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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 825.
∠DOA = 135/2 = 67.5 = ∠CBOSo OD // BC.By sharing the same base BC and same height, ΔBCD has the same area as ΔOCB.S = sector(OBC) = π *(36)*(1/8) = (9/2) π
See attached sketchhttp://img706.imageshack.us/img706/5949/problem825.pngLet Area(XYZ)= area of triangle XYZLet Sector(XYZ)= area of sector XYZLet R=OA=6We have Area of white area= Area(ADB)+2.S1But S1= Sector(AOD)-Area(AOD)And Area(ADB)=2.Area(AOD)So white area=2.Sector(AOD) => S= sector(AOB)-white area=(pi/2-3/8.pi)*R^2= 9/2.pi
Since ∠DOC=∠OCB=67.5°, so OD//BC. Hence, S = area of sector BOC (with angle 45°)= 1/2×6^2×π/4= 9π/2