2020年1月份宁德市九年级数学期末试卷及答案

宁德市2019-2020学年度第一学期期末九年级质量检测数学本试卷分第Ⅰ卷（选择题）和第Ⅱ卷（非选择题）两部分，第Ⅰ卷1至2页，第Ⅱ卷3至6页．考试时间120分钟，满分150分．注意事项：1．答题前，考生务必在试题卷、答题卡规定位置填写本人准考证号、姓名等信息．考生要认真核对答题卡上粘贴的条形码的“准考证号、姓名”与考生本人准考证号、姓名是否一致．2．选择题每小题选出答案后，用2B铅笔把答题卡上对应题目的答案标号涂黑．如需改动，用橡皮擦干净后，再选涂其他答案标号．非选择题答案用0.5毫米黑色签字笔在答题卡上相应位置书写作答，在试题卷上答题无效．3．作图可先使用2B铅笔画出，确定后必须用0.5毫米黑色签字笔描黑．4．考试结束，考生必须将试题卷和答题卡一并交回．第Ⅰ卷一、选择题：本题共10小题，每小题4分，共40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．sin30︒=A．12B．22C．32D．332．已知一个几何体如图所示，则该几何体的左视图是A B C D3．在一个不透明的袋子中放有若干个球，其中有6个白球，其余是红球，这些球除颜色外完全相同．每次把球充分搅匀后，任意摸出一个球记下颜色再放回袋子．通过大量重复试验后，发现摸到白球的频率稳定在0.25左右，则红球的个数约是A．2B．12C．18D．244．如图，在Rt△ABC中，∠C=90°，BC=4，AB=5，则sin B的值是A．23B．35C．34D．45AC B第4题图第2题图正方向数学试题第1页共6页数学试题第2页共6页5．如图，四边形ABCD 与四边形GBEF 是位似图形，则位似中心是A ．点A B ．点B C ．点FD ．点D6．如图，在四边形ABCD 中，对角线AC ，BD 相交于点O ，且OA =OC ，OB =OD ．若要使四边形ABCD 为菱形，则可以添加的条件是A ．AC =BD B ．AB ⊥BC C ．∠AOB =60°D ．AC ⊥BD7．对于反比例函数4y x=-，下列说法正确的是A ．y 的值随x 值的增大而增大B ．y 的值随x 值的增大而减小C ．当0x >时，y 的值随x 值的增大而增大D ．当0x <时，y 的值随x 值的增大而减小8．若方程240x x m -+=有两个不相等的实数根，则实数m 的值可能是A ．3B ．4C ．5D ．69．已知抛物线2y ax bx c =++经过点（-4，m ），（-3，n ），若1x ，2x 是关于x 的一元二次方程20ax bx c ++=的两个根，且-4＜1x ＜-3，2x ＞0，则下列结论一定正确的是A ．0m n +＞B ．0m n -＜C ．0m n ⋅＜D ．0m n＞10．如图，一根电线杆PO 垂直于地面，并用两根拉线PA ，PB 固定，量得PAO α∠=，PBO β∠=，则拉线PA ，PB 的长度之比PAPB=A ．tan tan αβB ．cos cos βαC ．sin sin αβD ．sin sin βα第5题图ABGDE CF 第10题图P第6题图A BD CO数学试题第3页共6页第Ⅱ卷注意事项：1．用0.5毫米黑色签字笔在答题卡上相应位置书写作答，在试题卷上答题无效．2．作图可先用2B 铅笔画出，确定后必须用0.5毫米黑色签字笔描黑．二、填空题：本题共6小题，每小题4分，共24分．11．若32x y =，则x y=．12．已知一个几何体的主视图与俯视图如图所示，则该几何体可能是．13．如图，公路AC 与BC 互相垂直，公路AB 的中点M 与点C被湖隔开．若测得AB 的长为2.4km ，则M ，C 两点间的距离为km ．14．中国古代数学著作《九章算术》中记载：“今有户高多于广六尺八寸，两隅相去适一丈．问户高、广各几何？”译文为：已知长方形门的高比宽多6.8尺，门的对角线长为10尺，那么门的高和宽各是多少尺？设长方形门的宽为x 尺，则可列方程为．15．如图，E ，F 分别为矩形ABCD 的边AD ，BC 的中点，若矩形ABCD 与矩形EABF 相似，则相似比等于．16．如图，正方形的顶点A ，C 分别在y 轴和x 轴上，边BC的中点F 在y 轴上，若反比例函数12y x=的图象恰好经过CD 的中点E ，则OA 的长为．三、解答题：本题有9小题，共86分．17．（本题满分8分）解方程：2650x x +-=．18．（本题满分8分）如图，D ，E 分别是△ABC 的边AB ，AC 上的点，//DE BC ，7AB =，2BD =，6AE =，求AC 的长．第16题图ADE B CC第13题图第15题图AC MB Axy OBDEF主视图俯视图第12题图EABCDF数学试题第4页共6页19．（本题满分8分）如图，点A 在y 轴正半轴上，点B (4，2)是反比例函数图象上的一点，且OAB ∠tan =1．过点A 作AC ⊥y 轴交反比例函数图象于点C ．（1）求反比例函数的表达式；（2）求点C 的坐标．20．（本题满分8分）如图，在菱形ABCD 中，点E 在对角线AC 上，延长BE 交AD 于点F ．（1）求证：EF FAEB BC=；（2）已知点P 在边CD 上，请以CP 为边，用尺规作一个△CPQ 与△AEF 相似，并使得点Q 在AC 上．（只须作出一个△CPQ ，保留作图痕迹，不写作法）21．（本题满分8分）某化肥厂2019年生产氮肥4000吨，现准备通过改进技术提升生产效率，计划到2021年生产氮肥4840吨．现技术攻关小组按要求给出甲、乙两种技术改进方案，其中运用甲方案能使每年产量增长的百分率相同，运用乙方案能使每年增长的产量相同．问运用哪一种方案能使2020年氮肥的产量更高？高多少？y ABOxCADE FP数学试题第5页共6页22．（本题满分10分）某地要建造一个圆形喷水池，在水池中央垂直于水面安装一个柱子OA ，点O 恰好在水面中心，安装在柱子顶端A 处的圆形喷头向外喷水，水流在各个方向上沿形状相同的抛物线路径落下，且在过OA 的任意平面上，水流喷出的高度(m)y 与水平距离(m)x 之间的关系如图所示，建立平面直角坐标系，右边抛物线的关系式为223y x x =-++．