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Let OC = R so FG = R & SQUARE HEFG = R^2.Further, OG = CG = R/√2 so inner circle =(∏R^2)/2 and the blue corner FCG = R^2/2 -(∏R^2)/8Thus, total blue area= 4[R^2/2 -(∏R^2)/8]+2[(∏R^2 -2R^2)/4]=R^2 on simplification. Hence etc.Ajit
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Let OC = R so FG = R & SQUARE HEFG = R^2.
ReplyDeleteFurther, OG = CG = R/√2 so inner circle =(∏R^2)/2 and the blue corner FCG = R^2/2 -(∏R^2)/8
Thus, total blue area= 4[R^2/2 -(∏R^2)/8]+2[(∏R^2 -2R^2)/4]=R^2 on simplification. Hence etc.
Ajit