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

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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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New post 26 Nov 2007, 23:33
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.

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New post 26 Nov 2007, 23:37
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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New post 27 Nov 2007, 02:45
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 M1-M2. No way. D is the only one 8-)

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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New post 29 May 2014, 04:46
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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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
Cube.png [ 14.44 KiB | Viewed 3608 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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New post 16 Jun 2014, 05:10
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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New post 16 Jun 2014, 05:42
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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New post 01 Jul 2015, 04:52
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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New post 14 Oct 2015, 09:09
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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New post 14 Oct 2015, 09:46
jayanthjanardhan wrote:


Bunuel,

Can u elaborate a bit more on how the line is perpendicular?


Planes ABCD and ADHE are perpendicular to each other. As such, any such lines drawn on these 2 mutually perpendicular planes will be perpendicular to each other.
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Square ABCD is the base of the cube while square EFGH is the [#permalink]

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New post 14 Oct 2015, 09:48
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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Re: Square ABCD is the base of the cube while square EFGH is the [#permalink]

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New post 22 Oct 2016, 04:41
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Re: Square ABCD is the base of the cube while square EFGH is the   [#permalink] 22 Oct 2016, 04:41
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