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inscribed coordinate geometry (m06q32)

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Re: inscribed coordinate geometry (m06q32) [#permalink] New post 26 Oct 2012, 22:17
Area of sector BOC - area of triangle BOC

pi*sq(sqrt(6))*[90/360] - [1/2]*sqrt(6)*sqrt(6)

= [3pi/2] - 3
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Re: inscribed coordinate geometry (m06q32) [#permalink] New post 28 Oct 2012, 04:51
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There is a slightly more convoluted approach.
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Re: inscribed coordinate geometry (m06q32) [#permalink] New post 28 Oct 2012, 08:32
Great problem Bunuel! This one was simpler - Thanks to your MATH NOTE on polygons.

1. As circle is circumscribed to the square - Area of circle = \frac{Pi}{2}*(Area of square). So this means we need to know area of sqaure first.
2. Area of sqaure = \frac{(diagonal)^2}{2} and diagonal (in this case) = 2 * radius i.e 2\sqrt{6}.Therefore, Area of square = 12 and Area of circle (using point 1) = \frac{Pi}{2}*12.
3. Now, area of all 4 sectors = area of circle - area of sqaure and area of 1 sector (say CD) = \frac{1}{4}*(AREA of 4 sectors)
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Re: inscribed coordinate geometry (m06q32)   [#permalink] 28 Oct 2012, 08:32
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