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06 Sep 2018, 01:03
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B
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87% (01:07) correct
13%
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If the area of the square inscribed in the circle shown above is 36, what is the area of the circle?
A. \(9\pi\)
B. \(18\pi\)
C. \(\sqrt{72} \pi\)
D. \(4\sqrt{18} \pi\)
E. \(36\pi\)
Attachment:
image009.jpg [ 1.76 KiB | Viewed 2336 times ]
06 Sep 2018, 01:13
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Bunuel
side of the square = 6
diagonal of the square = diameter of the circle = a\(\sqrt{2}\) = 6\(\sqrt{2}\)
area of the circle = \(\pi\) 3\(\sqrt{2}\)*3\(\sqrt{2}\) = 18\(\pi\)
06 Sep 2018, 03:23
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+1 for B.
Area of Square = S^2 = 36 --> S = 6
Diagonal of Circle = 6√2 (Applying Pythagoras Theorem) , and Radius = 3√2 (Half a diagonal)
Therefore, Area of the circle = πr^2 = π * (3√2)^2 = π * 9 * 2 = 18π
Hence, B.
Area of Square = S^2 = 36 --> S = 6
Diagonal of Circle = 6√2 (Applying Pythagoras Theorem) , and Radius = 3√2 (Half a diagonal)
Therefore, Area of the circle = πr^2 = π * (3√2)^2 = π * 9 * 2 = 18π
Hence, B.
06 Sep 2018, 07:57
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Bunuel
Diameter of the Circle is \(6√2\) , Hence Raidus of the Circle is \(3√2\)
So, Area of the circle is \(π*3√2*3√2 = 18π\), Answer must be (B)
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image001 %281%29.jpg [ 4.71 KiB | Viewed 2095 times ]
