GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 19 May 2019, 18:03

### 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

# There are two squares in a semi-circle as above figure, OC=CE. If the

Author Message
TAGS:

### Hide Tags

Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 7343
GMAT 1: 760 Q51 V42
GPA: 3.82
There are two squares in a semi-circle as above figure, OC=CE. If the  [#permalink]

### Show Tags

25 May 2016, 18:30
00:00

Difficulty:

35% (medium)

Question Stats:

81% (02:35) correct 19% (02:18) wrong based on 70 sessions

### HideShow timer Statistics

There are two squares in a semi-circle as above figure, OC=CE. If the radius of a semi-circle is 4√5, what is the area of the small square?
A. 12
B. 14
C. 15
D. 16
E. 27

*An answer will be posted in 2 days.

Attachments

1.png [ 25.26 KiB | Viewed 1381 times ]

_________________
MathRevolution: Finish GMAT Quant Section with 10 minutes to spare
The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy.
"Only $149 for 3 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Manager Joined: 05 Jan 2014 Posts: 69 Location: India GMAT 1: 610 Q47 V26 GPA: 3.76 WE: Information Technology (Computer Software) Re: There are two squares in a semi-circle as above figure, OC=CE. If the [#permalink] ### Show Tags 26 May 2016, 22:27 Let us assume x = OC = OE Side of larger square = BO + OC = x + x = 2*x or CD = 2*x Using pythagoras theorem (2*x)^2 + x^2 = (4*sqrt(5))^2 5*x^2 = 16*5 ==> x = 4 Area of smaller square = x*x = 4*4 = 16 Attachments File comment: Solution image 1.png [ 32.92 KiB | Viewed 1332 times ] Senior Manager Joined: 18 Jan 2010 Posts: 251 There are two squares in a semi-circle as above figure, OC=CE. If the [#permalink] ### Show Tags 26 May 2016, 23:42 Say OC = x. Then BC= 2x. This means that CD = 2x. Consider $$\triangle$$ OCD, $$OC^2$$ + $$DC^2$$ = Square of (4* $$\sqrt{5}$$) $$x^2$$+4$$x^2$$ = 80 5$$x^2$$=80 $$x^2$$ = 16. We want area of smaller square; which is $$x^2$$. So our answer is 16. D is the correct option. Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 7343 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: There are two squares in a semi-circle as above figure, OC=CE. If the [#permalink] ### Show Tags 28 May 2016, 19:39 If OC=GC=DG=a, then a^2+(2a)^2=(4√5)^2=80. From 5a^2=80, we get a^2=16. So the area of the small square is a^2=16. Hence, the correct answer is D. _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$149 for 3 month Online Course"
"Free Resources-30 day online access & Diagnostic Test"
"Unlimited Access to over 120 free video lessons - try it yourself"
Non-Human User
Joined: 09 Sep 2013
Posts: 10952
Re: There are two squares in a semi-circle as above figure, OC=CE. If the  [#permalink]

### Show Tags

26 Nov 2018, 07:06
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Re: There are two squares in a semi-circle as above figure, OC=CE. If the   [#permalink] 26 Nov 2018, 07:06
Display posts from previous: Sort by