Right triangle $ACD$ with right angle at $C$ is constructed outward on the hypotenuse $\overline{AC}$ of the $30^\circ$-$60^\circ$-$90^\circ$ triangle $ABC$, which has its right angle at $B$, $\angle BAC=30^\circ$ and $BC=2$, so that the two triangles have equal perimeters. What is $CD$?

1. In the $30^\circ$-$60^\circ$-$90^\circ$ triangle $ABC$ with $BC=2$ (opposite the $30^\circ$ angle at $A$), $AB=2\sqrt3$ and $AC=4$. Its perimeter is $2+2\sqrt3+4=6+2\sqrt3$.
2. Let $CD=x$ and $AD=y$. The right triangle $ACD$ has perimeter $4+x+y$, which must equal $6+2\sqrt3$, so $x+y=2+2\sqrt3$. Also $y^2-x^2=AC^2=16$.
3. Dividing, $y-x=\dfrac{16}{2+2\sqrt3}=\dfrac8{1+\sqrt3}=4(\sqrt3-1)$.
4. Subtracting, $2x=(2+2\sqrt3)-4(\sqrt3-1)=6-2\sqrt3$, so $CD=x=\boxed{3-\sqrt3}\approx1.27$ (and $AD=y=3\sqrt3-1$).