Last visit was: 20 Sep 2026, 23:12 It is currently 20 Sep 2026, 23:12
Home
Messages
Tests
Schools
Events
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.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel
Back to Forum
Create Topic
 1   2   3   
805+ (Hard)|   Geometry|      
User avatar
singh_amit19
User avatar
Joined: 11 Sep 2005
Last visit: 05 Aug 2012
Posts: 181
Own Kudos:
4,050
 [594]
Posts: 181
Kudos: 4,050
 [594]
In the figure, each side of square ABCD has length 1, the length of li Updated on: 26 Nov 2018, 06:39
Originally posted by singh_amit19 on 19 Oct 2007, 04:37.
Last edited by Bunuel on 26 Nov 2018, 06:39, edited 7 times in total.
Renamed the topic, edited the question and added the OA.
33
Kudos
Add Kudos
561
Bookmarks
Bookmark this Post
00:00
Start Timer
Pause Timer
Resume Timer
Show Answer
Hide
Show
History
B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

95% (hard)

Question Stats:

48% (02:54) correct 52% (02:27) wrong based on 2578 sessions

History

Date
Time
Result
Not Attempted Yet

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


Show Spoiler
Attachment:
squareabcd.jpg
squareabcd.jpg [ 17.9 KiB | Viewed 264347 times ]
ShowHide Answer
Official Answer
B
Every tricky question should trigger a follow up quiz. Build that habit with GMAT Club Forum Quiz →.
Most Helpful Reply
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 20 Sep 2026
Posts: 113,703
Own Kudos:
842,586
 [293]
Given Kudos: 111,704
Products:
GMAT Club Tests
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 113,703
Kudos: 842,586
 [293]
In the figure, each side of square ABCD has length 1, the length of li 25 Jan 2012, 10:53
133
Kudos
Add Kudos
159
Bookmarks
Bookmark this Post

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

Answer: B.

Show Spoiler
Attachment:
Square ABCD.JPG
Square ABCD.JPG [ 12.11 KiB | Viewed 268157 times ]
User avatar
LogicGuru1
User avatar
Joined: 04 Jun 2016
Last visit: 28 May 2024
Posts: 463
Own Kudos:
2,686
 [57]
Given Kudos: 36
GMAT 1: 750 Q49 V43
GMAT 1: 750 Q49 V43
Posts: 463
Kudos: 2,686
 [57]
In the figure, each side of square ABCD has length 1, the length of li 30 Jul 2016, 00:07
39
Kudos
Add Kudos
18
Bookmarks
Bookmark this Post
singh_amit19
Attachment:
squareabcd.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?

A. 1/3

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

C. 1/2

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

E. 3/4
Show more
My advise is not to use any fancy triangle properties or weird geometry.
This is a 2D geometry question and the shape of the object is a KITE.

This question can be solved under 20 seconds, The trick is use obscure properties of shapes to your advantage ; in this case the area of a KITE.
If you see the question properly it is asking us to find the area of half KITE and then subtract the area of a Half square
Mathematically
Area = 1/2*area of KITE-1/2* area of square
Area = 1/2 (area of KITE-area of square)
Area of the square is super easy .. its 1; half of it will be = 1/2

For finding the area of the KITE you will need its two diagonal. Smaller one is \(\sqrt{2}\). Bigger one is \(1+\sqrt{2}\)
FORMULA FOR KITE AREA= \(\frac{1}{2}\)*(diagonal1 * diagonal 2)

Solve it and you will get \(\frac{\sqrt{2}}{4}\)
General Discussion
User avatar
ronneyc
User avatar
Joined: 02 Aug 2007
Last visit: 30 Nov 2007
Posts: 12
Own Kudos:
145
 [9]
Posts: 12
Kudos: 145
 [9]
In the figure, each side of square ABCD has length 1, the length of li 19 Oct 2007, 06:56
4
Kudos
Add Kudos
5
Bookmarks
Bookmark this Post
singh_amit19
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
Show more


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
User avatar
saikarthikreddy
User avatar
Joined: 19 Apr 2011
Last visit: 14 Feb 2022
Posts: 128
Own Kudos:
1,044
 [11]
Given Kudos: 53
Concentration: Finance,Entrepreneurship,General Management
Schools: Carlson '16 (I) Kelley '16 (I) Tuck '16 (D) Darden '24 (II)
GMAT 1: 720 Q49 V38
GRE 1: Q167 V167
Posts: 128
Kudos: 1,044
 [11]
In the figure, each side of square ABCD has length 1, the length of li 10 Feb 2012, 10:53
6
Kudos
Add Kudos
5
Bookmarks
Bookmark this Post
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 .
User avatar
imhimanshu
User avatar
Joined: 07 Sep 2010
Last visit: 08 Nov 2013
Posts: 216
Own Kudos:
6,587
 [2]
Given Kudos: 136
GMAT 1: 650 Q49 V30
Posts: 216
Kudos: 6,587
 [2]
In the figure, each side of square ABCD has length 1, the length of li 22 Sep 2013, 07:44
2
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
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.

