jmc

The length of the hypotenuse of an isosceles right triangle is $\sqrt{2}$ times the length of each leg. Therefore, $BD=\frac{BC}{\sqrt{2}}=\frac{2}{\sqrt{2}}\cdot \left(\frac{\sqrt{2}}{\sqrt{2}}\right)=\frac{2\sqrt{2}}{2}=\sqrt{2}$ units. Applying the same rule to triangle $ABD$, we find that $AB=BD/\sqrt{2}=\sqrt{2}/\sqrt{2}=1$ unit. The perimeter of quadrilateral $ABCD$ is $AB+BD+CD+DA=1+2+\sqrt{2}+1=\boxed{4+\sqrt{2}}$ units.