如图已知三棱锥的底面是直角三角形直角边长分别为3和4过直角顶点的侧棱长为4且垂直于底面该三棱锥的主视图是abcd——青夏教育 Sketch Coloring Page

如图已知三棱锥的底面是直角三角形直角边长分别为3和4过直角顶点的侧棱长为4且垂直于底面该三棱锥的主视图是abcd——青夏教育 Template

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tomski如图已知三棱锥的底面是直角三角形直角边长分别为3和4过直角顶点的侧棱长为4且垂直于底面该三棱锥的主视图是abcd——青夏教育 coloring

正三棱锥的高是多� 百度经验 sketch template

如图已知三棱锥的底面是直角三角形直角边长分别为3和4过直角顶点的侧棱长为4且垂直于底面该三棱锥的主视图是ABCD——青夏教育 sketch template

初学讲义之高中数学二十：简单几何体知乎 sketch template

已知三棱锥的三视图如右图所示其中侧视图为直角三角形俯视图为等腰直角三角形则此三棱锥的体积等于 A． B． C． D．题目和参考 sketch template

如图，在三棱锥P ABC中，AB=BC=2√ 2，PA=PB=PC=AC=4，O为AC的中点（1）证明：PO⊥ 平面ABC；（2）若点M在棱 sketch template

一个三棱锥的各个面均为正三角形以每个面的中心为顶点构成一个新三棱锥若新三棱锥体积为2求原三棱锥体百度知道 sketch template

三棱锥专题 Word文档在线阅读与下载免费文档 sketch template

如图已知三棱锥的底面是直角三角形直角边长分别为3和4过直角顶点的侧棱长为4且垂直于底面该三棱锥的主视图是ABCD——青夏教育 sketch template

已知三棱锥P ABC中顶点P在底面的射影O是三角形ABC的内心关于这个三棱锥有三个�题�侧棱PA=PB=PC�侧棱PAPBPC两两 sketch template

下面有四个�题1各个侧面都是等腰三角形的棱锥是正棱锥2三条侧棱都相等的棱锥是正棱锥3底面是正三角形的棱锥是正三棱锥4 sketch template

14． 三棱锥P ABC的四个顶点点在同一球面上若PA⊥底面ABC底面ABC是直角三角形PA=2 AC=BC=1则此球的表面积为 sketch template

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——青夏教育精英家教�—— sketch template

如图，已知正三棱锥P ABC的侧面是直角三角形，PA=6，顶点P在平面ABC内的正投影为点D，D在平面PAB内的正投影为点E，连接PE并延长交 sketch template

如图，已知圆锥的顶点为，底面圆心为，高为，底面半径高途 sketch template

如图已知四棱锥P ABCD中底面ABCD是直角梯形ABCDABC=45°DC=1AB=2PA平面ABCDPA=1 1求证 sketch template

如图，在四棱锥中，底面是梯形，，侧面是等边三角形，侧面是等腰直角三角形， 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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

如图已知正三棱柱 的底面边长为2侧棱长为点E在侧棱上点F在侧棱上且 I 求证 II 求直线与平面的夹角题目和参考 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

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=1M是SB的中点 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

如图已知四棱锥P ABCD中底面ABCD是直角梯形ABCDABC=45°DC=1AB=2PA平面ABCDPA=1 1求证 sketch template

20．如图在四棱锥p ABCD中底面ABCD是一直角梯形∠BAD=90°AD∥BCAB=BC=aAD=2a且PA⊥底面ABCD 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=2EF分别为PCBD的中点．Ⅰ sketch template

如图已知四棱锥P ABCD的底面为直角梯形AB∥CD∠DAB=90°PA⊥底面ABCD且PA=AD=DC=12AB=1M是PB的 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已知正三棱柱 的底面边长为2侧棱长为点E在侧棱上点F在侧棱上且 I 求证 II 求直线与平面的夹角题目和参考 sketch template

四棱锥 360百� sketch template

题目如图四棱锥的底面是矩形点为的中点与交于点Ⅰ求异面直线与所成角的余弦值Ⅱ求证Ⅲ求直线与平面所成角的正弦值 sketch template

如图已知六棱锥的底面是正六边形平面则下列结论正确的是 直线∥平面平面平面直线与平面所成的角为——青夏教育精英家教�—— sketch template

如图已知四棱锥P ABCD中底面ABCD是直角梯形ABCDABC=45°DC=1AB=2PA平面ABCDPA=1 1求证 sketch template

如图斜三棱柱ABC A1B1C1的底面是Rt ABC∠A＝90 °且BC1⊥AC作C1 H⊥底面ABC垂足为H�试判断H点的位置 sketch template

如图，在平行六面体ABCD A 1B 1C 1D 1中，底面是边长为2的正方形，若∠ 1AB=∠ 1AD=60^° ，且A 1A sketch template

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