It is currently 28 Jun 2017, 11:10

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

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

In the figure, each side of square ABCD has length 1, the length of li

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

Hide Tags

4 KUDOS received
Senior Manager
Senior Manager
User avatar
Joined: 11 Sep 2005
Posts: 310
In the figure, each side of square ABCD has length 1, the length of li [#permalink]

Show Tags

New post 19 Oct 2007, 04:37
4
This post received
KUDOS
86
This post was
BOOKMARKED
00:00
A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

43% (02:57) correct 57% (01:30) wrong based on 1119 sessions

HideShow timer Statistics

Attachment:
squareabcd.jpg
squareabcd.jpg [ 17.9 KiB | Viewed 51156 times ]
In the figure, each side of square ABCD has length 1, the length of line Segment CE is 1, and the length of line segment BE is equal to the length of line segment DE. What is the area of the triangular region BCE?

A. 1/3

B. \(\frac{\sqrt{2}}{4}\)

C. 1/2

D. \(\frac{\sqrt{2}}{2}\)

E. 3/4
[Reveal] Spoiler: OA

Last edited by Bunuel on 30 Sep 2014, 00:30, edited 5 times in total.
Renamed the topic, edited the question and added the OA.
1 KUDOS received
Intern
Intern
User avatar
Joined: 02 Aug 2007
Posts: 36
Re: In the figure, each side of square ABCD has length 1, the length of li [#permalink]

Show Tags

New post 19 Oct 2007, 06:56
1
This post received
KUDOS
4
This post was
BOOKMARKED
singh_amit19 wrote:
In the figure (attchd), each side of square ABCD has length 1, the length of line Segment CE is 1, and the length of line segment BE is equal to the length Of line segment DE. What is the area of the triangular region BCE?

a. 1/3
b. (2^-2 )/4
c. 1/2
d. (2^-2)/2
e. 3/4


I got sqrt(2) / 4

I am wondering about ans. b and d in fact 2^-2 = 1/4 I think it should be sqrt(2) instead ... maybe I am wrong

Here is how I did it:
let's suppose O the intersection of ABCD diagonals. In tirangle BDE, the points A, C and O are alignes since C is the barycenter of the triangle (CE=CB=CD) and BED is isocel.

We can calculate the area of BCE by substracting the area of BOC from BOE.

area of BOE=BO*BE/2 = (1/sqrt(2)) * (1/sqrt(2) + 1) / 2 = (1 + sqrt(2))/4
area of BOC= 0.25 area of the square= 1/4

Thus area of BCE = (1 + sqrt(2))/4 - 1/4 = sqrt(2)/4
Expert Post
50 KUDOS received
Math Expert
User avatar
D
Joined: 02 Sep 2009
Posts: 39751
In the figure, each side of square ABCD has length 1, the length of li [#permalink]

Show Tags

New post 25 Jan 2012, 10:53
50
This post received
KUDOS
Expert's post
47
This post was
BOOKMARKED
Image
In the figure, each side of square ABCD has length 1, the length of line Segment CE is 1, and the length of line segment BE is equal to the length of line segment DE. What is the area of the triangular region BCE?

A. 1/3

B. \(\frac{\sqrt{2}}{4}\)

C. 1/2

D. \(\frac{\sqrt{2}}{2}\)

E. 3/4


Image

Note that as BE=DE the triangles BCE and CDE are congruent (all three sides are equal) --> \(\angle{BCE}=\angle{DCE}\). So if we extend the line segment CE it will meet with the crosspoint of the diagonals, let's call it point O. Also note that EO will be perpendicular to BD, as diagonals in square make a right angle.

(The area of BCE) = (The area of BOE) - (The area of BOC).

Area of BOC is one fourth of the square's =\(\frac{1}{4}\).

Area BOE, \(\frac{1}{2}*BO*EO\). \(BO=\frac{\sqrt{2}}{2}\), half of the diagonal of a square. \(EO=CE+CO=1+\frac{\sqrt{2}}{2}=\frac{2+\sqrt{2}}{2}\), CO is also half of the diagonal of a square.

