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inscribed coordinate geometry (m06q32)

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Re: inscribed coordinate geometry (m06q32) [#permalink] New post 26 Oct 2012, 05:54
The four parts of circle and 1 part inside the square?
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Re: inscribed coordinate geometry (m06q32) [#permalink] New post 26 Oct 2012, 05:57
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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] 26 Oct 2012, 22:17
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