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

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In the figure, each side of square ABCD has length 1, the [#permalink]  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. (2^-2 )/4
c. 1/2
d. (2^-2)/2
e. 3/4
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set25q31.GIF [ 2.57 KiB | Viewed 10035 times ]

Last edited by Bunuel on 25 Jan 2012, 09:57, edited 2 times in total.
None of the answer choices shown is correct. See correct answer in my post below.
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Height of triangle BCE - 0.5
Base - 1

So area of BCE - 1/4
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If CE = 1 and BC = 1

so why is (1*1)/2 = 1/2 not the answer ?

this is way too easy

am I missing something ?

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Re: PS-SET25 Q31 [#permalink]  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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KillerSquirrel wrote:
If CE = 1 and BC = 1

so why is (1*1)/2 = 1/2 not the answer ?

this is way too easy

am I missing something ?

Area of a triangle = (1/2)XbaseXaltitude

So, how u getting (1*1)/2 = 1/2? R u assuming BCE a right angled triangle? if YES y n how???
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KillerSquirrel wrote:
ronneyc wrote:
KillerSquirrel wrote:
singh_amit19 wrote:
KillerSquirrel wrote:
If CE = 1 and BC = 1

so why is (1*1)/2 = 1/2 not the answer ?

this is way too easy

am I missing something ?

Area of a triangle = (1/2)XbaseXaltitude

So, how u getting (1*1)/2 = 1/2? R u assuming BCE a right angled triangle? if YES y n how???

height = CE = 1
base = BC = 1

1*1*1/2

whats the OA ?

I dont see why height = CE...

What is the OA please ?

the angle BCE should be 90° to use this formula which is not the case here
=> CE is not a height
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singh_amit19 wrote:
OA is D....any takers????

if we calculate ans D which is (2^-2)/2 we get (1/4) * (1/2) = 1/8

Can you please provide the method ? I dont still dont see it ...
thx
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Re: In the figure (attchd), each side of square ABCD has length [#permalink]  25 Jan 2012, 08:15
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Re: In the figure (attchd), each side of square ABCD has length [#permalink]  25 Jan 2012, 09:53
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Expert's post
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Baten80 wrote:

Responding to a pm.
Attachment:

Square ABCD.JPG [ 12.11 KiB | Viewed 10768 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?

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.
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Re: In the figure, each side of square ABCD has length 1, the [#permalink]  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 [#permalink]  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 [#permalink]  08 Feb 2012, 01:18
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The answer choices given are wrong.
The correct ACs are:

A. 1/3
B. (sq.root 2)/4
C. 1/2
D. (sq.root 2)/2
E. 3/4

OA: B
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Re: In the figure, each side of square ABCD has length 1, the [#permalink]  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 [#permalink]  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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+1 if you like my explanation .Thanks

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Re: In the figure (attchd), each side of square ABCD has length [#permalink]  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 (attchd), each side of square ABCD has length [#permalink]  23 Sep 2013, 00:22
Expert's post
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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.
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Re: In the figure (attchd), each side of square ABCD has length [#permalink]  28 Sep 2013, 04:09
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.
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Re: In the figure (attchd), each side of square ABCD has length [#permalink]  11 Nov 2013, 11:46
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!
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Re: In the figure, each side of square ABCD has length 1, the [#permalink]  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 [#permalink]  13 Jan 2014, 16:18
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oh man. I thought it was a 3d figure. Its only 2d.
Re: In the figure, each side of square ABCD has length 1, the   [#permalink] 13 Jan 2014, 16:18
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