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Here is a quick way to solve. From the question, we can know that the area of BCE must be less than 1/2, so it leaves us with only two choices A or B.

B looks more likely to the answer because the it is more aligned with given numbers of the square side length.
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If someone is still finding it hard to visualize, the following diagram might help!



Hope this helps!

Ambarish

Attachment:
sq.jpg
sq.jpg [ 335.44 KiB | Viewed 2402 times ]
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singh_amit19

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:
squareabcd.jpg

We know this feels like 3rd image however we need to flatten which is achieved by connecting EC and CO
i would rather solve this through a simple formulae and plug in and check for the value

We know area of a triangle with side and angle =1/2 * r^2 *sin(theta)
r=1 and we know CB and EC are equal given
Therefore let us assume the triangle to be 45 - 45 - 90
Which gives 1/2 * r^2 *1/(rt2 )
=>rt2/ 4

Therefore IMO B
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Wow I really struggled just to visualise this image. As someone who uses 3d models in Chemistry all the time, this looked like a 3d square based pyramid sticking out of the page to me. I took the complete wrong approach because of it leading me to an incorrect answer as it just led me to an area of 1/2. I checked my answer later, still couldn't see the shape I am meant to see. Checked the explanations here, I couldn't follow them because I was still seeing a 3d image. Only after focusing for 5 mins telling myself it's a 2d was I able to finally see a different shape. Kind of an optical illusion playing tricks on me here. :dazed
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Triangle BCE is obtuse.

Area = 1/2 * base * height

Extend CE and draw perpendicular to it to get height of obtuse triangle.

Height = \(\sqrt{2}/2\)

Area = 1/2 * 1 * \(\sqrt{2}/2\) = \(\sqrt{2}\)/4
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Hi brunel, how did you calcaulate this part? Could you explain more please? Thanks
1/2∗√2/2∗2+√2/2=√2+1/4
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Kimberly77
Hi brunel, how did you calcaulate this part? Could you explain more please? Thanks
1/2∗√2/2∗2+√2/2=√2+1/4


Please check: Writing Mathematical Formulas on the Forum. Thank you!
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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.

Thank you Bunuel for providing this clear explanation!

I have practiced this question twice, and even though every time I strongly felt that extension of the line segment CE could coincide with one diagonal of the square, I could not really find a proof. I was aware that the triangles BCE and DCE are congruent, but I completely ignored that the bigger triangle BDE is an isosceles triangle, in which whatever bisects the vertex could be the height and bisect the base, and vice versa. Because the line segment CE bisects the angle E, the extension of CE, OE, could be the height of the isosceles triangle and the height bisects the base BD. Hence, we get that the length of the segment BO is half the length of BD and thus we will get the area.

By the way, since the square's diagonal AC is perpendicular to another diagonal BD and bisects BD, we can know that the segment OC is doing the same thing as does the diagonal AC. But we do not necessarily need to know this to solve this problem.

Thank you!
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Tough one! Here's an alternate method where you can calculate the area of the triangle directly using the 45/45/90 ratio to calculate the triangle's height.

We know that CE is an extension of the square’s diagonal (and hence 45 degrees) because of the fact that the bigger triangle is isosceles, so the angles that open up to the sides must also be equal (135 degrees each). 135 - 90 = 45



If you're still confused by this method, then just estimate and take a guess! All Quant questions on the GMAT are drawn to scale, and upon close inspection, triangle BCE looks to be a little less than 1/2 the area of square ABCD—which leaves only choices A and B.
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Here, BCE=135°, BC=1, CE=1

SO, AREA OF BCE TRIANGLE IS,
= 1/2×sinBCE×BC×CE
=1/2×sin135°×1×1
=√2/4
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