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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] New post   20 Nov 2018, 09:09
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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] 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] New post   17 Jan 2019, 03:49
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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] New post   19 Feb 2019, 01:01
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 \(BCE=BOE-BOC=\frac{\sqrt{2}+1}{4}-\frac{1}{4}=\frac{\sqrt{2}}{4}\)

Answer: B.

Show Spoiler
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] New post   19 Feb 2019, 01:14
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Shef08 wrote:
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 \(BCE=BOE-BOC=\frac{\sqrt{2}+1}{4}-\frac{1}{4}=\frac{\sqrt{2}}{4}\)


Answer: B.

Show Spoiler
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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?



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.

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In the figure, each side of square ABCD has length 1, the length of li [#permalink] New post   16 Mar 2019, 06:17
singh_amit19 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


Show Spoiler
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] New post   23 May 2019, 17:39
~45 SECOND "SOLUTION"
I estimated the height of triangle BCE by using my pen after measuring the side square ABCE. I estimated BCE height to be 3/4 as ABCE side length was 1.
So b*h/2 ... (1*3/4) / 2 = 3/8 = .375

Looking at answers starting with C) which is .5, we need something smaller.
Trying B) √2/4 = 1.4/4 = .7/2 = .35
Trying A) 1/3 = .33
Between these two, B) is closer to .375 so I picked it.

It's not foolproof but it's the best method for Geometry when you're running low on time and not seeing a quick solution, imo.
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Re: In the figure, each side of square ABCD has length 1, the length of li [#permalink] New post   22 Sep 2019, 19:44
Do we seriously always assume 2D figures????

Holy cow... i've only realised this now... this would have made my life so much easier!
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Re: In the figure, each side of square ABCD has length 1, the length of li [#permalink] New post   09 Nov 2019, 22:25
Hey, I dont understand this part of Bunnel's explanation.

I too tried to find the area of the two triangles and subtract them. But I got lost in the calculations to find the actual height.

12∗BO∗EO12∗BO∗EO . BO=2√2
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In the figure, each side of square ABCD has length 1, the length of li [#permalink] New post   Updated on: 04 Oct 2020, 09:31
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Since ⧠ABCD is a square, internal ∠BCD is 90
Thus, external ∠BCD is 270 (… … 360-90=270)

Now segment CE is an angle bisector for external ∠BCD (… … BE=DE)
Thus, ∠BCE = 135 (… … 270/2 = 135)

Now, Extending segment BC and EF to make a right angle triangle
so ∠CFE = 90

Because, ∠BCE = 135, ∠ECF = 45 (… … 180 – 135 = 45)
This make △CEF an isosceles right angle triangle.

So, CF:FE:EC = 1:1:√2
Now, we know, that CE=1
Thus, CF=FE= 1/√2

With this, we have height of the △BCF = 1/√2
And given is the base of the △BCF = 1

So Area of △BCE = ½ * (Base) * (Height)
= ½ * 1 * 1/√2
=1/2√2
=√2/4

Ans:B
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Originally posted by NiravUmaretiya on 01 Mar 2020, 06:46.
Last edited by NiravUmaretiya on 04 Oct 2020, 09:31, edited 1 time in total.
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In the figure, each side of square ABCD has length 1, the length of li [#permalink] New post   15 Jun 2020, 11:18
Hi,

I tried to find the length of CO considering COD as a right triangle.
so \(CD^2\) = \(CO^2\) + \(OD^2\)

Because CO = OD, then

\(CD^2\) = \(CO^2\) + \(CO^2\)
\(CD^2\) = 2(\(CO^2\))
\(1^2\) = 2(\(CO^2\))
\(\frac{1}{2}\) = \(CO^2\)
\(\sqrt{\frac{1}{2}}\) = CO

But this does not match with the actual CO length of \(\frac{\sqrt{2}}{2}\)

Can someone please explain what silly mistake I am making?
Thanks.
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Re: In the figure, each side of square ABCD has length 1, the length of li [#permalink] New post   21 Jun 2020, 08:31
ALthough i answered correctly by taking a bold guess; assuming the BED is an equilateral triangle

How to answer the question without knowing the length of AE?

