Question

如图,已知在梯形ABCD中,AD平行BC,AB=CD=5,AD=4 BC=10,P是边BC上的一个动点,∠APQ=∠B,

Answer 1

俊狼猎英团队为您解答

过A作AR⊥BC于R,则BR=3,∴AR=√(AB^2-BR^2)=4,

∴PR=|4-X|,

∴AP=√(AR^2-PR^2)=√(8X-X^2)

∵AD∥BC,∴∠PAD=∠APB,

又∠B=∠APQ,

∴ΔABP∽ΔQPA,

∴PB/PA=PA/AQ,

∴AQ=PA^2/PB,

∴4-Y=(8X-X^2)/X,

Y=-X+4.(0