由勾股定理知BC=4
过C向AB作垂线交AB于E
易知RT⊿ABC∽⊿RT⊿ACE∽⊿RT⊿BCE
则CE=AC*BC/AB=12/5
AE=AC^2/AB=9/5
BE=AB-AE=16/5
于是S四边形ABCD=S梯形BDCE+S直角⊿ACE
=(1/2)(CE+BD)*BE+(1/2)CE*AE
=6+16√3/5
由勾股定理知BC=4
过C向AB作垂线交AB于E
易知RT⊿ABC∽⊿RT⊿ACE∽⊿RT⊿BCE
则CE=AC*BC/AB=12/5
AE=AC^2/AB=9/5
BE=AB-AE=16/5
于是S四边形ABCD=S梯形BDCE+S直角⊿ACE
=(1/2)(CE+BD)*BE+(1/2)CE*AE
=6+16√3/5