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SVP  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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Question Stats: 59% (03:01) correct 41% (02:46) wrong based on 360 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
Math Expert V
Joined: 02 Sep 2009
Posts: 60627
Square ABCD is the base of the cube while square EFGH is the  [#permalink]

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6
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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: 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 9482 times ]

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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

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.
Director  Joined: 09 Aug 2006
Posts: 545

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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

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
Senior Manager  Joined: 09 Oct 2007
Posts: 342

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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 M1-M2. No way. D is the only one Now, I'll learn how to solve it properly!
Manager  Joined: 24 Mar 2013
Posts: 52
Re: Square ABCD is the base of the cube while square EFGH is the  [#permalink]

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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
Intern  Joined: 11 May 2013
Posts: 3
Re: Square ABCD is the base of the cube while square EFGH is the  [#permalink]

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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 $$\sqrt{2}$$.

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

Math Expert V
Joined: 02 Sep 2009
Posts: 60627
Re: Square ABCD is the base of the cube while square EFGH is the  [#permalink]

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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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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?
Intern  Joined: 03 Jul 2015
Posts: 27
Square ABCD is the base of the cube while square EFGH is the  [#permalink]

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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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Schools: Kellogg '18 (M)
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Re: Square ABCD is the base of the cube while square EFGH is the  [#permalink]

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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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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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Posts: 1193
Location: India
GMAT 1: 730 Q49 V41 GPA: 4
Re: Square ABCD is the base of the cube while square EFGH is the  [#permalink]

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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
Math Expert V
Joined: 02 Aug 2009
Posts: 8335
Re: Square ABCD is the base of the cube while square EFGH is the  [#permalink]

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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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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 4045 times ]

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

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Bunuel / chetan2u,

Regards
Math Expert V
Joined: 02 Aug 2009
Posts: 8335
Re: Square ABCD is the base of the cube while square EFGH is the  [#permalink]

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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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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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Joined: 30 Dec 2016
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Re: Square ABCD is the base of the cube while square EFGH is the  [#permalink]

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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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Posts: 60627
Re: Square ABCD is the base of the cube while square EFGH is the  [#permalink]

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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: $$XZ=\sqrt{AX^2+AZ^2}=\sqrt{(\frac{\sqrt{2}}{2})^2+(\frac{\sqrt{2}}{2})^2}=1$$?
_________________ Re: Square ABCD is the base of the cube while square EFGH is the   [#permalink] 05 Mar 2018, 07:14

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