连接AF,作GH⊥AE于点H,则有AE=EF=HG=4,FG=2,AH=2,根据矩形的性质及勾股定理即可求得其周长. 解:如图,连接AF,作GH⊥AE于点H,则有AE=EF=HG=4,FG=2,AH=2,
![](http://img.shitiku.com.cn/uploads/allimg/20191029/20191029224626-61186.png) ∵AG= ,AF= , ∴AF2=AD2+DF2=(AG+GD)2+FD2=AG2+GD2+2AG?GD+FD2,GD2+FD2=FG2 ∴AF2=AG2+2AG?GD+FG2∴32=20+2×2 ×GD+4, ∴GD= ,FD= , ∵∠BAE+∠AEB=90°=∠FEC+∠AEB, ∴∠BAE=∠FEC, ∵∠B=∠C=90°,AE=EF, ∴△ABE≌△ECF, ∴AB=CE,CF=BE, ∵BC=BE+CE=AD=AG+GD=2 + , ∴AB+FC=2 +![](http://img.shitiku.com.cn/uploads/allimg/20191029/20191029224627-69763.png) ∴矩形ABCD的周长=AB+BC+AD+CD=2BC+AB+CF+DF =2 + +2 + +2 + + =8 . 故答案为,8![](http://img.shitiku.com.cn/uploads/allimg/20191029/20191029224627-46473.png) |