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九年级数学试题和答案

发布时间：2019-12-23   来源：文档文库   
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九年级数学综合测试试卷
一、选择题（本题共16分，每小题2分）
1．下列图形中，不是轴对称图形的是
．．

(A (B (C (D
2．如图，直径为单位1 的圆从数轴上的原点沿着数轴无滑动地顺时针滚动一周到达点A，则点A表示的数是
DEC

(A2 (B2 (C π (D4
ABF3．如图，正五边形ABCDE，点F是AB延长线上的一点，则∠CBF的度数是 (A 60° (B72° (C108° (D120° 4．某颗人造地球卫星绕地球运行的速度是7．9×103 m／s，那么这颗卫星绕地球运行一年（一年以3．2×107 s计算）走过的路程约是 (A1.1×1010m (B7.9×1010m (C2.5×1010m (D2.5×1011m 5．如图，点A，B，C，D在⊙O上，AC是⊙O的直径，∠BAC=40°，D则∠D的度数是 C (A 40° (B50° (C60° (D90° 6．如果a+b=2，那么代数式1(A 2bab的值是 22aba2abbOAB1 (B1 (C 22 (D2
7．某非物质文化遗产共有16名传承艺人，为了了解每位艺人的日均生产能力，随机调查了某一天每位艺人的生产件数．获得数据如下表：
生产件数10 11 12 13 14 15 （件）
1 6 3 3 2 1 人数（人）
从这一天16名艺人中随意抽取1人，则他的这一天生产件数最可能的是
(A 11件 (B 12件 (C 13件 (D 15件
y28．如图，二次函数y＝ax＋bx＋c(a≠0的图象经过点A，B，C．现有5B下面四个推断：
4①抛物线开口向下； ②当x=－2时，y取最大值；
3C2③当m<4时，关于x的一元二次方程ax＋bx＋c=m必有两个不相等A2的实数根；
1④直线y=kx+c(k≠0经过点A，C，当kx+c> ax2＋bx＋c时，x的取值范围是－4<x<0；
–4–3–2–1O1其中推断正确的是 –1 (A ①② (B ①③ (C ①③④ (D ②③④
2x二、填空题(本题共16分，每小题2分

9．如图，该正方体的主视图是 形． 10．若分式1的值是正数，则x的取值范围是 ． ．．x111．某商场甲、乙、丙三名业务员5个月的销售额（单位：万元）如下表： 销售额 业务员
甲 乙 丙
第1月 7.2 7.8
9.2 第2月 9.6 9.7
5.8 第3月 9.6 9.8
8.5 第4月 8.0 5.8
9.9 第5月 9.3 9.9
9.9 AFDEC则甲、乙、丙三名业务员中销售额最稳定的是 ．
12．如图，在△ABC中，射线AD交BC于点D，BE⊥AD于E，CF⊥AD于F，请补充一个条件，使△BED≌△CFD，你补充的条件是 （填出一个即可）． 12．甲乙二人分别从相距20km的A，B两地出发，相向而行．下B图是小华绘制的甲乙二人运动两次的情形，设甲的速度是x km/h，乙的速度是y km/h，根据题意所列的方程组是 ．
甲走0.5h的路程甲走2h的路程相遇乙走2h的路程乙走1h的路程b第一次A第二次A甲走1h的路程Bab相距11kmBaAEBA，B两地相距20km14．如图，从一个边长为a的正方形的一角上剪去一个边长为b（a>b）的正方形，则剩余（阴影）部分正好能够表示一个乘法公式，则这个乘法公式是 （用含a，b的等式表示）．
15．如图，在Rt△ABC中，∠C=90°，AD平分∠BAC交BC于点D，过点D作DE⊥AB于点E，若CD=2，BD=4，则AE的长是 ．

16．小明家的客厅有一张直径为1.2米，高0.8米的圆桌BC，在距地面2米的A处有一盏灯，圆桌的影子为DE，依据题意建立平面直角坐标系，其中D点坐标为（2,0），则点E的坐标是 ．

