如图,在直角梯形ABCD中,AD‖BC,AD⊥DC,AD=4,BC=7,AB=5,AE=ED,P在AB上移动（不与A、B
1个回答

作AM⊥BC,垂足M,

AM²=AB²-BM²=5²-(BC-MC)²=25-(7-AD)²=25-(7-4)²=16,

AM=4=DC;

延长BA与CE的延长线交于G,AE=ED=AD/2=4/2=2;

AE//BC,AG:(AG+AB)=AE:BC,

AG:(AG+5)=2：7

7AG-2AG=10,

AG=2,

BG=AG+AB=2+5=7=BC,

PF//EC,BF=BP=x；

1,

△BFP底边BF上的高=h,h：AM=PB:AB,h=4x/5;

则△EAP底边EA上的高=AM-h=4-4x/5,

y=(AD+BC)DC/2-ED*DC/2-AE*(4-4x/5)/2-BF*h/2,

y=14+4x/5-2x²/5,(0

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