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P.o.t.W. #27 Solution

Since \({\textrm{AB}} = 1 then \({\textrm{BC}} = 1 ; so, the radius of the larger circle is 2 . This makes \({\textrm{CQ}} = 2 and triangle CQB a 30-60-90 triangle where angle BCQ \( = 60^\circ = \dfrac{\pi}{3} radians. It follows that triangle BCR is also a 30-60-90 triangle. Hence, QCR . The shaded area is found by subtracting half of the area of the smaller circle and the area of triangle QCR from the area of the...

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