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福建省泉州市2020年（春秋版）九年级上学期数学期中考试试卷（II）卷姓名:________ 班级:________ 成绩:________一、单选题 (共10题；共20分)1. （2分）是一个正整数，则n的最小正整数是（）A . 1B . 2C . 3D . 42. （2分）方程x2﹣9=0的两个根为（）A . x1=﹣3，x2=3B . x1=﹣9，x2=9C . x1=﹣1，x2=9D . x1=﹣9，x2=13. （2分）(2018·方城模拟) 下列计算正确的是（）A .B .C .D .4. （2分）对任意实数，多项式的值是一个（）A . 正数B . 负数C . 非负数D . 无法确定5. （2分） (2017八下·抚宁期末) 化简的值是（）A . ﹣3B . 3C . ±3D . 96. （2分）小丽要在一幅长为80cm，宽为50cm的矩形风景画的四周外围镶上一条宽度相同的金色纸边制成一幅矩形挂图，使整幅挂图的面积是5400，设金色纸边的宽度为xcm，则x满足的方程是（）A .B .C .D .7. （2分） (2018九上·恩阳期中) 下列四组线段中，能构成比例线段的一组是（）A . ，，，B . ，，，C . ，，，7cmD . ，，，8. （2分）下列各组条件中，不能判定△ABC与△A′B′C′相似的是（）A . ∠A=∠A′，∠B=∠B′B . ∠C=∠C′=90°，∠A=12°，∠B′=78°C . ∠A=∠B，∠B′=∠A′D . ∠A＋∠B=∠A′＋∠B′，∠A－∠B=∠A′－∠B′9. （2分）(2011·苏州) 如图，在四边形ABCD中，E、F分別是AB、AD的中点，若EF=2，BC=5，CD=3，则tanC等于（）A .B .C .D .10. （2分）(2018·惠山模拟) 如图，在△ABC中，D为AB边上一点，E为CD中点，AC= ，∠ABC=30°，∠A=∠BED=45°，则BD的长为（）A .B . +1﹣C . ﹣D . ﹣1二、填空题 (共5题；共6分)11. （1分）(2019·慈溪模拟) 二次根式有意义，则x的取值范围是________ 。
2020学年第一学期九年级期中检测数学参考答案一、选择题：本题有10个小题，每小题3分，共30分． 题号 1 2 3 4 5 6 7 8 9 10 答案BCACBCBDCB二、填空题：本题有6个小题，每小题4分，共24分． 11．4112．22 13． 35；3 14．m ≥－1． 15． 1或716． ①③三、解答题：本题有7个小题，共66分．解答应写出文字说明、证明过程或演算步骤． 17．(本题6分)解：∵抛物线y ＝a (x －3)2＋2经过点(1，－2)∴－2＝a (x －3)2＋2∴a ＝x －1 ∴y ＝－(x －3)2＋2 ……3分∴此函数的图象开口向下，当x ＜3时，y 随x 的增大而增大，当x ＞3时，y 随x 的增大而减小，∵点A (m ，s )，B (n ，t )(m ＜n ＜3)都在该抛物线上 ∴s ＜t……3分18．(本题8分)解：如图，连接OC ，∵∠AOC ＝2∠B ，∠DAC ＝2∠B ， ∴∠AOC ＝∠DAC ， ∴AO ＝AC ， ……3分 又∵OA ＝OC ，∴△AOC 是等边三角形，……2分 ∴AC ＝AO ＝21AD ＝3cm ． ……3分19．(本题8分) 解：选择方案二；……1分 ∵方案一获奖的概率为31，…… 3分方案二中出现的可能性如下表所示：第一次第二次红 黄 蓝 红 (红，红) (黄，红) (蓝，红) 黄 (红，黄) (黄，黄) (蓝，黄) 蓝(红，蓝)(黄，蓝)(蓝，蓝)共有9种不同的情况，其中指针落在不同颜色区域的可能性为96＝32……3分 ∵32＞31∴选择方案二 ……1分20．(本小题满分10分)解：(1)如图所示，⊙O 即为所求．……4分(2)如图，连接OB ，由题意知CD ＝8m ，AB ＝40m ， ∵OD ⊥AB ，∴BC ＝AC ＝21AB ＝20m ， ……3分设圆的半径为r ，则OC ＝r －8，在Rt △BOC 中，由BO 2＝BC 2＋OC 2可得r 2＝(r －8)2＋202， 解得：r ＝29，答：这钢梁圆弧的半径为29米．……3分21．(本小题满分10分)解：(1)设抛物线解析式为y ＝a (x －6)2＋5(a ≠0)，∵A (0，2)在抛物线上，∴代入得a ＝121-， ∴抛物线的解析式为y ＝121-(x －6)2＋5． ……5分(2)∵令y ＝0，即121-(x －6)2＋5＝0，解得x 1＝6－215(舍去)，x 2＝6＋215 ∴OC ＝6＋215．答：该同学把实心球扔出(6＋215)m ． ……5分22．(本小题满分12分)(1)110101+-=x y……2分 (2)解：当200=x 时，9011020=+-=y1800090200=⨯(元)答：零售商一次性批发200件，需要支付18000元． ……3分(3)解：当300100≤≤x 时5.3802)195(101)39101()71(2+--=+-=-=x x x x y w ……2分 ∵0101<-=a ，抛物线开口向下 当 195<x 时，w 随x 的增大而增大 又 x 为10的正整数倍∴x ＝190时，w 最大，最大值是3800 当 195>x 时，w 随x 的增大而减小 又 x 为10的正整数倍∴x ＝200 时，w 最大，最大值是3800 ……2分当 400300≤<x 时，w ＝(80－71)x ＝9x∵k ＝9>0 ∴ w 随x 的增大而增大 ∴x ＝400 时， w 最大，最大值是3600……2分综上所述，当 x ＝190或 x ＝200时，w 最大，最大值是3800……1分23．(本小题满分12分)(1)解：把A (－1，0)，C (0，3)代入抛物线解析式c x ax y ++=492得抛物线的解析式为349432++-=x x y ……4分 (2)解：令0349432=++-x x ，解得B (4，0) ∴343+-=x y CB 令3)2(43343)343(34943222+--=+-=+--++-=x x x x x x h∴x ＝2时，h 有最大值为3，此时Q 的坐标为)29,2( ……2分 此时S △BCQ ＝4321⨯⨯＝6 ……2分(3)解：已有343+-=x y CB ∵∠DCB ＝2∠ABC ∴CD 所在直线与CB 所在直线关于y 轴对称∴343+=x y CD又连列⎪⎪⎩⎪⎪⎨⎧++-=+=349433432x x y x y 01=x (舍去)，22=x∴则点D 坐标为)29,2(……4分。
