In the figure, each side of square ABCD has length 1, the length of li

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




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 of BCE) = (The area of BOE) - (The area of 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.


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My advise is not to use any fancy triangle properties or weird geometry.
This is a 2D geometry question and the shape of the object is a KITE.

This question can be solved under 20 seconds, The trick is use obscure properties of shapes to your advantage ; in this case the area of a KITE.
If you see the question properly it is asking us to find the area of half KITE and then subtract the area of a Half square
Mathematically
Area = 1/2*area of KITE-1/2* area of square
Area = 1/2 (area of KITE-area of square)
Area of the square is super easy .. its 1; half of it will be = 1/2

For finding the area of the KITE you will need its two diagonal. Smaller one is \(\sqrt{2}\). Bigger one is \(1+\sqrt{2}\)
FORMULA FOR KITE AREA= \(\frac{1}{2}\)*(diagonal1 * diagonal 2)

Solve it and you will get \(\frac{\sqrt{2}}{4}\)

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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

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 .
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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.
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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.

Area = 1/2 * 1 * 1/(sqrt)2 = 1/2*(sqrt)2 = sqrt(2)/4.

HTH

Hi All,

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.


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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..

whats wrong..plz xplain.
thanks

You are overthinking it. On the GMAT, all figures lie in a plane unless otherwise indicated.
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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.

Thanks in advance!

This figure has a symmetry through the segment \(ACE\) (since \(BE=DE\)). Thus \(CA\) is the extension of the side \(EC\) of the \(\triangle BCE\). Since in squares the diagonals are perpendicular, the height of the \(\triangle BCE\), relative to the side \(EC\), is half of the diagonal of the square \(ABCD\) (draw the segment \(BO\), where \(O\) is the intersection of the diagonals). Therefore the area of the \(\triangle BCE=\frac{\frac{\sqrt{2}}{2}*1}{2}=\frac{\sqrt{2}}{4}\). Letter B.

Just in case something does not click during exam , you can use some intelligent guess work the base of BCE given as one
so area = 1/2 * 1 * height .

Now CE is 1 since in PS problems the figure is drawn to scale height is surely less than CE is not 90 degree so the area is less than 1/2 .
This leaves you with 2 possible options sqrt2/ 4 and 1/3 , since the sqrt2 is the diagonal of ABCA it is more probable to figure in the answer.

guess as B ...

This Q is can be solved if we extend the line from E to touch one of the diagonals of the square.

Diagonals of a square intersect at right angles. This implies angle BPD will be 90.

here ir my colorful contribution

Bunuel

Hi Bunuel, Can you please explain what property of isosceles triangle is this? I mean if the triangle BED is isosceles and EO is angle bisector, why is it that EO is height of this triangle? I am trying to understand why EO meets at point O, making a 90 degree angle, and not at any other point on the diagonal of square?

Thanks for your help!

In an isosceles triangle angle bisector from top vertex to the base coincides with altitude (height) and median.
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Hi Brunuel, how do you get the length of CO to be \frac{\sqrt{2}}{2}?

I thought BC is the hypotenuse. Thus, by pythagoras thm, CO = square root of [1^2 - (sq root 2/2)^2]?

Please help!

CO is half the length of the diagonal of the square, which is \(\sqrt{1^2+1^2}=\sqrt{2}\)
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