It is currently 27 Jun 2017, 21:01

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
TAGS:

### Hide Tags

CEO
Joined: 21 Jan 2007
Posts: 2739
Location: New York City
Square ABCD is the base of the cube while square EFGH is the [#permalink]

### Show Tags

27 Nov 2007, 00:24
2
KUDOS
16
This post was
BOOKMARKED
00:00

Difficulty:

65% (hard)

Question Stats:

64% (03:17) correct 36% (01:48) wrong based on 370 sessions

### HideShow timer Statistics

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
[Reveal] Spoiler: OA
SVP
Joined: 29 Aug 2007
Posts: 2473

### Show Tags

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

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: 755

### Show Tags

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
Senior Manager
Joined: 09 Oct 2007
Posts: 466

### Show Tags

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!
Manager
Joined: 24 Mar 2013
Posts: 61
Re: Square ABCD is the base of the cube while square EFGH is the [#permalink]

### Show Tags

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
Math Expert
Joined: 02 Sep 2009
Posts: 39723
Re: Square ABCD is the base of the cube while square EFGH is the [#permalink]

### Show Tags

29 May 2014, 06:45
2
KUDOS
Expert's post
3
This post was
BOOKMARKED
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 $$\sqrt{2}$$.

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

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

### Show Tags

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 $$\sqrt{2}$$.

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

Math Expert
Joined: 02 Sep 2009
Posts: 39723
Re: Square ABCD is the base of the cube while square EFGH is the [#permalink]

### Show Tags

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?
_________________
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 16014
Re: Square ABCD is the base of the cube while square EFGH is the [#permalink]

### Show Tags

28 Jun 2015, 11:13
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Current Student
Joined: 29 May 2013
Posts: 119
Location: India
Concentration: Technology, Marketing
WE: Information Technology (Consulting)
Re: Square ABCD is the base of the cube while square EFGH is the [#permalink]

### Show Tags

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?
Intern
Joined: 03 Jul 2015
Posts: 36
Square ABCD is the base of the cube while square EFGH is the [#permalink]

### Show Tags

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?
Math Forum Moderator
Joined: 20 Mar 2014
Posts: 2647
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)
Re: Square ABCD is the base of the cube while square EFGH is the [#permalink]

### Show Tags

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.
_________________

Thursday with Ron updated list as of July 1st, 2015: http://gmatclub.com/forum/consolidated-thursday-with-ron-list-for-all-the-sections-201006.html#p1544515
Inequalities tips: http://gmatclub.com/forum/inequalities-tips-and-hints-175001.html
Debrief, 650 to 750: http://gmatclub.com/forum/650-to-750-a-10-month-journey-to-the-score-203190.html

Math Forum Moderator
Joined: 20 Mar 2014
Posts: 2647
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)
Square ABCD is the base of the cube while square EFGH is the [#permalink]

### Show Tags

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.
_________________

Thursday with Ron updated list as of July 1st, 2015: http://gmatclub.com/forum/consolidated-thursday-with-ron-list-for-all-the-sections-201006.html#p1544515
Inequalities tips: http://gmatclub.com/forum/inequalities-tips-and-hints-175001.html
Debrief, 650 to 750: http://gmatclub.com/forum/650-to-750-a-10-month-journey-to-the-score-203190.html

GMAT Club Legend
Joined: 09 Sep 2013
Posts: 16014
Re: Square ABCD is the base of the cube while square EFGH is the [#permalink]

### Show Tags

22 Oct 2016, 05:41
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Re: Square ABCD is the base of the cube while square EFGH is the   [#permalink] 22 Oct 2016, 05:41
Similar topics Replies Last post
Similar
Topics:
3 A cube and a pyramid of square base share the same dimensions: length, 1 04 Sep 2016, 06:22
11 A cube and a pyramid of square base share the same dimensions length, 4 28 Feb 2017, 00:11
8 Square PQRS is inscribed in the square ABCD whose 13 07 Jan 2016, 10:43
9 ABCD is a square picture frame (see figure). EFGH is a 14 14 Aug 2016, 00:42
15 In the figure above, square AFGE is inside square ABCD such 9 29 Nov 2016, 00:07
Display posts from previous: Sort by