题目内容
正三角形,B1C1∥BC,B1C1=
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| 2 |
(Ⅰ)求证:面A1AC⊥面ABC;
(Ⅱ)求该几何体的体积.
考点:平面与平面垂直的判定,棱柱、棱锥、棱台的体积
专题:空间位置关系与距离
分析:(Ⅰ)由已知得A1C=A1B=
,A1A=AC=1,从而A1A⊥AC,由此能证明面A1AC⊥面ABC.
(Ⅱ)依题意得:V=VC-A1B1BA+VC-A1B1C1而VC-A1B1BA=
×SA1B1BA×CA=
×1×1=
,VC-A1B1C1=
×SA1B1C1×A1A=
×(
×
×
)×1=
,由此能求出该几何体的体积.
| 2 |
(Ⅱ)依题意得:V=VC-A1B1BA+VC-A1B1C1而VC-A1B1BA=
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| 3 |
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| 3 |
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解答:
(Ⅰ)证明:∵在多面体ABC-A1B1C1中,四边形ABB1A1是正方形,AC=AB=1,
△A1BC是正三角形,B1C1∥BC,B1C1=
BC,
∴A1C=A1B=
,A1A=AC=1,
∴A1A2+AC2=A1C2,
∴A1A⊥AC,
又A1A⊥AB,∴A1A⊥平面ABC,
∴面A1AC⊥面ABC.
(Ⅱ)解:依题意得:V=VC-A1B1BA+VC-A1B1C1
而VC-A1B1BA=
×SA1B1BA×CA=
×1×1=
,
VC-A1B1C1=
×SA1B1C1×A1A=
×(
×
×
)×1=
,
故:V=
+
=
.
△A1BC是正三角形,B1C1∥BC,B1C1=
| 1 |
| 2 |
∴A1C=A1B=
| 2 |
∴A1A2+AC2=A1C2,
∴A1A⊥AC,
又A1A⊥AB,∴A1A⊥平面ABC,
∴面A1AC⊥面ABC.
(Ⅱ)解:依题意得:V=VC-A1B1BA+VC-A1B1C1
而VC-A1B1BA=
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| 3 |
| 1 |
| 3 |
| 1 |
| 3 |
VC-A1B1C1=
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| 3 |
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故:V=
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| 12 |
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| 12 |
点评:本题考查面面垂直的证明,考查几何体的体积的求法,解题时要认真审题,注意空间思维能力的培养.
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