Show more
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
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 20 Sep 2026
Posts: 113,703
Own Kudos:
842,586
 [11]
Given Kudos: 111,704
Products:
GMAT Club Tests
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 113,703
Kudos: 842,586
 [11]
In the figure, each side of square ABCD has length 1, the length of li 23 Sep 2013, 01:22
7
Kudos
Add Kudos
4
Bookmarks
Bookmark this Post
imhimanshu
Bunuel
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
Show more

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.
avatar
sb1
avatar
Joined: 05 Jul 2011
Last visit: 27 Dec 2015
Posts: 2
Own Kudos:
11
 [10]
Posts: 2
Kudos: 11
 [10]
In the figure, each side of square ABCD has length 1, the length of li 02 Jan 2014, 17:46
9
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
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
User avatar
AtlanticGMAT
User avatar
Joined: 27 Jan 2013
Last visit: 09 Jul 2026
Posts: 255
Own Kudos:
646
 [8]
Given Kudos: 38
GMAT 1: 760 Q47 V48
GMAT 2: 770 Q49 V47
GMAT 3: 780 Q49 V51
Expert
Expert reply
GMAT 3: 780 Q49 V51
Posts: 255
Kudos: 646
 [8]
In the figure, each side of square ABCD has length 1, the length of li 11 May 2014, 09:56
7
Kudos
Add Kudos
1
Bookmarks
Bookmark this 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.

avatar
shreygupta3192
avatar
Joined: 08 Dec 2013
Last visit: 13 Apr 2020
Posts: 22
Own Kudos:
8
 [1]
Given Kudos: 23
Posts: 22
Kudos: 8
 [1]
In the figure, each side of square ABCD has length 1, the length of li 14 Oct 2014, 06:35
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 20 Sep 2026
Posts: 113,703
Own Kudos:
842,586
 [7]
Given Kudos: 111,704
Products:
GMAT Club Tests
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 113,703
Kudos: 842,586
 [7]
In the figure, each side of square ABCD has length 1, the length of li 14 Oct 2014, 06:42
5
Kudos
Add Kudos
2
Bookmarks
Bookmark this Post
shreygupta3192
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
Show more

You are overthinking it. On the GMAT, all figures lie in a plane unless otherwise indicated.
avatar
Chandni170
avatar
Joined: 29 Sep 2012
Last visit: 31 Mar 2015
Posts: 7
Own Kudos:
20
 [4]
Given Kudos: 1
Posts: 7
Kudos: 20
 [4]
In the figure, each side of square ABCD has length 1, the length of li 26 Oct 2014, 22:41
3
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
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!
avatar
Hugoba
avatar
Joined: 27 Dec 2011
Last visit: 01 Feb 2017
Posts: 29
Own Kudos:
25
 [3]
Given Kudos: 71
Location: Brazil
Concentration: Entrepreneurship, Strategy
Schools: Insead Sept '17 IESE '18 Haas '18 CBS '18 Wharton '18
GMAT 1: 620 Q48 V27
GMAT 2: 680 Q46 V38
GMAT 3: 750 Q50 V41
GPA: 3.5
Schools: Insead Sept '17 IESE '18 Haas '18 CBS '18 Wharton '18
GMAT 3: 750 Q50 V41
Posts: 29
Kudos: 25
 [3]
In the figure, each side of square ABCD has length 1, the length of li 11 Oct 2015, 19:02
2
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
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.
User avatar
shekhar_2017
User avatar
Joined: 27 Apr 2015
Last visit: 31 Oct 2023
Posts: 45
Own Kudos:
493
 [8]
Given Kudos: 4
Location: India
Concentration: Organizational Behavior, Technology
Schools: Tuck '19 (WL) Kellogg '19 (D) Booth '19 (D) Duke '19 (WL) Ross '19 (D) Kenan-Flagler '19 (A$)
GMAT 1: 600 Q42 V32
GMAT 2: 770 Q50 V45
GPA: 3.7
WE:Engineering (Telecommunications)
Products:
My Reviews
Schools: Tuck '19 (WL) Kellogg '19 (D) Booth '19 (D) Duke '19 (WL) Ross '19 (D) Kenan-Flagler '19 (A$)
GMAT 2: 770 Q50 V45
Posts: 45
Kudos: 493
 [8]
In the figure, each side of square ABCD has length 1, the length of li 28 Aug 2016, 07:23
6
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
Just in case something does not click during exam , you can use some intelligent guess work the base of BCE given as one
so area = 1/2 * 1 * height .

