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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 [#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. \(2\sqrt{3}\)

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Re: PS Cube [#permalink]

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New post 27 Nov 2007, 00:33
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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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Re: PS Cube [#permalink]

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New post 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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 [#permalink]

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New post 27 Nov 2007, 03:45
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, 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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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:

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 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.

[Reveal] Spoiler:
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Cube.png
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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 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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New post 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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New post 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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New post 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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New post 14 Oct 2015, 10:46
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, 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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Re: Square ABCD is the base of the cube while square EFGH is the [#permalink]

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New post 18 Feb 2018, 03:52
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.


Hello Bunuel / chetan2u,

The answer will be different if EH were the diagonal of the square. I got this wrong because of that only.
Either the diagram should be given with the question or it must be mentioned that EH is not a diagonal. I can place E above A and etc in any manner right?

Please correct me if i am wrong.

Regards
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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 18 Feb 2018, 07:25
gmatexam439 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.


Hello Bunuel / chetan2u,

The answer will be different if EH were the diagonal of the square. I got this wrong because of that only.
Either the diagram should be given with the question or it must be mentioned that EH is not a diagonal. I can place E above A and etc in any manner right?

Please correct me if i am wrong.

Regards



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


this specifies that ABCD has EFGH over them in same sequence.
yes if it was not given E is over A, F over B etc, you would be correct..
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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 18 Feb 2018, 12:08
chetan2u wrote:
gmatexam439 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:
The attachment Cube.png is no longer available
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.


Hello Bunuel / chetan2u,

The answer will be different if EH were the diagonal of the square. I got this wrong because of that only.
Either the diagram should be given with the question or it must be mentioned that EH is not a diagonal. I can place E above A and etc in any manner right?

Please correct me if i am wrong.

Regards



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


this specifies that ABCD has EFGH over them in same sequence.
yes if it was not given E is over A, F over B etc, you would be correct..


Hello chetan2u,

I meant such a figure. In this E is over A, F over B etc, but still the answer will be different.

Am I missing anything here or is the question flawed?

Regards
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Untitled.png
Untitled.png [ 4.85 KiB | Viewed 756 times ]


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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 20 Feb 2018, 02:12
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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 20 Feb 2018, 07:31
gmatexam439 wrote:
Bunuel / chetan2u,

Awaiting your response on my above doubt. Please reply.

Regards


hi..

when a square is given as ABCD, it would mean the corners would be named accordingly ..
AB
DC
or
BA
CD..
when you read in clockwise or anticlockwise, it should read ABCD..
in the example given by you, the square is ABDC
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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 20 Feb 2018, 08:29
chetan2u wrote:
gmatexam439 wrote:
Bunuel / chetan2u,

Awaiting your response on my above doubt. Please reply.

Regards


hi..

when a square is given as ABCD, it would mean the corners would be named accordingly ..
AB
DC
or
BA
CD..
when you read in clockwise or anticlockwise, it should read ABCD..
in the example given by you, the square is ABDC


Thank you for replying chetan2u. Can I take your above statements as a standard followed by GMAT?

Regards
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7. GMAT Prep Challenge RC

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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 05 Mar 2018, 07:08
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.


How do we know that XZ is 1 ?
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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 05 Mar 2018, 07:14
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 \(\sqrt{2}\).

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

Answer: D.


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] 05 Mar 2018, 07:14

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