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

It is currently 25 Apr 2019, 16:59

Close

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
Your Progress

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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel

The side of square ABCD is 6. The midpoints of AB and BC are E and F,

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

 
Retired Moderator
avatar
Joined: 29 Apr 2015
Posts: 838
Location: Switzerland
Concentration: Economics, Finance
Schools: LBS MIF '19
WE: Asset Management (Investment Banking)
GMAT ToolKit User
The side of square ABCD is 6. The midpoints of AB and BC are E and F,  [#permalink]

Show Tags

New post 09 Aug 2015, 08:33
1
4
00:00
A
B
C
D
E

Difficulty:

  25% (medium)

Question Stats:

80% (02:29) correct 20% (03:09) wrong based on 85 sessions

HideShow timer Statistics

Attachment:
T7914.png
T7914.png [ 4.52 KiB | Viewed 4893 times ]


The side of square ABCD is 6. The midpoints of AB and BC are E and F, respectively. If square HGID intersects EF at G, what is the length of DI?

A. 3
B. 4
C. 4.5
D. 5
E. 5.5

Economist says ballpark, but I would like to know the calculations for this. What is wrong with my approach:

square HGID intersects EF at G. Triangle EBF is a 90-45-45 triangle and therefore the hypothenuse is: 3*sqrt(2). Half of 3*sqrt(2) would be a little more than 2. No lets subtract a little more than 2 from 6, which would equal a little less than 4. It does not make sense, because none of the answers are in this region.

Thanks

_________________
Saving was yesterday, heat up the gmatclub.forum's sentiment by spending KUDOS!

PS Please send me PM if I do not respond to your question within 24 hours.
Most Helpful Community Reply
Retired Moderator
User avatar
Joined: 06 Jul 2014
Posts: 1229
Location: Ukraine
Concentration: Entrepreneurship, Technology
GMAT 1: 660 Q48 V33
GMAT 2: 740 Q50 V40
GMAT ToolKit User
Re: The side of square ABCD is 6. The midpoints of AB and BC are E and F,  [#permalink]

Show Tags

New post 09 Aug 2015, 12:11
4
1
reto wrote:
Attachment:
T7914.png


The side of square ABCD is 6. The midpoints of AB and BC are E and F, respectively. If square HGID intersects EF at G, what is the length of DI?

A. 3
B. 4
C. 4.5
D. 5
E. 5.5

Economist says ballpark, but I would like to know the calculations for this. What is wrong with my approach:

square HGID intersects EF at G. Triangle EBF is a 90-45-45 triangle and therefore the hypothenuse is: 3*sqrt(2). Half of 3*sqrt(2) would be a little more than 2. No lets subtract a little more than 2 from 6, which would equal a little less than 4. It does not make sense, because none of the answers are in this region.

Thanks



Hello reto

This task is built on isosceles triangle and its ratio \(1:1:\sqrt{2}\)

\(EB=BF=3\) so this is isosceles triangle and we can infer that \(EF = 3\sqrt{2}\)
So \(EG = GF = GB = 3\sqrt{2}/2\)

Triangle DAB is isosceles so its side \(DB = 6\sqrt{2}\)

\(DG = DB - GB\) --> \(6\sqrt{2} - 3\sqrt{2}/2 = 4.5\sqrt{2}\)

Triangle DGH is isosceles so its side \(HG = 4.5\)

--------------------

In your solution you are correctly find length of EG but we should subtract from AB not EG but IC which much less than EG.
_________________
General Discussion
CEO
CEO
avatar
S
Joined: 20 Mar 2014
Posts: 2624
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)
GMAT ToolKit User Reviews Badge
Re: The side of square ABCD is 6. The midpoints of AB and BC are E and F,  [#permalink]

Show Tags

New post 09 Aug 2015, 08:55
reto wrote:
Attachment:
T7914.png


The side of square ABCD is 6. The midpoints of AB and BC are E and F, respectively. If square HGID intersects EF at G, what is the length of DI?

