解法一 I直三棱柱abc A1b1c1中cc1⊥平面abc 1分底面三边长ac=3bc=4ab=5∴ ac⊥bc 2分且 Sketch Coloring Page

解法一 I直三棱柱abc A1b1c1中cc1⊥平面abc 1分底面三边长ac=3bc=4ab=5∴ ac⊥bc 2分且 Template

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tomski解法一 i直三棱柱abc a1b1c1中cc1⊥平面abc coloring

解法一 I直三棱柱ABC A1B1C1中CC1⊥平面ABC 1分底面三边长AC=3BC=4AB=5∴ AC⊥BC 2分且 sketch template

如图，在直三棱柱ABC—A1B1C1中，∠ACB＝90°，∠BAC＝30°，BC＝1，AA1＝，M是棱CC1的中点． 1 求证：A1B⊥ sketch template

已知直三棱柱ABC A1B1C1中，侧面AA1B1B为正方形 sketch template

19．解1在直三棱柱ABC A1B1C1中 1∵ ∴为异面直线AD与A1B1所成的角 2分连结BD 在中 sketch template

【题文�如图，在直三棱柱ABC－A1B1C1中，，AC＝3，BC＝4，AA1＝4，点D是AB的中点，（1）求证：AC⊥BC1；（2）求证 sketch template

如图所示在直三棱柱ABC A1B1C1中AB＝1AC＝AA1＝∠ABC＝60° 1证明AB⊥A1C 2求二面角A A1C sketch template

直三棱柱ABC A1B1C1的底面是直角三角形其中斜边BC= C 二面角B A1C A的大小为45°求这棱柱的侧面积广水一中 sketch template

如图在三棱柱ABC A1B1C1中B1B⊥平面A1B1C1AC=CB=CC1=2∠ACB=90°DE分别是A1B1CC1的中点 sketch template

如图在直三棱柱ABC A1B1C1中三角形ABC为等腰直角三角形AC=BC=$sqrt{2}$AA1=1点D是AB的中点．1 sketch template

如图在三棱柱ABC A1B1C1中B1B⊥平面A1B1C1AC=CB=CC1=2∠ACB=90°DE分别是A1B1CC1的中点 sketch template

如图已知正三棱柱ABC A1B1C1的各条棱长均为2M是BC的中点．Ⅰ求证A1C∥平面AB1MⅡ求证在棱CC1上找一点N使得 sketch template

如图已知三棱柱ABC A1B1C1的侧棱与底面垂直 AA1=AB=AC=1AB⊥AC M是CC1的中点 N是BC的中点点P在线段 sketch template

如图，已知三棱柱ABC A1B1C1 ， 平面A1AC1C⊥平面ABC，∠ABC=90°∠BAC=30°，A1A=A1C=AC，E，F分别是 sketch template

如图，直三棱柱ABC A1B1C1中，CA=CB=1，∠BCA=90°，棱AA1=2，M，N分别是A1B1，A1A的中点（1）求BN−→−的 sketch template

17．在三棱柱ABC A1B1C1中A1B1是A1C和B1C1的公垂线段A1B与平面ABC成60°角AB=A1A=AC=2 1 sketch template

7．如图在正三棱柱ABC A1B1C1中AB=3AA1=4M为AA1的中点P是BC上一点且由P沿棱柱侧面经过棱CC1到M点的最短 sketch template

正三棱柱正三棱锥正三棱锥直观图大山谷图库 sketch template

如图在直三棱柱ABC A1B1C1中∠ACB=90°AC=BC=CC1=2．1求证AB1⊥BC1．2求二面角C AC1 B的 sketch template

百度教育 sketch template

斜三棱柱ABC A1B1C1的底面是边长为2的正三角形顶点A1在底面的射影O是 ABC的中心异面直线AB与CC1所成的角为45° 1 sketch template

如图所示直三棱柱ABC A1B1C1中AC=BC=1∠ACB=90°点D为AB的中点．1求证BC1∥面A1DC2求棱AA1的长 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

11 直三棱柱ABC A1B1C1中 ∠A1B1C1=90° 且AB=BC=BB1 E F分别是AB CC1的中点那么A1C sketch template

如图，在直三棱柱ABC A1B1C1中，AC⊥BC，AC=BC=CC1=2，点D，E，F分别为棱A1C1，B1C1，BB1的中点．（Ⅰ）求证 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

18．如图在直三棱柱中为中点 Ⅰ求证Ⅱ求证平面平面 Ⅲ求三棱锥的体——青夏教育精英家教�—— sketch template

9．在斜三棱柱ABC A1B1C1中A0B0分别为侧棱AA1BB1上的点且知BB0B0B1=32过A0B0C1的截面将三 sketch template

三棱柱ABC A1B1C1的底面是边长为4的正三角形侧面AA1C1C是菱形 PA⊥BC点P是A1C1的中点∠C1CA=60° 1 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

在直三棱柱ABC A1B1C1 中AC=4CB=2AA1=2∠ACB=60°EF分别是A1C1BC的中点．1证明C1F∥ sketch template

如图所示在直三棱柱ABC A1B1C1中底面是等腰三角形∠ACB=90°侧棱AA1=2CA=2D是CC1的中点试问在线段A1B上 sketch template

已知直三棱柱ABC A1B1C1中，侧面AA1B1B为正方形 sketch template

如图在直三棱锥ABC A1B1C1中AC=BC=AA1∠ACB=90°EF分别是ABBC的中点G是AA1上的点若一个三棱锥的三 sketch template

百度教育 sketch template

如图直三棱柱ABC A1B1C1中AB⊥BCD为AC的中点AA1=AB=2．1求证AB1∥平面BC1D2若四棱锥B sketch template

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