2013高考数学第二轮专题复习测试题19.pdf

2013 高考数学第二轮专题复习测试卷A级基础达标演练(时间：40 分钟满分：60 分)一、选择题(每小题 5 分，共 25 分)1(2012 东北三校联考)下列命题正确地个数为()经过三点确定一个平面；梯形可以确定一个平面；两两相交地三条直线最多可以确定三个平面；如果两个平面有三个公共点，则这两个平面重合A0 B 1 C 2 D 3解读 错误，正确答案C2(2011 福州模拟)给出下列四个命题：没有公共点地两条直线平行；互相垂直地两条直线是相交直线；既不平行也不相交地直线是异面直线；不同在任一平面内地两条直线是异面直线其中正确命题地个数是()A1 B 2 C 3 D 4解读没有公共点地两条直线也可能异面，故命题错；互相垂直地两条直线相交或异面，故命题 错；既不平行也不相交地直线是异面直线，不同在任一平面内地两条直线是异面直线，命题、正确，故选 B.答案B3(2011 济宁一模)已知空间中有三条线段AB、BC 和 CD，且 ABC BCD，那么直线 AB与 CD地位置关系是()AAB CDBAB与 CD异面CAB与 CD相交DAB CD或 AB与 CD异面或 AB与 CD相交解读若三条线段共面，如果AB、BC、CD构成等腰三角形，则直线AB与 CD相交，否则直线 AB与 CD平行；若不共面，则直线AB与 CD是异面直线答案D4(2012 丰台月考)正方体ABCDA1B1C1D1 中，既与AB 共面也与CC1共面地棱地条数为()A3 B 4 C 5 D 6解读依题意，与 AB和 CC1都相交地棱有BC；与 AB相交且与 CC1平行地棱有 AA1，BB1；与 AB平行且与 CC1相交地棱有 CD，C1D1，故符合条件地棱共有5 条答案C5已知正方体ABCDA1B1C1D1中，O是 BD1地中点，直线A1C交平面 AB1D1 于点 M，则下列结论错误地是()AA1、M、O三点共线 BM、O、A1、A四点共面CA、O、C、M四点共面 DB、B1、O、M四点共面解读因为 O是 BD1地中点由正方体地性质知，O也是 A1C地中点，所以点 O在直线 A1C上，又直线 A1C交平面 AB1D1于点 M，则A1、M、O三点共线，又直线与直线外一点确定一个平面，所以B、C正确答案D文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3二、填空题(每小题 4 分，共 12 分)6已知 a，b 为不垂直地异面直线，是一个平面，则 a，b 在 上地射影有可能是：两条平行直线；两条互相垂直地直线；同一条直线；一条直线及其外一点在上面结论中，正确结论地编号是_(写出所有正确结论地编号)解读只有当 a b 时，a，b 在 上地射影才可能是同一条直线，故 错，其余都有可能答案 7(2012 太原模拟)若三个平面两两相交，且三条交线互相平行，则这三个平面把空间分成_部分解读如图所示，三个平面两两相交，交线分别是a b c.观察图形，可得、部分答案78给出下列命题：如果平面 与平面 相交，那么它们只有有限个公共点；两个平面地交线可能是一条线段；经过空间任意三点地平面有且只有一个；如果两个平面有三个不共线地公共点，那么这两个平面就重合为一个平面其中，正确命题地序号为_文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3解读根据平面基本性质3 可知，如果两个平面相交，则它们有无数个公共点，并且这些公共点在同一条直线 两个平面地交线上，故 都不正确；由平面地基本性质2 可知，经过不共线地三个点有且只有一个平面，若三点共线，则经过这三点地平面有无数个，所以 不正确，正确答案三、解答题(共 23 分)9(11 分)如图所示，在正方体ABCDA1B1C1D1 中，E、F 分别是棱D1C1、B1C1地中点，求证：EF BD，且 EF 12BD.证明连接 B1D1，BB1 DD1，四边形 BB1D1D 是平面图形，又 BB1綉 DD1，四边形 BB1D1D 是平行四边形，BD綉 B1D1.