We know CE =1 and BE =DE; However, AE is not given.

AE can be longer or shorter than 1cm and <EBD or <EDB is not necessary 60%
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Re: In the figure, each side of square ABCD has length 1, the length of li [#permalink] New post   13 Jul 2020, 02:03
singh_amit19 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


Please find attached the explanation for the question.



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The attachment squareabcd.jpg is no longer available

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Re: In the figure, each side of square ABCD has length 1, the length of li [#permalink] New post   16 Jul 2020, 23:11
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 \(BCE=BOE-BOC=\frac{\sqrt{2}+1}{4}-\frac{1}{4}=\frac{\sqrt{2}}{4}\)

Answer: B.

Show Spoiler
Attachment:
Square ABCD.JPG


Hey Bunuel - how is CE=1?
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Re: In the figure, each side of square ABCD has length 1, the length of li [#permalink] New post   17 Jul 2020, 00:13
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KevinMaloneSr wrote:
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 \(BCE=BOE-BOC=\frac{\sqrt{2}+1}{4}-\frac{1}{4}=\frac{\sqrt{2}}{4}\)

Answer: B.

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Attachment:
Square ABCD.JPG


Hey Bunuel - how is CE=1?


It's given in the stem. Check the highlighted part above.
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Re: In the figure, each side of square ABCD has length 1, the length of li [#permalink] New post   17 Jul 2020, 00:15
Bunuel wrote:
KevinMaloneSr wrote:
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 \(BCE=BOE-BOC=\frac{\sqrt{2}+1}{4}-\frac{1}{4}=\frac{\sqrt{2}}{4}\)

Answer: B.

Show Spoiler
Attachment:
Square ABCD.JPG


Hey Bunuel - how is CE=1?


It's given in the stem. Check the highlighted part above.


Wow- I am sorry for wasting your time... Bunuel

Posted from my mobile device
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Re: In the figure, each side of square ABCD has length 1, the length of li [#permalink] New post   03 Jan 2021, 04:42
Dear Bunnel,

Thank you for your detailed explanation. Cant we assume that it is a cube, i was thinking in that direction while solving the question.

Happy to hear your thoughts.

Regards,
Gaurav
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Re: In the figure, each side of square ABCD has length 1, the length of li [#permalink] New post   03 Jan 2021, 06:03
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Gaurav25993 wrote:
Dear Bunnel,

Thank you for your detailed explanation. Cant we assume that it is a cube, i was thinking in that direction while solving the question.

Happy to hear your thoughts.

Regards,
Gaurav


It' a square: In the figure, each side of square ABCD has length 1...
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In the figure, each side of square ABCD has length 1, the length of li [#permalink] New post   04 Apr 2021, 16:48
Bunuel wrote:

Considering your figure with O as the middle of the square.

I tried to get the length of BO by using BC as hypotenuse. Why is this not possible?

BO = CO -->

BO^2 + BO^2 = 1^2

2*BO^2 = 1^2

BO = (1/2)^0,5 = 1/(2^0,5)

Thanks!
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Re: In the figure, each side of square ABCD has length 1, the length of li [#permalink] New post   04 Apr 2021, 22:09
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Bambi2021 wrote:
Bunuel wrote:

Considering your figure with O as the middle of the square.

I tried to get the length of BO by using BC as hypotenuse. Why is this not possible?

BO = CO -->

BO^2 + BO^2 = 1^2

2*BO^2 = 1^2

BO = (1/2)^0,5 = 1/(2^0,5)

Thanks!


That's correct: \(BO=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}\) (we are getting \(\frac{\sqrt{2}}{2}\) by multiplying \(\frac{1}{\sqrt{2}}\) by \(\frac{\sqrt{2}}{\sqrt{2}}\)).
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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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