请完成下列问题：（1）将223y x x =-++化为2()y a x h k =-+的形式，并写出喷出的水流距水平面的最大高度是多少米；（2）写出左边那条抛物线的表达式；（3）不计其他因素，若要使喷出的水流落在池内，水池的直径至少要多少米？23．（本题满分10分）4月23日，为迎接“世界读书日”，某书城开展购书有奖活动．顾客每购书满100元获得一次摸奖机会，规则为：一个不透明的袋子中装有4个小球，小球上分别标有数字1，2，3，4，它们除所标数字外完全相同，摇匀后同时从中随机摸出两个小球，则两球所标数字之和与奖励的购书券金额的对应关系如下：两球所标数字之和34567奖励的购书券金额（元）306090（1）通过列表或画树状图的方法计算摸奖一次获得90元购书券的概率；（2）书城规定：如果顾客不愿意参加摸奖，那么可以直接获得30元的购书券．在“参加摸奖”和“直接获得购书券”两种方式中，你认为哪种方式对顾客更合算？请通过求平均数的方法说明理由．y OxA数学试题第6页共6页24．（本题满分12分）如图，已知平行四边形ABCD 中，sin ∠DBC =513，BD =24，∠BDC =60°．平行四边形MPNQ 的顶点P ，Q 在线段BD 上（点P 在Q 的左边），顶点M ，N 分别在线段AD 和BC 上．（1）求证：DM =BN ；（2）如图1，将△BCD 沿直线BD 折叠得到△BC ʹD ，当BC ʹ恰好经过点M 时，求证：四边形MPNQ 是菱形；（3）如图2，若四边形MPNQ 是矩形，且MP ∥AB ，求BP 的长．（结果中的分母可保留根式）25．（本题满分14分）已知二次函数2y ax bx c =++(a ＞0)的图象经过点A （1，2）．（1）当c =4时，若点B （3，10）在该二次函数的图象上，求该二次函数的表达式；（2）已知点M （3t -，5），N （3t +，5）在该二次函数的图象上，求t 的取值范围；（3）当a =1时，若该二次函数的图象与直线31y x =-交于点P ，Q，且PQ =，求b 的值．图2MA CBDQN PA MC BDQN PCʹ图1数学试题参考答案及评分说明第1页共6页宁德市2019-2020学年度第一学期期末九年级质量检测数学试题参考答案及评分标准⑴本解答给出了一种或几种解法供参考，如果考生的解法与本解答不同，可参照本答案的评分标准的精神进行评分．⑵对解答题，当考生的解答在某一步出现错误时，如果后续部分的解答未改变该题的立意，可酌情给分．⑶解答右端所注分数表示考生正确作完该步应得的累加分数．⑷评分只给整数分，选择题和填空题均不给中间分．一、选择题：（本大题有10小题，每小题4分，满分40分）1．A2．B3．C4．B5．B6．D7．C8．A9．C10．D二、填空题：（本大题有6小题，每小题4分，满分24分）11．23；12．三棱柱；13． 1.2；14．222( 6.8)10x x ++=；15．2∶1（或2）；16．26．三、解答题（本大题共9小题，共86分）17．（本题满分8分）解法一：265x x +=．················································································1分26+959x x +=+．···············································································3分2(3)14x +=．··················································································5分∴314x +=±．··············································································7分即1314x =-+，2314x =--．···························································8分解法二：1,6,5a b c ===-．········································································1分224641(5)560b ac -=-⨯⨯-=>．·························································3分∴6566214322x -±-±===-±．··················································7分即13x =-+，23x =--···························································8分18．（本题满分8分）解：∵AB =7，BD=2，∴AD=AB -BD=5．·············································································1分∵//DE BC ，数学试题参考答案及评分说明第2页共6页∴AD AEAB AC =．··················································································5分∵AE =6∴567AC=．·····················································································6分∴425AC =．······················································································8分19．（本题满分8分）解：（1）设反比例函数的表达式为ky x=，∵点B (4，2)在反比例函数图象上，∴24k=．·······················································································2分解得8k =．