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M14-23

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M14-23  [#permalink]

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New post 16 Sep 2014, 00:53
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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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Re M14-23  [#permalink]

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New post 16 Sep 2014, 00:53
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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:

Image

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 \(\sqrt{2}\).

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

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


Answer: D
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Re M14-23  [#permalink]

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New post 01 Sep 2016, 07:32
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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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Re: M14-23  [#permalink]

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New post 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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Re: M14-23  [#permalink]

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New post 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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Re: M14-23  [#permalink]

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New post 23 Feb 2017, 21:04
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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)
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M14-23  [#permalink]

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New post 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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Re M14-23  [#permalink]

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New post 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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