GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 23 Mar 2019, 05:47

### 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

# M14-23

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 53795

### Show Tags

16 Sep 2014, 00:53
00:00

Difficulty:

75% (hard)

Question Stats:

57% (03:08) correct 43% (03:17) wrong based on 47 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. $$2\sqrt{3}$$

_________________
Math Expert
Joined: 02 Sep 2009
Posts: 53795

### Show Tags

16 Sep 2014, 00:53
2
Official Solution:

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 $$\sqrt{2}$$.

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

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

_________________
Intern
Joined: 16 Sep 2012
Posts: 1

### Show Tags

01 Sep 2016, 07:32
1
I agree with explanation. easier way to answer this Q. can be
We know that distance is more than an edge which is 2^1/2 but less than longest diagonal which is 3*2^1/2 so it has to be 3^1/2
Current Student
Joined: 08 Jan 2015
Posts: 76

### Show Tags

23 Oct 2016, 07:19
kashyap001 wrote:
I agree with explanation. easier way to answer this Q. can be
We know that distance is more than an edge which is 2^1/2 but less than longest diagonal which is 3*2^1/2 so it has to be 3^1/2

Agree, but I guess you have a typo. It should be (3*2)^1/2
Intern
Joined: 16 Nov 2016
Posts: 1

### Show Tags

26 Nov 2016, 06:47
X and Y are the midpoints of their respective sides and as can be seen from the figure above XY is also half the length of the space diagonal.

Formula for Space Diagonal: d^2= 3(a)^2

So, Formula for Half Space Diagonal:
(XY)^2= 3(a)^2/ 2
(XY)^2= 3.(√2)^2/2
(XY)^2= 3
XY= √3

Hope that helps. Cheers!
>> !!!

You do not have the required permissions to view the files attached to this post.

Current Student
Joined: 23 Nov 2016
Posts: 74
Location: United States (MN)
GMAT 1: 760 Q50 V42
GPA: 3.51

### Show Tags

23 Feb 2017, 21:04
1
A shortcut for 3 dimensional distances would be to take the square root of the sum of the squares of the distances for each dimension.

AB= sqrt(2)

distance between the two points: [(distance in X dimension)^2+(distance in Y dimension)^2+(distance in Z dimension)^2)^.5

=[(sqrt(2)/2)^2 + (sqrt(2))^2 + (sqrt(2)/2)^2]^.5

=(2/4+2+2/4)^.5 = sqrt(3)
Intern
Joined: 26 May 2016
Posts: 20
GMAT 1: 640 Q49 V30

### Show Tags

04 May 2017, 04:34
brooklyndude wrote:
A shortcut for 3 dimensional distances would be to take the square root of the sum of the squares of the distances for each dimension.

AB= sqrt(2)

distance between the two points: $$\sqrt{(distance In X dimension)^2+(distance In Y dimension)^2+(distance In Z dimension)^2)}$$

=sq root of ($$(\sqrt{2}/2)^2 + (\sqrt{2})^2 + (\sqrt{2}/2)^2$$)

=$$\sqrt{(2/4+2+2/4)} =\sqrt{3}$$

brooklyndude's answer with proper symbols to make it easy to understand as many will find this useful.
Manager
Joined: 22 Jun 2017
Posts: 178
Location: Argentina
Schools: HBS, Stanford, Wharton

### Show Tags

27 Feb 2019, 12:21
I think this is a high-quality question and I agree with explanation.
_________________

The HARDER you work, the LUCKIER you get.

Re M14-23   [#permalink] 27 Feb 2019, 12:21
Display posts from previous: Sort by

# M14-23

Moderators: chetan2u, Bunuel

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.