····················································································3分∴反比例函数的表达式为xy 8=．························································4分（2）过点B 作BD ⊥AO 于点D ．∵点B 的坐标为(4，2)，∴BD ＝4，DO ＝2．································5分在Rt △ABD 中，OAB ∠tan =ADBD=1，∴AD =BD=4．·················································································6分∴AO =AD+DO=6．∵AC ⊥y 轴，∴点C 的纵坐标为6．·······································································7分将y =6代入x y 8=，得x =34．∴点C 的坐标为（34，6）．·································································8分20．（本题满分8分）解：（1）∵四边形ABCD 是菱形，∴AD ∥BC ．∴∠FAE ＝∠ACB ．又∵∠AEF ＝∠CEB ，·······································································2分∴△AEF ∽△CEB ．··········································································3分∴EF FAEB BC=．·················································································4分D y ABOxC数学试题参考答案及评分说明第3页共6页（2）尺规作图如图所示：或·····································································································7分∴△CPQ 就是所求作的三角形．·························································8分21．（本题满分8分）解：设甲方案的平均增长率为x ，依题意得························································1分24000(1)4840x += ．··································································4分解得，10.1x =，2 2.1x =-（不合题意，舍去）．····································5分甲方案2020年产量：4000(10.1)4400 ⨯+= ，···············································6分乙方案2020年产量：14000(48404000)44202 +⨯ - = ．··································7分44004420 < ，4420440020 - =（吨）．答：乙方案能使2020年氮肥的产量更高，高20吨．··········································8分22．（本题满分10分）解：（1）∵223y x x =-++=214x --+（），∴抛物线的顶点式为214y x =--+（）．·············································3分∴喷出的水流距水平面的最大高度是4米；··········································4分（2）左边抛物线的表达式为214y x =-++（）．············································7分（3）将y =0代入223y x x =-++，则得2230x x -++=，·········································································8分解得x 1=3，x 2=-1(不合题意，舍去)．···················································9分∵326⨯=（米）∴水池的直径至少要6米．··························································10分数学试题参考答案及评分说明第4页共6页23．（本题满分10分）解：（1）列表如下：第1球第2球12341（1，2）（1，3）（1，4）2（2，1）（2，3）（2，4）3（3，1）（3，2）（3，4）4（4，1）（4，2）（4，3）······················································································3分由上表可知，共有12种等可能的结果.其中“两球数字之和等于7”有2种，∴P （获得90元购书券）=21126=．·······················································5分（2）由（1）中表格可知，两球数字之和的各种情况对应的概率如下：······················································································7分∴摸球一次平均获得购书券金额为2242200306090351212121212⨯+⨯+⨯+⨯+⨯=·····································9分∵35＞30，∴在“参加摸球”和“直接获得购书券”两种方式中，我认为选择“参加摸球”对顾客更合算.