解答：解：设B′C′和CD的交点是O，连接OA，∵AD=AB′，AO=AO，∠D=∠B′=90°，∴Rt△ADO≌Rt△AB′O，∴∠OAD=∠OAB′=30°，∴OD=OB′= 2 ，S四边形AB′OD=2S△AOD=2× 1 2 2 × 6 =2 3 ，∴S阴影部分=S正方形-S四边形AB′OD=6-2 3 ．