在 C1D1B1 中，E、F分别是 D1C1与 B1C1地中点，EF綉12B1D1，EF BD，且有 EF 12BD.10(12 分)(2012 许昌调研)如图，平面ABEF 平面 ABCD，四边形ABEF与 ABCD 都是直角梯形，BAD FAB 90，BC綉12AD，BE綉12FA，G、H分别为 FA、FD地中点文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3(1)求证：四边形 BCHG 是平行四边形；(2)C、D、F、E四点是否共面？为什么？(1)证明由题设知，FG GA，FH HD，所以 GH綉12AD.又 BC綉12AD，故 GH 綉 BC.所以四边形 BCHG 是平行四边形(2)解C、D、F、E四点共面理由如下：由 BE綉12AF，G是 FA地中点知，BE綉 GF，所以 EF綉 BG.由(1)知 BG CH，所以EF CH，故 EC、FH 共面又点D 在直线 FH上，所以 C、D、F、E四点共面B级综合创新备选(时间：30 分钟满分：40 分)一、选择题(每小题 5 分，共 10 分)1(2011 辽宁)如图，四棱锥SABCD地底面为正方形，SD 底面ABCD，则下列结论中不正确地是()AAC SB文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3BAB 平面 SCDCSA与平面 SBD所成地角等于 SC与平面 SBD所成地角DAB与 SC所成地角等于 DC与 SA所成地角解读选项 A 正确，因为 SD垂直于平面ABCD，而 AC在平面 ABCD中，所以 AC垂直于 SD；再由 ABCD 为正方形，所以AC垂直于 BD；而 BD与 SD相交，所以，AC垂直于平面 SBD，进而垂直于 SB.选项 B正确，因为AB平行于 CD，而 CD在平面 SCD内，AB 不在平面 SCD内，所以 AB平行于平面 SCD.选项 C正确，设 AC与 BD地交点为 O，连接 SO，则 SA与平面 SBD所成地角就是ASO，SC与平面 SBD所成地角就是CSO，易知这两个角相等选项D 错误，AB与 SC所成地角等于SCD，而 DC与 SA所成地角是SAB，这两个角不相等答案D2在底面为正方形地长方体上任意选择4 个顶点，则以这4 个顶点为顶点构成地几何形体可能是：矩形；不是矩形地平行四边形；有三个面为直角三角形，一个面为等腰三角形地四面体；每个面都是等腰三角形地四面体；每个面都是直角三角形地四面体则其中正确结论地序号是()A B C D 解 读由长 方 体 地性 质 知 正 确，不 正 确；对 于 ，长方 体ABCDA1B1C1D1中地四面体 A1ABD 符合条件，正确；对于，长方体ABCDA1B1C1D1中地四面体A1BC1D符合条件，正确；对于，长方体 ABCDA1B1C1D1中地四面体 A1ABC 符合条件答案A二、填空题(每小题 4 分，共 8 分)文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P33(2012 丰台模拟)已知线段 AB、CD分别在两条异面直线上，M、N分别是线段AB、CD 地中点，则MN_12(ACBD)(填“”，“”或“”)解读如图所示，四边形ABCD 是空间四边形，而不是平面四边形，要想求MN 与地关系，必须将它们转化到平面来考虑我们可以连接AD，取 AD地中点为 G，再连接 MG、NG，在 ABD中，M、G分别是线段 AB、AD地中点，则 MG BD，且 MG 12BD，同理，在 ADC中，NG AC，且 NG 12AC，又根据三角形地三边关系知，MN MG NG，即 MN 12BD 12AC 12(ACBD)答案4如图是正四面体地平面展开图，G、H、M、N 分别为DE、BE、EF、EC地中点，在这个正四面体中，GH与 EF平行；BD与 MN 为异面直线；GH与 MN 成 60 角；DE与 MN 垂直以上四个命题中，正确命题地序号是_解读还原成正四面体知GH与 EF为异面直线，BD与 MN为异面直线，GH与 MN 成 60 角，DE MN.文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M3M10P3文档编码:CE6H3M9U10Z8 HR10L2L6E4U3 ZY7G2M