济南市槐荫区2017.1北师大版八年级数学期末考试题（含答案）

2016～2017 学年度第一学期槐荫区八年级数学调研测试题( 2017.1) 本试题分试卷和答题卡两部分．第 1 卷共 2 页，满分为 48 分；第 1I 卷共 4 页，满分为 102 分．本试题共 6 页，满分为 150 分．考试时间为 120 分钟． 第 I 卷（选择题共 48 分） 一、选择题（本大题共 12 个小题，每小题 4 分，共 48 分．在每小题给出的四个选项中，只 有一项是符合题目要求的．） 1. 25 的平方根是 A．5 B．－5 C．± 5 D．±5 2．下列图形中，是中心对称图形的是 3．某射击小组有 20 人，教练根据他们某次射击的数据绘制成如图所示的统计图，则这组数 据的众数和中位数分别是 A. 7, 7 B. 8, 7.5 C. 7, 7.5 D. 8, 6.5 4．如图，两个较大正方形的面积分别为 225、289，则字母 A 所代表的正方形的面积为 A．4 B．8 C．16 D．64 5．化简 2 x2－1÷ 1 x－1 的结果是 A． 2 x－1 B.2 x C. 2 x＋1 D. 2(x＋1) 6.不等式组 x－1≤0 2x＋4＞0的解集在数轴上表示为 7．如果关于 x 的不等式（a＋1）x＞a＋1 的解集为 x＜1，则 a 的取值范围是 A.a＜0 B.a＜－1 C.a＞1 D.a＞－1 8．实数 a 在数轴上的位置如图所示，则 (a－4)2＋ (a－11)2化简后为 A． 7 B． －7 C．2a－15 D．无法确定 9．若方程 A x－3 ＋ B x＋4 ＝ 2x＋1 (x－3)(x＋4) 那么 A、B 的值 A.2,1 B.1,2 C.1,1 D.－1, －1 10．已知长方形 ABCD 中，AB＝3，AD＝9，将此长方形折叠，使点 B 与点 D 重合，折痕为 EF,则△ABE 的面积为 A．6 B．8 C．10 D．12 11．如图，△ABC 绕点 A 顺时针旋转 45°得到△AB′C′，若∠BAC＝90°,AB＝AC＝ 2， 则图中阴影部分的面积等于 A．2- 2 B.1 C． 2 D. 2-l 12．如图，△ABC 中，∠ACB＝90°，AC＞BC，分别以△ABC 的边 AB、BC、CA 为一边内 △ABC 外作正方形 ABDE、BCMN、CAFG，连接 EF、GM、ND，设△AEF、△BND、△ CGM 的面积分别为 S1、S2、S3，则下列结论正确的是 A．Sl＝S2＝S3 B．S1＝S2＜S3 C．Sl＝S3＜S2 D．S2＝S3＜Sl 第 II 卷（非选择题共 102 分） 二、填空题（本大题共 6 个小题．每小题 4 分，共 24 分．把答案填在答题卡的横线上．） 13．计算： 8一 2＝______________. 14．分解因式：a2－6a＋9＝______________. 15．当 x＝______时，分式 x2－9 (x－1)(x－3) 的值为 0. 16.已知 a＋b＝3，a2b＋ab2＝1，则 ab＝____________· 17．如图，一只蚂蚁沿着边长为 2 的正方体表面从点 4 出发，经过 3 个面爬到点 B，如果它 运动的路径是最短的，则最短路径的是长为__________________. 18．如图，在四边形 ABCD 中，AD＝4，CD＝3, ∠ABC＝∠ACB＝∠ADC＝45°，则 BD 的 长为______________. 三、解答题（本大题共 9 个小厦，共 78 分，解答应写出文字说明，证明过程或演算步骤．） 19．(本小题满分 6 分）计算： (1) 18＋ 2 2 －3 (2)a＋2 a－2 ÷ 1 a2—2a 20．（本小题满分 6 分） (1)因式分解：m3n―9mn. (2)求不等式x－2 2 ≤7－x 3 的正整数解 21．（本小题满分 8 分） (1)解方程：1－2x x－2 ＝2＋ 3 2－x (2)解不等式组 4x―3＞x x＋4＜2x 一 1，并把解集在数轴上表示出来 22.（本小题满分 10 分） (1)如图 1，△ABC 是边长为 2 的等边三角形，将△ABC 沿直线 BC 向右平移，使点 B 与 点 C 重合，得到△DCE，连接 BD，交 AC 于点 F．求线段 BD 的长． (2)一次环保知识竞赛共有 25 道题，规定答对一道题得 4 分，答错或不答一道题扣 1 分. 在这次竞赛中，小明被评为优秀（85 分或 85 分以上），小明至少答对了几道题？ 23．（本小题满分 8 分） 济南与北京两地相距 480 千米，乘坐高铁列车比乘坐普通快车能提前 4 小时到达．已知 高铁列车的平均行驶速度是普通快车的 3 倍，求高铁列车的平均行驶速度. 24．（本小题满分 6 分） 先化简再求值：(x＋1 一 3 x－1 )×x－1 x－2 ，其中 x＝－ 2 2＋2 25．（本小题满分 10 分） 某公司需招聘一名员工，对应聘者甲、乙、丙从笔试、面试、体能三个方面进行量化考 核，甲、乙、丙各项得分如下表： 笔试 面试 体能 甲 83 79 90 乙 85 80 75 丙 80 90 73 (1)根据三项得分的平均分，从高到低确定三名应聘者的排名顺序． (2)该公司规定：笔试，面试、体能得分分别不得低于 80 分，80 分，70 分，并按 60%， 30%，10%的比例计入总分，根据规定，请你说明谁将被录用． 26．（本小题满分 12 分） 如图，在四边形 ABCD 中，对角线 AC，BD 交于点 E，∠BAC＝90°，∠CED＝45°， ∠DCE＝30°，DE＝ 2，BE＝2 2． (1)求 CD 的长： (2)求四边形 ABCD 的面积 27．（本小题满分 12 分） 已知，点 D 是等边△ABC 内的任一点，连接 OA，OB，OC. (1)如图 1，己知∠AOB＝150°，∠BOC＝120°，将△BOC 绕点 C 按顺时针方向旋转 60° 得△ADC． ①∠DAO 的度数是_______________ ②用等式表示线段 OA，OB，OC 之间的数量关系，并证明； (2)设∠AOB＝α，∠BOC＝β. ①当α，β满足什么关系时，OA＋OB＋OC 有最小值？请在图 2 中画出符合条件的图形， 并说明理由； ②若等边△ABC 的边长为 1，直接写出 OA＋OB＋OC 的最小值． 八年级数学试题参考答案与评分标准 一、选择题 题号 1 2 3 4 5 6 7 8 9 10 11 12 答案 D B C D C B B A C A D A 二、填空题 13. 2 14. ( a－3) 2 15. －3 16. 1 3 17. 2 10 18. 41 三．解答题： 19.解： (1) 18 2 3 2   = 3 2 2 3 2   ···················································································· 1 分 = 4 2 3 2  ··························································································2 分 =1·····································································································3 分 (2) 2 2 1 2 2 a a a a    = 2 ( 2) 2 1 a a a a   ·················································································· 5 分 = 2 2a a ····························································································6 分 20.