如图,已知四棱柱ABCD-A1B1C1D1的棱长都为a,底面ABCD是菱形,且∠BAD=60°,侧棱A1A⊥平面ABCD,F为棱B1

2025-06-20 11:17:11
推荐回答(1个)
回答1:

(1)证明:(如上图)连结BD,由直四棱柱ABCD-A1B1C1D1,可知:A1A⊥平面ABCD,
又∵BD?平面ABCD,
∴A1A⊥BD.
∵四边形ABCD为菱形,
∴AC⊥BD.
又∵AC∩A1A=A,AC、A1A?平面ACC1A1
∴BD⊥平面ACC1A1.…(7分)
而NA∥BD,
∴NA⊥平面ACC1A1
又∵NA?平面AFC1
∴平面AFC1⊥平面ACC1A1                                           …(9分)
(2)解:∵∠DAB=60°,∴C到AB的距离为:asin60°=

3
2
a,就是C1到平面ABF的距离,AD=AA1=a,
∴三棱锥A1-AC1F的体积:
1
3
×
1
2
AB?BF?
3
2
a
=
1
3
×
1
2
×a×
1
2
3
2
a
=
3
24
a3
…(12分)