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In the figure, each side of square ABCD has length 1, the [#permalink]
19 Oct 2007, 03:37

1

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A

B

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E

Difficulty:

5% (low)

Question Stats:

61% (02:27) correct
38% (02:09) wrong based on 13 sessions

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

Attachments

set25q31.GIF [ 2.57 KiB | Viewed 5814 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.

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

Re: In the figure (attchd), each side of square ABCD has length [#permalink]
25 Jan 2012, 09:53

12

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Expert's post

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 [ 12.11 KiB | Viewed 6221 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. _________________

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 .
_________________

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
_________________

Re: In the figure (attchd), each side of square ABCD has length [#permalink]
23 Sep 2013, 00:22

Expert's post

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.
_________________

Re: In the figure (attchd), each side of square ABCD has length [#permalink]
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.

Re: In the figure (attchd), each side of square ABCD has length [#permalink]
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!

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.