已知斜三棱柱ABC-A1B1C1中,四边形A1ACC1为菱形,∠ACB=90°,AC=BC=2,点D为AC的中点,A1D⊥平面ABC.(

2025-05-09 17:44:03
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回答1:

(Ⅰ)∵A1D⊥平面ABC,BC?平面ABC,
∴A1D⊥BC;
又∵BC⊥AC,且A1D∩AC=D,
∴BC⊥平面A1ACC1
∴BC⊥AC1;①
又∵四边形A1ACC1为菱形,
∴A1C⊥AC1;②
由①②得,AC1⊥平面A1BC,
且A1B?平面A1BC,
∴AC1⊥A1B;
(Ⅱ)∵D是线段AC的中点,∴

AD
A1C1
1
2

AM
MC1
1
2
,即
C1M
C1A
2
3

V三棱锥C1?MBC=V三棱锥C1?ABC-V三棱锥M-ABC
=3V三棱锥M-ABC-V三棱锥M-ABC=2V三棱锥M-ABC
=2×
1
3
×S△ABC?MD
S△ABC
1
2
×2×2=2

MD=
1
3
×A1D=
1
3
×
3
3
3

V三棱锥C1?MBC=2×
1
3
×2×
3
3
=
4
9