Now CE is 1 since in PS problems the figure is drawn to scale height is surely less than CE is not 90 degree so the area is less than 1/2 .
This leaves you with 2 possible options sqrt2/ 4 and 1/3 , since the sqrt2 is the diagonal of ABCA it is more probable to figure in the answer.

guess as B ...
User avatar
Manonamission
User avatar
Joined: 11 Jul 2016
Last visit: 22 Mar 2017
Posts: 72
Own Kudos:
209
 [8]
Given Kudos: 87
Posts: 72
Kudos: 209
 [8]
In the figure, each side of square ABCD has length 1, the length of li 23 Oct 2016, 03:51
5
Kudos
Add Kudos
3
Bookmarks
Bookmark this Post
This Q is can be solved if we extend the line from E to touch one of the diagonals of the square.

Diagonals of a square intersect at right angles. This implies angle BPD will be 90.
Attachments

File comment: Geometry Question
Geo Q.PNG
Geo Q.PNG [ 288.2 KiB | Viewed 133163 times ]

User avatar
felippemed
User avatar
Joined: 23 Jun 2009
Last visit: 27 Oct 2020
Posts: 122
Own Kudos:
897
 [2]
Given Kudos: 138
Location: Brazil
GMAT 1: 470 Q30 V20
GMAT 2: 620 Q42 V33
GMAT 2: 620 Q42 V33
Posts: 122
Kudos: 897
 [2]
In the figure, each side of square ABCD has length 1, the length of li 10 Dec 2016, 09:36
2
Kudos
Add Kudos
Bookmarks
Bookmark this Post
here ir my colorful contribution :-D
Attachments

deletar.png
deletar.png [ 348.74 KiB | Viewed 132723 times ]

User avatar
sdlife
User avatar
Joined: 21 Jul 2015
Last visit: 05 Nov 2025
Posts: 185
Own Kudos:
69
Given Kudos: 489
GMAT 1: 730 Q50 V38
Products:
My Reviews
GMAT 1: 730 Q50 V38
Posts: 185
Kudos: 69
In the figure, each side of square ABCD has length 1, the length of li 13 May 2017, 12:44
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
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.
Show more

Bunuel

Hi Bunuel, Can you please explain what property of isosceles triangle is this? I mean if the triangle BED is isosceles and EO is angle bisector, why is it that EO is height of this triangle? I am trying to understand why EO meets at point O, making a 90 degree angle, and not at any other point on the diagonal of square?

Thanks for your help!
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 20 Sep 2026
Posts: 113,703
Own Kudos:
842,586
 [2]
Given Kudos: 111,704
Products:
GMAT Club Tests
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 113,703
Kudos: 842,586
 [2]
In the figure, each side of square ABCD has length 1, the length of li 13 May 2017, 13:13
1
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
yt770
Bunuel
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.

Bunuel

Hi Bunuel, Can you please explain what property of isosceles triangle is this? I mean if the triangle BED is isosceles and EO is angle bisector, why is it that EO is height of this triangle? I am trying to understand why EO meets at point O, making a 90 degree angle, and not at any other point on the diagonal of square?

Thanks for your help!
Show more

In an isosceles triangle angle bisector from top vertex to the base coincides with altitude (height) and median.
User avatar
roastedchips
User avatar
Joined: 18 Apr 2013
Last visit: 11 Jun 2020
Posts: 25
Own Kudos:
182
Given Kudos: 3
Posts: 25
Kudos: 182
In the figure, each side of square ABCD has length 1, the length of li 09 Sep 2017, 05:34
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hi Brunuel, how do you get the length of CO to be \frac{\sqrt{2}}{2}?

I thought BC is the hypotenuse. Thus, by pythagoras thm, CO = square root of [1^2 - (sq root 2/2)^2]?

Please help!


Bunuel

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

Answer: B.

Show Spoiler
Attachment:
Square ABCD.JPG
Show more
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 20 Sep 2026
Posts: 113,703
Own Kudos:
842,586
Given Kudos: 111,704
Products:
GMAT Club Tests
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 113,703
Kudos: 842,586
In the figure, each side of square ABCD has length 1, the length of li 09 Sep 2017, 05:38
Kudos
Add Kudos
Bookmarks
Bookmark this Post
roastedchips
Hi Brunuel, how do you get the length of CO to be \frac{\sqrt{2}}{2}?

I thought BC is the hypotenuse. Thus, by pythagoras thm, CO = square root of [1^2 - (sq root 2/2)^2]?

Please help!


Bunuel

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

Answer: B.

Show Spoiler
Attachment:
Square ABCD.JPG
Show more

CO is half the length of the diagonal of the square, which is \(\sqrt{1^2+1^2}=\sqrt{2}\)
 1   2   3   
Moderator:
Bunuel
Math Expert
113703 posts