如图在斜棱柱abc A1b1c1中底面为正三角形侧棱长等于底面边长且侧棱与底面所成的角为60°顶点b1在底面abc上的射影o恰好是 Sketch Coloring Page

如图在斜棱柱abc A1b1c1中底面为正三角形侧棱长等于底面边长且侧棱与底面所成的角为60°顶点b1在底面abc上的射影o恰好是 Template

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sketch #929 tree stump

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anonymous

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sketch #3575 Stitch by Laura Gutierrez Hinojosa

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anonymous

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sketch #3554 Dr. House by Carrieann Benthem

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anonymous

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anonymous

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tomski如图在斜棱柱abc a1b1c1中底面为正三角形侧棱长等于底面边长且侧棱与底面所成的角为60°顶点b1在底面abc上的射影o恰好是 coloring

4．如图在斜三棱柱ABC A1B1C1中侧面AA1B1B⊥底面ABC侧棱AA1与底面ABC成60°的角AA1=2底面ABC是边长为2 sketch template

如图在斜三棱柱ABC A1B1C1中侧面ABB1A1⊥底面ABC侧棱AA1与底面ABC成60°的角AA1=2．底面ABC是边长为2的正 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在三棱柱ABC A1B1C1中侧面AA1C1C⊥底面ABCAA1=A1C=AC=2AB=BCAB⊥BCO为AC的中点1 sketch template

如图在斜棱柱ABC A1B1C1中底面为正三角形侧棱长等于底面边长且侧棱与底面所成的角为60°顶点B1在底面ABC上的射影O恰好是 sketch template

如图在三棱柱ABC A1B1C1中侧面AA1C1C⊥底面ABCAA1=A1C=AC=2AB=BCAB⊥BCO为AC的中点1 sketch template

如图在斜棱柱ABC A1B1C1中底面为正三角形侧棱长等于底面边长且侧棱与底面所成的角为60°顶点B1在底面ABC上的射影O恰好是 sketch template

如图，在三棱柱ABC A1B1C1中，底面ABC是以AC为斜边的等腰直角三角形，侧面AA1C1C为菱形，点A1在底面上的投影为AC的中点D，且 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

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

如图在斜棱柱ABC A1B1C1中底面为正三角形侧棱长等于底面边长且侧棱与底面所成的角为60°顶点B1在底面ABC上的射影O恰好是 sketch template

题目如图已知三棱柱ABC A1B1C1的底面是正三角形侧面BB1C1C是矩形MN分别为BCB1C1的中点P为AM上一点过 sketch template

在体积为15的斜三棱柱ABC A1B1C1中S是C1C上的一点S ABC的体积为3则三棱锥S A1B1C1的体积为 A．1 B 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三棱柱ABC A1B1C1中B1B⊥平面A1B1C1AC=CB=CC1=2∠ACB=90°DE分别是A1B1CC1的中点 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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

18．如图已知斜三棱柱ABC A1B1C1中∠BCA=90°AC=BC=a点A1在底面ABC上的射影 恰为AC的中点DBA1⊥AC1 sketch template

如图在三棱柱ABC A1B1C1中B1B⊥平面A1B1C1AC=CB=CC1=2∠ACB=90°DE分别是A1B1CC1的中点 sketch template

百度教育 sketch template

如图在三棱柱ABC A1B1C1中 ABC为等边三角形侧棱AA1⊥平面ABCAB=2AA1=23DE分别为AA1BC1的中点 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

如图已知三棱柱ABC A1B1C1的侧棱与底面垂直 AA1=AB=AC=1AB⊥AC M是CC1的中点 N是BC的中点点P在线段 sketch template

如图，在直四棱柱中，底面为直角梯形，，，，为上一点高途 sketch template

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

18．如图已知斜三棱柱ABC A1B1C1中∠BCA=90°AC=BC=a点A1在底面ABC上的射影 恰为AC的中点DBA1⊥AC1 sketch template

如图在正三棱柱ABC 中AB=1点D在棱上BD=1若AD与平面BC所成的角为a则sina= ——青夏教育精英家教�—— sketch template

如图在正三棱柱ABC A1B1C1中点DE分别为BCB1C1的中点且AB=AA1=2．1求证A1E⊥C1D2求证A1E 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

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正三棱柱ABC A1B1C1的九条棱均相等D是BC上一点AD⊥C1D 1求证截面ABC1⊥侧面BCC1B1 2求 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

已知斜三棱柱ABC A1B1C1的底面为正三角形侧面A1ACC1为菱形∠A1AC=60°且平面A1ACC1⊥平面ABCM是C1C的中点 sketch template

如图在三棱柱ABC A1B1C1中∠BAC=90°AB=AC=2AA1=4A1在底面ABC的射影为BC的中点ED是B1C1的中点 sketch template

在正三棱柱ABC A1B1C1中若AB=2BB1则AB1与C1B所成的角的大小．题目和参考答案——青夏教育精英家教�—— sketch template

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