?$?解:設AE=x,則AB^2=2^2+x^2,AD^2=6^2+x^2$?
?$在Rt△ABD中,∵ AD^2+AB^2=BD^2$?
?$∴ 36+{x}^{2}+4+{x}^{2}={8}^{2}$?
?$解得,x=±2{\sqrt {3}}$?
?$∵ x>0$?
?$∴ x=2{\sqrt {3}}$?
?$∴ AB={\sqrt {{2}^{2}+{x}^{2}}}=4,AD={\sqrt {{6}^{2}+{x}^{2}}}=4{\sqrt {3}}$?
?$∴ 矩形ABCD的長為4\sqrt {3},寬為4$?