Jul 13 10:00 PM PDT  11:00 PM PDT Take 20% off the plan of your choice, now through midnight on 7/13 Jul 14 08:00 PM PDT  09:00 PM PDT A better GRE prep experience. We believe GRE prep should be real education that prepares you not just for the test, but for the challenges you’ll face as you earn your graduate degree and advance your career. Jul 15 08:00 PM PDT  09:00 PM PDT Regardless of whether you choose to study with Greenlight Test Prep, I believe you'll benefit from my many free resources. Jul 22 07:30 AM PDT  08:30 AM PDT Learn the basics of quadratics to demystify one of the most commonly feared concepts on the GRE and GMAT.
Author 
Message 
TAGS:


Founder
Joined: 18 Apr 2015
Posts: 12093
Followers: 256
Kudos [?]:
3016
[1]
, given: 11281

In the figure above, ABCD is a square with sides equal to [#permalink]
04 Aug 2017, 07:52
1
This post received KUDOS
Question Stats:
50% (03:21) correct
49% (03:03) wrong based on 59 sessions
Attachment:
#GRepracticequestion In the figure above, ABCD is a square with sides equal to 1,.jpg [ 26.23 KiB  Viewed 2742 times ]
In the figure above, ABCD is a square with sides equal to 1, AFC is an arc of a circle centered at B, and AEC is an arc of a circle centered at D. What is the area of rhombus AECF? A. 2  \(\sqrt{2}\) B. \(\sqrt{2}\)  1 C. \(\sqrt{2}(2  \sqrt{2})\) D. \(\sqrt{2}\) E. 1 + \(\sqrt{2}\)
_________________
Need Practice? 20 Free GRE Quant Tests available for free with 20 Kudos GRE Prep Club Members of the Month: Each member of the month will get three months free access of GRE Prep Club tests.




Director
Joined: 03 Sep 2017
Posts: 518
Followers: 2
Kudos [?]:
420
[3]
, given: 66

Re: In the figure above, ABCD is a square with sides equal to [#permalink]
19 Sep 2017, 08:10
3
This post received KUDOS
If the side of the square is equal to 1 its diagonal is \(sqrt(2)\), which coincides with the rhombus main diagonal.
To find the smaller one, I exploit the fact that DE = 1 since it is a radius of the semicircle so as AD, side of the square. Thus, the smaller diagonal is equal to the diagonal of the square minus twice DF, i.e. \(EF=sqrt(2)2*(sqrt(2)1)\).
Given the two diagonals we can compute the area of the rhombus as \(\frac{sqrt(2)*[sqrt(2)2*(sqrt(2)1]}{2}=sqrt(2)1\). Thus, the answer is B!



Intern
Joined: 21 Apr 2017
Posts: 3
Followers: 0
Kudos [?]:
1
[1]
, given: 3

Re: In the figure above, ABCD is a square with sides equal to [#permalink]
30 Sep 2017, 22:34
1
This post received KUDOS
fraction[sqrt(2)*[sqrt(2)2*(sqrt(2)1]]/[2]=sqrt(2)1



Intern
Joined: 07 Jan 2019
Posts: 30
Followers: 0
Kudos [?]:
2
[0], given: 0

Re: In the figure above, ABCD is a square with sides equal to [#permalink]
09 Jan 2019, 13:23
done thank you



Manager
Joined: 18 Jun 2019
Posts: 123
Followers: 1
Kudos [?]:
27
[0], given: 62

Re: In the figure above, ABCD is a square with sides equal to [#permalink]
07 Sep 2019, 17:30
Hi,
Can somebody please post a better solution to this?
Thanks!



Manager
Joined: 04 Apr 2020
Posts: 91
Followers: 0
Kudos [?]:
33
[0], given: 22

Re: In the figure above, ABCD is a square with sides equal to [#permalink]
04 May 2020, 01:07
The key to this problem is finding the length of Rhombus diagonal EF, which is the base of both upper and lower triangles of the Rhombus.
The diagonal of the square is root(2). This length of root(2) is comprised of 2 segments of 1 (radius of the circle) "merged partially". So consider this as a Venn Diagram problem.
Full length = Sum of overlapping lengths  common length. Diagonal = (Radius1 + Radius2)  EF. root(2) = (1 + 1)  EF. EF = 2  root(2).
Now we have the base of the 2 triangles. The height of the triangles are half the length of the diagonal, which is root(2) / 2 = 1 / root(2).
Area of one triangle = 1/2 * base * height = 1/2 (EF) (half diagonal) = 1/2 (2  root(2)) (1 / root(2)) = (2root(2))/2root(2)
Area of rhombus = sum of areas of two triangles = 2 * Area of one triangle = 2 * (2  root(2)) / 2root(2) = (1  root(2)).
While this solution is very simple, it took me long enough to come up with it, sadly. The main point of this solution is using the Venn Diagram principle to find the length of the overlapping section of 2 line segments.




Re: In the figure above, ABCD is a square with sides equal to
[#permalink]
04 May 2020, 01:07