A. 3
B. 4
C. 4.5
D. 5
E. 5.5

Economist says ballpark, but I would like to knoNw the calculations for this. What is wrong with my approach:

square HGID intersects EF at G. Triangle EBF is a 90-45-45 triangle and therefore the hypothenuse is: 3*sqrt(2). Half of 3*sqrt(2) would be a little more than 2. No lets subtract a little more than 2 from 6, which would equal a little less than 4. It does not make sense, because none of the answers are in this region.

Thanks


You are assuming in your calculations that EG = GF .
Retired Moderator
avatar
Joined: 29 Apr 2015
Posts: 838
Location: Switzerland
Concentration: Economics, Finance
Schools: LBS MIF '19
WE: Asset Management (Investment Banking)
GMAT ToolKit User
Re: The side of square ABCD is 6. The midpoints of AB and BC are E and F,  [#permalink]

Show Tags

New post 09 Aug 2015, 11:10
Engr2012 wrote:
reto wrote:
Attachment:
T7914.png


The side of square ABCD is 6. The midpoints of AB and BC are E and F, respectively. If square HGID intersects EF at G, what is the length of DI?

A. 3
B. 4
C. 4.5
D. 5
E. 5.5

Economist says ballpark, but I would like to knoNw the calculations for this. What is wrong with my approach:

square HGID intersects EF at G. Triangle EBF is a 90-45-45 triangle and therefore the hypothenuse is: 3*sqrt(2). Half of 3*sqrt(2) would be a little more than 2. No lets subtract a little more than 2 from 6, which would equal a little less than 4. It does not make sense, because none of the answers are in this region.

Thanks


You are assuming in your calculations that EG = GF . In triangle GFM with GM perpendicular to BF.


The pain of non-native. I confused intersect with bisect. Mh, what should I say, pretty important :-) Thanks, lessons learned.
_________________
Saving was yesterday, heat up the gmatclub.forum's sentiment by spending KUDOS!

PS Please send me PM if I do not respond to your question within 24 hours.
CEO
CEO
avatar
S
Joined: 20 Mar 2014
Posts: 2624
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)
GMAT ToolKit User Reviews Badge
Re: The side of square ABCD is 6. The midpoints of AB and BC are E and F,  [#permalink]

Show Tags

New post 09 Aug 2015, 11:15
reto wrote:
Engr2012 wrote:
reto wrote:
Attachment:
T7914.png


The side of square ABCD is 6. The midpoints of AB and BC are E and F, respectively. If square HGID intersects EF at G, what is the length of DI?

A. 3
B. 4
C. 4.5
D. 5
E. 5.5

Economist says ballpark, but I would like to knoNw the calculations for this. What is wrong with my approach:

square HGID intersects EF at G. Triangle EBF is a 90-45-45 triangle and therefore the hypothenuse is: 3*sqrt(2). Half of 3*sqrt(2) would be a little more than 2. No lets subtract a little more than 2 from 6, which would equal a little less than 4. It does not make sense, because none of the answers are in this region.

Thanks


You are assuming in your calculations that EG = GF .


The pain of non-native. I confused intersect with bisect. Mh, what should I say, pretty important :-) Thanks, lessons learned.


I am not a big fan of ball parking in Quant as is done by Economist GMAT . I will ballpark/simplify my calculations but only based on some common sense rules. GNAT tutor's ballpaking methods are a bit more 'forced', IMO.
Retired Moderator
User avatar
Joined: 06 Jul 2014
Posts: 1229
Location: Ukraine
Concentration: Entrepreneurship, Technology
GMAT 1: 660 Q48 V33
GMAT 2: 740 Q50 V40
GMAT ToolKit User
Re: The side of square ABCD is 6. The midpoints of AB and BC are E and F,  [#permalink]

Show Tags

New post 09 Aug 2015, 12:05
Engr2012 wrote:
reto wrote:
Attachment:
T7914.png


The side of square ABCD is 6. The midpoints of AB and BC are E and F, respectively. If square HGID intersects EF at G, what is the length of DI?

