Question

矩形ABCD中,AD等于5,AB等于8.点E在AD边上,且AE等于2,点F,H分别是AB,CD上的动点,以EF,EH

Answer 1

1、∵ABCD是矩形∴∠D=90°,在RT△DEH中：DH=4,DE=AD-AE=5-2=3∴EH²=DH²+DE²=4²+3²=5²EH=5∴菱形EFGH周长=4EH=202、∵EFGH是正方形∴EH=EF,∠HEF=90°∴∠DEH+∠AEF=90°∵∠AEF+∠AFE=90°∴∠DEH=∠AFE∵∠D=∠A=90°EH=EF∴△AEF≌△DHE(AAS)∴AF=DE3、得△DEH≌△BFG,那么BG=DE=AD-AE=3∴在RT△AEF和RT△BFG中：EF²=AE²+AF²=2²+(8-BF)²FG²=BG²+BF²=3²+BF²∴4+(8-BF)²=9+BF²68-16BF=916BF=59BF=59/16