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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, 18:41

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.

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

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30 Jul 2016, 00:07

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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 \(\frac{\sqrt{2}}{4}\) _________________

Posting an answer without an explanation is "GOD COMPLEX". The world doesn't need any more gods. Please explain you answers properly. FINAL GOODBYE :- 17th SEPTEMBER 2016. .. 16 March 2017 - I am back but for all purposes please consider me semi-retired.

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

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28 Aug 2016, 07:23

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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.

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

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11 Oct 2016, 11:40

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?

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

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13 May 2017, 12:44

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.

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?

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.

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.
_________________

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

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03 Aug 2017, 01:22

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 \(1:1:\sqrt{2}\)

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

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09 Sep 2017, 05:34

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]?

Please help!

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

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!

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

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

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11 Sep 2017, 01:53

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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