1． Abc的三边长分别为ac=3 Bc=4 Ab=5 以ab所在直线为轴将此三角形旋转一� 求所得旋转体的表面积——青夏 Sketch Coloring Page

1． Abc的三边长分别为ac=3 Bc=4 Ab=5 以ab所在直线为轴将此三角形旋转一� 求所得旋转体的表面积——青夏 Template

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sketch #2996 American Flag

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sketch #2895 Hulk by Dmitry Cherepanin

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sketch #4264 oh yeah

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sketch #4264 oh yeah

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sketch #4834 Aayla Secura by Nînâ Yämâhä

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tomski1． abc的三边长分别为ac=3 bc=4 coloring

1． ABC的三边长分别为AC=3 BC=4 AB=5 以AB所在直线为轴将此三角形旋转一� 求所得旋转体的表面积——青夏 sketch template

百度教育 sketch template

如图以 ABC的三边为边长在BC的同侧作三个等边三角形即 ABD BCE ACF．1四边形ADEF是什么四边形试说明理由．2 sketch template

5．斜三棱柱ABC A1B1C1的底面是边长为a的正三角形侧棱长等于b一条侧棱AA1与底面相邻两边ABAC都成450角求这个三棱柱的侧 sketch template

23．已知 ABC的三边长分别为．．且．．满足等式试说明该三角形是等边三角形．——青夏教育精英家教�—— 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

如图所示，已知边长为4的正三角形ABC中，E，F分别为BC和AC的中点，PA=2，且PA⊥平面ABC，设Q是CE的中点（1）求证：AE∥平面 sketch template

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 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

正三棱锥S ABC的侧棱长和底面边长相等如果EF分别为SCAB的中点那么异面直线EF与SA所成角为A．90°B．60°C．45°D sketch template

如图 ABC是边长为2的等边三角形D是AB边的中点P是BC边上的动点Q是AC边上的动点当PQ的位置在何处时才能使 DPQ的�长最小 sketch template

如图已知正三角形ABC的边长为1EFG分别是ABBCCA上的点且AE=BF=CG设 EFG的面积为yAE的长为x则y关于x sketch template

如图在Rt ABC中∠ACB=90°AC=3BC=4将边AC沿CE翻折使点A落在AB上的点D处再将边BC沿CF翻折使点B落在 sketch template

如图 ABC是等边三角形分别延长CAABBC到A′B′C′使AA′=BB′=CC′=AC若 ABC的面积为1则 A′B′C sketch template

如图所示，分别以 ABC的边AB、AC为边，�三角形的外侧作正方形ABDE和正方形ACFG，点M为BC中点，1求证：AM⊥EG．2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

如图在直三棱锥ABC A1B1C1中AC=BC=AA1∠ACB=90°EF分别是ABBC的中点G是AA1上的点若一个三棱锥的三 sketch template

如图 ABC是边长为4的等边三角形以BC为底边作一个顶角为120°的等腰三角形 DBC以D为顶点作∠EDF=60°使点EF分别在边 sketch template

18．已知如图1正三角形ABC 的边长为2aCD 是AB 边上的高EF分别是AC 和BC 边上的点且满足现将 ABC 沿CD 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 ABC中D为BC的中点G为AD的中点过点G任作一直线MN分别交ABAC于MN两点若AM=xABAN=yAC．试问1x sketch template

如图以 ABC的三边为边在BC的同一侧分别作三个等边三角形即 ABD BCE ACF为等边三角形．请回答下列问题1四边形ADEF sketch template

已知 ABC是边长为3的等边三角形以BC为底边作一个顶角为120°等腰 BDC．点M点N分别是AB边与AC边上的点并且满足∠MDN sketch template

斜三棱柱ABC A1B1C1的底面是边长为2的正三角形侧棱长为2侧棱AA1和底面相邻两边ABAC都成45°角求全面积和体积题目和 sketch template

百度教育 sketch template

已知 ABC是边长为4的等边三角形BC在x轴上点D为BC的中点点A在第一象限内AB与y轴的正半轴相交于点E点BP是AC上的一个动点 sketch template

如图，在等边 ABC中，D为BC边上一点，E为AC边上一点，且∠ADE=60°，BD=3，CE=2，则 ABC的边长为（） A 9 B sketch template

15．三棱锥P ABC的四个顶点在同一球面上若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——青夏 sketch template

如图 ABC为等边三角形边长为1DEF分别为ABBCAC上的动点且AD=BE=CF．若AD=x DEF的面积为y．1求y sketch template

如果一个三角形的三条中线分别是3，4，5，求这个三角形的面积？知乎 sketch template

百度教育 sketch template

已知空间四边形ABCD的各边长分别为AB=AC=AD=BC=BD=CD=2EF分别是ABCD的中点 1求异面直线AB与CD间的距离 sketch template

如图1是边长分别为4和3的两个等边三角形纸片ABC和CD′E′叠放在一起．1操作固定 ABC将 CD′E′绕点C顺时针旋转得到 sketch template

如图以 ABC的ABAC为一边�外作正方形ABGF和ACDE两个正方形对角线交点分别是MNP是BC中点求证PM＝PN．题目和 sketch template

三角形三边长分别为√13、√10、√17，则面积是多�？知乎 sketch template

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