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Square ABCD is the base of the cube while square EFGH is the [#permalink]
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27 Nov 2007, 00:24
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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. \(2\sqrt{3}\) m14 q23
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Re: PS Cube [#permalink]
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27 Nov 2007, 00:33
bmwhype2 wrote: Square ABCD is the base of the cube while square EFGH is the cube's top face 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?
1/sqrt2 1 sqrt2 sqrt3 2sqrt3
Please explain your answer.
distance from mid point of AB to AD = sqrt [(1/sqrt2)^2+(1/sqrt2)^2] = 1
the distance between the midpoint of edge AB and the midpoint of edge EH = sqrt [1^2+(sqrt2)^2] = sqrt3.
D.



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Re: PS Cube [#permalink]
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27 Nov 2007, 00:37
bmwhype2 wrote: Square ABCD is the base of the cube while square EFGH is the cube's top face 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?
1/sqrt2 1 sqrt2 sqrt3 2sqrt3
Please explain your answer.
sqrt 3.
let midpoint of EH = M
let midpoint of AB = N
Drop perpendicular from M to side AD on F
AN = (sqrt 2)/2
AF = (sqrt 2)/2
FN = [(sqrt 2)/2]^2 + [(sqrt 2)/2]^2 = 1
MN^2 = 1^2 + (sqrt 2)^2 = 3
MN = sqrt 3



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haha! I guesstimated D.
Midpoint AB=M1
Midpoint EF=M2
Midpoint EH=M3
Segment M1M2=sqrt2, hence, M1M3 had to be bigger than that.
Eliminate A, B, C.
Down to D, E. E = 2*1.7*=3.4, which is more than 2 times the segment between M1M2. No way. D is the only one
Now, I'll learn how to solve it properly!



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Re: Square ABCD is the base of the cube while square EFGH is the [#permalink]
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29 May 2014, 05:46
I used deluxe pythag to solve....
We know that the area of the square is 2, therefore the side = sqrt2 or 2^1/2. We know the midpoints are (2^1/2)/2.
So in reality, we are just finding the main diagonal of a rectangular solid with lengths (2^1/2)/2, (2^1/2)/2, and (2^1/2); apply deluxe pythag theorum.
Find x  Main diagonal ((2^1/2)/2)^2 + ((2^1/2)/2)^2 + (2^1/2)^2 = x^2 (2/2)+(2/2)+2=x^2 1/2+1/2+2=x^2 3=x^2 3^1/2=x



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Re: Square ABCD is the base of the cube while square EFGH is the [#permalink]
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29 May 2014, 06:45
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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. \(2\sqrt{3}\) Look at the diagram below: Attachment:
Cube.png [ 14.44 KiB  Viewed 3975 times ]
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 \(\sqrt{2}\). \(XZ=\sqrt{AX^2+AZ^2}=\sqrt{(\frac{\sqrt{2}}{2})^2+(\frac{\sqrt{2}}{2})^2}=1\); \(XY=\sqrt{XZ^2+YZ^2}=\sqrt{1^2+(\sqrt{2})^2}=\sqrt{3}\). Answer: D.
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Re: Square ABCD is the base of the cube while square EFGH is the [#permalink]
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16 Jun 2014, 06:10
Why angle Z is the right angle? 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 \(\sqrt{2}\). \(XZ=\sqrt{AX^2+AZ^2}=\sqrt{(\frac{\sqrt{2}}{2})^2+(\frac{\sqrt{2}}{2})^2}=1\); \(XY=\sqrt{XZ^2+YZ^2}=\sqrt{1^2+(\sqrt{2})^2}=\sqrt{3}\). Answer: D.



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Re: Square ABCD is the base of the cube while square EFGH is the [#permalink]
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16 Jun 2014, 06:42
amar13 wrote: Why angle Z is the right angle? 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 \(\sqrt{2}\). \(XZ=\sqrt{AX^2+AZ^2}=\sqrt{(\frac{\sqrt{2}}{2})^2+(\frac{\sqrt{2}}{2})^2}=1\); \(XY=\sqrt{XZ^2+YZ^2}=\sqrt{1^2+(\sqrt{2})^2}=\sqrt{3}\). Answer: D. YZ is perpendicular to plane ABCD. Hence any line segment which is on the same plane and originates from Z will also be perpendicular to YZ. Does this make sense?
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Re: Square ABCD is the base of the cube while square EFGH is the [#permalink]
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28 Jun 2015, 11:13
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Re: Square ABCD is the base of the cube while square EFGH is the [#permalink]
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01 Jul 2015, 05:52
Bunuel wrote: amar13 wrote: Why angle Z is the right angle? 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 \(\sqrt{2}\). \(XZ=\sqrt{AX^2+AZ^2}=\sqrt{(\frac{\sqrt{2}}{2})^2+(\frac{\sqrt{2}}{2})^2}=1\); \(XY=\sqrt{XZ^2+YZ^2}=\sqrt{1^2+(\sqrt{2})^2}=\sqrt{3}\). Answer: D. YZ is perpendicular to plane ABCD. Hence any line segment which is on the same plane and originates from Z will also be perpendicular to YZ. Does this make sense? Bunuel, Can u elaborate a bit more on how the line is perpendicular?



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Square ABCD is the base of the cube while square EFGH is the [#permalink]
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14 Oct 2015, 10:09
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 \(\sqrt{2}\). \(XZ=\sqrt{AX^2+AZ^2}=\sqrt{(\frac{\sqrt{2}}{2})^2+(\frac{\sqrt{2}}{2})^2}=1\); \(XY=\sqrt{XZ^2+YZ^2}=\sqrt{1^2+(\sqrt{2})^2}=\sqrt{3}\). Answer: D. abcd looks like rombos and it is not like a square then how <xaz= 90?



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Re: Square ABCD is the base of the cube while square EFGH is the [#permalink]
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Square ABCD is the base of the cube while square EFGH is the [#permalink]
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14 Oct 2015, 10:48
anik19890 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 \(\sqrt{2}\). \(XZ=\sqrt{AX^2+AZ^2}=\sqrt{(\frac{\sqrt{2}}{2})^2+(\frac{\sqrt{2}}{2})^2}=1\); \(XY=\sqrt{XZ^2+YZ^2}=\sqrt{1^2+(\sqrt{2})^2}=\sqrt{3}\). Answer: D. abcd looks like rombos and it is not like a square then how <xaz= 90? ABCD can not be a hombus as it a face of a cube. All 6 faces in a cube are squares. The 1st few words of the question itself mention that ABCD is a square face of the cube.Additionally, you are basing your observation on a 3D figure drawn on a 2D surface. This is going to lead to a bit of distortion and hence the square 'looks like' a rhombus. In reality, all 6 faces of a cube are square, by definition.
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