80分太少
算了,没人不说我心地不好,先给你答案好了.
(1)BP=3-t,BQ=t,且作过A的△ABC中线,有AQ=4,则sin∠ABC=4/5,cos∠ABC=3/5,因为PQ⊥AB则∠BPQ=90,所以BP/BQ=cos∠ABC=3/5即(3-t)/t=3/5,自己解t
(2)以B为原点BC为X轴正方向作直角坐标系,有线AB:y=4/3x,根据1S1CM且sin∠ABC=4/5,cos∠ABC=3/5,可得P(3-3/5t,4-4/5t),Q(t,0)则S△PBQ=(4-4/5t)*t/2所以S四边形APQC=S△ABC-S△PBQ=12-(4-4/5t)*t/2(0