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

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Joined: 21 Jan 2007
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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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58% (03:05) correct 42% (02:45) wrong based on 463 sessions

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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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Posts: 49968
Square ABCD is the base of the cube while square EFGH is the  [#permalink]

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29 May 2014, 06:45
3
5
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:

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 $$\sqrt{2}$$.

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

Attachment:

Cube.png [ 14.44 KiB | Viewed 6430 times ]

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27 Nov 2007, 00:33
2
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

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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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

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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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

Now, I'll learn how to solve it properly!
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Posts: 56
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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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 $$\sqrt{2}$$.

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

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Posts: 49968
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 $$\sqrt{2}$$.

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

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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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 $$\sqrt{2}$$.

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

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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Joined: 03 Jul 2015
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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 $$\sqrt{2}$$.

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

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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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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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 $$\sqrt{2}$$.

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

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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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 $$\sqrt{2}$$.

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

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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6. Challange OG RC
7. GMAT Prep Challenge RC

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Posts: 6956
Re: Square ABCD is the base of the cube while square EFGH is the  [#permalink]

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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 $$\sqrt{2}$$.

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

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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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 $$\sqrt{2}$$.

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

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
Attachments

File comment: My Figure

Untitled.png [ 4.85 KiB | Viewed 1406 times ]

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My study resources:
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2. e-GMAT's ALL SC Compilation
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4. Actual LSAT CR collection by Broal
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6. Challange OG RC
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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20 Feb 2018, 02:12
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Posts: 6956
Re: Square ABCD is the base of the cube while square EFGH is the  [#permalink]

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

Regards

hi..

when a square is given as ABCD, it would mean the corners would be named accordingly ..
AB
DC
or
BA
CD..
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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20 Feb 2018, 08:29
chetan2u wrote:
gmatexam439 wrote:
Bunuel / chetan2u,

Regards

hi..

when a square is given as ABCD, it would mean the corners would be named accordingly ..
AB
DC
or
BA
CD..
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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Preparing for RC my way

My study resources:
1. Useful Formulae, Concepts and Tricks-Quant
2. e-GMAT's ALL SC Compilation
3. LSAT RC compilation
4. Actual LSAT CR collection by Broal
5. QOTD RC (Carcass)
6. Challange OG RC
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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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 $$\sqrt{2}$$.

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

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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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 $$\sqrt{2}$$.

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

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 &nbs [#permalink] 05 Mar 2018, 07:14

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