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 Coloring Page

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 Template

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tomski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 coloring

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

已知菱形ABCD的边长为2∠BAD=60°．将三角形ABD沿对角线BD折到ABD使得二面角A BD C的大小为60°则AD与平面 sketch template

已知菱形ABCD的边长为2∠BAD=60°．将三角形ABD沿对角线BD折到ABD使得二面角A BD C的大小为60°则AD与平面 sketch template

菱形对折对折折页第5页大山谷图库 sketch template

如图，菱形的边长为，，为的中点．将沿折起，使到达，高途 sketch template

十四平面几何之四边形知乎 sketch template

矩形、菱形、正方形知识点汇总 5大题型解析例题知乎 sketch template

如图在菱形ABCD中AB=2∠A=120°过菱形ABCD的对称中心O分别作边ABBC的垂线交各边于点EFGH则四边形 sketch template

已知菱形的一个内角为60度，一条对角线的长为2倍根号3，则另一条对角线的长为？百度知道 sketch template

如图已知菱形ABCD的边长为2∠BAD=60°若DE⊥AB垂足为点E则DE的长为33．题目和参考答案——青夏教育精英家教�—— sketch template

百度教育 sketch template

2．如图所示已知菱形ABCD的边长为2cm∠BAD＝120°求这个菱形的两条对角线的长．——青夏教育精英家教�—— sketch template

在边长为1的菱形ABCD中∠ABC＝60°将菱形沿对角线AC折起使折起后BD＝1则二面角B AC D的余弦值为 A B C sketch template

如图�已知点O为菱形ABCD的对称中心∠A=60°将等边 OEF的顶点放在点O处OEOF分别交ABBC于点MN．1求证OM sketch template

如图已知菱形ABCD中∠ABC=60°AB=6点E是AB的中点F是边BC上的任意一点将 BEF沿EF折叠B点的对应点为B′连接 sketch template

已知菱形ABCD的边长为2∠BAD=120°则菱形的对角线长是．题目和参考答案——青夏教育精英家教�—— sketch template

菱形ABCD的边长为2∠A=60°则它的面积为 ．题目和参考答案——青夏教育精英家教�—— sketch template

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已知菱形，，将沿对角线折起，使以，，，四点为顶点的 高途 sketch template

已知菱形ABCD中∠A=72°如图�可以用3条直线将菱形ABCD分割成4个等腰三角形．你能在图��中再作出符合条件的两种分割方法吗 sketch template

已知如图在菱形ABCD中∠B=60°点EF分别是ABAD上的两个动点且始终保持∠ECF=60°1试判断 ECF的形状并说明 sketch template

如图1，已知O为正方形ABCD对角线的交点，点E在边CB的延长线上，连结EO，OF⊥OE交BA延长线于点F，连结EF．1求证：EO＝FO sketch template

题目如图已知正方形ABCD的边长为6EF分别是ABBC边上的点且∠EDF=45°将 DAE绕点D逆时针旋转90°得到 DCM sketch template

如图所示菱形ABCD的边长为6厘米∠B=60度．从初始时刻开始点PQ同时从A点出�点P以1厘米/秒的速度沿A→C→B的方�运动点Q sketch template

如图已知正方形ABCD的边长为2沿对角线AC将三角形ADC折起使平面ADC与平面ABC垂直折叠后BD两点的距离是 B sketch template

$已知菱形ABCD边长为2．∠BAD=$frac{π}{3}$．将 ABD沿BD折起．折成二面角A1 BD C．则下列说法正确的是 A．当 sketch template$

已知如图10菱形ABCD的�长为16 Cm ∠ABC＝60°对角线AC和BD相交于点O 求AC和BD的长——青夏教育精英家教�—— sketch template

如图，在四边形ABCD中，对角线AC与BD交于点O，已知OA =OC，OB=OD，过点O作EF⊥ BD，分别交AB、DC于点E，F，连接DE sketch template

已知边长为10的菱形ABCD对角线BD=16过线段BD上的一个动点P分别�直线ABAD作垂线垂足分别为EF．1如图1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1将由5个边长为1的小正方形组成的十字形纸板沿虚线剪拼成一个大正方形需剪4 刀思考�现大正方形的面积等于5个小正方形的面积和大 sketch template

利用高中数学知识解初中题（降维打击）知乎 sketch template

如图，在平行六面体ABCD A 1B 1C 1D 1中，底面是边长为2的正方形，若∠ 1AB=∠ 1AD=60^° ，且A 1A 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

如图1已知菱形ABCD的边长为23点A在x轴负半轴上点B在坐标原点．点D的坐标为 33抛物线y=ax2 B经过ABCD两边的 sketch template

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