It is currently 24 Jan 2020, 05:14

### 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

# In the figure above, ABCD is a square with sides equal to

Author Message
TAGS:
Founder
Joined: 18 Apr 2015
Posts: 9396
Followers: 195

Kudos [?]: 2288 [1] , given: 8892

In the figure above, ABCD is a square with sides equal to [#permalink]  04 Aug 2017, 07:52
1
KUDOS
Expert's post
00:00

Question Stats:

44% (02:50) correct 55% (02:59) wrong based on 43 sessions

Attachment:

#GRepracticequestion In the figure above, ABCD is a square with sides equal to 1,.jpg [ 26.23 KiB | Viewed 1794 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

_________________

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: 519
Followers: 2

Kudos [?]: 396 [3] , given: 66

Re: In the figure above, ABCD is a square with sides equal to [#permalink]  19 Sep 2017, 08:10
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 [1] , given: 3

Re: In the figure above, ABCD is a square with sides equal to [#permalink]  30 Sep 2017, 22:34
1
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: 124
Followers: 0

Kudos [?]: 23 [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!
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