试题分析:(1)方法一:连接
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824044753874562.png)
交于菱形的中心
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824044753889292.png)
,过
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824044753889292.png)
作
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824044753921592.png)
,
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824044753936316.png)
为垂足,连接
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824044753952516.png)
,根据定义可知
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824044753967516.png)
为二面角
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824044753983577.png)
的平面角,在三角形
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824044753999456.png)
中求出此角即可;
方法二:设
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824044754030402.png)
与
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824044754045380.png)
交点为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824044753889292.png)
,以
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824044753889292.png)
为坐标原点,分别以
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824044754295559.png)
所在直线为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824044754311266.png)
轴
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824044754326310.png)
轴建立如图所示的空间直角坐标系, 设平面
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824044754342475.png)
,平面
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824044754373474.png)
的法向量分别为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824044754389462.png)
,利用
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824044754404654.png)
的公式进行计算.
(2)连接
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824044754420629.png)
,设直线
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824044754451400.png)
与直线
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824044754467414.png)
相交于点
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824044754482303.png)
,则四棱锥
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824044754513642.png)
与四棱锥
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824044754529625.png)
的公共部分为四棱锥
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824044754545600.png)
,过
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824044754482303.png)
作
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824044754576383.png)
平面
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824044754591534.png)
,
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824044754607289.png)
为垂足,然后求出
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824044754623359.png)
,利用体积公式
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824044754654919.png)
求解即可.
试题解析:(1)方法一:如图(1)连结AC、BD交于菱形的中心O,过O
作OG⊥AF,G为垂足. 连结BG、DG.
由BD⊥AC,BD⊥CF,得BD⊥平面ACF, 故BD⊥AF. 于是AF⊥平面BGD,
所以BG⊥AF,DG⊥AF,∠BGD为二面角B-AF-D的平面角. 3分
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408240447546696241.jpg)
由FC⊥AC,FC=AC=2,得∠FAC
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824044754701387.png)
,
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824044754716593.png)
.
由OB⊥OG,OB=OD=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824044754732387.png)
,得∠BGD=2∠BGO
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824044754747405.png)
.
即二面角B-AF-D的大小为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824044754763381.png)
. 6分
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408240447547947600.png)
方法二:设AC与BD交点为O,以O为坐标原点,分别以BD 、AC所在直线为x轴
y轴建立如图所示的空间直角坐标系
则A(0,-1,0),B(
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824044754810415.png)
,0,0),D(
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824044754732387.png)
,0,0),F(0,1,2)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824044754903656.png)
,
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824044754919807.png)
,
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824044754950802.png)
2分
设平面ABF,平面ADF的法向量分别为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824044754389462.png)
设
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824044754997647.png)
由
令
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824044755028643.png)
4分
同理可得
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824044755075649.png)
∴
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824044755091707.png)
∴
∴二面角B-AF-D的大小为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824044754763381.png)
6分
(2)如图(2)连EB、EC、ED,设直线AF与直线CE相交于点H,
则四棱锥E-ABCD与四棱锥F-ABCD的公共部分为四棱锥H-ABCD.
过H作HP⊥平面ABCD,所以平面ACFE⊥平面ABCD,
从而
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824044755137650.png)
. 7分
由
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408240447551531020.png)
,得
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824044755169541.png)
. 9分
又因为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408240447552001172.png)
故四棱锥
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824044755200604.png)
的体积
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408240447552311229.png)
. 12分