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欢迎光临中学数学信息网下载资料 浙江省瓯海中学 徐进光新课标高一数学同步期中测试一、选择题:在每小题给出的四个选项中,只有一项是符合题目要求的,请把正确答案的代号填在题后的括号内(每小题5分,共50分)1一个棱锥所有的棱长都相等,则该棱锥一定不是( )A三棱锥B四棱锥C五棱锥D六棱锥2面积为Q的正方形,绕其一边旋转一周,则所得几何体的侧面积为( )AQ B2Q C 3Q D 4Q3已知高与底面的直径之比为2:1的圆柱内接于球,且圆柱的体积为500,则球的体积 为 ( )A B C D4到空间四点距离相等的平面的个数为( )A4 B7 C4或7 D7或无穷多5在阳光下一个大球放在水平面上, 球的影子伸到距球与地面接触点10米处, 同一时刻, 一根长1米一端接触地面且与地面垂直的竹竿的影子长为2米, 则该球的半径等于( )A10(2)米 B(6)米 C(94)米 D5 米6已知ABCD 是空间四边形,M 、N 分别是AB 、CD 的中点,且AC 4,BD 6,则( )A1MN 5 B2MN 10 C1MN 5 D2MN 57空间一个角的两边分别垂直于另一角的两边,则这两个角( )A相等B互补C相等或互补D 不确定8已知平面a 平面b ,m 是a 内一条直线,n 是b 内一条直线,且m n 那么,甲:m b ;乙:n a ;丙:m b 或n a ;丁:m b 且n a 这四个结论中,不正确的三个是( )A甲、乙、丙 B甲、乙、丁 C甲、丙、丁 D乙、丙、丁9如图,ABCDE 是一个四棱锥,AB 平面BCDE ,且四边 形BCDE为矩形,则图中互相垂直的平面共有( )A4组 B5组 C6组 D7组10棱台的两底面积分别为S上、S下、平行于底面的戴面把棱台的高自上而下分为两段之比 为mn则截面面S0为( )A BC()2 D()2二、填空题:请把答案填在题中横线上(每小题6分,共24分)11半径为a的球放在墙角,同时与两墙面和地面相切,那么球心到墙角顶点的距离为 12a 、b 是两个不同的平面,m 、n 是平面a 及b 之外的两条不同直线,给出四个论断:(1)m n (2)a b (3)n b (4)m a 以其中三个论断作为条件,余下一个论断作为结论,写出你认为正确的一个命题_13如图,三棱柱ABCA1B1C1中,若E、F分别为AB、AC 的中点,平面EB1C1将三棱柱分成体积为V1、V2的两部分,那么V1V2= _14下图都是正方体的表面展开图,还原成正方体后,其中两个完全一样的是_(1) (2) (3) (4)三、解答题:解答应写出文字说明、证明过程或演算步骤(共76分)15(12分)如图,长方体ABCDA1B1C1D1中被截去一部分,其中EFA1D1剩下的几何体是什么?截取的几何体是什么?若FHEG,但FHEG,截取的几何体是什么?16(12分)有一正三棱锥和一个正四棱锥,它们的所有棱长都相等,把正三棱锥和正四棱锥的一个全等的面重合说明组合体是什么样的几何体?证明你的结论17(12分)正四棱台的高,侧棱,对角线长分别为7cm,9cm,11cm,求它的侧面积18(12分)三棱锥S-ABC的三条侧棱两两垂直,SA=5,SB=4,SC=3,D为AB中点,E为AC中点,求四棱锥S-BCED的体积19(14分)如图,在正方体 (1)证明:; (2)求所成的角; (3)证明:20(14分)如图,ABC 为正三角形,EC 平面ABC ,BD CE ,CE CA 2 BD ,M 是EA 的中点,求证: (1)DE DA ; (2)平面BDM 平面ECA ; (3)平面DEA 平面ECA 高一新数学期中测试题参考答案一、DBDDA ADBCDMNCBADSE二、11;12;1375;14;三、15 五棱柱,三棱柱,三棱台。16解:(1)是斜三棱柱。(2)正三棱锥为SAED,正四棱锥为SABCD,重合的面为ASD,如图示, 设AD,BC中点分别为M、N,由AD平面MNS知平面MES重合;因为SE=AB=MN,EM=SN,MNSE为平行四边行。ESMN,又ABMN,ESAB,ABSE为平行四边形,同理,CDES为平行四边形。面SBC面EAD,ABCDSE,且AB不垂直平面SBC,组合体为斜三棱柱。17解:如图,在中过A作于E,则AE=OO1=7cm18解: 19 (1)(2)(3)20证明:(1)如图,取EC 中点F ,连结DF EC 平面ABC ,BD CE ,得DB 平面ABC DB AB ,EC BC BD CE ,BD CE FC ,则四边形FCBD 是矩形,DF EC 又BA BC DF , RtDEF RtABD ,所以DE DA (2)取AC 中点N ,连结MN 、NB , M 是EA 的中点, MN EC 由BD EC ,且BD 平面ABC ,可得四边形MNBD 是矩形,于是DM MN DE DA ,M 是EA 的中点, DM EA 又EA MN M , DM 平面ECA ,而DM 平面BDM ,则平面ECA 平面BDM (3) DM 平面ECA ,DM 平面DEA , 平面DEA 平面ECA 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