Question

如图,某学校校园内有一块形状为直角梯形的空地ABCD.其中AB平行于DB,角B=90°,AB=100m,BC=80m,C

Answer 1

1）过点D作DE⊥AB于E则DE∥BC且DE=BC,CD=BE,DE∥PMRt△ADE中,DE=80m∴AE=AB-BE=100-40=60m∴AD=√( AE^2+DE^2)=√(3600+6400)=100m（2）∵DE∥PM∴△APM∽△ADE∴ AP/AD=PM/DE=AM/AE即 x/100=PM/80=AM/60∴PM= 4x/5,AM= 3x/5即MB=AB-AM=100- 3x/5S=PM•MB= 4x/5•（100- 3/5）x= 12x*2/15+80x由PM= 45x≥36,得x≥45∴自变量x的取值范围为45≤x≤100（3）当S=3300m2时,80x- 12x^2/25=330012x^2-2000x+82500=03x^2-500x+20625=0 x={ 500±√ [ (-500)^2-4×3×20625 ] }/ 6=(500±50) / 6∴x1= 550/6≈91.7（m）,x2= 450/6=75（m）即当s=3300m2时,PA的长为75m,或约为91.7m．向左转|向右转