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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 822.
∵ D, O are mid-points of AC, AB resp.∴ OD//BCS = area of sector OBC (with angle 45°)= π×6^2×1/8= 9π/2
http://img812.imageshack.us/img812/2623/problem822.pngDraw lines per attached sketchWe have Area S=Area(DCB)+ Area S1But Area(DCB)= Area ( OBC) …… ( Same base BC and altitude)So S= Area of sector COB = pi x OB^2/8 = 4.5 pi
Join BC, OC(i) Area bounded by chord BC and arc BC = Area of sector OBC - Area of ΔOBC= (1/2).6.6. (π /4)- (1/2).6.6.sin (π/4) = 9 π/2 - 9√2(ii) Area of Δ BDC = Area of Δ ADB = (1/2) Area of Δ ABC= (1/2)(twice the area of Δ AOC)= area of Δ AOC= (1/2) OA.OC.sin (3 π /4)= (1/2).6.6.(1/ √ 2)= 9 √ 2So S = Area (i) + area (ii) = 9 π /2
Jacob Ha's proof is nice.I realized that if T denotes the area cut off by arc BC and chord BC,S = T + area of ΔBDC = T + area of ΔBOC = area of sector BOC