解： (1) m3n－9mn． = 2( 9)mn m  ······················································································· 1 分 = 2 2( 3 )mn m  ······················································································ 2 分 = ( 3)( 3)mn m m  ················································································ 3 分 (2)解：3(x－2)≤2(7－x) ····································································4 分 3x－6≤14－2x 5x≤20 x≤4·············································································· 5 分 ∴这个不等式的正整数解为 1、2、3、4.·················································· 6 分 21.(1) 1 2 322 2 x x x     1 2 2( 2) 3x x    ······································································1 分 1 2 2 4 3x x    ········································································2 分 4 8x   2x  ················································································· 3 分 经检验 2x  是增根，原方程无解······················································4 分 （2） 4 3 4 2 1 x x x x     ＞ ＜ ， 解：解不等式①得：x＞1，··································································· 5 分 解不等式②得：x＞5，········································································· 6 分 ∴不等式组的解集为 x＞5，···································································7 分 在数轴上表示不等式组的解集为： ．··············································8 分 22. (1)解：∵正△ABC 沿直线 BC 向右平移得到正△DCE ∴ BE=2BC=4, BC=CD,DE=AC=2,∠E=∠ACB=∠DCE=∠ABC=60°················· 2 分 ∴∠DBE= 1 2 ∠DCE =30°······································································· 3 分 ∴∠BDE=90°······················································································4 分 在 Rt△BDE 中，由勾股定理得 2 2 2 24 2 2 3BD BE DE     ························································· 5 分 (2)解：设小明答对了 x 道题，································································6 分 4x－(25－x) ≥85·················································································8 分 x≥22················································································· 9 分 所以，小明至少答对了 22 道题.···························································· 10 分 23. 解：设普通快车的速度为 xkm/h，由题意得：······································ 1 分 480 480 43x x   ····················································································· 3 分 480 160 4x x   320 x =4······························································································ 4 分 x=80··································································································5 分 经检验 x=80 是原分式方程的解·······························································6 分 3x=3×80=240 ····················································································· 7 分 答：高铁列车的平均行驶速度是 240km/h．···············································8 分 24.解： 3 11 1 2 xx x x        = ( 1)( 1) 3 1[ ]1 1 2 x x x x x x       ·························································· 1 分 = 2 4 1 1 2 x x x x    ·············································································2 分 = ( 2)( 2) 2 x x x    ·············································································3 分 = 2x  ······················································································· 4 分 当 2 2 2 x    = 2( 2 2) 2 2 ( 2 2)( 2 2)      时············································ 5 分 原式= 2 2 2  = 2 ············································································6 分 25. 