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

Question

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

Answer 1

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．