题目
题型:不详难度:来源:
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021024518-20342.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021024519-60063.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021024519-25871.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021024519-16305.gif)
(1)求证:
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021024519-88305.gif)
(2)设线段
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021024520-99321.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021024520-82154.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021024520-60028.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021024520-16926.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021024520-34905.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021024520-16926.gif)
(3)求二面角
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021024521-29968.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021024521-60074.gif)
答案
(2)略
(3)二面角
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021024521-29968.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021024521-87587.gif)
解析
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021024521-15471.gif)
所以BC⊥平面ABEF.
所以BC⊥EF. ……………………………………2分
因为⊿ABE为等腰直角三角形,AB=AE,
所以∠AEB=45°,
又因为∠AEF=45,
所以∠FEB=90°,即EF⊥BE. …………………3分
因为BC
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021024521-15471.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021024521-15471.gif)
BC∩BE=B
所以
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021024519-88305.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021024522-73639.jpg)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021024522-10016.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021024522-66391.jpg)
∴PMNC为平行四边形,所以PM∥CN. ………6分
∵CN在平面BCE内,PM不在平面BCE内,PM∥平面BCE ………8分
(III)由EA⊥AB,平面ABEF⊥平面ABCD,易知EA⊥平面ABCD.
作FG⊥AB,交BA的延长线于G,则FG∥EA.从而FG⊥平面ABCD,
作GH⊥BD于H,连结FH,则由三垂线定理知BD⊥FH.
∴ ∠FHG为二面角F-BD-A的平面角. …………………10分
∵ FA=FE,∠AEF=45°,∠AEF=90°, ∠FAG=45°.
设AB=1,则AE=1,AF=
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021024522-80045.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021024522-64239.gif)
在Rt⊿BGH中, ∠GBH=45°,BG=AB+AG=1+
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021024523-91495.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021024523-41944.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021024523-70852.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021024523-76343.jpg)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021024524-37707.jpg)
在Rt⊿FGH中,
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021024524-50926.gif)
∴ 二面角
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021024521-29968.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021024521-87587.gif)
核心考点
试题【 (本题满分12分) 如图,正方形所在平面与平面四边形所在平面互相垂直,△是等腰直角三角形(1)求证:;(2)设线段的中点为,在直线 上是否存在一点,使得?若】;主要考察你对空间几何体的结构特征等知识点的理解。[详细]
举一反三
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021024508-69397.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021024508-62822.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021024508-79354.gif)
(1)求直线
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021024509-82291.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021024509-43541.gif)
(2)求三棱锥
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021024509-31764.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021024509-86441.gif)
如图所示,在长方体
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021024500-58253.gif)
(1)求异面直线A1M和C1D1所成的角的正切值;
(2)证明:直线BM⊥平面A1B1M1
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021024500-33363.jpg)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021024453-62035.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021024453-93566.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021024453-81764.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021024454-46638.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021024454-60204.gif)
A.![]() | B.![]() | C.![]() | D.![]() |
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021024455-41980.jpg)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021024449-36326.gif)
A.![]() | B.![]() | C.![]() | D.![]() |
如图,在四棱锥
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021024438-86544.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021024438-48391.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021024439-17002.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021024439-16925.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021024439-82462.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021024439-18374.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021024438-48391.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021024440-44339.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021024440-86766.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021024440-44569.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021024440-42727.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021024440-84314.gif)
(1)求四棱锥
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021024438-86544.gif)
(2)求
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021024441-45177.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021024441-84375.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021024441-70999.jpg)
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