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19 Jul 2018, 21:55
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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
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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 4989 times ]
19 Jul 2018, 22:08
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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
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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= \(\pi *\) \({Radius}^2=\pi*16\)