三、解答题(本题共68分，第17-21题，每小题5分，第22-27题，每小题6分，第28题7分解答应写出文字说明、演算步骤或证明过程. 17．下面是小元设计的“作已知角的角平分线”的尺规作图过程．
已知：如图，∠AOB．
求作：∠AOB的角平分线OP． 作法：如图，
①在射线OA上任取点C； ②作∠ACD=∠AOB；
③以点C为圆心CO长为半径画圆，交射线CD于点P； ④作射线OP；
所以射线OP即为所求．
DCACDOB
根据小元设计的尺规作图过程，完成以下任务． （1）补全图形；
（2）完成下面的证明：
证明：∵ ∠ACD=∠AOB，
∴ CD∥OB（____________）（填推理的依据）． ∴∠BOP=∠CPO． 又∵ OC=CP，
∴∠COP=∠CPO（____________）（填推理的依据）． ∴∠COP=∠BOP． ∴ OP平分∠AOB． 18．计算：2sin60312

031．

2x13x1,19．解不等式组：x1

1. 2


20．已知关于x的一元二次方程x(k1xk20 （1）求证：方程总有两个实数根；
（2）若方程有一根为正数，求实数k的取值范围．

21．如图，在平面直角坐标系xOy中，函数2
ykx0的图象经过点A，作AC⊥x轴于点C．
x（1）求k的值；
（2）直线AB：yaxba0图象经过点A交x轴于点B．横、纵坐标都是整数的点叫做整点．线段AB，AC，BC围成的区域（不含边界）为W． ①直线AB经过0,1时，直接写出区域W内的整点个数； ②若区域W内恰有1个整点，结合函数图象，求a的取值范围．

22．如图，在△ABC中，AB=AC，点D是BC边的中点，连接AD，分别过点A，C作AE∥BC，CE∥AD交于点E，连接DE，交AC于点O． （1）求证：四边形ADCE是矩形； （2）若AB=10，sin∠COE=AEO4，求CE的长．
5BDC

23．费尔兹奖是国际上享有崇高荣誉的一个数学奖项，每4年评选一次，在国际数学家大会上颁给有卓越贡献的年龄不超过40岁的年轻数学家，美籍华人丘成桐1982年获得费尔兹奖．为了让学生了解费尔兹奖得主的年龄情况，我们查取了截止到2018年60名费尔兹奖得
主获奖时的年龄数据，并对数据进行整理、描述和分析.下面给出了部分信息. a．截止到2018年费尔兹奖得主获奖时的年龄数据的频数分布直方图如下
（数据分成5组，各组是28≤x＜31，31≤x＜34，34≤x＜37，37≤x＜40，x≥40）：
b．如图，在a的基础上，画出扇形统计图；


c．截止到2018年费尔兹奖得主获奖时的年龄在34≤x＜37这一组的数据是：
36 35 34 35 35 34 34 35 36 36 36 36 34 35 d．截止到2018年时费尔兹奖得主获奖时的年龄的平均数、中位数、众数如下：
年份 截止到2018 平均数
35.58 中位数
m 众数 37 ，38 根据以上信息，回答下列问题： （1）依据题意，补全频数直方图；
（2）31≤x＜34这组的圆心角度数是 度，并补全扇形统计图； （3）统计表中中位数m的值是 ；
（4）根据以上统计图表试描述费尔兹奖得主获奖时的年龄分布特征．

24．如图，AB是⊙O的直径，AC切⊙O于点A，连接BC交⊙O于点CD，点E是BD的中点，连接AE交BC于点F．
DFAEOB
（1）求证：AC=CF；
（2）若AB=4，AC=3，求∠BAE的正切值．

