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In the figure below, vertex Q of square OPQR is on a circle
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02 Feb 2019, 09:09
02 Feb 2019, 09:09
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Question Stats:
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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 7309 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 7309 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
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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
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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
