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# Square ABCD is the base of the cube while square EFGH is the

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Joined: 30 Dec 2016
Posts: 16
Location: India
GMAT 1: 480 Q22 V22
Re: Square ABCD is the base of the cube while square EFGH is the  [#permalink]

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05 Mar 2018, 07:18
Bunuel wrote:
ayush98 wrote:
Bunuel wrote:
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:
Attachment:
Cube.png
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 to $$\sqrt{2}$$.

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

How do we know that XZ is 1 ?

Please tell me what to elaborate here: $$XZ=\sqrt{AX^2+AZ^2}=\sqrt{(\frac{\sqrt{2}}{2})^2+(\frac{\sqrt{2}}{2})^2}=1$$?

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Re: Square ABCD is the base of the cube while square EFGH is the  [#permalink]

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18 Jun 2019, 00:20
bmwhype2 wrote:
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. $$2\sqrt{3}$$

m14 q23

Please find teh solution as attached

Attachments

Screenshot 2019-06-18 at 12.49.48 PM.png [ 265.37 KiB | Viewed 81 times ]

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Re: Square ABCD is the base of the cube while square EFGH is the   [#permalink] 18 Jun 2019, 00:20

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