Let a triangle ABC be inscribed in the circle x² -$\sqrt{2}(\mathrm{x}+\mathrm{y})+{\mathrm{y}}^2=0$ such that$\mathrm{\angle }\mathrm{BAC}=\frac{\mathrm{\pi }}{2}$. If the length of side AB is $\sqrt{2}$, then the area of the $\mathrm{△}\mathrm{ABC}$ is equal to