GRE Prep 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.
In the figure below, vertex Q of square OPQR is on a circle
[#permalink]
02 Feb 2019, 09:09
02 Feb 2019, 09:09
Expert Reply
00:00
Question Stats:
75% (01:05) correct
24% (01:10) wrong based on 103 sessions
Hide Show timer Statistics
In the figure below, vertex Q of square OPQR is on a circle with center O. If the area of the square is 8, what is the area of the circle?
#GREpracticequestion In the figure below, vertex Q of square OPQR.png [ 10.4 KiB | Viewed 7918 times ]
A) \(8 \pi\)
B) \(8\pi \sqrt{2}\)
C) \(16 \pi\)
D) \(32 \pi\)
E) \(64 \pi\)
Attachment:
#GREpracticequestion In the figure below, vertex Q of square OPQR.png [ 10.4 KiB | Viewed 7918 times ]
A) \(8 \pi\)
B) \(8\pi \sqrt{2}\)
C) \(16 \pi\)
D) \(32 \pi\)
E) \(64 \pi\)
Re: In the figure below, vertex Q of square OPQR is on a circle
[#permalink]
02 Feb 2019, 09:58
02 Feb 2019, 09:58
1
1
Area of the square = 8, so side =√8 = 2√2.
Diagonal of the square = s√2
= 2√2 * 2 = 4
Diagonal of the square = radius of the circle
Area of the circle = 16π
Answer - C
Diagonal of the square = s√2
= 2√2 * 2 = 4
Diagonal of the square = radius of the circle
Area of the circle = 16π
Answer - C
Re: In the figure below, vertex Q of square OPQR is on a circle
[#permalink]
08 Feb 2021, 00:18
08 Feb 2021, 00:18
1
Area of circle = Radios^2* pi.
Given that the area of square is 8 = A of square = S^2.
Clearly that the side is 2√2.
To find the Radios we can use the Pythagorean theorem to find the diagonal.
(2√2)^2+(2√2)^2=(d)^2
16=d^2= d=4
Finally, Area of circle = (4)^2*pi = 16pi
Given that the area of square is 8 = A of square = S^2.
Clearly that the side is 2√2.
To find the Radios we can use the Pythagorean theorem to find the diagonal.
(2√2)^2+(2√2)^2=(d)^2
16=d^2= d=4
Finally, Area of circle = (4)^2*pi = 16pi
