Events & Promotions
|
|
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.
Sep 0907:00 AM PDT
-10:00 AM PDT
Meet AdComs and explore top Master’s programs - MiM, MiF, MSc, MSBA and more. - Application Fee Waivers - Free 1-Week of GMAT Club Tests: - Master's Application Toolkit - Grand Prize Giveaway- Sep 09
10:00 AM PDT
-11:00 AM PDT
Three MBA applications. Three rejections. No interview invites. A year later, Aman was admitted to Cambridge Judge, SMU, and IMD. In this episode of MBA Admit Stories, Aman shares how he rebuilt his MBA application... 
Sep 0410:00 AM EDT
-11:59 PM EDT
We’ve worked incredibly hard to build TTP into the best test prep experience possible, and it would mean a lot to us to win Newsweek’s 2026 Readers’ Choice Award for Best Test Prep. If TTP has helped you, we’d be incredibly grateful for your vote.
19 Jul 2018, 21:55
Kudos
Bookmarks
C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
73% (01:00) correct
28%
(01:10)
wrong
based on 40
sessions
History
Date
Time
Result
Not Attempted Yet
In the figure, vertex Q of square OPQR is on circle with centre O. If the area of the square is 8, what is the area of the circle?
A. \(8 \pi\)
B. \(8 \sqrt{2} \pi\)
C. \(16 \pi\)
D. \(32 \pi\)
E. \(64 \pi\)
Attachment:
figure 11.jpg [ 14.68 KiB | Viewed 5075 times ]
19 Jul 2018, 22:08
Kudos
Bookmarks
vertex Q of square OPQR is on circle with centre O
=> radius of the circle = diagonal of the square
area of square is \(a^2\) = 8
=> a = \(2\sqrt{2}\)
diagonal of the square = \(\sqrt{a^2 + a^2}\)
=> diagonal = \(\sqrt{8 + 8}\)
=> diagonal = \(\sqrt{16}\) = 4
=> radius of the circle = 4
=> Area of the circle = \(4^2 \pi\) = \(16 \pi\)
Hence option C
=> radius of the circle = diagonal of the square
area of square is \(a^2\) = 8
=> a = \(2\sqrt{2}\)
diagonal of the square = \(\sqrt{a^2 + a^2}\)
=> diagonal = \(\sqrt{8 + 8}\)
=> diagonal = \(\sqrt{16}\) = 4
=> radius of the circle = 4
=> Area of the circle = \(4^2 \pi\) = \(16 \pi\)
Hence option C
19 Jul 2018, 23:28
Kudos
Bookmarks
Bunuel
OA:C
As per figure,Diagonal (\(diagonal_{square}\)) of Square is equal to Radius of circle.
Area of Square = \(\frac{1}{2}*Diagonal_{square}^2=8\)
\(Diagonal_{square}^2=16\);
As \(Diagonal_{square}\) = Radius of circle
\({Radius}^2 =16\)
Area of circle= \(\pi *\) \({Radius}^2=\pi*16\)
