqrt{2}$$
D. $$\sqrt{3}$$
E. $$2\sqrt{3}$$

Look at the diagram below:

Notice that $$Z$$ is the midpoint of $$AD$$. We need to find the length of line segment $$XY$$.

Now, since the area of $$ABCD$$ is 2, then each edge of the cube equals $$\sqrt{2}$$.

$$XZ=\sqrt{AX^2+AZ^2}=\sqrt{(\frac{\sqrt{2}}{2})^2+(\frac{\sqrt{2}}{2})^2}=1$$;

$$XY=\sqrt{XZ^2+YZ^2}=\sqrt{1^2+(\sqrt{2})^2}=\sqrt{3}$$.

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Joined: 16 Sep 2012
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01 Sep 2016, 07:32
1
I agree with explanation. easier way to answer this Q. can be
We know that distance is more than an edge which is 2^1/2 but less than longest diagonal which is 3*2^1/2 so it has to be 3^1/2
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Joined: 08 Jan 2015
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23 Oct 2016, 07:19
kashyap001 wrote:
I agree with explanation. easier way to answer this Q. can be
We know that distance is more than an edge which is 2^1/2 but less than longest diagonal which is 3*2^1/2 so it has to be 3^1/2

Agree, but I guess you have a typo. It should be (3*2)^1/2
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Joined: 16 Nov 2016
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26 Nov 2016, 06:47
X and Y are the midpoints of their respective sides and as can be seen from the figure above XY is also half the length of the space diagonal.

Formula for Space Diagonal: d^2= 3(a)^2

So, Formula for Half Space Diagonal:
(XY)^2= 3(a)^2/ 2
(XY)^2= 3.(√2)^2/2
(XY)^2= 3
XY= √3

Hope that helps. Cheers!
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Joined: 23 Nov 2016
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GMAT 1: 760 Q50 V42
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23 Feb 2017, 21:04
1
A shortcut for 3 dimensional distances would be to take the square root of the sum of the squares of the distances for each dimension.

AB= sqrt(2)

distance between the two points: [(distance in X dimension)^2+(distance in Y dimension)^2+(distance in Z dimension)^2)^.5

=[(sqrt(2)/2)^2 + (sqrt(2))^2 + (sqrt(2)/2)^2]^.5

=(2/4+2+2/4)^.5 = sqrt(3)
Intern
Joined: 26 May 2016
Posts: 20
GMAT 1: 640 Q49 V30

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04 May 2017, 04:34
brooklyndude wrote:
A shortcut for 3 dimensional distances would be to take the square root of the sum of the squares of the distances for each dimension.

AB= sqrt(2)

distance between the two points: $$\sqrt{(distance In X dimension)^2+(distance In Y dimension)^2+(distance In Z dimension)^2)}$$

=sq root of ($$(\sqrt{2}/2)^2 + (\sqrt{2})^2 + (\sqrt{2}/2)^2$$)

=$$\sqrt{(2/4+2+2/4)} =\sqrt{3}$$

brooklyndude's answer with proper symbols to make it easy to understand as many will find this useful.
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Joined: 22 Jun 2017
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Location: Argentina
Schools: HBS, Stanford, Wharton
GMAT 1: 630 Q43 V34

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27 Feb 2019, 12:21
I think this is a high-quality question and I agree with explanation.
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Re M14-23   [#permalink] 27 Feb 2019, 12:21
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# M14-23

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