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abcd——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abcd——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abcd——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ABCD——青夏教育 sketch template

正三棱锥的高是多� 百度经验 sketch template

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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

已知四棱锥P ABCD它的底面是边长为a的菱形且∠ABC＝120°PC⊥平面ABCD又PC＝aE为PA的中点． 1求证平面 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

11．底面边长为2的正三棱锥P ABC中E是BC的中点若 PAE的面积为则侧棱PA与底面所成角的正切值是 A．1 B． C． D．——青 sketch template

已知正三棱柱ABC A1B1C1的底面边长为2高为4过BC作一截面截面与底面ABC成600角则截面面积是 A．4 B． C． D 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

2．如图正三棱锥P ABC顶点P在底面上的射影在ΔABC内部M是侧面PAB上的点且M到点P的距离等于M到底面的距离则点M的轨迹是 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

详解三角面片：实现现实场景的数字化 CSDN博客 sketch template

在四棱锥中其底面是正方形一条侧棱垂直于底面不通过此棱的一个侧面与底面所成的二面角为45°且最长的侧棱长为15cm则棱锥的高为 sketch template

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

如图已知四棱锥P ABCDPA⊥平面ABCD底面ABCD为直角梯形∠A=90°且AB//CDAB=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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

6．如图在底面为直角梯形的四棱锥 ABCDBC=6 Ⅰ求证 Ⅱ求二面角的平面角的余弦值解Ⅰ如图建立坐标系则 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⊥ADCD⊥ADCD=2AD=2AB=2平面PAD⊥ sketch template

四棱锥的顶点在底面上的投影恰好是其正视图与侧视图都是腰长为的等腰直角三角形则在四棱锥的任意两个顶点的连线中互相垂直的异面直线共有对 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° B45° C60° D90 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ABCDSD=DC点E是SC的中点作EF⊥SB交SB于点F．1 sketch template

如图已知四棱锥的底面是边长为4的正方形在底面上的射影落在正方形内的长为到的距离分别为和 是的中点． Ⅰ求证平面底面 Ⅱ设 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 ABCDPA⊥平面ABCD底面ABCD为直角梯形∠A=90°且AB//CDAB=CD 1点F在线段PC上运动 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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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⊥底面ABCDE为PC的中点 ——青夏 sketch template

如图已知四棱锥S ABCD的侧面SCD⊥底面ABCDBC∥ADBC⊥SC且SC=SD=CD=BC=2AD =2P为SB的中点求 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

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