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九年级数学期中模拟试题 （考试时间：120分钟；满分：120分） 温馨提示：亲爱的同学，欢迎你参加本次考试，祝你答题成功！
一、选择题（本题满分24分，共有8道小题，每小题3分） 下列每小题都给出标号为A、B、C、D的四个结论，其中只有一个是正确的。每小题选对得分；不选、选错或者选出的标号超过一个的不得分。
1．方程xx32的解是（ ）
A．3x B．3x C．0x D．3x或0x
2.如图，DE是△ABC的中位线，延长DE至F使EF=DE，连接CF，则S△CEF：S四边形BCED
的值为（ ）
A.1:3 B.2:3 C.1:4 D.2:5 3.点C是线段AB的黄金分割点（AC>BC），若AB=10㎝，则AC等于（ ） A.6 cm B.(55+1)㎝ C.(515)㎝ D.(515)cm 4.关于x的一元二次方程(a-1)x2-2x+3=0有实数根，则整数a的最大值是( ) A.2 B.1 C.0 D.-1 5.如图，点E、F、G、H分别是任意四边形ABCD中AD、BD、BC、CA的中点，若四边形EFGH是菱形，则四边形ABCD的边需满足的条件是( ) A.AB∥DC B.AC=BD C.AC⊥BD D.AB=DC 6．在一次篮球联赛中，每个小组的各队都要与同组的其他队比赛两场，然后决定小组出线的球队．如果某一小组共有x个队，该小组共赛了90场，那么列出正确的方程是（ ） A． B．x（x﹣1）=90 C． D. x（x+1）=90 7.一个布袋内只装有1个黑球和2个白球，这些球除颜色外其余都相同，随机摸出一个球后放回搅匀，再随机摸出一个球，则两次摸出的球都是黑球的概率是（ ）
A．94 B．91 C．61 D．31 最新 文档 可修改 欢迎下载 1
8.如图，正方形ABCD中，点E、F分别在BC、CD上，△AEF是等边三角形，连接AC交EF于G，下列结论：①BE = DF；②∠DAF =15°；③AC垂直平分EF；④BE + DF = EF；⑤S△CEF=2S△ABE。其中正确结论有（ ） A．4 B.3 C.2 D.1 二、填空题（本题满分18分，共有6道小题，每小题3分）
第 1 页 共 8 页 2020版九年级（上）期中数学试卷（word版含答案解析）B卷 姓名:________ 班级:________ 成绩:________ 一、单选题 1 . 如图，将Rt△ABC绕直角顶点A，沿顺时针方向旋转后得到Rt△AB1C1，当点B1恰好落在斜边BC的中点时，则∠B1AC＝（ ）
A．25° B．30° C．40° D．60° 2 . 如图，AB∥CD，FH平分∠BFG，∠EFB＝58°，则下列说法错误的是（ ）
A．∠EGD＝58° B．GF＝GH C．∠FHG＝61° D．FG＝FH 3 . 两个不相等的实数m，n满足m2-6m=4，n2-6n=4，则mn的值为（ ） A．6 B．-6 C．4 D．-4 4 . 下列图标是轴对称图形的是（ ）
A． B． C．
D．
5 . 已知关于x的一元二次方程的一个根是0，则m的值为（ ） A．1 B．0 C．-1 D．1或-1 第 2 页 共 8 页
6 . 对于二次函数y=2(x﹣1)2﹣8，下列说法正确的是（ ） A．图象的开口向下 B．当x=﹣1时，取得最小值为y=﹣8 C．当x＜1时，y随x的增大而减小 D．图象的对称轴是直线x=﹣1
7 . 抛物线的对称轴是（ ） A．x=1 B．x=-1 C．x=2 D．x=-2 8 . 下列图形中，把△ABC平移后，能得到△DEF的是（ ）
A． B． C． D． 9 . 如图，把一个球放在长方体纸盒内，球的一部分露出盒外，其截面如图所示．已知EF=CD=4 cm，则球的半径长是（ ）
A．2 cm B．2.5 cm C．3 cm D．4 cm 10 . 对于任意实数h，抛物线y=（x﹣h）2与抛物线y=x2（ ） A．开口方向相同 B．对称轴相同 C．顶点相同 D．都有最高点 二、填空题
11 . 二次函数y=+2的顶点坐标为 ． 12 . 哈尔滨市某楼盘以每平方米10000元的均价对外销售，经过连续两次上调后，均价为每平方米12100元，则平均每次上调的百分率为_____． 第 3 页 共 8 页
2023年初中学业质量检查数学试题参考答案及评分标准说明：（一）考生的正确解法与“参考答案”不同时，可参照“参考答案及评分标准”的精神进行评分．（二）如解答的某一步出现错误，这一错误没有改变后续部分的考查目的，可酌情给分，但原则上不超过后面应得的分数的二分之一；如属严重的概念性错误，就不给分．（三）以下解答各行右端所注分数表示正确做完该步应得的累计分数．一、选择题（每小题4分，共40分）1.B2.D3.A4.C5.A6.C7.D8.C9.B 10.D 二、填空题（每小题4分，共24分）11.如1(答案不唯一)12.()0,1-13.不可能14.415.1216.321x x x <<三、解答题（共86分）17.（8分）解：原式27=- (6)分9=-············································································································8分(其它解法，请参照以上评分标准)18.（8分）解：原式()()()2222111a a a a a a a +--=÷-+++····················································································2分()()()2221211a a a a a a a +-+=⋅--++······················································································3分211a aa a +=-++·········································································································4分21a =+.