Author 
Message 
TAGS:


Moderator
Joined: 18 Apr 2015
Posts: 2501
Followers: 37
Kudos [?]:
361
[1]
, given: 1494

AB = DE, BC = CD, BE is parallel to CD, and BC is par [#permalink]
02 Aug 2017, 14:49
1
This post received KUDOS
Question Stats:
0% (00:00) correct
0% (00:00) wrong based on 0 sessions
Attachment:
box.jpg [ 18.52 KiB  Viewed 954 times ]
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.
_________________
Get the 17 FREE GREPrepclub Tests




Senior Manager
Joined: 20 Apr 2016
Posts: 264
Followers: 1
Kudos [?]:
162
[1]
, given: 54

Re: AB = DE, BC = CD, BE is parallel to CD, and BC is par [#permalink]
26 Sep 2017, 22:20
1
This post received KUDOS
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}*xTherefore Area of Triangle ABE = x^2  ( \frac{1}{4}*x^2 + \frac{1}{4}*x^2) = \frac{1}{2} *x^2Now 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^2Hence option C. All queries are welcome!!!
Attachments
box (1).jpg [ 13.66 KiB  Viewed 570 times ]
_________________
If you found this post useful, please let me know by pressing the Kudos Button



Senior Manager
Joined: 20 Apr 2016
Posts: 264
Followers: 1
Kudos [?]:
162
[1]
, given: 54

Re: AB = DE, BC = CD, BE is parallel to CD, and BC is par [#permalink]
26 Sep 2017, 22:31
1
This post received KUDOS
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
Attachments
box .jpg [ 16.57 KiB  Viewed 567 times ]
_________________
If you found this post useful, please let me know by pressing the Kudos Button



Intern
Joined: 26 Sep 2017
Posts: 31
Followers: 0
Kudos [?]:
2
[1]
, given: 19

Re: AB = DE, BC = CD, BE is parallel to CD, and BC is par [#permalink]
27 Sep 2017, 05:57
1
This post received KUDOS
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



Senior Manager
Joined: 20 Apr 2016
Posts: 264
Followers: 1
Kudos [?]:
162
[1]
, given: 54

Re: AB = DE, BC = CD, BE is parallel to CD, and BC is par [#permalink]
27 Sep 2017, 09:17
1
This post received KUDOS
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 306090 or 459045? Can you show in details
_________________
If you found this post useful, please let me know by pressing the Kudos Button



Intern
Joined: 28 Feb 2017
Posts: 7
Followers: 0
Kudos [?]:
0
[0], given: 3

Re: AB = DE, BC = CD, BE is parallel to CD, and BC is par [#permalink]
27 Sep 2017, 11:35
Thank you.



Intern
Joined: 21 Aug 2017
Posts: 2
Followers: 0
Kudos [?]:
0
[0], given: 0

Re: AB = DE, BC = CD, BE is parallel to CD, and BC is par [#permalink]
15 Nov 2017, 12:14
Can Someone Pls explain how we know that BA = BD...or just how BDAE is a square.



Intern
Joined: 14 Oct 2017
Posts: 5
Followers: 0
Kudos [?]:
2
[0], given: 0

Re: AB = DE, BC = CD, BE is parallel to CD, and BC is par [#permalink]
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}*xTherefore Area of Triangle ABE = x^2  ( \frac{1}{4}*x^2 + \frac{1}{4}*x^2) = \frac{1}{2} *x^2Now 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^2Hence 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.



Senior Manager
Joined: 20 Apr 2016
Posts: 264
Followers: 1
Kudos [?]:
162
[1]
, given: 54

Re: AB = DE, BC = CD, BE is parallel to CD, and BC is par [#permalink]
06 Dec 2017, 22:37
1
This post received KUDOS
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.
Attachments
box %281%29.jpg [ 14.57 KiB  Viewed 67 times ]
_________________
If you found this post useful, please let me know by pressing the Kudos Button



Intern
Joined: 03 Dec 2017
Posts: 16
Followers: 0
Kudos [?]:
5
[0], given: 5

Re: AB = DE, BC = CD, BE is parallel to CD, and BC is par [#permalink]
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