··································································································10分24．（本题满分12分）解：（1）如图1，∵四边形ABCD 是平行四边形，∴AD ∥BC ．∴∠ADB ＝∠DBC ．∵四边形MPNQ 是平行四边形，∴MQ ∥PN ，MQ =PN ．∴∠MQP ＝∠NPQ ．·······································································2分数字之和34567获奖金额（元）00306090相应的概率212212412212212图1AMCBDQP数学试题参考答案及评分说明第5页共6页∴∠MQD ＝∠NPB ．∴△MQD ≌△NPB ．·······································································3分∴DM =BN ．····················································································4分（2）证法一：如图2，∵△BC ʹD 与△BCD关于BD 对称，∴∠DBC ʹ＝∠DBC ．由（1）得∠ADB ＝∠DBC ，∴∠DBC ʹ＝∠ADB ．∴BM ＝DM ．·······························5分由（1）得DM ＝BN ，∴BM ＝BN ．······························6分∵BP ＝BP ，∴△MBP ≌△NBP ．····················7分∴PM =PN ．∴ MPNQ 是菱形．·····················8分证法二：如图3，连接MN ，MN 交BD 于点O ．（同证法一得）BM ＝BN ．··········6分∵∠DBC ʹ＝∠DBC ，∴BO ⊥MN ．······························7分∴ MPNQ 是菱形．·····················8分（3）如图4，过点N 作NH ⊥BD 于点H ，连接MN 交PQ 于O ．设OP =x ，∵四边形PMQN 是矩形，MP ∥AB ，∴∠PQN =∠BDC =60°，OQ =OP =ON =QN =x ，∴12HQ HO x ==，32HN x =，1122BH x =+．························9分在Rt △BHN 中，由5sin 13DBC ∠=，得5tan 12HBN ∠=··································10分52112122xx =+，解得x =.∴12BP =-········································································12分MAC B DQNPHO图4AMCBDQNP Cʹ图2图3AMCBDQNPCʹO数学试题参考答案及评分说明第6页共6页25．（本题满分14分）解：（1）∵c =4，∴二次函数的表达式为24y ax bx =++．∵点A （1，2），B （3，10）在二次函数的图象上，∴4293410a b a b ++=⎧⎨++=⎩．······································································2分解得24a b =⎧⎨=-⎩．·······································································3分∴该抛物线的函数表达式为2244y x x =-+．········································4分（2）∵点M (3,5)t -，N (3,5)t +在该二次函数的图象上，∴该二次函数的对称轴是直线tt t x =++-=2)3()3(．······························5分∵抛物线2y ax bx c =++(a ＞0)开口向上，A (1，2)，M (3,5)t -，N (3,5)t +在该二次函数图象上，且5＞2，由图象得，点M ，N 分别落在点A 的左侧和右侧，∴313-<<+t t .································································8分解得t 的取值范围是24t -<<.························································9分（3）当a =1时，2y x bx c =++.∵二次函数2y x bx c =++的图象经过点A (1，2),∴21b c =++，即1=-c b .∴二次函数表达式为21=++-y x bx b .················································10分根据二次函数的图象与直线31y x =-交于点P ，Q ，由2=131++--x bx b x ，解得1212x x b ==-,.∴点P ，Q 的横坐标分别是1，2-b .····················································12分不妨设点P 的横坐标是1，则点P 与点A 重合，即P 的坐标是(1，2).∴点Q 的坐标是（2-b ,1)2(3--b ），即Q 的坐标是（2b -,53b -）.∵10=PQ ,∴根据勾股定理，22221532[()][()]--+--=b b .解得b =0或2.················································································14分。