25．如图，点P是AB 所对弦AB上一动点，点Q是AB与弦AB所围成的图形的内部的一定点，作射线PQ交AB于点C，连接BC．已知AB=6cm，设A，P两点间的距离为xcm，P，C两点间的距离为y1cm，B，C两点间的距离为y2cm．（当点P与点A重合时，x的值为0）．


小平根据学习函数的经验，分别对函数y1，y2随自变量x的变化而变化的规律进行了探究． 下面是小平的探究过程，请补充完整：
（1）按照下表中自变量x的值进行取点、画图、测量，分别得到了y与x的几组对应值；
x/cm 0 1 2 3 4 5 6 y1/cm 5.37 4.06 2.83 m 3.86 4.83 5.82 y2/cm 2.68 3.57 4.90 5.54 5.72 5.79 5.82 经测量m的值是 （保留一位小数）．
（2）在同一平面直角坐标系xOy中，描出补全后的表中各组数值所对应的点(x，y1， (x，y2，并画出函数y1，y2的图象；



（3）结合函数图象，解决问题：当△BCP为等腰三角形时，AP的长度约为 cm．

26．平面直角坐标系xOy中，抛物线yx22mxm23与y轴交于点A，过A作AB∥x轴与直线x=4交于B点．
（1）抛物线的对称轴为x= （用含m的代数式表示）； （2）当抛物线经过点A，B时，求此时抛物线的表达式； （3）记抛物线在线段AB下方的部分图象为G（包含A，B两点），点P（m,0）是x轴上一动点，过P作PD⊥x轴于P，交图象G于点D，交AB于点C，若CD≤1，求m的取值范围．

y654321–6–5–4–3–2–1O–1–2–3–4–5–6123456x

27．在△ABC中，∠ABC=120°，线段AC绕点A逆时针旋转60°得到线段AD，连接CD，BD交AC于P．
（1）若∠BAC=α，直接写出∠BCD的度数 （用含α的代数式表示）； （2）求AB，BC，BD之间的数量关系；
（3）当α=30°时，直接写出AC，BD的关系．
DPA

CB

28．对于平面直角坐标系xoy中的图形P，Q，给出如下定义：M为图形P上任意一点，N为图形Q上任意一点，如果M，N两点间的距离有最小值，那么称这个最小值为图形P，Q间的“非常距离”，记作d（P,Q）．已知点A（4,0），B（0,4），连接AB． （1）d（点O，AB）=

（2）⊙O半径为r，若d（⊙O，AB）=0，求r的取值范围； （3）点C（－3，－2），连接AC，BC，⊙T的圆心为T（t，0），半径为2，d（⊙T，△ABC），且0<d <2，求t的取值范围．



数学试卷参考答案及评分标准
一、选择题（本题共16分，每小题2分）
题号 答案
1 C 2 C 3 B 4 D 5 B 6 A 7 A 8 B 二、填空题（本题共16分，每小题2分）
9．正方； 10．x>-1； 11．甲； 12．答案不唯一，如BD=DC； 13．2.5x2y2022； 14．ababab； 15．23； 16．(4,0．
xy1120三、解答题(本题共68分，第17-21题，每小题5分，第22-27题，每小题6分，第28题7分解答应写出文字说明、演算步骤或证明过程. 17．（1）如图；····························································································· 1 ACPOBD
（2）同位角相等，两直线平行； ································································ 3 等边对等角． ·················································································· 5 18．解：原式=2312331 ··························································· 4 2 =0． ························································································· 5 19．解：由①得x<3 ······················································································· 1 由①得x+1>2， ················································································· 2 x>1． ················································································ 3 ∴1<x<3． ················································································ 5 20．解：(1k22k14k8 ································································· 1 k3 ················································································ 2