··················································································································5分当1a =时，原式= (6)分=8分(其它解法，请参照以上评分标准)19.（8分）证明：∵C ACB AB ∠=∠，∴AB AC =，··········································································2分∵AB ∥CE ，∴BAD ACE ∠=∠，·············································································································4分(第19题图)ABCED在ABD △与CAE △中，,,AB AC BAD ACE AD CE =⎧⎪∠=∠⎨⎪=⎩·················································································5分∴ABD △≌CAE △()S.A.S ，································································································6分∴AE BD =.·······················································································································8分(其它解法，请参照以上评分标准)20．（8分）证明：∵四边形ABCD 是矩形，∴90ADC DAB ∠=∠=︒，AD ∥BC .······································2分∵30DAC ∠=︒，∴903060ACD ∠=︒-︒=︒.····················································3分由旋转的性质可得30DAC FAE ∠=∠=︒，60ACD E ∠=∠=︒，·····················································4分∴303060DAE ∠=︒+︒=︒，··································································································5分∵AD ∥BC ，∴60PQE DAE ∠=∠=︒，·····································································································6分∴180606060QPE ∠=︒-︒-︒=︒，··························································································7分∴PQE △是等边三角形.·······································································································8分(其它解法，请参照以上评分标准)21.（8分）解：(1)8040%200÷=(人)；··································································································3分(2)补全图形如图所示，3036054200⨯︒=︒，∴扇形统计图中喜欢跳绳的扇形圆心角的度数为54︒.(3)法一：画出树状图如下：·······································································································································6分ABCDA B C D A B C D A B C D A B C D小张小王项目A B C D人数(单位：名)受调查学生喜欢运动项目条形统计图80304050(第20题图)ABCDEF PQ共有16种等可能结果，其中同时选中相同项目的共有4种，故41()164P ==项目相同.·······································································································································8分法二：列表如下：ABCDA ()A,A ()A,B ()A,C ()A,D B ()B,A ()B,B ()B,C ()B,D C ()C,A ()C,B ()C,C ()C,D D()D,A ()D,B ()D,C ()D,D ·······································································································································6分共有16种等可能结果，其中同时选中相同项目的共有4种，故41()164P ==项目相同.·······································································································································8分(其它解法，请参照以上评分标准)22.