解:$(1)$此矩形的周长为
$ (\frac {1}{2}\sqrt {32}+\frac {1}{3}\sqrt {18})×2=(2\sqrt {2}+\sqrt {2})×2=6\sqrt {2}$
$ (2)$此矩形的面积$=\frac {1}{2}\sqrt {32}×\frac {1}{3}\sqrt {18}=2\sqrt {2}×\sqrt {2}=4$,
$ $故与此矩形面积相等的正方形的边长为$2$,
∴对角线的长为$\sqrt {2^2+2^2}=2\sqrt {2}$