A. 3
B. 4
C. 4.5
D. 5
E. 5.5

Economist says ballpark, but I would like to knoNw the calculations for this. What is wrong with my approach:

square HGID intersects EF at G. Triangle EBF is a 90-45-45 triangle and therefore the hypothenuse is: 3*sqrt(2). Half of 3*sqrt(2) would be a little more than 2. No lets subtract a little more than 2 from 6, which would equal a little less than 4. It does not make sense, because none of the answers are in this region.

Thanks


You are assuming in your calculations that EG = GF .


Hello Engr2012
And why this assuming is wrong?
_________________
CEO
CEO
avatar
S
Joined: 20 Mar 2014
Posts: 2624
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)
GMAT ToolKit User Reviews Badge
Re: The side of square ABCD is 6. The midpoints of AB and BC are E and F,  [#permalink]

Show Tags

New post 09 Aug 2015, 13:00
Harley1980 wrote:
Engr2012 wrote:
reto wrote:
Attachment:
T7914.png


The side of square ABCD is 6. The midpoints of AB and BC are E and F, respectively. If square HGID intersects EF at G, what is the length of DI?

A. 3
B. 4
C. 4.5
D. 5
E. 5.5

Economist says ballpark, but I would like to knoNw the calculations for this. What is wrong with my approach:

square HGID intersects EF at G. Triangle EBF is a 90-45-45 triangle and therefore the hypothenuse is: 3*sqrt(2). Half of 3*sqrt(2) would be a little more than 2. No lets subtract a little more than 2 from 6, which would equal a little less than 4. It does not make sense, because none of the answers are in this region.

Thanks


You are assuming in your calculations that EG = GF .


Hello Engr2012
And why this assuming is wrong?


Harley1980, my bad. Yes, it is infact correct that EG=GF.
Manager
Manager
avatar
Joined: 20 Jul 2011
Posts: 80
GMAT 1: 660 Q49 V31
GMAT ToolKit User
Re: The side of square ABCD is 6. The midpoints of AB and BC are E and F,  [#permalink]

Show Tags

New post 10 Aug 2015, 05:54
reto wrote:
Attachment:
T7914.png


The side of square ABCD is 6. The midpoints of AB and BC are E and F, respectively. If square HGID intersects EF at G, what is the length of DI?

A. 3
B. 4
C. 4.5
D. 5
E. 5.5

Economist says ballpark, but I would like to know the calculations for this. What is wrong with my approach:

square HGID intersects EF at G. Triangle EBF is a 90-45-45 triangle and therefore the hypothenuse is: 3*sqrt(2). Half of 3*sqrt(2) would be a little more than 2. No lets subtract a little more than 2 from 6, which would equal a little less than 4. It does not make sense, because none of the answers are in this region.

Thanks



Given: DC = 6, we need to find DI => DI = DC - IC

We know that EB = BF = 3(Isosceles, 45:45:90), hence EF =3\(\sqrt{2}\).

Extend HG to meet BC at O.

Since angle BGF = 90, angle FGO = 45. Also GOF = 90, Hence angle OFG = 45. Traingle OGF is also a (45:45:90 triangle)

Since GF = 3\(\sqrt{2}\) /2 => 1.5 \(\sqrt{2}\) => GO = OF = 1.5

We know that GO = IC = 1.5

Hence DI = 6 -1.5 = 4.5

Option D
Retired Moderator
avatar
Joined: 29 Apr 2015
Posts: 838
Location: Switzerland
Concentration: Economics, Finance
Schools: LBS MIF '19
WE: Asset Management (Investment Banking)
GMAT ToolKit User
Re: The side of square ABCD is 6. The midpoints of AB and BC are E and F,  [#permalink]

Show Tags

New post 18 Aug 2015, 11:28
Harley1980 wrote:
reto wrote:
Attachment:
T7914.png


The side of square ABCD is 6. The midpoints of AB and BC are E and F, respectively. If square HGID intersects EF at G, what is the length of DI?

A. 3
B. 4
C. 4.5
D. 5
E. 5.5

Economist says ballpark, but I would like to know the calculations for this. What is wrong with my approach:

square HGID intersects EF at G. Triangle EBF is a 90-45-45 triangle and therefore the hypothenuse is: 3*sqrt(2). Half of 3*sqrt(2) would be a little more than 2. No lets subtract a little more than 2 from 6, which would equal a little less than 4. It does not make sense, because none of the answers are in this region.

