∵AA1∥BB1,∴∠CB1B为异面直线A1A与B1C所成的角,
∵AB=BC=2,异面直线A1A与B1C所成的角的大小为arctan
1 |
2 |
∴BB1=4,
∴正四棱柱ABCD-A1B1C1D1的侧面积S=4×2×4=32.
故答案是32.
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