如图正四棱锥s Abcd 的底面是边长为a正方形o为底面对角线交点侧棱长是底面边长的2倍p为侧棱sd上的点．Ⅰ求证ac⊥sd Sketch Coloring Page

如图正四棱锥s Abcd 的底面是边长为a正方形o为底面对角线交点侧棱长是底面边长的2倍p为侧棱sd上的点．Ⅰ求证ac⊥sd Template

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tomski如图正四棱锥s abcd 的底面是边长为a正方形o为底面对角线交点侧棱长是底面边长的2倍p为侧棱sd上的点．Ⅰ求证ac⊥sd coloring

正四棱锥图册 360百� 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

$底面边长和侧棱长均为2的正四棱锥的体积为$frac{4sqrt{2}}{3}$．题目和参考答案——青夏教育精英家教�—— sketch template$

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是正方形侧面PAD⊥底面ABCD且PA=PD=AD=2EF分别为PCBD的中点．Ⅰ sketch template

如图在四棱锥P—ABCD中PA⊥底面ABCDAB⊥ADAC⊥CD∠ABC=60°PA=AB=BCE是PC的中点1证明AE sketch template

如图正四棱锥S ABCD 的底面是边长为a正方形O为底面对角线交点侧棱长是底面边长的2倍P为侧棱SD上的点．Ⅰ求证AC⊥SD sketch template

正四棱锥的直观图画法千图� sketch template

正四棱锥S ABCD的侧棱长为底面边长为E是SA的中点则异面直线BE和SC所成的角等于 A．30° B45° C60° D90 sketch template

如图，正四棱锥S ABCD，SA=SB=SC=SD=4，，P为侧棱SD上的点，且SP=3PD．（1）求正四棱锥S ABCD的表面积；（2）求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

20．（12分）如图，四棱锥P ABCD的底面为正方形，PD⊥底面ABCD．设平面PAD与平面PBC的交线为l．（1）证明：l⊥平面PDC 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

14．已知正四棱锥P ABCD的高为4侧棱与底面所成的角为600则该正四棱锥的侧面积是 ．——青夏教育精英家教�—— sketch template

如图在四棱锥P ABCD中PA⊥底面ABCDAB⊥ADBC∥ADPA=AB=BC=2AD=4．1求四棱锥P ABCD的体积2 sketch template

如图已知四棱锥S ABCD的侧面SCD⊥底面ABCDBC∥ADBC⊥SC且SC=SD=CD=BC=2AD =2P为SB的中点求 sketch template

如图，在四棱锥P ABCD中，底面ABCD是平行四边形，∠BCD=120°，侧面PAB⊥底面ABCD，PB=2√6，AB=AC=PA=2√3 sketch template

$如图四棱锥P ABCD中侧面PAD为等边三角形且垂直于底面ABCDAB=BC=$frac{1}{2}$AD∠BAD=∠ABC=90 sketch template$

如图在底面是菱形的四棱锥 P ABCD中∠ABC=60°PA⊥平面ABCD点EFG分别为CDPDPB的中点．PA=AD=2 sketch template

如图正四棱锥S ABCD的所有棱长都相等过底面对角线 AC作平行于侧棱SB的截面交SD于E 1求AB与SC所成角的大小 2求 sketch template

如图，在四棱锥P ABCD中，侧面PAD⊥ 底面ABCD，侧棱PA=PD=√2，底面ABCD为直角梯形，其中BC//AD，AB⊥ AD，AD sketch template

如图正四棱锥S ABCD的所有棱长都相等过底面对角线 AC作平行于侧棱SB的截面交SD于E 1求AB与SC所成角的大小 2求 sketch template

如图已知两个正四棱锥P ABCD与Q ABCD的高分别为1和2AB=4 � 证明PQ⊥平面ABCD � 求异面直线AQ与PB所成的角 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PCB2求点C sketch template

如图已知四棱锥P ABCDPA⊥平面ABCD底面ABCD为直角梯形∠A=90°且AB//CDAB=CD 1点F在线段PC上运动 sketch template

四棱锥 360百� 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在四棱锥P ABCD中底面ABCD是矩形侧棱PA垂直于底面EF分别是ABPC的中点 1求证CD⊥PD 2求证 sketch template

19．如图四棱锥S ABCD的底面是直角梯形∠ABC＝∠BCD＝90°AB＝BC＝SB＝SC＝2CD＝2侧面SBC⊥底面ABCD sketch template

如图已知两个正四棱锥P ABCD与Q ABCD的高分别为1和2AB=4 � 证明PQ⊥平面ABCD � 求异面直线AQ与PB所成的角 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正四棱锥，，，为侧棱上的点，且．高途 sketch template

高中数学第一册上棱锥 Word文档在线阅读与下载无忧文档 sketch template

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

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