2k320，
······························································· 3 ∴方程总有两个实数根． ·(2 ∵xk1k322∴x11，x22k． ······························································· 4 ，
∵方程有一个根为正数， ∴2k0 k2．··············································································· 5
21．（1）k=4；····························································································· 1 （2）①1个； ························································································· 2
②当直线AB经过点A（2，﹣2），（0,1）时区域W内恰有1个整点， ∴a1．
2当直线AB经过点A（2，﹣2），（1,1）时区域W内没有整点， ∴a=1． ······················································································ 3 ∴当1········································ 5 a1时区域W内恰有1个整点．
·222．（1）证明：∵AB=AC，点D是BC边的中点，
∴AD⊥BC于点D． ··································································· 1 ∵AE∥BC，CE∥AD，
∴四边形ADCE是平行四边形． ··················································· 2 ∴平行四边形ADCE是矩形． ······················································ 3 （2）解： 过点E作EF⊥AC于F．
A∵AB=10， ∴AC=10．
∵对角线AC，DE交于点O， O∴DE=AC=10． F∴OE=5． ···················· 4 BD∵sin∠COE=E4，
5C∴EF=4 ··················································································· 5 ∴OF=3．
∵OE=OC=5， ∴CF=2．
∴CE=25． ········································································· 6 23．（1）如图； ··························································································· 1 （2）31≤x＜34这组的圆心角度数是 78 度， ·················································· 2 如图（画图1分，数据1分）； ································································· 4 （3）统计表中中位数m的值是 36 ； ··························································· 5 （4）答案不唯一，如：费尔兹奖得主获奖时年龄集中在37岁至40岁． ··················· 6 24．（1）证明：∵AC切⊙O于点A，

∴∠BAC=90°． ········································································· 1 连接AD．
∵点E是BD的中点，
CDE∴∠BAE=∠DAE． F∵AB是⊙O的直径， ∴∠ADB=90°． BAO∵∠CAD+∠DAB=∠DAB+∠B=90°， ∴∠CAD=∠B．
∵∠CAD+∠DAE =∠B+∠BAE， ∴∠CAF=∠CFA． ···································································· 2 ∴AC=CF． ············································································· 3 （2）解：∵AB=4，AC=3，
∴BC=5． ················································································ 4 ∵AC=CF=3， ∴BF=2．
∵cosB∴BD=BDAB4， ABBC516． ············································································· 5 5126∴AD=，DF=．
551∴tan∠BAE= tan∠DAE = ························································· 6 225．（1）3.0； ····························································································· 1 （2）如图； ··························································································· 3 （3）1.2或1.6或3.0． ············································································· 6
26．（1）m； ······························································································· 1 （2）∵yx22mxm23xm3，
∴抛物线顶点坐标为（m,－3）． ························································ 2 ∵抛物线经过点A，B时，且AB∥x轴， ∴抛物线对称轴为x=m=2． ······························································· 3 ∴抛物线的表达式为yx4x1； ················································· 4 （3）0m1． ···················································································· 6 27．（1）∠BCD=120°－α． ············································································· 1 22
（2）解：
方法一：延长BA使AE=BC，连接DE． ······ 2 由（1）知△ADC是等边三角形，
∴AD=CD．
∵∠DAB+∠DCB=∠DAB+∠DAE=180°， ∴∠DAB=∠DAE． ∴△ADE≌△CDB． ···················· 3 ∴BD=BE．
∴BD=AB+BC． ·························· 4
方法二：延长AB使AF=BC，连接CF． ······ 2
∠BDC=∠ADE． ∵∠ABC=120°， ∴∠CBF=60°．
∴△BCF是等边三角形． ∴BC=CF．
∵∠DCA=∠BCF=60°，
∴∠DCA+∠ACB=∠BCF+∠ACB． 即∠DCB=∠ACF． ∵CA=CD，
∴△ACF≌△DCB．····················· 3 ∴BD=AF．
∴BD=AB+BC． ·························· 4 （3）AC，BD的数量关系是：ACDPEABCDPABCF3BD； ········································· 5 2位置关系是：AC⊥BD于点P． ······································· 6 28．（1）22； ·························································································· 1 （2）22r4； ················································································ 3 （3）252t52或6<r<8． ····················································· 7


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