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2020年山东省威海市初中升学考试数学试题及答案(word版)

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2020年山东省威海市初中升学考试数学试题及答案

(word版)

威海市二○○九年初中升学考试

数 学

友爱的同学:

你好!答题前,请认真阅读以下讲明:

1. 本试卷共12页,分第一卷和第二卷两部分.第一卷〔1-4页〕为选择题,第二卷〔5-12

页〕为非选择题.试卷总分值120分.考试时刻120分钟.

2. 请清点试卷,然后将考生信息填涂在答题卡上,并将第二卷密封线内的项目填写清晰. 3. 将选择题答案用2B铅笔涂在答题卡对应题目的标号上,将非选择题答案用蓝色或黑色

钢笔、圆珠笔填写在试卷上.不要求保留精确度的题目,运算结果保留准确值. 期望你能愉快地度过这120分钟,祝你成功!

第一卷〔选择题,共36分〕

一、选择题〔本大题共12小题,每题3分,共36分,在每题给出的四个选项中,只有一个是正确的.每题选对得3分,选错、不选或多项选择,均不得分〕 1.327的绝对值是〔 〕 A.3

B.3

C.

B

1 3D.

132.如图,ABAC,BDBC,假设A40,那么

A D

C

ABD的度数是〔 〕

A.20

B.30

C.35

D.40

3.实数a,b在数轴上的位置如下图,那么以下结论正确的选项是〔 〕 A. ab0 B. ab0

0 1 a aC. ab0 D.0

1 b

b4.形状相同、大小相等的两个小木块放置于桌面,其俯视图如以下图所示,那么其主视图是〔 〕 A. B. C. D. 〔俯视图〕 5.化简y11x的结果是〔 〕 xyA.y xB. x y C.

x yD.

y x6.某公司职员的月工资如下表: 职员 月工资/元 经理 4800 副经理 3500 职员A 2000 职员B 1900 职员C 1800 职员D 1600 职员E 1600 职员F 1600 职员G 1000 那么这组数据的平均数、众数、中位数分不为〔 〕 A.2200元 1800元 1600元 B.2000元 1600元 1800元 C.2200元 1600元 1800元 D.1600元 1800元 1900元 7.二次函数y3x6x5的图象的顶点坐标是〔 〕 A.(18), B.(18),

C.(1,2)

D.(1,4)

28.在梯形ABCD中,AB//CD,A60,B30,ADCD6,那么AB的长度为〔 〕 A.9

B.12

C.18

D.633 y 9.如图,A,B的坐标为〔2,0〕,〔0,1〕假设将线段AB平移至A1B1,那么ab的值为〔 〕

A.2 B.3 C.4 D.5

10.如图,在四边形ABCD中,E是BC边的中点,连结DE并延长,交AB的延长线于F点,ABBF.添加一个条件,使四边形ABCD是平行四边形.你认为下面四个条件中可选择的是〔 〕

A.ADBC B.CDBF C.AC D.FCDE

11.O是△ABC的外接圆,假设AB=AC=5,BC=6,那么O的半径为〔 〕 A.4 B.3.25 C.3.125 D.2.25 12.如图,△ABC和的△DEF是等腰直角三角形,

B1(a,2) B(0,1) O D A1(3,bx )

A(2,0) C E

A

F

B

CF90,AB2,DE4.点B与点D重合,点

A,(BD),E在同一条直线上,将△ABC沿DE方向平移,至点A与点E重合时停

止.设点B,D之间的距离为x,△ABC与△DEF重叠部分的面积为y,那么准确反映y与x之间对应关系的图象是〔 〕

二、填空题〔本大题共6分,每题3分,共18分.只要求填出最后结果〕

1013.运算(23)(21)的结果是_________.

l 1

a

2

b

14.如图,直线l与直线a,b相交.假设a∥b,170,那么2的度数是_________.

15.分解因式:(x3)(x3)___________.

16.如图,△ABC与△ABC是位似图形,点O是位似中心,假设OA2AA,S△ABC8,那么S△ABC________. 17.假设关于x的一元二次方程x(k3)xk0的一个根是

22〔第14题图〕

C

C O 2,那么另一个根是______.

18.如图,

A A B B

〔第16题图〕

O1和O2的半径为1和3,连接O1O2,交O2于

O1绕点P按顺时针方向旋转360,那么O1与O2共相切

o1 第18题图P 点P,O1O28,假设将

_______次.

三、解答题〔本大题共7小题,共66分〕 19.〔7分〕

o2 22先化简,再求值:(ab)(ab)(2ab)3a,其中a23,b32.

20.〔7分〕

除颜色外完全相同的六个小球分不放到两个袋子中,一个袋子中放两个红球和一个白球,另一个袋子中放一个红球和两个白球.随机从两个袋子中分不摸出一个小球,试判定摸出两个异色小球的概率与摸出两个同色小球的概率是否相等,并讲明理由.

21.〔9分〕

如图,一巡逻艇航行至海面B处时,得知其正北方向上C处一渔船发生故障.港口A处在B处的北偏西37方向上,距B处20海里;C处在A处的北偏东65方向上. 求B,C之间的距离〔结果精确到0.1海里〕.

