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

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In the figure, each side of square ABCD has length 1, the length of li [#permalink] New post 19 Oct 2007, 03:37
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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 29 Sep 2014, 23:30, edited 5 times in total.
Renamed the topic, edited the question and added the OA.
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Re: In the figure, each side of square ABCD has length 1, the length of li [#permalink] New post 19 Oct 2007, 05:56
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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
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In the figure, each side of square ABCD has length 1, the length of li [#permalink] New post 25 Jan 2012, 09:53
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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
Square ABCD.JPG [ 12.11 KiB | Viewed 12148 times ]

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}

Answer: B.
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Re: In the figure, each side of square ABCD has length 1, the length of li [#permalink] New post 25 Jan 2012, 10:03
very hard one for me
thanx for explanation
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Re: In the figure, each side of square ABCD has length 1, the length of li [#permalink] New post 26 Jan 2012, 05:34
Bunnel thanks for the explanation +1 :)
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Re: In the figure, each side of square ABCD has length 1, the length of li [#permalink] New post 08 Feb 2012, 05:07
Could not solve it. But thanks for the explanation. One level to hard for me.
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Re: In the figure, each side of square ABCD has length 1, the length of li [#permalink] New post 10 Feb 2012, 09:53
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 .
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Re: In the figure, each side of square ABCD has length 1, the length of li [#permalink] New post 22 Sep 2013, 06:44
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
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Re: In the figure, each side of square ABCD has length 1, the length of li [#permalink] New post 23 Sep 2013, 00:22
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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.
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Re: In the figure, each side of square ABCD has length 1, the length of li [#permalink] New post 28 Sep 2013, 04:09
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.
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Re: In the figure, each side of square ABCD has length 1, the length of li [#permalink] New post 11 Nov 2013, 11:46
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!
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Re: In the figure, each side of square ABCD has length 1, the length of li [#permalink] New post 02 Jan 2014, 16:46
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
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Re: In the figure, each side of square ABCD has length 1, the length of li [#permalink] New post 13 Jan 2014, 16:18
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oh man. I thought it was a 3d figure. Its only 2d.
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Re: In the figure, each side of square ABCD has length 1, the length of li [#permalink] New post 11 May 2014, 08:56
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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.

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Re: In the figure, each side of square ABCD has length 1, the length of li [#permalink] New post 24 Jul 2014, 17:35
Took me 3 minutes, but this is a very doable problem.
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Re: In the figure, each side of square ABCD has length 1, the length of li [#permalink] New post 14 Oct 2014, 05:35
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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
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Re: In the figure, each side of square ABCD has length 1, the length of li [#permalink] New post 14 Oct 2014, 05:42
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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.
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Re: In the figure, each side of square ABCD has length 1, the length of li   [#permalink] 14 Oct 2014, 05:42
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