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

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

正三棱锥的高是多� 百度经验 sketch template

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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

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

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

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

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

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

如图在三棱锥P ABC中PA=PB=PC=AC=4AB=BC=2．1若点P在底面ABC内的射影是点O试指出点O的位置并说明理由 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

81 基本立体图形1 棱柱、棱锥、棱台湛江贵哥讲数学博客园 sketch template

如图，平面四边形 ，将沿折起到的位置，此时二面角的大小为 连接则三棱锥外接球的表面积为；三棱锥的体积为 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

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

7．一个四棱锥的底面为正方形其三视图如图所示则这个四棱锥的体积是 A． B． C． D．——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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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

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

如图已知在四棱锥P ABCD中底面ABCD是边长为4的正方形 PAD是正三角形EFG分别是PDPCBC的中点．1求证直线 sketch template

如图在三棱柱ABC A1B1C1中侧棱垂直于底面AB⊥BCEF分别是A1BAC1的中点．1求证平面AEF⊥平面AA1B1B 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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六棱锥P ABCDEF其中底面ABCDEF是正六边形点P在底面的投影是正六边形的中心底面边长为2cm侧棱长为3cm求六棱锥P sketch template

已知斜三棱柱ABC A1B1C1的侧面A1ACC1与底面ABC垂直∠ABC＝90°BC＝2AC＝2且AA1⊥A1CAA1＝A1C sketch template

如图，已知三棱柱ABC A1B1C1的底面是正三角形，侧面BB1C1C是矩形，M，N分别为BC，B1C1的中点，P为AM上一点，过B1C1和P 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

六棱柱有12个顶点18条棱8个面．题目和参考答案——青夏教育精英家教�—— sketch template

19． 已知斜三棱柱ABC A1B1C1的高考资^源�侧面A1ACC1与底面ABC垂直∠ABC＝90°BC＝2AC＝2且AC 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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

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