（10分）解：(1)由表中数据猜想：y 是x 的反比例函数.··············································································1分设()0k y k x =≠，将225x =，40y =，代入ky x=，得225409000k =⨯=，∴故所求函数关系式为9000y x=.································································································4分(2)由题意得()1304500x y -=，·································································································6分把9000y x =代入，得()90001304500x x-⋅=，···············································································7分解得260x =，························································································································8分经检验，260x =是原方程的根，且符合题意.·····································9分∴若直销店计划每天的销售利润为4500元，则其售价应定为260元．·····10分(其它解法，请参照以上评分标准)23.（10分）(1)解：如图1，⊙O 是所求作的圆.·············································3分(2)延长GO 交⊙O 于点F ，连接OB 、BF ，则180F BEG ∠+∠=︒，又∵180GEM BEG ∠+∠=︒，45GEM ∠=︒，∴45F GEM ∠=∠=︒，∴90BOG ∠=︒，∵⊙O 与AM 相切于点G ，∴AG OG ⊥，90AGD ∠=︒ABCGMODE FABC M(第23题图1)OG∴AGD BOG ∠=∠，∴OB ∥AG ，························································································································4分∴BOD △∽AGD △，∴BO OD AG GD =，即BO AGOD GD=，····································································································5分∵G 是ABC △的重心，∴2AG GM =，······················································································································6分又∵1DG GM ==，∴2AG =，···························································································································7分∴221BO OD ==，2BO OD =，·····································································································8分设半径OB OG r ==，则1OD r =-，∴()21r r =-，解得2r =，······································································································9分∴290212223602S ππ⨯⨯=-⨯⨯=-弓形.··························································································10分(其它解法，请参照以上评分标准)24.（13分）(1)证明：∵CD 与CE 都是⊙O 的切线，∴CD CE =，即CDE △是等腰三角形.·························································································3分(2)如图1，作AM ⊥直线DE 于点M ，作BN ⊥直线DE 于点N ，则90AMD BNE ∠=∠=︒.由(1)证得CD CE =，∴CDE CED ∠=∠，又∵CDE ADM ∠=∠，CED BEN ∠=∠，∴ADM BEN ∠=∠,∴AMD △∽BNE △,················································································································5分∴AM ADBN BE=，∵AD 与AF 都是⊙O 的切线，∴AD AF =，同理可证BF BE =，·····························································································6分∵AM DE ⊥，BN DE ⊥，FG DE ⊥，∴AM ∥FG ∥BN ，∴MG AF ADGN BF BE ==，∴AM MGBN GN=，又∵90AMD BNE ∠=∠=︒，∴AMG △∽BNG △,················································································································7分∴AGM BGN ∠=∠，∵FG DE ⊥,∴90FGM FGN ∠=∠=︒,∴FGM AGM FGN BGN ∠-∠=∠-∠，∴AGF BGF ∠=∠，即FG 平分AGB ∠.