So \(AreaBOE=\frac{1}{2}*BO*EO=\frac{1}{2}*\frac{\sqrt{2}}{2}*\frac{2+\sqrt{2}}{2}=\frac{\sqrt{2}+1}{4}\).

Area \(BCE=BOE-BOC=\frac{\sqrt{2}+1}{4}-\frac{1}{4}=\frac{\sqrt{2}}{4}\)

Answer: B.

[Reveal] Spoiler:
Attachment:
Square ABCD.JPG
Square ABCD.JPG [ 12.11 KiB | Viewed 58085 times ]

_________________

New to the Math Forum?
Please read this: All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Senior Manager
Senior Manager
User avatar
Joined: 19 Apr 2011
Posts: 277
Schools: Booth,NUS,St.Gallon
Re: In the figure, each side of square ABCD has length 1, the length of li [#permalink]

Show Tags

New post 10 Feb 2012, 10:53
2
This post was
BOOKMARKED
the figure shown above is a square pyramid .Only then BE and DE will be the same .Now we are asked to find the are of one of the slant faces .
Area of a pyramid =1/2 * perimeter of the base * slant height
=1/2 *(4)*1

We need the area of one of the sides only .Hence Area = 1/2.
Option B is the answer .Sorry if i had brought in some new formulas and concepts .Hope you are aware .
_________________

+1 if you like my explanation .Thanks :)

Senior Manager
Senior Manager
avatar
Joined: 07 Sep 2010
Posts: 327
Re: In the figure, each side of square ABCD has length 1, the length of li [#permalink]

Show Tags

New post 22 Sep 2013, 07:44
Bunuel wrote:
Three sides are equal) --> \(\angle{BCE}=\angle{DCE}\). So if we continue the line segment CE it will meet with the crosspoint of the diagonals, let's call it point O. Also note that EO will be perpendicular to BD, as diagonals in square make a right angle.


Hello Bunuel
Could you please explain the underlying logic behind highlighted portion. I failed to understand why it would meet at point O. Please explain.
Thanks
_________________

+1 Kudos me, Help me unlocking GMAT Club Tests

Expert Post
Math Expert
User avatar
D
Joined: 02 Sep 2009
Posts: 39751
Re: In the figure, each side of square ABCD has length 1, the length of li [#permalink]

Show Tags

New post 23 Sep 2013, 01:22
Expert's post
2
This post was
BOOKMARKED
imhimanshu wrote:
Bunuel wrote:
Three sides are equal) --> \(\angle{BCE}=\angle{DCE}\). So if we continue the line segment CE it will meet with the crosspoint of the diagonals, let's call it point O. Also note that EO will be perpendicular to BD, as diagonals in square make a right angle.


Hello Bunuel
Could you please explain the underlying logic behind highlighted portion. I failed to understand why it would meet at point O. Please explain.
Thanks


Since triangles BCE and CDE are congruent, then \(\angle{BEC}=\angle{DEC}\), which means that CE is the bisector of angle E. But triangle BED is an isosceles, thus bisector of angle E must also be the height and the median.
_________________

New to the Math Forum?
Please read this: All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Intern
Intern
avatar
Joined: 21 Sep 2013
Posts: 30
Location: United States
Concentration: Finance, General Management
GMAT Date: 10-25-2013
GPA: 3
WE: Operations (Mutual Funds and Brokerage)
Re: In the figure, each side of square ABCD has length 1, the length of li [#permalink]

Show Tags

New post 28 Sep 2013, 05:09
Bunuel wrote:
Baten80 wrote:
I still in trouble with this problem. My document shows OA is B. Please help.

Responding to a pm.
Attachment:
Square ABCD.JPG
In the figure, each side of square ABCD has length 1, the length of line Segment CE is 1, and the length of line segment BE is equal to the length of line segment DE. What is the area of the triangular region BCE?

