 It is currently 15 Sep 2019, 09:57 ### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here. # In the figure above, ABCD is a square with sides equal to  Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS: Founder  Joined: 18 Apr 2015
Posts: 8061
Followers: 155

Kudos [?]: 1682  , given: 7386

In the figure above, ABCD is a square with sides equal to [#permalink]
1
KUDOS
Expert's post 00:00

Question Stats: 38% (02:56) correct 61% (03:04) wrong based on 34 sessions

Attachment: #GRepracticequestion In the figure above, ABCD is a square with sides equal to 1,.jpg [ 26.23 KiB | Viewed 1036 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 + $$\sqrt{2}$$
[Reveal] Spoiler: OA

_________________ Director Joined: 03 Sep 2017
Posts: 520
Followers: 1

Kudos [?]: 369  , given: 66

Re: In the figure above, ABCD is a square with sides equal to [#permalink]
3
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  , given: 3

Re: In the figure above, ABCD is a square with sides equal to [#permalink]
1
KUDOS
fraction[sqrt(2)*[sqrt(2)-2*(sqrt(2)-1]]/=sqrt(2)-1
Intern Joined: 07 Jan 2019
Posts: 30
Followers: 0

Kudos [?]: 2 , given: 0

Re: In the figure above, ABCD is a square with sides equal to [#permalink]
done thank you
Manager Joined: 18 Jun 2019
Posts: 83
Followers: 0

Kudos [?]: 4 , given: 37

Re: In the figure above, ABCD is a square with sides equal to [#permalink]
Hi,

Can somebody please post a better solution to this?

Thanks! Re: In the figure above, ABCD is a square with sides equal to   [#permalink] 07 Sep 2019, 17:30
Display posts from previous: Sort by

# In the figure above, ABCD is a square with sides equal to  Question banks Downloads My Bookmarks Reviews Important topics Powered by phpBB © phpBB Group Kindly note that the GRE® test is a registered trademark of the Educational Testing Service®, and this site has neither been reviewed nor endorsed by ETS®.