解：(1) x甲 =（83+79+90）÷3=84， x乙 =（85+80+75）÷3=80， x丙 =（80+90+73）÷3=81．·································································· 3 分 从高到低确定三名应聘者的排名顺序为：甲，丙，乙；·······························4 分 (2)∵该公司规定：笔试，面试、体能得分分别不得低于 80 分，80 分，70 分， ∴甲淘汰，······················································································· 5 分 乙成绩=85×60%+80×30%+75×10%=82.5，················································ 7 分 丙成绩=80×60%+90×30%+73×10%=82.3，··············································· 9 分 ∴乙将被录取． ··············································································10 分 26 解： (1)过点 D 作 DH⊥AC，·····························································1 分 ∵∠CED=45°， ∴∠EDH=45°， ∴∠HED=∠EDH， ∴EH=DH，······················································································· 3 分 ∵EH2+DH2=DE2，DE= 2 ， ∴EH2=1， ∴EH=DH=1，···················································································· 5 分 又∵∠DCE=30°，∠DHC=90°， ∴DC=2 ····························································································6 分 (2)∵在 Rt△DHC 中， 2 2 2DH HC DC  ··················································7 分 ∴12+HC2=22， ∴HC= 3 ，······················································································· 8 分 ∵∠AEB=∠CED=45°，∠BAC=90°，BE=2 2 ， ∴AB=AE=2，····················································································· 9 分 ∴AC=2+1+ 3 =3+ 3 ，······································································10 分 ∴S 四边形 ABCD =S△BAC+S△DAC···················································································· 11 分 = 1 2 ×2×（3+ 3 ）+ 1 2 ×1×（3+ 3 ） = 3 3 9 2  ··························································································12 分 27. 解：（1）①90°. ·············································································2 分 ②线段 OA，OB，OC 之间的数量关系是 2 2 2OA OB OC  . ························· 3 分 如图 1，连接 OD.················································································ 4 分 ∵△BOC 绕点 C 按顺时针方向旋转 60°得△ADC， ∴△ADC≌△BOC，∠OCD=60°. ∴CD = OC，∠ADC =∠BOC=120°，AD= OB. ∴△OCD 是等边三角形，····································································· 5 分 ∴OC=OD=CD，∠COD=∠CDO=60°， ∵∠AOB=150°，∠BOC=120°， ∴∠AOC=90°， ∴∠AOD=30°，∠ADO=60°. ∴∠DAO=90°.·····················································································6 分 在 Rt△ADO 中，∠DAO=90°， ∴ 2 2 2OA AD OD  . ∴ 2 2 2OA OB OC  .···············································································7 分 (2)①如图 2，当α=β=120°时，OA+OB+OC 有最小值. ·································8 分 作图如图 2，······················································································ 9 分 如图 2，将△AOC 绕点 C 按顺时针方向旋转 60°得△A’O’C，连接 OO’. ∴△A′O′C≌△AOC，∠OCO′=∠ACA′=60°. ∴O′C= OC, O′A′ = OA，A′C = BC, ∠A′O′C =∠AOC. ∴△OC O′是等边三角形.····································································· 10 分 ∴OC= O′C = OO′，∠COO′=∠CO′O=60°. ∵∠AOB=∠BOC=120°， ∴∠AOC =∠A′O′C=120°. ∴∠BOO′=∠OO′A′=180°. ∴四点 B，O，O′，A′共线. ∴OA+OB+OC= O′A′ +OB+OO′ =BA′ 时值最小.········································ 11 分 ②当等边△ABC 的边长为 1 时，OA+OB+OC 的最小值 A′B= 3 . ·················12 分 不用注册，免费下载！