Thanks



Hello reto

This task is built on isosceles triangle and its ratio \(1:1:\sqrt{2}\)

\(EB=BF=3\) so this is isosceles triangle and we can infer that \(EF = 3\sqrt{2}\)
So \(EG = GF = GB = 3\sqrt{2}/2\)

Triangle DAB is isosceles so its side \(DB = 6\sqrt{2}\)

\(DG = DB - GB\) --> \(6\sqrt{2} - 3\sqrt{2}/2 = 4.5\sqrt{2}\)

Triangle DGH is isosceles so its side \(HG = 4.5\)

--------------------

In your solution you are correctly find length of EG but we should subtract from AB not EG but IC which much less than EG.


Hello!

I am now reviewing this question. Why do you infer that EG = GF = GB ... I mean it is clear that EG = GF but why should it be also equal to the height of the triangle GB?

Thank you Sergii!
_________________
Saving was yesterday, heat up the gmatclub.forum's sentiment by spending KUDOS!

PS Please send me PM if I do not respond to your question within 24 hours.
CEO
CEO
avatar
S
Joined: 20 Mar 2014
Posts: 2624
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)
GMAT ToolKit User Reviews Badge
Re: The side of square ABCD is 6. The midpoints of AB and BC are E and F,  [#permalink]

Show Tags

New post 18 Aug 2015, 11:51
reto wrote:
Harley1980 wrote:
reto wrote:
Attachment:
T7914.png


The side of square ABCD is 6. The midpoints of AB and BC are E and F, respectively. If square HGID intersects EF at G, what is the length of DI?

A. 3
B. 4
C. 4.5
D. 5
E. 5.5

Economist says ballpark, but I would like to know the calculations for this. What is wrong with my approach:

square HGID intersects EF at G. Triangle EBF is a 90-45-45 triangle and therefore the hypothenuse is: 3*sqrt(2). Half of 3*sqrt(2) would be a little more than 2. No lets subtract a little more than 2 from 6, which would equal a little less than 4. It does not make sense, because none of the answers are in this region.

Thanks



Hello reto

This task is built on isosceles triangle and its ratio \(1:1:\sqrt{2}\)

\(EB=BF=3\) so this is isosceles triangle and we can infer that \(EF = 3\sqrt{2}\)
So \(EG = GF = GB = 3\sqrt{2}/2\)

Triangle DAB is isosceles so its side \(DB = 6\sqrt{2}\)

\(DG = DB - GB\) --> \(6\sqrt{2} - 3\sqrt{2}/2 = 4.5\sqrt{2}\)

Triangle DGH is isosceles so its side \(HG = 4.5\)

--------------------

In your solution you are correctly find length of EG but we should subtract from AB not EG but IC which much less than EG.


Hello!

I am now reviewing this question. Why do you infer that EG = GF = GB ... I mean it is clear that EG = GF but why should it be also equal to the height of the triangle GB?

Thank you Sergii!


In the text in red above, you mean why in triangle EBF, the height GB = EG = GF ?

If this is the case then look below,

You know in triangle EBF, we have 45-45-90 triangle and angle EFB = 45. Draw the height GB such that angle BGF = 90.

Thus in triangle BGF, angle BGF = 180-angle GFB - angle BGF = 180-90-45 = 45 ----> BG = GF
Intern
Intern
User avatar
Joined: 13 Nov 2014
Posts: 44
GMAT 1: 590 Q42 V29
GMAT 2: 630 Q47 V29
Re: The side of square ABCD is 6. The midpoints of AB and BC are E and F,  [#permalink]

Show Tags

New post 18 Aug 2015, 19:12
Engr2012 wrote:
reto wrote:
Attachment:
T7914.png


The side of square ABCD is 6. The midpoints of AB and BC are E and F, respectively. If square HGID intersects EF at G, what is the length of DI?