参考数据:sin370.60, cos370.80,tan370.75,北 sin650.91,cos650.42,tan652.14.

北 C 65°

A

37°

B

22.〔10分〕

响应〝家电下乡〞的惠农,某商场决定从厂家购进甲、乙、丙三种不同型号的电冰箱80台,其中甲种电冰箱的台数是乙种电冰箱台数的2倍,购买三种电冰箱的总金额不超过...132 000元.甲、乙、丙三种电冰箱的出厂价格分不为:1 200元/台、1 600元/台、2 000元/台.

(1) 至少购进乙种电冰箱多少台?

(2) 假设要求甲种电冰箱的台数不超过丙种电冰箱的台数,那么有哪些购买方案?

23.〔10分〕

如图1,在正方形ABCD中,E,F,G,H分不为边AB,BC,CD,DA上的点,

交点为O. HAEBFCGD,连接EG,FH,〔1〕如图2,连接EF,FG,GH,HE,试判定四边形EFGH的形状,并证明你的结论;

G G C C D D F F

H H O

A A B E B E

〔第23题图1〕 〔第23题图2〕 〔第23题图3〕

〔2〕将正方形ABCD沿线段EG,HF剪开,再把得到的四个四边形按图3的方式拼接成一个四边形.假设正方形ABCD的边长为3cm,HAEBFCGD1cm,那么图3中阴影部分的面积为_________cm2. 24.〔11分〕

如图,在直角坐标系中,点A,B,C的坐标分不为(1,,,,,0)(30)(03),过A,B,C三点的抛物线的对称轴为直线l,D为对称轴l上一动点. (1) 求抛物线的解析式;

(2) 求当ADCD最小时点D的坐标; (3) 以点A为圆心,以AD为半径作A.

①证明:当ADCD最小时,直线BD与A相切.

②写出直线BD与A相切时,D点的另一个坐标:___________. 25.〔12分〕

y C l A O B x 一次函数yaxb的图象分不与x轴、y轴交于点M,N,与反比例函数yk的图象相x交于点A,B.过点A分不作ACx轴,AEy轴,垂足分不为C,E;过点B分不作

BFx轴,BDy轴,垂足分不为F,D,AC与BD交于点K,连接CD.

〔1〕假设点A,B在反比例函数y①S四边形AEDKS四边形CFBK; ②ANBM.

〔2〕假设点A,B分不在反比例函数y还相等吗?试证明你的结论.

y

N E D A(x1,y1)k的图象的同一分支上,如图1,试证明: xk的图象的不同分支上,如图2,那么AN与BMxy

F M B(x,y)B(x2,y2)K O C F M E N A(x1,y1)x O C D K x

威海市2018年初中升学考试 数学试题参考解答及评分意见

评卷讲明:

1. 第一大题〔选择题〕和第二大题〔填空题〕的每题,只有总分值和零分两个评分档,不

给中间分.

2. 第三大题〔解答题〕每题的解答中所对应的分数,是指考生正确解答到该步骤所应得的

累计分数.部分试题有多种解法,对考生的其他解法,请参考评分意见进行评分. 3. 假如考生在解答的中间过程显现运算错误,但并没有改变试题的实质和难度,其后续部

分酌情给分,但最多不超过正确解答分数的一半;假设显现严峻的逻辑错误,后续部分就不再给分.

一、选择题〔本大题共12小题,每题3分,共36分〕 题号 答案 1 A 2 B 3 A 4 D 5 D 6 C 7 A 8 C 9 A 10 D 11 C 12 B 二、填空题〔本大题共6小题,每题3分,共18分〕

13.2; 14.110°; 15.(x2)(x3) 16.18; 17.1; 18.3. 三、解答题〔本大题共7小题,共66分〕 19.〔本小题总分值7分〕

解:(ab)(ab)(2ab)3aa2abb2aabb3a ················· 3分 ······································································································· 5分 ab. ·

当a23,b222222232时,

22原式(23)(32)(2)(3)1 ······················································ 7分

20.〔本小题总分值7分〕

解:摸出两个异色小球的概率与摸出两个同色小球的概率不相等. ··························· 1分

画树状图如下〔画出一种情形即可〕:

开始 或 开始

红 红 白 白 白 红

红 白 白 红 白 白 红 白 白 红 红 白 红 红 白 红 红 白

································· 4分 ∴摸出两个异色小球的概率为摸出两个同色小球的概率

5, ····································································· 5分 94. ··········································································· 6分 9即摸出两个异色小球的概率与摸出两个同色小球的概率不相等. ······························ 7分 21.〔本小题总分值9分〕

北 解:过点A作ADBC,垂足为D.·····················1分 北 在Rt△ABD中,AB20,B37°,

C 65° ∴ADAB················3分 ·sin37°20sin37°≈12. ·

A D ··················5分 BDAB·cos37°20cos37°≈16. ·

在Rt△ADC中,ACD65°,

AD1237° ···························8分 ≈≈5.61 ·

tan65°2.14B BCBDCD≈5.611621.61≈21.6〔海里〕 答:B,C之间的距离约为21.6海里. ······························································· 9分

