在△ABC中,BC=12,AB=10,sinB=五分之三,动点D从点A出发,以每秒1个单位的速度沿线段AB向点B运动,D

2个回答

  • 1)正方形DEFG的边GF在BC上时,BD=AB-AD=10-t

    ∵sinB=DG/BD=3/5; ∴DG= (3/5)BD = (3/5)(10-t)

    又∵△ADE相似△ABC

    ∴AD/AB=DE/BC ∴DE=(AD*BC)/AB=6t/5

    又∵DG=DE ∴ (3/5)(10-t)=6t/5 ∴t=10/3

    (2)由(1)可知:S△CEP+S△BDQ=(1/2)CP*EP+=(1/2)BQ*DQ

    =(1/2)(BC-PQ)*PQ

    =(1/2)[12 - ( 6t/5)]* (6t/5)

    = - 18t²/25+ 36t/5

    S△ABC=(1/2)BC*ABsinB=(1/2)12*10*3/5=36

    令:

    - 18t²/25+ 36t/5=36/4

    2t²- 20t-25=0

    t1=5+5√6/2 ; t2=5-5√6/2