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.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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

Show Tags

25 Sep 2008, 03:05

6

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

(N/A)

Question Stats:

58% (02:47) correct
42% (01:44) wrong based on 107 sessions

HideShow timer Statistics

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?

Hello, I have attached a pic as well to clraify the problem.

The problem:

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?

Triangle BCE is congruent to triangle CDE. Also triangle BED is an isosceles triangle, since BE=DE. Now draw a diagonal BD to form triangle BCD. You realize angle CBE IS 45 and so is angle CEB. Therefore, triangle BEC is a rt-angled triangle with base 1 and ht. 1. Hence its area is 1/2 *1*1= 1/2.

Triangle BCE is congruent to triangle CDE. Also triangle BED is an isosceles triangle, since BE=DE. Now draw a diagonal BD to form triangle BCD. You realize angle CBE IS 45 and so is angle CEB. Therefore, triangle BEC is a rt-angled triangle with base 1 and ht. 1. Hence its area is 1/2 *1*1= 1/2.

how did you get angle CBE = 45 ? Can you explain
_________________

Your attitude determines your altitude Smiling wins more friends than frowning

Triangle BCE is congruent to triangle CDE. Also triangle BED is an isosceles triangle, since BE=DE. Now draw a diagonal BD to form triangle BCD. You realize angle CBE IS 45 and so is angle CEB. Therefore, triangle BEC is a rt-angled triangle with base 1 and ht. 1. Hence its area is 1/2 *1*1= 1/2.

Angle CBE is definitely not 45 degrees. We know:

-if you add the three angles around C, you must get 360; -one of these angles is 90, so the two large angles in triangles BCE and CED must add to 270; -triangles BCE and CED are congruent (identical sides), so angle BCE and angle DCE must both be 135.

So in BCE, the other two angles must add to 180-135 = 45, and since BCE is isosceles, angle CBE must be 22.5 degrees (not 45 degrees).

The diagram is pretty misleading, since it disguises the symmetry in the picture- in an accurate diagram, triangles BCE and DCE should appear to be the same size, and the square should look like a square. Even with the diagram above, though, it should be clear that BCE is not a right triangle. Regardless, I posted four different solutions to this problem on another forum:

None of the answer choices is correct, incidentally- B would be correct if the exponent on 2 in the numerator were 1/2 instead of -2.
_________________

GMAT Tutor in Toronto

If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com

Now make draw your BCE Triangle (make it look like an equilateral).. BC = 1 and and BE = 1 sqrt (2) .. Now let's find the area by finding the height. We know BC = 1 so lets divide the triangle so that we have 1/2 the base (1/2). So your base is 1/2 and your BE hypotneuse = 1 sqrt (2).. looks like we have 30-60-90 triangle making the height (1/2) sqrt (2).

so the Area = (1/2) * (1) * (1/2) *sqrt(2) = 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?

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

No correct answer in answer choices.
_________________

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. .

I am not sure about this point. Is it because both BOC+BDE=ECD+DC0?

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. .

I am not sure about this point. Is it because both BOC+BDE=ECD+DC0?

Drawing might help:

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

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

Show Tags

08 Sep 2014, 04:57

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

After days of waiting, sharing the tension with other applicants in forums, coming up with different theories about invites patterns, and, overall, refreshing my inbox every five minutes to...

I was totally freaking out. Apparently, most of the HBS invites were already sent and I didn’t get one. However, there are still some to come out on...

In early 2012, when I was working as a biomedical researcher at the National Institutes of Health , I decided that I wanted to get an MBA and make the...