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

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

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14 Dec 2015, 17:03
Bunuel Can someone please explain my question above?

" I get that ∠BCE=∠DCE, but how do you know that CE meets at the diagonal's midpoint which is the crux of the rest of the problem?"

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

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21 Dec 2015, 22:59
I could nt think of point ), So i tried to find area of triangle BED by using cosine formula.

Then, I subtracted area of triangle BCD.

But it is too complicated.

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

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20 Jun 2016, 17:41
I should have scrolled down ... I spent 20 minutes trying to understand how one is supposed to extend the line to this imaginary origin. I actually considered making this shape out of paper...

Lesson learned, it is not some sort of 3-D prism. One of those face palm moments.

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

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29 Jul 2016, 23:07
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singh_amit19 wrote:
Attachment:
squareabcd.jpg
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

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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FINAL GOODBYE :- 17th SEPTEMBER 2016. .. 16 March 2017 - I am back but for all purposes please consider me semi-retired.

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

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28 Aug 2016, 06:23
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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 ...

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

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11 Oct 2016, 10:40
HerrGrau wrote:
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.

Thank you for your explanation. I am not sure I understand how you manipulated the diagram. How did you put E right into the square?

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

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23 Oct 2016, 02:51
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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.

Attachments

File comment: Geometry Question

Geo Q.PNG [ 288.2 KiB | Viewed 1238 times ]

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

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10 Dec 2016, 08:36
here ir my colorful contribution
Attachments

deletar.png [ 348.74 KiB | Viewed 1078 times ]

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

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13 May 2017, 11:44
Bunuel wrote:
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.

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

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13 May 2017, 12:13
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yt770 wrote:
Bunuel wrote:
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.

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

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13 May 2017, 13:13
Thanks a lot Bunuel
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Re: In the figure, each side of square ABCD has length 1, the length of li [#permalink]

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03 Aug 2017, 00:22
AREA of BCE =$$\frac{1}{2}$$ * BC * height EK
Angle ECK = angle CAD = $$45^o$$ as E is equidistant from B and D.
So if EC = 1, then height EK = $$\frac{1}{\sqrt{2}}$$ as the ratio of sides is $$1:1:\sqrt{2}$$

Area = $$1 * \frac{1}{2} * \frac{1}{\sqrt{2}} = \frac{1}{2*\sqrt{2}}=\sqrt{2} * \frac{1}{4}$$
Attachments

squareabcd.jpg [ 43.52 KiB | Viewed 632 times ]

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

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09 Sep 2017, 04:34
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]?

Bunuel wrote:

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 $$BCE=BOE-BOC=\frac{\sqrt{2}+1}{4}-\frac{1}{4}=\frac{\sqrt{2}}{4}$$

[Reveal] Spoiler:
Attachment:
Square ABCD.JPG

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

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09 Sep 2017, 04:38
roastedchips wrote:
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]?

Bunuel wrote:

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 $$BCE=BOE-BOC=\frac{\sqrt{2}+1}{4}-\frac{1}{4}=\frac{\sqrt{2}}{4}$$

[Reveal] Spoiler:
Attachment:
Square ABCD.JPG

CO is half the length of the diagonal of the square, which is $$\sqrt{1^2+1^2}=\sqrt{2}$$
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Re: In the figure, each side of square ABCD has length 1, the length of li [#permalink]

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11 Sep 2017, 00:53
Area of triangle BCE is: 1/2*CE*height
we know CE=1,
For height, we extend CE inwards upto the diagonal of square. It lands midway on diagonal BD. This means height of triangle BCE=1/2* Diagonal BD= 1/2*Sqrt2
Area BCE: 1/2*1*(1/2)*sqrt2=sqrt2/4

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Re: In the figure, each side of square ABCD has length 1, the length of li   [#permalink] 11 Sep 2017, 00:53

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