题目内容
(本小题满分14分)如图,在四棱锥
中,平面
平面
,
为等边三角形,底面
为菱形,
,
为
的中点,
。
(1)求证:
平面
;
(2) 求四棱锥
的体积
(3)在线段
上是否存在点
,使
平面
; 若存在,求出
的值。
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(1)求证:
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(2) 求四棱锥
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(3)在线段
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(1)见解析;(2)
;
(3)存在,当
时,
平面
。
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(3)存在,当
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本试题主要是考查了空间几何体中线面的垂直问题,以及锥体的体积,和线面平行的判定综合运用。
(1)连BD,四边形ABCD菱形, ∵AD⊥AB, ∠BAD=60°
△ABD为正三角形, Q为AD中点, ∴AD⊥BQ
∵PA=PD,Q为AD的中点,AD⊥PQ 又BQ∩PQ=Q ∴AD⊥平面PQB.
(2)因为
平面
,那么
是四棱锥
的高,
利用锥体的体积公式得到。
(3)因为AQ//BC,那么结合PA//MN,得到判定定理,从而得到证明。
解:(1)连BD,四边形ABCD菱形, ∵AD⊥AB, ∠BAD=60°
△ABD为正三角形, Q为AD中点, ∴AD⊥BQ…………………………2分
∵PA=PD,Q为AD的中点,AD⊥PQ……………………………3分
又BQ∩PQ=Q ∴AD⊥平面PQB. ………………………………5分
(2)平面
平面
平面
平面
=
平面
,
所以
平面
…………………………………7分
是四棱锥
的高,
…………………………………9分
(3)存在,当
时,
平面
由
可得,
,
……………………11分
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………………………………………………………12分
平面
,
平面
,
平面
………………14分
(1)连BD,四边形ABCD菱形, ∵AD⊥AB, ∠BAD=60°
△ABD为正三角形, Q为AD中点, ∴AD⊥BQ
∵PA=PD,Q为AD的中点,AD⊥PQ 又BQ∩PQ=Q ∴AD⊥平面PQB.
(2)因为
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利用锥体的体积公式得到。
(3)因为AQ//BC,那么结合PA//MN,得到判定定理,从而得到证明。
解:(1)连BD,四边形ABCD菱形, ∵AD⊥AB, ∠BAD=60°
△ABD为正三角形, Q为AD中点, ∴AD⊥BQ…………………………2分
∵PA=PD,Q为AD的中点,AD⊥PQ……………………………3分
又BQ∩PQ=Q ∴AD⊥平面PQB. ………………………………5分
(2)平面
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平面
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所以
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(3)存在,当
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由
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