解:矩形的长为$\frac {1}{2}\sqrt {32}=\frac {1}{2}×4\sqrt {2}=2\sqrt {2} ,$
宽为$\frac {1}{3}\sqrt {18}=\frac {1}{3}×3\sqrt {2}=\sqrt {2} 。$
周长:$2×(2\sqrt {2}+\sqrt {2})=2×3\sqrt {2}=6\sqrt {2}。$
对角线长:$\sqrt {(2\sqrt {2})^2+(\sqrt {2})^2}=\sqrt {8+2}=\sqrt {10} 。$
答:矩形的周长为$6\sqrt {2},$对角线长为$\sqrt {10} 。$