This proble can be solved in many ways. One of the approaches:

Note that as BE=DE the triangles BCE and CDE are congruent (all three sides are equal) --> \(\angle{BCE}=\angle{DCE}\). So if we continue the line segment CE it will meet with the crosspoint of the diagonals, let's call it point O. Also note that EO will be perpendicular to BD, as diagonals in square make a right angle.

The area BCE=BOE-BOC.

Area of BOC is one fourth of the square's =\(\frac{1}{4}\).

Area BOE, \(\frac{1}{2}BO*EO\). \(BO=\frac{\sqrt{2}}{2}\), half of the diagonal of a square. \(EO=CE+CO=1+\frac{\sqrt{2}}{2}=\frac{2+\sqrt{2}}{2}\), CO is also half of the diagonal of a square.
So \(AreaBOE=\frac{1}{2}BO*EO=\frac{1}{2}*\frac{\sqrt{2}}{2}*\frac{2+\sqrt{2}}{2}=\frac{\sqrt{2}+1}{4}\).

Area \(BCE=BOE-BOC=\frac{\sqrt{2}+1}{4}-\frac{1}{4}=\frac{\sqrt{2}}{4}\)

None of the answer choices shown is correct.



can we expect such kind of question on the exam ... damn they are tough. moreover completing it in less than 2 min is out question.
Intern
Intern
avatar
Joined: 12 Sep 2012
Posts: 7
Location: India
Concentration: International Business, Entrepreneurship
GMAT 1: 650 Q46 V33
GPA: 4
WE: Information Technology (Computer Software)
Re: In the figure, each side of square ABCD has length 1, the length of li [#permalink]

Show Tags

New post 11 Nov 2013, 12:46
Yash12345 wrote:
Bunuel wrote:
Baten80 wrote:
I still in trouble with this problem. My document shows OA is B. Please help.

Responding to a pm.
Attachment:
Square ABCD.JPG
In the figure, each side of square ABCD has length 1, the length of line Segment CE is 1, and the length of line segment BE is equal to the length of line segment DE. What is the area of the triangular region BCE?

This proble can be solved in many ways. One of the approaches:

Note that as BE=DE the triangles BCE and CDE are congruent (all three sides are equal) --> \(\angle{BCE}=\angle{DCE}\). So if we continue the line segment CE it will meet with the crosspoint of the diagonals, let's call it point O. Also note that EO will be perpendicular to BD, as diagonals in square make a right angle.

The area BCE=BOE-BOC.

Area of BOC is one fourth of the square's =\(\frac{1}{4}\).

Area BOE, \(\frac{1}{2}BO*EO\). \(BO=\frac{\sqrt{2}}{2}\), half of the diagonal of a square. \(EO=CE+CO=1+\frac{\sqrt{2}}{2}=\frac{2+\sqrt{2}}{2}\), CO is also half of the diagonal of a square.
So \(AreaBOE=\frac{1}{2}BO*EO=\frac{1}{2}*\frac{\sqrt{2}}{2}*\frac{2+\sqrt{2}}{2}=\frac{\sqrt{2}+1}{4}\).

Area \(BCE=BOE-BOC=\frac{\sqrt{2}+1}{4}-\frac{1}{4}=\frac{\sqrt{2}}{4}\)

None of the answer choices shown is correct.



can we expect such kind of question on the exam ... damn they are tough. moreover completing it in less than 2 min is out question.



FYI, I encountered the same question in GMAT Prep Exam Pack 1! So, yes you can expect such questions in the real test too!
1 KUDOS received
Intern
Intern
avatar
Joined: 05 Jul 2011
Posts: 6
Re: In the figure, each side of square ABCD has length 1, the length of li [#permalink]

Show Tags

New post 02 Jan 2014, 17:46
1
This post received
KUDOS
1
This post was
BOOKMARKED
another way to do this problem

extend BC to the right, let's say this point is X. Connect E to X so that CXE is a right angle.

