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