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

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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?

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

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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

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 [ 12.11 KiB | Viewed 49921 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 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 \(BCE=BOE-BOC=\frac{\sqrt{2}+1}{4}-\frac{1}{4}=\frac{\sqrt{2}}{4}\)

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

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10 Feb 2012, 09:53

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

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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
_________________

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

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

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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 length of li [#permalink]

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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.

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

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

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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..

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.
_________________

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

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26 Oct 2014, 21:41

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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.

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

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05 Aug 2015, 00:35

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

Answer: B.

hi bunuel,

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

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