Now, for triangle BCE base BC = 1 and height EX = 1/(sqrt)2.
How length of EX is derived:
angle ECX = 180 - angle BCE = 180 - 135 = 45 deg.
so, triangle ECX is a 45-45-90 right angle triangle and EC is 1.

Area = 1/2 * 1 * 1/(sqrt)2 = 1/2*(sqrt)2 = sqrt(2)/4.

HTH
6 KUDOS received
Intern
Intern
avatar
Joined: 28 Oct 2013
Posts: 10
Re: In the figure, each side of square ABCD has length 1, the length of li [#permalink]

Show Tags

New post 13 Jan 2014, 17:18
6
This post received
KUDOS
oh man. I thought it was a 3d figure. Its only 2d.
Expert Post
1 KUDOS received
Manager
Manager
User avatar
Joined: 27 Jan 2013
Posts: 230
GMAT 1: 780 Q49 V51
Re: In the figure, each side of square ABCD has length 1, the length of li [#permalink]

Show Tags

New post 11 May 2014, 09:56
1
This post received
KUDOS
Expert's post
Hi All,

Lots of great information here on this tricky question. I think there may be one more way to look at the solution that might be easier for some people. The basic idea is: flatten the shape. Put point E right onto the square. Then, do all the normal things that you should always be doing in GMAT geometry. For a square you should always be thinking about the diagonal. For the area of a triangle you should always draw a height. Look for special triangles. In this case you have a 45-45-90 which allows you to get the measure of the height. Plug in all of the numbers into the area of a triangle formula. Rationalize the denominator. Simplify and you're there.

I added a diagram. I hope that it's clear. Feel free to follow up with any questions!

Happy Studies,

A.

Image
_________________

"It is a curious property of research activity that after the problem has been solved the solution seems obvious. This is true not only for those who have not previously been acquainted with the problem, but also for those who have worked over it for years." -Dr. Edwin Land

GMAT vs GRE Comparison

If you found my post useful KUDOS are much appreciated.

IMPROVE YOUR READING COMPREHENSION with the ECONOMIST READING COMPREHENSION CHALLENGE:

Here is the first set along with some strategies for approaching this work: http://gmatclub.com/forum/the-economist-reading-comprehension-challenge-151479.html

Intern
Intern
avatar
Joined: 08 Dec 2013
Posts: 47
Re: In the figure, each side of square ABCD has length 1, the length of li [#permalink]

Show Tags

New post 14 Oct 2014, 06:35
1
This post was
BOOKMARKED
hey ..
why can`t we assume this a 3-d figure where extended part "bedc" seems to be coming out ..
and as a result angle bce=dce=90..and then 1/2 bh
area comes as 1/2..

whats wrong..plz xplain.
thanks
Expert Post
1 KUDOS received
Math Expert
User avatar
D
Joined: 02 Sep 2009
Posts: 39751
Re: In the figure, each side of square ABCD has length 1, the length of li [#permalink]

Show Tags

New post 14 Oct 2014, 06:42
1
This post received
KUDOS
Expert's post
1
This post was
BOOKMARKED
shreygupta3192 wrote:
hey ..
why can`t we assume this a 3-d figure where extended part "bedc" seems to be coming out ..
and as a result angle bce=dce=90..and then 1/2 bh
area comes as 1/2..

whats wrong..plz xplain.
thanks


You are overthinking it. On the GMAT, all figures lie in a plane unless otherwise indicated.
_________________

New to the Math Forum?
Please read this: All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

2 KUDOS received
Intern
Intern
avatar
Joined: 29 Sep 2012
Posts: 7
Re: In the figure, each side of square ABCD has length 1, the length of li [#permalink]

Show Tags

New post 26 Oct 2014, 22:41
2
This post received
KUDOS
1
This post was
BOOKMARKED
Hi,

I received this question towards the end of my mock test very recently. Since I didn't have enough time to literally solve it, I made an educated guess, used POE and moved on. Fortunately it worked.

