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.

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]
25 Jan 2010, 09:05

1

This post received KUDOS

00:00

A

B

C

D

E

Difficulty:

5% (low)

Question Stats:

100% (16:24) correct
0% (00:00) wrong based on 10 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?

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. _________________

in triangle BCE , draw a perpendicular to base BE which meets BE at Z

Now in Triangle BCZ , BC = 1 Angle EBC = 45/2

=> BZ = cos(45/2) CZ = sin(45/2)

Area of triangle BCE = \frac{1}{2}* BE * CZ = \frac{1}{2}* 2*cos(45/2) * Sin(45/2) =\frac{1}{2}* sin45 using -> sin(2A) = 2 * SinA * cosA =\frac{1}{2\sqrt{2}}

which should be 2^ (-3/2)

I think you have wrongly written your ans.. Please check if D is 2^ (-3/2)
_________________

Basically adding a picture for more clarity... original one is throwing bit off...! OB=OC=(sq. rt 2)/2 rest is simplification... agree with the answers above of sqrt2/4.

Attachments

Triangle89651.jpg [ 8.45 KiB | Viewed 1750 times ]