作DH⊥AB于H,相交EF于G,根据条件：
梯形ABFE与EFCD的面积相等,有
(CD+EF)*DG/2=(EF+AB)*GH/2  即  (2+EF)*DG=(EF+4)*GH
又
DG:(EF-CD)=DH:(AB-CD)=GH:(AB-EF)  有  DG*(4-EF)=GH*(EF-2)
所以
(EF+2)*(EF-2)=(4+EF)*(4-EF)
EF=√10≈3.16   即EF的长是根号10,约等于3.16