合聚咖

合聚咖

(2013

admin

设正方形EFGC的边长为a,即EC=EF=CG=FG=a,

∴ED=EC-DC=a-5,BG=BC+CG=a+5,

∴S△EFD=

1
2
a(a-5),

∴S四边形DCGF=a2-

1
2
a(a-5),

∵S△BCD=

1
2
×52=12.5,S△BCF=
1
2
a(a+5),

∴S△BDF=S△BCD+S四边形DCGF-S△BCF=12.5+a2-

1
2
a(a-5)-
1
2
a(a+5)=12.5.

故答案为:12.5.