题目内容
如图,三棱柱ABC-A1B1C1的侧棱AA1⊥底面ABC,∠ACB=90°,E是棱CC1的中点,F是AB的中点,AC=BC=1,AA1=2.
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/201408240400289123750.jpg)
(1)求证:CF∥平面AB1E;
(2)求三棱锥C-AB1E在底面AB1E上的高.
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/201408240400289123750.jpg)
(1)求证:CF∥平面AB1E;
(2)求三棱锥C-AB1E在底面AB1E上的高.
(1)见解析(2)![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824040028928419.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824040028928419.png)
(1)证明:取AB1的中点G,连接EG,FG,
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/201408240400289283986.jpg)
∵F、G分别是AB、AB1的中点,
∴FG∥BB1,FG=
BB1.
∵E为侧棱CC1的中点,
∴FG∥EC,FG=EC,
∴四边形FGEC是平行四边形,
∴CF∥EG,∵CF?平面AB1E,EG?平面AB1E,
∴CF∥平面AB1E.
(2)∵三棱柱ABC-A1B1C1的侧棱AA1⊥底面ABC,
∴BB1⊥平面ABC.
又AC?平面ABC,∴AC⊥BB1,∵∠ACB=90°,∴AC⊥BC,
∵BB1∩BC=B,∴AC⊥平面EB1C,∴AC⊥CB1,
∴VA-EB1C=
S△EB1C·AC
=
×
×1=
.
∵AE=EB1=
,AB1=
,∴S△AB1E=
,
∵VC-AB1E=VA-EB1C,∴三棱锥C-AB1E在底面AB1E上的高为
.
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/201408240400289283986.jpg)
∵F、G分别是AB、AB1的中点,
∴FG∥BB1,FG=
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824040028944338.png)
∵E为侧棱CC1的中点,
∴FG∥EC,FG=EC,
∴四边形FGEC是平行四边形,
∴CF∥EG,∵CF?平面AB1E,EG?平面AB1E,
∴CF∥平面AB1E.
(2)∵三棱柱ABC-A1B1C1的侧棱AA1⊥底面ABC,
∴BB1⊥平面ABC.
又AC?平面ABC,∴AC⊥BB1,∵∠ACB=90°,∴AC⊥BC,
∵BB1∩BC=B,∴AC⊥平面EB1C,∴AC⊥CB1,
∴VA-EB1C=
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824040028959327.png)
=
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824040028959327.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824040028990708.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824040029022336.png)
∵AE=EB1=
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824040029037344.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824040029037341.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824040029053453.png)
∵VC-AB1E=VA-EB1C,∴三棱锥C-AB1E在底面AB1E上的高为
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/201408240400290681224.png)
![](http://thumb.zyjl.cn/images/loading.gif)
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