Here's how I approached it:

Given:

- ABCD are all equal to 1; therefore BC=1
- CE=1
- BE=DE

Area of BCE = 1/2 x BC x height (ht)

Since ht is the perpendicular distance and CE is the hypotenuse, height<1

Therefore area of BCE = 1/2 x 1 x something less than 1 = LT 1/2------- (this narrows down ur options to A and B)

Now A and B are very close: A= 0.33 and B=0.32.

I chose B (Not much logic here: honestly because every other answer choice had an even denominator and root 2 appeared as a split in 2 ans choices).

It would be great if any expert can help me figure out whether there's any other solid way to eliminate A (other than the super difficult approach already mentioned in the post)

My exam is around the corner and any quick help would be highly appreciated.

Thanks in advance!
Manager
Manager
avatar
B
Joined: 10 Mar 2014
Posts: 245
Premium Member
Re: In the figure, each side of square ABCD has length 1, the length of li [#permalink]

Show Tags

New post 05 Aug 2015, 01:35
Bunuel wrote:
Image
In the figure, each side of square ABCD has length 1, the length of line Segment CE is 1, and the length of line segment BE is equal to the length of line segment DE. What is the area of the triangular region BCE?
A. 1/3

B. \(\frac{\sqrt{2}}{4}\)

C. 1/2

D. \(\frac{\sqrt{2}}{2}\)

E. 3/4

Attachment:
Square ABCD.JPG

Note that as BE=DE the triangles BCE and CDE are congruent (all three sides are equal) --> \(\angle{BCE}=\angle{DCE}\). So if we extend the line segment CE it will meet with the crosspoint of the diagonals, let's call it point O. Also note that EO will be perpendicular to BD, as diagonals in square make a right angle.

The area BCE=BOE-BOC.

Area of BOC is one fourth of the square's =\(\frac{1}{4}\).

Area BOE, \(\frac{1}{2}BO*EO\). \(BO=\frac{\sqrt{2}}{2}\), half of the diagonal of a square. \(EO=CE+CO=1+\frac{\sqrt{2}}{2}=\frac{2+\sqrt{2}}{2}\), CO is also half of the diagonal of a square.
So \(AreaBOE=\frac{1}{2}BO*EO=\frac{1}{2}*\frac{\sqrt{2}}{2}*\frac{2+\sqrt{2}}{2}=\frac{\sqrt{2}+1}{4}\).

Area \(BCE=BOE-BOC=\frac{\sqrt{2}+1}{4}-\frac{1}{4}=\frac{\sqrt{2}}{4}\)

Answer: B.


hi bunuel,

I have a query here. Cant we use directly 1/2 * BC*CE

= 1/2*1*1 = 1/2.

Thanks
1 KUDOS received
Intern
Intern
avatar
Joined: 21 Jul 2015
Posts: 3
Re: In the figure, each side of square ABCD has length 1, the length of li [#permalink]

Show Tags

New post 11 Aug 2015, 23:03
1
This post received
KUDOS
holy smoke I thought that was a 3D picture, like a cone or something.
Intern
Intern
avatar
Joined: 06 Aug 2015
Posts: 5
Re: In the figure, each side of square ABCD has length 1, the length of li [#permalink]

Show Tags

New post 11 Sep 2015, 05:19
Same here I saw it as a 3D figure. Luckily it was just a Gmat Prep question but on a real exam I would bite mysel in the a** for this. IMO it's really misleading but to be honest I knew I did it wrong becaus A=1/2 is just too easy for a 700 Level question (since BC and CE are already given in the text).
Expert Post
Math Forum Moderator
avatar
B
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)
GMAT ToolKit User Premium Member Reviews Badge
Re: In the figure, each side of square ABCD has length 1, the length of li [#permalink]

Show Tags

New post 11 Sep 2015, 05:28
Kekki wrote:
Same here I saw it as a 3D figure. Luckily it was just a Gmat Prep question but on a real exam I would bite mysel in the a** for this. IMO it's really misleading but to be honest I knew I did it wrong becaus A=1/2 is just too easy for a 700 Level question (since BC and CE are already given in the text).


