Vẽ AH vuông với BC.
Ta có: [tex]\widehat{BAC}=45^o\Rightarrow \widehat{BOC}=90^o\Rightarrow BC=\sqrt{2}OC=\sqrt{2}R[/tex]
Lại có:[tex]OH=\frac{1}{2}BC=\frac{\sqrt{2}}{2}R\Rightarrow AH=AO+OH=(1+\frac{\sqrt{2}}{2})R\Rightarrow S_{ABC}=\frac{1}{2}.AH.BC=\frac{1}{2}.(1+\frac{\sqrt{2}}{2})R.\sqrt{2}R=\frac{\sqrt{2}+1}{2}R^2[/tex]