(1)证明:连结AD1,则EF//AD1.
∵AD1//BC1,∴EF//BC1
∵BC1⊂平面A1BC1,BC1⊄平面A1BC1,∴EF//平面A1BC1.
(2)设AA1=h,则
V(ABCD-A1C1D1)=V(ABCD-A1B1C1D1)-V(B-A1B1C1)
=2×2h-(1/3)×(1/2)×2×2h
=4h-2h/3
=10h/3=40/3
∴h=4,即AA1=4
(1)证明:连结AD1,则EF//AD1.
∵AD1//BC1,∴EF//BC1
∵BC1⊂平面A1BC1,BC1⊄平面A1BC1,∴EF//平面A1BC1.
(2)设AA1=h,则
V(ABCD-A1C1D1)=V(ABCD-A1B1C1D1)-V(B-A1B1C1)
=2×2h-(1/3)×(1/2)×2×2h
=4h-2h/3
=10h/3=40/3
∴h=4,即AA1=4