|
|
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
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
25 May 2016, 18:30
Kudos
Bookmarks
D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
75% (02:34) correct
25%
(02:30)
wrong
based on 75
sessions
History
Date
Time
Result
Not Attempted Yet
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.
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 3490 times ]
26 May 2016, 22:27
Kudos
Bookmarks
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
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 3418 times ]
26 May 2016, 23:42
Kudos
Bookmarks
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.
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.
