已知平行六面体ABCD-A1B1C1D1中AB=4,AD=3,AA1=5,∠BAD=90,∠BAA1=∠DAA1=60,则|AC1|=_.

问题描述:

已知平行六面体ABCD-A1B1C1D1中AB=4,AD=3,AA1=5,∠BAD=90,∠BAA1=∠DAA1=60,则|

AC1
|=______.

连接AC,∵AB=4,AD=3,∠BAD=90°
∴AC=5
根据cos∠A'AB=cos∠A'AC•cos∠CAB

1
2
=cos∠A'AC•
2
2

∴∠A'AC=45°则∠C'CA=135°
而AC=5,AA′=5,
根据余弦定理得AC′=
85

故答案为:
85