························································································8分(3)AT BF =，理由如下：(第24题图1)DEFAB GC NMO过点Q 作⊙O 的切线HK ，分别交AC 、BC 于点H 、K ，连接OA 、OB 、OK 、OH .∵AB 与HK 都是⊙O 的切线，∴AB FQ ⊥，HK FQ ⊥，∴AB ∥HK ，90AFO OQH ∠=∠=︒，∴180BAH AHK ∠+∠=︒，∵AB 与AC 都是⊙O 的切线，∴OA 平分BAC ∠，即12OAH BAH ∠=∠，同理可证12OHA AHK ∠=∠，∴()111809022OAH OHA BAH AHK ∠+∠=∠+∠=⨯︒=︒，∴()18090AOH OAH OHA ∠=︒-∠+∠=︒，∴90HOQ AOF ∠+∠=︒，∵90FAO AOF ∠+∠=︒，∴HOQ FAO ∠=∠，∴AFO △∽OQH △，··············································································································9分∴FO AFQH OQ=，又∵FO OQ =，∴2FO AF QH =⋅，同理可证得BFO △∽OQK △，∴FO BFQK OQ=，即2FO BF QK =⋅，∴AF QH BF QK ⋅=⋅，即AF QKBF QH=，························································································11分∵AB ∥HK ，∴CHQ △∽CAT △，∴HQ CQ AT CT =，同理可证CQ QKCT TB=，∴HQ QK AT TB =，QK TBHQ AT=，∴AF TBBF AT=，························································································································12分BF AT AF BF AT TB =++，即BF ATAB AB=，∴AT BF =.···························································································································13分(其它解法，请参照以上评分标准)25.（13分）解：(1)由题意，得()2247y x n n =-+--+，令0x =，则23y n =-+，∴点F 的坐标为()20,3n -+.·····································································································2分(2)抛物线()2247y x n n =-+--+的对称轴为直线2x n =-，顶点E 的坐标为()2,47n n --+，∵当2x >时，y 随x 的增大而减小，∴22n -≤，解得0n ≥，·········································································································3分∵点F 在y 轴的正半轴，∴230n -+>，解得n <<，····························································································4分(第24题图2)DEF ABQ CT O H K。
一、选择题（每小题 3 分，满分 27 分．在每小题给出的四个选项中， 只有一项是正确的；每小题选出答案后，用 2B 铅笔将答题卡上对应题目的答案标号的 小框涂黑）1 、下列二次根式中，与 3 是同类二次根式的是（ ）A. 18B. 6C. 30D. 3002 、在下列图形中，既是中心对称图形又是轴对称图形的是（ ）A. 等边三角形B. 平行四边形C. 圆D. 等 腰梯形3、方程 x 2 x 的根为（ ）A. x 1 1,x 2 1B. x 1 0,x 2 1C. x 0D. x=14 、随机掷一枚均匀的硬币两次，落地后有两次正面朝上的概率是7 、若一元二次方程 x2+2x-3=0 的两个根为 x1 ，x2,则 x1+x2 的值为 （）A.14B. 12C.D. 1 5 、如图，⊙O 是△ABC 的外接圆， ∠OCB ＝40 °则∠ A A ． 60 B ． 50 C ． 40 D ． 30 6、在平面直角坐标系中，抛物线 y （x 1） 2 1的顶点坐标是 （ A ．（ 1,0 ） B ．（-1,0 ） C ．（1,-1 ） D ．（ 1， 1） 的度数等A. 2B. 2C.3D. 3 或38、把抛物线y 5x2向上平移2 个单位后，所得抛物线的解析式是( ) A. y 5x2 2 B. y 5x2 2 C. y 5x2 2 D. y 5x2 29、在Rt △ABC 中，∠C=90 °，AC=12 ，BC=5 ，将△ABC 绕边AC 所在直线旋转一周得到圆锥，则该圆锥的侧面积是( ) A．25πB．65πC．90πD．130 π二、填空题(细心填一填，本大题共6 小题，每小题3 分，满分18 分)10、已知式子1 x有意义，则x 的取值范围是11、已知两个圆的半径分别为2 和7，两个圆的圆心之A间的距离是5 ，则这两个圆的位置关系是.12、如图，△ABC以点A为旋转中心，按逆时针方向旋转B6C0 ，得△ABC ，则△ABB 是三角形。
