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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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New post 20 Nov 2018, 09:09
hi guys,

I have prepared a detailed solution of this video.

Click here to see the explanation
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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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New post 17 Jan 2019, 03:44
What is the level of this question?
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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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New post 17 Jan 2019, 03:49
SushiVoyage wrote:
What is the level of this question?



You can check difficulty level of a question in the tags above original post. For this one is 700. The difficulty level of a question is calculated automatically based on the timer stats from the users which attempted the question.
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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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New post 19 Feb 2019, 01:01
Bunuel wrote:
Image
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


Image

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.

Attachment:
Square ABCD.JPG







Hi Bunuel,

Please, can you explain this in lame terms as to how the area of BOC is one forth and the diagonal part?


Is there any other way to solve it?
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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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New post 19 Feb 2019, 01:14
Shef08 wrote:
Bunuel wrote:
Image
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


Image

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.

Attachment:
The attachment Square ABCD.JPG is no longer available







Hi Bunuel,

Please, can you explain this in lame terms as to how the area of BOC is one forth and the diagonal part?


Is there any other way to solve it?


Image
Diagonals of a square are perpendicular bisectors of each other. That is, they cut each other in half and at 90 degrees. As you can see above diagonals of a square cut the square into four equal parts.

For other solutions please check the discussion. There are many different solutions there.

Attachment:
Untitled.png
Untitled.png [ 27.17 KiB | Viewed 153 times ]

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

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New post 16 Mar 2019, 06:17
singh_amit19 wrote:
Image
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:
The attachment squareabcd.jpg is no longer available



Hi Bunuel,

I tried solving this problem in the following way.

\(\triangle\) BCD and \(\triangle\) CDE are similar triangles (Side-Side-Side)

So \(\angle\) DCE =\(\angle\) BCE

\(\angle\)DCE+\(\angle\)BCE=360-90= 270 degrees

So \(\angle\) DCE=\(\angle\)BCE =135 degree

Area of any triangle = 1/2 ab sin (angle between the sides)
= 1/2 *1 * 1 Sin(90+45)
=\(\frac{1}{2\sqrt[]{2}}\)
=\(\frac{\sqrt{2}}{4}\)

This way of solving would eliminate drawing extra sides in the figure. i hope you find this helpful.
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In the figure, each side of square ABCD has length 1, the length of li   [#permalink] 16 Mar 2019, 06:17

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