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Bunuel

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.

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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Bunuel

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.

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?


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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singh_amit19

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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~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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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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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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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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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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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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singh_amit19

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.



Attachment:
The attachment squareabcd.jpg is no longer available

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Bunuel

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.

Attachment:
Square ABCD.JPG

Hey Bunuel - how is CE=1?
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Bunuel

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.

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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Bunuel
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Bunuel

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.

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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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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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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Bunuel
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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Bunuel
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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