A. 3
B. 4
C. 4.5
D. 5
E. 5.5

Economist says ballpark, but I would like to knoNw the calculations for this. What is wrong with my approach:

square HGID intersects EF at G. Triangle EBF is a 90-45-45 triangle and therefore the hypothenuse is: 3*sqrt(2). Half of 3*sqrt(2) would be a little more than 2. No lets subtract a little more than 2 from 6, which would equal a little less than 4. It does not make sense, because none of the answers are in this region.

Thanks


You are assuming in your calculations that EG = GF .


Assumption EG = GF looks valid because HGID is square and thus HD must be equal to DI, and hence AH and CI must be equal. so EG = GF must be true.
Please correct me if i am wrong
_________________
-----------------------------------------
Consider Cudos if you like this post.
-----------------------------------------
Retired Moderator
avatar
Joined: 29 Apr 2015
Posts: 838
Location: Switzerland
Concentration: Economics, Finance
Schools: LBS MIF '19
WE: Asset Management (Investment Banking)
GMAT ToolKit User
Re: The side of square ABCD is 6. The midpoints of AB and BC are E and F,  [#permalink]

Show Tags

New post 19 Aug 2015, 09:43
Engr2012 wrote:
In the text in red above, you mean why in triangle EBF, the height GB = EG = GF ?

If this is the case then look below,

You know in triangle EBF, we have 45-45-90 triangle and angle EFB = 45. Draw the height GB such that angle BGF = 90.

Thus in triangle BGF, angle BGF = 180-angle GFB - angle BGF = 180-90-45 = 45 ----> BG = GF


Hey! Thanks I got it now. But for me, impossible to see in 2min... now. Maybe after a lot more of practice. Better to rely on ballparking ...
_________________
Saving was yesterday, heat up the gmatclub.forum's sentiment by spending KUDOS!

PS Please send me PM if I do not respond to your question within 24 hours.
Intern
Intern
User avatar
Joined: 02 Jun 2015
Posts: 33
Location: United States
Concentration: Operations, Technology
GMAT Date: 08-22-2015
GPA: 3.92
WE: Science (Other)
Re: The side of square ABCD is 6. The midpoints of AB and BC are E and F,  [#permalink]

Show Tags

New post 19 Aug 2015, 10:13
3
Everyone is making this problem way harder than it actually is. This problem does not require using sqrt(2) or any ballpark "guessing". If you split up the top triangle into a square and two triangles it should be easy to see the square has a side of 3/2 as well as the neighbor triangles. DI simply has length 6-3/2.
CEO
CEO
avatar
S
Joined: 20 Mar 2014
Posts: 2624
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)
GMAT ToolKit User Reviews Badge
The side of square ABCD is 6. The midpoints of AB and BC are E and F,  [#permalink]

Show Tags

New post 19 Aug 2015, 11:20
PolarCatastrophe wrote:
Everyone is making this problem way harder than it actually is. This problem does not require using sqrt(2) or any ballpark "guessing". If you split up the top triangle into a square and two triangles it should be easy to see the square has a side of 3/2 as well as the neighbor triangles. DI simply has length 6-3/2.


Yes, I missed the fact that HGID is a square as well.
CEO
CEO
avatar
S
Joined: 20 Mar 2014
Posts: 2624
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)
GMAT ToolKit User Reviews Badge
Re: The side of square ABCD is 6. The midpoints of AB and BC are E and F,  [#permalink]

Show Tags

New post 19 Aug 2015, 11:23
dav90 wrote:
Engr2012 wrote:
reto wrote:
Attachment:
T7914.png


The side of square ABCD is 6. The midpoints of AB and BC are E and F, respectively. If square HGID intersects EF at G, what is the length of DI?

A. 3
B. 4
C. 4.5
D. 5
E. 5.5

Economist says ballpark, but I would like to knoNw the calculations for this. What is wrong with my approach:

square HGID intersects EF at G. Triangle EBF is a 90-45-45 triangle and therefore the hypothenuse is: 3*sqrt(2). Half of 3*sqrt(2) would be a little more than 2. No lets subtract a little more than 2 from 6, which would equal a little less than 4. It does not make sense, because none of the answers are in this region.