Official guide mentions this about figures in GMAT Quant : " All figures lie in a plane unless otherwise stated" . Thus you should not assume that a given figure is 3D unless specifically mentioned as such. For this question, it was not mentioned that the given figure is 3D.
_________________

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
Rules for Posting in Quant Forums: http://gmatclub.com/forum/rules-for-posting-please-read-this-before-posting-133935.html
Writing Mathematical Formulae in your posts: http://gmatclub.com/forum/rules-for-posting-please-read-this-before-posting-133935.html#p1096628
GMATCLUB Math Book: http://gmatclub.com/forum/gmat-math-book-in-downloadable-pdf-format-130609.html
Everything Related to Inequalities: http://gmatclub.com/forum/inequalities-made-easy-206653.html#p1582891
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

Intern
Intern
avatar
Joined: 27 Dec 2011
Posts: 46
Location: Brazil
Concentration: Entrepreneurship, Strategy
GMAT 1: 620 Q48 V27
GMAT 2: 680 Q46 V38
GMAT 3: 750 Q50 V41
GPA: 3.5
Premium Member
Re: In the figure, each side of square ABCD has length 1, the length of li [#permalink]

Show Tags

New post 11 Oct 2015, 19:02
This figure has a symmetry through the segment \(ACE\) (since \(BE=DE\)). Thus \(CA\) is the extension of the side \(EC\) of the \(\triangle BCE\). Since in squares the diagonals are perpendicular, the height of the \(\triangle BCE\), relative to the side \(EC\), is half of the diagonal of the square \(ABCD\) (draw the segment \(BO\), where \(O\) is the intersection of the diagonals). Therefore the area of the \(\triangle BCE=\frac{\frac{\sqrt{2}}{2}*1}{2}=\frac{\sqrt{2}}{4}\). Letter B.
Intern
Intern
avatar
Joined: 08 Oct 2015
Posts: 11
In the figure, each side of square ABCD has length 1, the length of li [#permalink]

Show Tags

New post 17 Nov 2015, 19:09
Bunuel

Perhaps you can further explain why and how you know this is a one-dimensional picture. It looked 3D as well to me and I made the same assumptions and arrived at 1/2 for the area. Also, in your explanation, how do you know that CE extends to meet directly at the midpoint ("O") of ABCD's diagonal? I get that ∠BCE=∠DCE, but how do you know that CE meets at the diagonal's midpoint which is the crux of the rest of the problem?

Last edited by dubyap on 14 Dec 2015, 18:04, edited 1 time in total.
In the figure, each side of square ABCD has length 1, the length of li   [#permalink] 17 Nov 2015, 19:09

Go to page    1   2    Next  [ 31 posts ] 

    Similar topics Author Replies Last post
Similar
Topics:
2 Experts publish their posts in the topic If each square in the preceding figure has a side of length 3, what is Bunuel 2 09 Apr 2016, 14:42
Experts publish their posts in the topic In the figure above, the square LMNO has a side of length 2x + 1 and Bunuel 4 09 Apr 2016, 14:47
16 Experts publish their posts in the topic In the figure shown, ABCD is a square with side length 5 sqr sudhir18n 10 01 Jun 2017, 18:24
12 Experts publish their posts in the topic In the figure, each side of square ABCD has length 1, the length of li vaivish1723 8 30 Sep 2014, 00:40
18 Experts publish their posts in the topic In the figure, each side of square ABCD has length 1, the jugolo1 10 08 Sep 2014, 04:57
Display posts from previous: Sort by

In the figure, each side of square ABCD has length 1, the length of li

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


cron

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

Powered by phpBB © phpBB Group and phpBB SEO

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®.