19．如图所示四棱锥p abcd的底面为直角梯形⊥平面e为ab中点 1求证平面⊥平面 2求直线与平面所成角的正弦值 Sketch Coloring Page

19．如图所示四棱锥p abcd的底面为直角梯形⊥平面e为ab中点 1求证平面⊥平面 2求直线与平面所成角的正弦值 Template

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tomski19．如图所示四棱锥p abcd的底面为直角梯形⊥平面e为ab中点 1求证平面⊥平面 coloring

如图在四棱锥P ABCD中底面ABCD为等腰梯形AB∥CDAD=BC=2对角线AC⊥BD于O∠DAO=60°且PO⊥平面ABCD sketch template

如图所示四棱锥P ABCD的底面ABCD是边长为a的正方形侧棱PA=a PB=PD=a则它的5个面中互相垂直的面有对——青夏教育 sketch template

19．如图所示四棱锥P ABCD中底面ABCD为直角梯形点E在棱PA上且PE=2EA 1 求异面直线PA与CD所成的角 2 sketch template

如图在四棱锥P—ABCD中PA⊥底面ABCDAB⊥ADAC⊥CD∠ABC=60°PA=AB=BCE是PC的中点1证明AE sketch template

如图所示在四棱锥P ABCD中底面四边形ABCD是菱形AC∩BD=O PAC是边长为2的等边三角形PB=PD=6AP=4AF．Ⅰ sketch template

如图在四棱锥P ABCD中侧面底面ABCD底面ABCD是菱形侧面PAD是等边三角形且PB与面PAD所成角为1求四棱锥P sketch template

一个四棱锥的底面为正方形其直观图与三视图如图所示则二面角P DC A的正弦值为 ABCD——青夏教育精英家教�—— sketch template

如图所示，四棱锥的底面是正方形，每条侧棱的长都是底 高途 sketch template

如图在四棱锥P ABCD中底面ABCD为直角梯形AD∥BC∠ADC=90°平面PAD⊥底面ABCDQ为AD的中点PA=PD=2 sketch template

如图已知四棱锥P ABCD中底面ABCD是直角梯形ABCDABC=45°DC=1AB=2PA平面ABCDPA=1 1求证 sketch template

19．如图所示四棱锥P ABCD的底面为直角梯形⊥平面E为AB中点 1求证平面⊥平面 2求直线与平面所成角的正弦值 sketch template

已知四棱锥如图5 1所示，其三视图如图5 2所示，其中正视图和侧视图都是直角三角形，俯视图是矩形 1求此四棱锥的体积； 2若E是PD sketch template

百度教育 sketch template

如图所示在四棱锥S ABCD中底面ABCD是正方形侧棱SD⊥平面ABCDSD=DC点E是SC的中点作EF⊥SB交SB于点F．1 sketch template

19．如图所示四棱锥P ABCD的底面为直角梯形⊥平面E为AB中点 1求证平面⊥平面 2求直线与平面所成角的正弦值 sketch template

19．如图在四棱锥P ABCD中底面ABCD是矩形MN分别为PABC的中点 PD⊥平面ABCD且PD=AD=CD=1 1 sketch template

如图所示，四棱锥的底面是正方形，每条侧棱的长都是底 高途 sketch template

如图已知四棱锥S ABCD的底面是直角梯形AB∥DC∠DAB=90°SA⊥底面ABCD且SA=AD=DC=12AB=1M是SB的中点 sketch template

如图所示四棱锥P ABCD中底面ABCD是正方形PA⊥底面ABCDPA=AB=2点F是PB的中点点E在棱BC上移动．1当E为 sketch template

如图所示正四棱锥P ABCD的各棱长均为13MN分别为PABD上的点且PM∶MA=BN∶ND=5∶8 1求证直线MN∥平面 sketch template

如图四棱锥P ABCD中PD⊥平面ABCD底面ABCD是正方形PD=AB=2E为PC中点．1求证DE⊥平面PCB2求点C sketch template

如图所示在四棱锥P ABCD中底面ABCD为菱形且∠DAB=60°PA=PDM为CD的中点BD⊥PM．1求证平面PAD⊥平面 sketch template

如图所示的四棱锥中，平面，，．高途 sketch template

17． 如图所示在四棱锥P ABCD中∠ABC＝∠ACD＝90°∠BAC＝∠CAD＝60°PA⊥平面ABCDE为PD的中点PA sketch template

17． 如图所示在四棱锥P ABCD中∠ABC＝∠ACD＝90°∠BAC＝∠CAD＝60°PA⊥平面ABCDE为PD的中点PA sketch template

17． 如图所示在四棱锥P ABCD中∠ABC＝∠ACD＝90°∠BAC＝∠CAD＝60°PA⊥平面ABCDE为PD的中点PA sketch template

19． 如图四棱锥P ABCD中底面四边形ABCD是正方形侧面PDC是边长为a的正三角形且平面PDC⊥底——青夏教育精英家教�—— sketch template

19．已知四棱锥P ABCD的底面ABCD的边长为1的正方形 PD⊥底面ABCD且PD=1 1求证BC//平面PAD 2若 sketch template

如图所示，底面边长为的正四棱锥被平行于其底面的平面 高途 sketch template

17． 如图所示在四棱锥P ABCD中∠ABC＝∠ACD＝90°∠BAC＝∠CAD＝60°PA⊥平面ABCDE为PD的中点PA sketch template

如图所示侧面是全等的等腰三角形的棱锥是 A正棱锥 B侧棱长相等的棱锥C斜高长相等的棱锥 D以上均不对题目和参考答案——青夏教育精英 sketch template

如图所示正四棱锥P ABCD的各棱长均为13MN分别为PABD上的点且PM∶MA=BN∶ND=5∶8 1求证直线MN∥平面 sketch template

如图所示在四棱锥P ABCD中底面ABCD是正方形侧棱PD⊥底面ABCDPD=DCE是PC的中点作EP⊥PB交PB于点F1证明 sketch template

百度教育 sketch template

如图所示在四棱锥P ABCD中底面ABCD是矩形PA⊥平面ABCDPA=AD=2AB=1BM⊥PD垂足为M．Ⅰ求证AM⊥ sketch template

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