11．如图正四面体a Bcd的棱长为a点ef分别是棱bdbc的中点则平面aef截该正四面体的内切球所得截面的面积为 A． B Sketch Coloring Page

11．如图正四面体a Bcd的棱长为a点ef分别是棱bdbc的中点则平面aef截该正四面体的内切球所得截面的面积为 A． B Template

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tomski11．如图正四面体a bcd的棱长为a点ef分别是棱bdbc的中点则平面aef截该正四面体的内切球所得截面的面积为 a． coloring

深入浅出数学系列立体几何（3）：正四面体的性质知乎 sketch template

深入浅出数学系列立体几何（3）：正四面体的性质知乎 sketch template

正四面体ABCD棱长为2，E、F分别为BC、AD中点，则EF的长为 百度教育 sketch template

如图，正四面体的长为1，， 则． sketch template

已知正四面体ABCD的棱长为1求1该四面体的内切球的表面积2与该四面体各条棱均相切的球的体积3该四面体的外接球上AB两点间 sketch template

求棱长为a的正四面体的内切球、棱切球的半径百度教育 sketch template

高中数学：几何体表面积与体积的教学设计知乎 sketch template

15．如图2正四面体ABCD的棱长为1H是顶点D的底面 ABC上的射影M在DH上且使得∠ABC=90°则DM的长 为 ——青夏教育 sketch template

7．如图正四面体的棱长均为且平面于A点BCD均在平面外且在平面同一侧则点B到平面的距离是 A． B． C． D．——青夏教育精英 sketch template

11．如图正四面体A BCD的棱长为a点EF分别是棱BDBC的中点则平面AEF截该正四面体的内切球所得截面的面积为 A． B sketch template

正四面体 中心到顶点距离为a，求四个顶点坐标百度知道 sketch template

任意四面体的外接球半径公式知乎 sketch template

在棱长为a的正四面体中，相对两条棱间的距离为 百度教育 sketch template

正四面体怎么建立空间直角坐标系�？知乎 sketch template

一个正四面体的各条棱长都是a那么这个正四面体的高百度经验 sketch template

已知菱形的边长为2，将沿折起，使得点至点的位置，得到四面体当二面角的大小为120°时，四面体的体积为；当四面体的体积为1时，以为球心，的长 sketch template

正四面体一个顶点处的立体角是多大？知乎 sketch template

15．如图2正四面体ABCD的棱长为1H是顶点D的底面 ABC上的射影M在DH上且使得∠ABC=90°则DM的长 为 ——青夏教育 sketch template

5．如图A BCD是一体积为的正四面体连结两个面的重心EF则线段EF的长是 ——青夏教育精英家教�—— sketch template

如图已知正四面体A BCD的棱长为1EF分别为棱ABCD的中点 1建立适当的空间直角坐标系写出顶点ABCD的坐标 2 sketch template

由四个全等的等边三角形的封面几何体称为正四面体如图正四面体ABCD中EF分别是棱BCAD的中点CF与DE是一对异面直线在图形中 sketch template

为什么正四面体顶点在底面的投影在底面三角形的中心�？知乎 sketch template

正四面体一个顶点处的立体角是多大？知乎 sketch template

正四面体的中心到底面的距离与这四面体的高的比是（）A12B13C14D16 百度教育 sketch template

如图四面体ABCD的每条棱长都等于2点EF分别为棱ABAD的中点则AB BC= BC EF= EF与AC所成的角为 sketch template

如图，已知正三棱柱ABC A 1B 1C 1的所有棱长都为2，D为CC 1中点，E为BC的中点（1）求证：BD⊥ 平面AB 1E；（2）求三 sketch template

正四面体 360百� sketch template

截角四面体是一种半正八面体可由四面体经过适当的截角即截去四面体的四个顶点所产生的多面体如图将棱长为3的正四面体沿棱的三等分点作平行于底 sketch template

百度教育 sketch template

百度教育 sketch template

$如图已知正四面体D ABC所有棱长均相等的三棱锥PQR分别为ABBCCA上的点AP=PB$frac{BQ}{QC sketch template$

题目半正多面体亦称“阿基米德多面体 是由边数不全相同的正多边形为面组成的多面体如将正四面体所有棱各三等分沿三等分点从原几何体割去四个 sketch template

如图是一个正四面体它的四个面都是正三角形现沿它的三条棱ACBCCD剪开展成平面图形则所得的展开图是 ——青夏教育精英家教�—— sketch template

如图，已知正四面体的棱长为，，分别为棱，的中点．若高途 sketch template

11．如图四棱锥P ABCD的底面ABCD是边长为2的正方形侧面PAB是正三角形且与底面垂直E为PD中点过直线BC和点E的平面与侧棱 sketch template

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