$\begin{align}  & \text{Using Pythagoras theorem} \\ & {{a}^{2}}+{{a}^{2}}={{2}^{2}} \\ & 2{{a}^{2}}=4 \\ & {{a}^{2}}=2c{{m}^{2}} \\ & a=\sqrt{2}cm \\ & \text{Area of triangle}=\tfrac{1}{2}\cdot b\cdot h \\ & A=\tfrac{1}{2}\times \sqrt{2}\times \sqrt{2}c{{m}^{2}}=\frac{2}{2}c{{m}^{2}}=1c{{m}^{2}} \\\end{align}$