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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] New post 25 Sep 2008, 02:05
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A
B
C
D
E

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30% (05:57) correct 70% (01:41) 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?

OPEN DISCUSSION OF THIS QUESTION IS HERE: in-the-figure-each-side-of-square-abcd-has-length-1-the-54152.html
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Last edited by Bunuel on 16 Jun 2013, 22:11, edited 1 time in total.
Edited the question.
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Re: Area of Triangle [#permalink] New post 25 Sep 2008, 02:58
jugolo1 wrote:
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?

I hope someone could help in this.

thanks


Answer is C i.e. 1/2

Please confirm.
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Re: Area of Triangle [#permalink] New post 25 Sep 2008, 05:51
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.
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Re: Area of Triangle [#permalink] New post 25 Sep 2008, 06:00
KASSALMD wrote:
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
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Re: Area of Triangle [#permalink] New post 25 Sep 2008, 06:20
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KASSALMD wrote:
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:

http://www.beatthegmat.com/area-of-tria ... 15073.html

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.
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Re: Area of Triangle [#permalink] New post 25 Sep 2008, 09:39
I am getting an answer of (1/4)*sqrt2. But, I do not see that in the answer choice.

My apprpach:
I my diagram (sorry for poor drawing), since BC = CE = CD, hence, angle CBE = angle BEC = CED = angle EDC.

Now, in triangle BCD, angle CDB = angle CBD = 45.

Hence, EF will be perpendicular bisector of BD.

Now, area of triangle BCE = 1/2(area of triangle BED - area of triangle BCD).

=1/2(1/2*(1+1/sqrt2)*sqrt2 - 1/2*1/sqrt2*sqrt2)
=1/2(1/2*sqrt2) = (1/4)*sqrt2
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Re: Area of Triangle [#permalink] New post 30 Sep 2010, 21:40
This is how I did it and got the Answer: B.

BC = 1 and CE = 1 so BE = 1 sqrt (2)

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
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Re: Area of Triangle [#permalink] New post 30 Sep 2010, 22:23
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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?

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.
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Re: Area of Triangle [#permalink] New post 16 Jun 2013, 15:57
Bunuel wrote:
[b]

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.
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I am not sure about this point. Is it because both BOC+BDE=ECD+DC0?
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Re: Area of Triangle [#permalink] New post 16 Jun 2013, 22:16
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gmatfighter12 wrote:
Bunuel wrote:


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:
Image

[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}

None of the answer choices shown is correct.

Open discussion of this question is here: in-the-figure-each-side-of-square-abcd-has-length-1-the-54152.html In case of any questions pleas post in that thread.
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Re: Area of Triangle   [#permalink] 16 Jun 2013, 22:16
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