题目内容：

如图，长方形ABCD的面积是60，若EB=2AE，AF=FD，则S_{四边形AEOF}=___．

优质解答

7


连接AO、BD，因AF=FD，所以S_△AFO=S_△DFO
因EB=2AE，长方形ABCD的面积是60
S_△ABF=

1

2

S_△ABD
=

1

2

×

1

2

×S_{长方形ABCD}
=

1

2

×

1

2

×60
=15
EB=2AE
S_△ADE=

1

3

S_△ABD
=

1

3

×

1

2

×60
=10
S_△EBO=S_△DFO+（15-10）=S_△DFO+5=S_△AFO+5
S_△AEO=

1

2

S_△EBO
                  S_△AFO+S_△AEO+S_△EBO=15
                S_△AFO+

1

2

S_△EBO+S_△EBO=15
S_△AFO+

1

2

×（S_△AFO+5）+S_△AFO+5=15
      S_△AFO+

1

2

S_△AFO+2.5+S_△AFO+5=15
                                       

5

2

S_△AFO=7.5
                                           S_△AFO=3
S_{四边形AEOF}=S_△ADE-S_△AFO=10-3=7
答：S_{四边形AEOF}的面积是7．
故答案为：7．