Question

如图,梯形ABCD中,AD平行BC,角ABC=90度,AD=6,BC=8,AB=根号m,（m>0)

Answer 1

⑴过D作DH⊥BC于H,则四边形ABHD是矩形,∴BH=AD=6,

∴CH=2,

当CP=2时,P与H重合,∴E与B重合.

⑵∵∠DPE=90°,∴∠BPE+∠DPH=90°,

∵∠BPE+∠BEP=90°,∴∠BEP=∠DPH,

∴RTΔBEP∽RTΔHPD,∴BE/CH=BP/DH,

Y/(X-2)=(8-X)/√m,

Y=-1/√m(X^2-10X+16).(2