∵ABCD-A1B1C1D1是正方体,
E,F为AA1 CC1中点
∴S平行四边形EBFD1=2SΔBED1
∴VA1-EBFD1=2VA1-BED1
而VA1-BED1=VB-A1D1E
=1/3*SΔA1D1E*AB
=1/3*(1/2*a/2*a)*a
=a³/12
∴VA1-EBFD1=2VA1-BED1=a³/6
故1/6a³为正解
∵ABCD-A1B1C1D1是正方体,
E,F为AA1 CC1中点
∴S平行四边形EBFD1=2SΔBED1
∴VA1-EBFD1=2VA1-BED1
而VA1-BED1=VB-A1D1E
=1/3*SΔA1D1E*AB
=1/3*(1/2*a/2*a)*a
=a³/12
∴VA1-EBFD1=2VA1-BED1=a³/6
故1/6a³为正解