如图四棱锥p Abcd它的正视图如图1是等腰三角形侧视图如图2是等腰直角三角形俯视图如图3是正方形abcd各长度如图 Sketch Coloring Page

如图四棱锥p Abcd它的正视图如图1是等腰三角形侧视图如图2是等腰直角三角形俯视图如图3是正方形abcd各长度如图 Template

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tomski如图四棱锥p abcd它的正视图如图1是等腰三角形侧视图如图2是等腰直角三角形俯视图如图3是正方形abcd各长度如图 coloring

如图在四棱锥P ABCD中PA⊥平面ABCD底面ABCD为平行四边形且AB=1BC=2EF分别为ADBC的中点．I求证 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PA⊥底面ABCDAB⊥ADAC⊥CD∠ABC=60°PA=AB=BCE是PC的中点1证明AE sketch template

四棱锥俯视图正四棱锥的俯视图正视图侧视图俯视图 大山谷图库 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

如图，在四棱锥P ABCD中，底面ABCD是边长为1的正方形，PA⊥底面ABCD，点M是棱PC的中点，AM⊥平面PBD．（1）求四棱锥P sketch template

百度教育 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

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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

如图四棱锥P ABCD的底面ABCD为一直角梯形侧面PAD是等边三角形其中BA⊥ADCD⊥ADCD=2AD=2AB=2平面PAD⊥ 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}$AD∠BAD=∠ABC=90 sketch template$

四棱锥的直观图正四棱锥的直观图画正四棱锥直观图第2页大山谷图库 sketch template

2019年高考真题理�数学北京卷百度题库 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

如图四棱锥P ABCD中底面ABCD为矩形PA⊥平面ABCDE为PD的中点．1证明PB∥平面AEC2设AP=AB=1AD sketch template

如图四棱锥P ABCD中PD⊥平面ABCDPD=DC=BC=1AB=2AB∥DC∠BCD=900 1求证PC⊥BC 2求 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

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

【题目� 如图，在四棱锥 P ABCD 中，底面为正方形， PD ⊥平面，＝ AD 百度教育 sketch template

如图四棱锥P ABCD它的正视图如图1是等腰三角形侧视图如图2是等腰直角三角形俯视图如图3是正方形ABCD各长度如图 sketch template

如图四棱锥P ABCD中底面ABCD是边长为a的正方形PD=aPA=PC=a 1求证PD⊥平面ABCD 2求异面直线PB sketch template

如图四棱锥P ABCD的底面是直角梯形AB∥CDAB⊥AD PAB和 PAD是两个边长为2的正三角形DC=4O为BD的中点．1 sketch template

如图在四棱锥P ABCD中PA⊥平面ABCD四边形ABCD为直角梯形AD//BC且AD&gtBC∠DAB=∠ABC=90°PA=AB sketch template

如图在四棱锥P ABCD中PA⊥底面ABCDAD∥BCAB⊥ADAC与BD交于点OPA=3AD=2AB=23BC=6．Ⅰ 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的底面为正方形PA平面ABCDPA=AB=2F为PA上的点 1求证无论F点在PA上如何运动都有BDFC 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

四棱锥P ABCD的底面是边长为3的正方形PD⊥平面ABCD．异面直线AD与PB所成角为60°E为线段PC上一点PE=2EC．求二面角P 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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在四棱锥P ABCD中PA⊥平面ABCD底面ABCD为直角梯形AB⊥ADAB∥CDCD=AD=2AB=2AP．1求证平面 sketch template

如图四棱锥P ABCD中底面ABCD是边长为2的正方形PB⊥BCPD⊥CD且PA=2E为PD中点．1求二面角B EC A的正弦 sketch template

如图四棱锥P ABCD中PA⊥平面ABCD底面ABCD是直角梯形且AB∥CD∠BAD=90°PA=AD=DC=2AB=4．1 sketch template

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