∴CD22.〔本小题总分值10分〕 解:〔1〕设购买乙种电冰箱x台,那么购买甲种电冰箱2x台,

丙种电冰箱(803x)台,依照题意,列不等式: ·················································· 1分 ·············································· 3分 12002x1600x(803x)2000≤132000. ·

解那个不等式,得x≥14. ············································································· 4分 至少购进乙种电冰箱14台. ·········································································· 5分 〔2〕依照题意,得2x≤803x. ··································································· 6分 解那个不等式,得x≤16. ············································································· 7分 由〔1〕知x≥14. 14≤x≤16. 又x为正整数,

··························································································· 8分 x141516,,. ·

因此,有三种购买方案:

方案一:甲种电冰箱为28台,乙种电冰箱为14台,丙种电冰箱为38台; 方案二:甲种电冰箱为30台,乙种电冰箱为15台,丙种电冰箱为35台; 方案三:甲种电冰箱为32台,乙种电冰箱为16台,丙种电冰箱为32台. ··············· 10分 23.〔本小题总分值10分〕 解:〔1〕四边形EFGH是正方形. ··················· 1分 证明:四边形ABCD是正方形,

G

C D

F

H A

O E B

ABCD90°,ABBCCDDA. HAEBFCGD,

····························· 2分 AEBFCGDH. ·

····· 3分 △AEH≌△BFE≌△CGF≌△DHG. ·

····························· 4分 EFFGGHHE. ·

四边形EFGH是菱形. ······························· 5分 由△DHG≌△AEH知DHGAEH. AEHAHE90°, DHGAHE90°.

························································································· 6分 GHE90°. ·

四边形EFGH是正方形. ············································································ 7分

〔2〕1. ····································································································· 10分 24.〔本小题总分值11分〕 解:〔1〕设抛物线的解析式为ya(x1)(x3). ··············································· 1分 将(0,3)代入上式,得3a(01)(03).

解,得a1. ····························································································· 2分

抛物线的解析式为y(x1)(x3).

即yx2x3. ······················································································ 3分 〔2〕连接BC,交直线l于点D. 点B与点A关于直线 l对称,

····················································· 4分 ADBD. ·

ADCDBDCDBC.

由〝两点之间,线段最短〞的原理可知:

现在ADCD最小,点D的位置即为所求. ·············· 5分 设直线BC的解析式为ykxb,

2y C l D E B x A O 由直线BC过点(3,0),(0,3),得03kb,

3b.解那个方程组,得k1,

b3.直线BC的解析式为yx3. ··································································· 6分

由〔1〕知:对称轴l为x21,即x1.

2(1)将x1代入yx3,得y132.

点D的坐标为〔1,2〕. ··············································································· 7分

讲明:用相似三角形或三角函数求点D的坐标也可,答案正确给2分. 〔3〕①连接AD.设直线l与x轴的交点记为点E. 由〔1〕知:当ADCD最小时,点D的坐标为〔1,2〕. DEAEBE2.

············································································ 8分 DABDBA45°. ·

ADB90°. AD⊥BD.

······················································································· 9分 BD与⊙A相切. ·

②(1································································································· 11分 ,2). 25.〔本小题总分值12分〕 解:〔1〕①

y N E D A B AC⊥x轴,AE⊥y轴,

四边形AEOC为矩形.

BF⊥x轴,BD⊥y轴,

四边形BDOF为矩形.

K O C F M x AC⊥x轴,BD⊥y轴,

四边形AEDK,DOCK,CFBK均为矩形. ·········· 1分

图1 OCx1,ACy1,x1y1k, S矩形AEOCOCACx1y1k

OFx2,FBy2,x2y2k, S矩形BDOFOFFBx2y2k. S矩形AEOCS矩形BDOF.

S矩形AEDKS矩形AEOCS矩形DOCK,

S矩形CFBKS矩形BDOFS矩形DOCK,

S矩形AEDKS矩形CFBK. ················································································· 2分

②由〔1〕知S矩形AEDKS矩形CFBK.

AKDKBKCK. AKBK. ····························································································· 4分 CKDKAKBCKD90°, △AKB∽△CKD. ··················································································· 5分 CDKABK.

AB∥CD. ······························································································ 6分

AC∥y轴,

四边形ACDN是平行四边形. ANCD. ······························································································ 7分 同理BMCD.

····························································································· 8分 ANBM.·

〔2〕AN与BM仍旧相等. ············································································ 9分

S矩形AEDKS矩形AEOCS矩形ODKC, S矩形BKCFS矩形BDOFS矩形ODKC,

y 又

S矩形AEOCS矩形BDOFk,

E N F M A S矩形AEDKS矩形BKCF. ···························· 10分

x O C AKDKBKCK.

D K CKDK B. AKBK图2 KK,

△CDK∽△ABK. CDKABK. AB∥CD. ···························································································· 11分

AC∥y轴,

四边形ANDC是平行四边形. ANCD. 同理BMCD. ANBM. ··························································································· 12分

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