Thanks


You are assuming in your calculations that EG = GF .


Assumption EG = GF looks valid because HGID is square and thus HD must be equal to DI, and hence AH and CI must be equal. so EG = GF must be true.
Please correct me if i am wrong


Yes, you are correct.
Retired Moderator
avatar
Joined: 29 Apr 2015
Posts: 838
Location: Switzerland
Concentration: Economics, Finance
Schools: LBS MIF '19
WE: Asset Management (Investment Banking)
GMAT ToolKit User
Re: The side of square ABCD is 6. The midpoints of AB and BC are E and F,  [#permalink]

Show Tags

New post 20 Aug 2015, 03:47
PolarCatastrophe wrote:
Everyone is making this problem way harder than it actually is. This problem does not require using sqrt(2) or any ballpark "guessing". If you split up the top triangle into a square and two triangles it should be easy to see the square has a side of 3/2 as well as the neighbor triangles. DI simply has length 6-3/2.


Wow, respect. Yes that seems like a useful approach. So expanding triangle EBF into a Square with side length of 6/2. Following this, half of the newly built square will be 3/2. KUDOS for that :)
_________________
Saving was yesterday, heat up the gmatclub.forum's sentiment by spending KUDOS!

PS Please send me PM if I do not respond to your question within 24 hours.
Intern
Intern
avatar
Joined: 01 Jul 2015
Posts: 14
Re: The side of square ABCD is 6. The midpoints of AB and BC are E and F,  [#permalink]

Show Tags

New post 20 Aug 2015, 08:58
Even though if we make the traiangle to form a small square, where did 3/2 logic came from. Are you assuming "G" as the mid point of "EF".
Pls explain.
Retired Moderator
User avatar
Joined: 06 Jul 2014
Posts: 1229
Location: Ukraine
Concentration: Entrepreneurship, Technology
GMAT 1: 660 Q48 V33
GMAT 2: 740 Q50 V40
GMAT ToolKit User
Re: The side of square ABCD is 6. The midpoints of AB and BC are E and F,  [#permalink]

Show Tags

New post 20 Aug 2015, 11:01
Shreeya24 wrote:
Even though if we make the traiangle to form a small square, where did 3/2 logic came from. Are you assuming "G" as the mid point of "EF".
Pls explain.



Hello Shreeya24
We are not assuming: G is the mid point of EF because DHGI is a square and E and F are midpoints of AB and BC
Try to move point G and you will see that if G is not a midpoint of EF then DHGI is not a square.

It is proved here:
the-side-of-square-abcd-is-6-the-midpoints-of-ab-and-bc-are-e-and-f-203250.html#p1558985
But you can see it visually without any proves.
_________________
Intern
Intern
avatar
Joined: 01 Jul 2015
Posts: 14
Re: The side of square ABCD is 6. The midpoints of AB and BC are E and F,  [#permalink]

Show Tags

New post 21 Aug 2015, 09:32
Understood
Thanks
Intern
Intern
avatar
B
Joined: 20 May 2017
Posts: 10
The side of square ABCD is 6. The midpoints of AB and BC are E and F,  [#permalink]

Show Tags

New post 21 May 2017, 13:51
Here's a faster explanation with almost no calculation involved, or what I thought:

GI splits EB in halves because of nature of the square which is symmetrical , therefor HG = 6 - 1/2 EB = 4.5.

Also, G must be the mid point of EF because if it's not, then you cannot fold the 2 squares into a bigger triangle with the biggest side being the diagonal.
GMAT Club Bot
The side of square ABCD is 6. The midpoints of AB and BC are E and F,   [#permalink] 21 May 2017, 13:51

Go to page    1   2    Next  [ 21 posts ] 

Display posts from previous: Sort by

The side of square ABCD is 6. The midpoints of AB and BC are E and F,

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  


Copyright

GMAT Club MBA Forum Home| About| Terms and Conditions and Privacy Policy| GMAT Club Rules| Contact| Sitemap

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®.