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AB = DE, BC = CD, BE is parallel to CD, and BC is par

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AB = DE, BC = CD, BE is parallel to CD, and BC is par [#permalink] New post 02 Aug 2017, 14:49
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AB = DE, BC = CD, BE is parallel to CD, and BC is parallel to DF.

Quantity A
Quantity B
The area of triangle ABE
The area of quadrilateral BCDF


A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
[Reveal] Spoiler: OA

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Re: AB = DE, BC = CD, BE is parallel to CD, and BC is par [#permalink] New post 26 Sep 2017, 22:20
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pclawong wrote:
explain please


This is a tough ques, let me try

consider AB=DE =x

now join Band D

in the quadrilateral ABDE we have angle A =90 degree

angle E = 90 degree

and AB=DE = x

So quadrilateral ABDE is a square.

and Area of square ABDE = x^2

Now we need the area of triangle ABE = Area of square ABDE - area of triangle BDE - area of triangle DEF

Area of triangle BDF = 1/2 * base * altitude =\frac{1}{2} * x *\frac{1}{2}*x(since side BD =x and 1/2 * total distance = altitude of triangle BDF, as BDEA is square)

and similarly Area of Triangle DEF =\frac{1}{2} *x * \frac{1}{2}*x

Therefore Area of Triangle ABE = x^2 - (\frac{1}{4}*x^2 + \frac{1}{4}*x^2)

=\frac{1}{2} *x^2

Now in quadrilaterl BCDF we have
BC=CD , BC parallel to FD and BE parallel to CD and angle F =90degree (since the diagonal cuts at 90 degree )

Therefore we can consider BCDF is a square

Area of square BCDF=\frac{diagonal ^2}{2}

= \frac{1}{2} *x^2

Hence option C.

All queries are welcome!!!
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Re: AB = DE, BC = CD, BE is parallel to CD, and BC is par [#permalink] New post 26 Sep 2017, 22:31
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pranab01 wrote:
pclawong wrote:
explain please


This is a tough ques, let me try


All queries are welcome!!!


Based on the above solution if the reasoning is correct then we can also have the diagram (attached)

Now if we divide to 10 small triangles (considering all the assumption given and proven in previous response)

Then Area of Triangle = ABE = sum of 4 triangles

and Area of quadrialteral = sum of 4 triangles

This also led to option C.


Kindly let me know if my reasoning are correct.

Thanks in advance. and all queries are welcome
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Re: AB = DE, BC = CD, BE is parallel to CD, and BC is par [#permalink] New post 27 Sep 2017, 05:57
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the fastest approach is by 30:60:90 angle formed by BAE
do it quick by 1:root3:2
you get root3/2 in quantity A as well as in quantity B
so c is the answer
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Re: AB = DE, BC = CD, BE is parallel to CD, and BC is par [#permalink] New post 27 Sep 2017, 09:17
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bim1946 wrote:
the fastest approach is by 30:60:90 angle formed by BAE
do it quick by 1:root3:2
you get root3/2 in quantity A as well as in quantity B
so c is the answer



is it 30-60-90 or 45-90-45?

Can you show in details
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Re: AB = DE, BC = CD, BE is parallel to CD, and BC is par [#permalink] New post 27 Sep 2017, 11:35
Thank you.
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Re: AB = DE, BC = CD, BE is parallel to CD, and BC is par [#permalink] New post 15 Nov 2017, 12:14
Can Someone Pls explain how we know that BA = BD...or just how BDAE is a square.
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Re: AB = DE, BC = CD, BE is parallel to CD, and BC is par [#permalink] New post 06 Dec 2017, 15:14
pranab01 wrote:
pclawong wrote:
explain please


This is a tough ques, let me try

consider AB=DE =x

now join Band D

in the quadrilateral ABDE we have angle A =90 degree

angle E = 90 degree

and AB=DE = x

So quadrilateral ABDE is a square.

and Area of square ABDE = x^2

Now we need the area of triangle ABE = Area of square ABDE - area of triangle BDE - area of triangle DEF

Area of triangle BDF = 1/2 * base * altitude =\frac{1}{2} * x *\frac{1}{2}*x(since side BD =x and 1/2 * total distance = altitude of triangle BDF, as BDEA is square)

and similarly Area of Triangle DEF =\frac{1}{2} *x * \frac{1}{2}*x

Therefore Area of Triangle ABE = x^2 - (\frac{1}{4}*x^2 + \frac{1}{4}*x^2)

=\frac{1}{2} *x^2

Now in quadrilaterl BCDF we have
BC=CD , BC parallel to FD and BE parallel to CD and angle F =90degree (since the diagonal cuts at 90 degree )

Therefore we can consider BCDF is a square

Area of square BCDF=\frac{diagonal ^2}{2}

= \frac{1}{2} *x^2

Hence option C.

All queries are welcome!!!


and AB=DE = x

So quadrilateral ABDE is a square.

To be a square AB = AE has to be given. Not AB=DE.
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Re: AB = DE, BC = CD, BE is parallel to CD, and BC is par [#permalink] New post 06 Dec 2017, 22:37
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LethalMonkey wrote:

and AB=DE = x

So quadrilateral ABDE is a square.

To be a square AB = AE has to be given. Not AB=DE.


Plz see the attached diagram,


In the quadrilateral ABDE,

we have AB = DE and angle A = angle E = 90 degree and the diagonal bisect each other at equal length . SO angle b = angle d = 90

Now the triangle ABF and triangle BDF are equal, since they both have the common same height

i.e AB = BD.
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Re: AB = DE, BC = CD, BE is parallel to CD, and BC is par [#permalink] New post 07 Dec 2017, 19:18
bim1946 wrote:
the fastest approach is by 30:60:90 angle formed by BAE
do it quick by 1:root3:2
you get root3/2 in quantity A as well as in quantity B
so c is the answer


Dear friend,
how do you get 30:60:90, please explain, really need your help.
thanks
Re: AB = DE, BC = CD, BE is parallel to CD, and BC is par   [#permalink] 07 Dec 2017, 19:18
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