Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack

 It is currently 23 May 2017, 06:36

### 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

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

Author Message
TAGS:

### Hide Tags

Senior Manager
Joined: 11 Sep 2005
Posts: 315
Followers: 2

Kudos [?]: 268 [4] , given: 0

In the figure, each side of square ABCD has length 1, the length of li [#permalink]

### Show Tags

19 Oct 2007, 04:37
4
KUDOS
81
This post was
BOOKMARKED
00:00

Difficulty:

95% (hard)

Question Stats:

43% (02:58) correct 57% (01:32) wrong based on 1082 sessions

### HideShow timer Statistics

Attachment:

squareabcd.jpg [ 17.9 KiB | Viewed 48292 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.
Intern
Joined: 02 Aug 2007
Posts: 36
Followers: 0

Kudos [?]: 35 [1] , given: 0

Re: In the figure, each side of square ABCD has length 1, the length of li [#permalink]

### Show Tags

19 Oct 2007, 06:56
1
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
Math Expert
Joined: 02 Sep 2009
Posts: 38822
Followers: 7716

Kudos [?]: 105858 [50] , given: 11593

In the figure, each side of square ABCD has length 1, the length of li [#permalink]

### Show Tags

25 Jan 2012, 10:53
50
KUDOS
Expert's post
46
This post was
BOOKMARKED

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

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 $$BCE=BOE-BOC=\frac{\sqrt{2}+1}{4}-\frac{1}{4}=\frac{\sqrt{2}}{4}$$

[Reveal] Spoiler:
Attachment:

Square ABCD.JPG [ 12.11 KiB | Viewed 55311 times ]

_________________
Senior Manager
Joined: 19 Apr 2011
Posts: 279
Schools: Booth,NUS,St.Gallon
Followers: 5

Kudos [?]: 310 [0], given: 51

Re: In the figure, each side of square ABCD has length 1, the length of li [#permalink]

### Show Tags

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
Joined: 07 Sep 2010
Posts: 329
Followers: 9

Kudos [?]: 767 [0], given: 136

Re: In the figure, each side of square ABCD has length 1, the length of li [#permalink]

### Show Tags

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

Math Expert
Joined: 02 Sep 2009
Posts: 38822
Followers: 7716

Kudos [?]: 105858 [0], given: 11593

Re: In the figure, each side of square ABCD has length 1, the length of li [#permalink]

### Show Tags

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.
_________________
Intern
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)
Followers: 0

Kudos [?]: 25 [0], given: 82

Re: In the figure, each side of square ABCD has length 1, the length of li [#permalink]

### Show Tags

28 Sep 2013, 05:09
Bunuel wrote:
Baten80 wrote:

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
Joined: 12 Sep 2012
Posts: 7
Location: India
GMAT 1: 650 Q46 V33
GPA: 4
WE: Information Technology (Computer Software)
Followers: 0

Kudos [?]: 16 [0], given: 2

Re: In the figure, each side of square ABCD has length 1, the length of li [#permalink]

### Show Tags

11 Nov 2013, 12:46
Yash12345 wrote:
Bunuel wrote:
Baten80 wrote:

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!
Intern
Joined: 05 Jul 2011
Posts: 6
Followers: 0

Kudos [?]: 2 [1] , given: 0

Re: In the figure, each side of square ABCD has length 1, the length of li [#permalink]

### Show Tags

02 Jan 2014, 17:46
1
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
Intern
Joined: 28 Oct 2013
Posts: 10
Followers: 0

Kudos [?]: 6 [6] , given: 1

Re: In the figure, each side of square ABCD has length 1, the length of li [#permalink]

### Show Tags

13 Jan 2014, 17:18
6
KUDOS
oh man. I thought it was a 3d figure. Its only 2d.
Manager
Joined: 27 Jan 2013
Posts: 230
GMAT 1: 780 Q49 V51
Followers: 51

Kudos [?]: 222 [1] , given: 32

Re: In the figure, each side of square ABCD has length 1, the length of li [#permalink]

### Show Tags

11 May 2014, 09:56
1
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.

_________________

"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.

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
Joined: 08 Dec 2013
Posts: 47
Followers: 0

Kudos [?]: 1 [0], given: 23

Re: In the figure, each side of square ABCD has length 1, the length of li [#permalink]

### Show Tags

14 Oct 2014, 06:35
1
This post was
BOOKMARKED
hey ..
why cant 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
Math Expert
Joined: 02 Sep 2009
Posts: 38822
Followers: 7716

Kudos [?]: 105858 [1] , given: 11593

Re: In the figure, each side of square ABCD has length 1, the length of li [#permalink]

### Show Tags

14 Oct 2014, 06:42
1
KUDOS
Expert's post
1
This post was
BOOKMARKED
shreygupta3192 wrote:
hey ..
why cant 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.
_________________
Intern
Joined: 29 Sep 2012
Posts: 7
Followers: 0

Kudos [?]: 6 [2] , given: 1

Re: In the figure, each side of square ABCD has length 1, the length of li [#permalink]

### Show Tags

26 Oct 2014, 22:41
2
KUDOS
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.

Manager
Joined: 10 Mar 2014
Posts: 244
Followers: 2

Kudos [?]: 79 [0], given: 13

Re: In the figure, each side of square ABCD has length 1, the length of li [#permalink]

### Show Tags

05 Aug 2015, 01:35
Bunuel wrote:

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 $$BCE=BOE-BOC=\frac{\sqrt{2}+1}{4}-\frac{1}{4}=\frac{\sqrt{2}}{4}$$

hi bunuel,

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

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

Thanks
Intern
Joined: 21 Jul 2015
Posts: 3
Followers: 0

Kudos [?]: 1 [1] , given: 0

Re: In the figure, each side of square ABCD has length 1, the length of li [#permalink]

### Show Tags

11 Aug 2015, 23:03
1
KUDOS
holy smoke I thought that was a 3D picture, like a cone or something.
Intern
Joined: 06 Aug 2015
Posts: 5
Followers: 0

Kudos [?]: 0 [0], given: 10

Re: In the figure, each side of square ABCD has length 1, the length of li [#permalink]

### Show Tags

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).
Math Forum Moderator
Joined: 20 Mar 2014
Posts: 2644
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)
Followers: 128

Kudos [?]: 1470 [0], given: 789

Re: In the figure, each side of square ABCD has length 1, the length of li [#permalink]

### Show Tags

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
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
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
Followers: 1

Kudos [?]: 10 [0], given: 71

Re: In the figure, each side of square ABCD has length 1, the length of li [#permalink]

### Show Tags

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
Joined: 08 Oct 2015
Posts: 11
Followers: 0

Kudos [?]: 1 [0], given: 0

In the figure, each side of square ABCD has length 1, the length of li [#permalink]

### Show Tags

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 Replies Last post
Similar
Topics:
2 If each square in the preceding figure has a side of length 3, what is 2 09 Apr 2016, 14:42
In the figure above, the square LMNO has a side of length 2x + 1 and 4 09 Apr 2016, 14:47
12 In the figure shown, ABCD is a square with side length 5 sqr 9 27 Apr 2016, 20:14
12 In the figure, each side of square ABCD has length 1, the length of li 8 30 Sep 2014, 00:40
17 In the figure, each side of square ABCD has length 1, the 10 08 Sep 2014, 04:57
Display posts from previous: Sort by