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Question Number 9594 by tawakalitu last updated on 19/Dec/16

Answered by sandy_suhendra last updated on 20/Dec/16

Commented by tawakalitu last updated on 20/Dec/16

God bless you sir.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

Commented by tawakalitu last updated on 20/Dec/16

Thanks but please what is the answer ?

$$\mathrm{Thanks}\:\mathrm{but}\:\mathrm{please}\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{answer}\:? \\ $$

Commented by sandy_suhendra last updated on 20/Dec/16

OA=OB=OC=OD=R OE=AE=EB=R sin 45°=(1/2)R(√2) AB=2AE=R(√2) CE=OC−OE=R−(1/2)R(√2) Area ΔABC=(1/2)AB×CE =(1/2)R(√2) (R−(1/2)R(√2)) =(1/2)R^2 ((√2)−1) Area ΔAOD=(1/2)×R×R×sin 135° =(1/4)R^2 (√2) ((Area ΔABC)/(Area ΔAOD))=(((1/2)R^2 ((√2)−1))/((1/4)R^2 (√2)))=((2((√2)−1))/(√2))×((√2)/(√2))=2−(√2)

$$\mathrm{OA}=\mathrm{OB}=\mathrm{OC}=\mathrm{OD}=\mathrm{R} \\ $$$$\mathrm{OE}=\mathrm{AE}=\mathrm{EB}=\mathrm{R}\:\mathrm{sin}\:\mathrm{45}°=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{R}\sqrt{\mathrm{2}} \\ $$$$\mathrm{AB}=\mathrm{2AE}=\mathrm{R}\sqrt{\mathrm{2}} \\ $$$$\mathrm{CE}=\mathrm{OC}−\mathrm{OE}=\mathrm{R}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{R}\sqrt{\mathrm{2}} \\ $$$$\mathrm{Area}\:\Delta\mathrm{ABC}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{AB}×\mathrm{CE} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{R}\sqrt{\mathrm{2}}\:\left(\mathrm{R}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{R}\sqrt{\mathrm{2}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{R}^{\mathrm{2}} \left(\sqrt{\mathrm{2}}−\mathrm{1}\right) \\ $$$$\mathrm{Area}\:\Delta\mathrm{AOD}=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{R}×\mathrm{R}×\mathrm{sin}\:\mathrm{135}° \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{4}}\mathrm{R}^{\mathrm{2}} \sqrt{\mathrm{2}} \\ $$$$\frac{\mathrm{Area}\:\Delta\mathrm{ABC}}{\mathrm{Area}\:\Delta\mathrm{AOD}}=\frac{\frac{\mathrm{1}}{\mathrm{2}}\mathrm{R}^{\mathrm{2}} \left(\sqrt{\mathrm{2}}−\mathrm{1}\right)}{\frac{\mathrm{1}}{\mathrm{4}}\mathrm{R}^{\mathrm{2}} \sqrt{\mathrm{2}}}=\frac{\mathrm{2}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)}{\sqrt{\mathrm{2}}}×\frac{\sqrt{\mathrm{2}}}{\sqrt{\mathrm{2}}}=\mathrm{2}−\sqrt{\mathrm{2}}\:\:\: \\ $$

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