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徐州市2018届中考第一次模拟考试数学试题(带答案)

徐州市2018届中考第一次模拟考试数学试题(带答案)
徐州市2018届中考第一次模拟考试数学试题(带答案)

2018年江苏省徐州中考数学模拟试卷

(满分:140分时间:120分钟)

一、选择题:本大题共8个小题,每小题3分,共24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. -2的倒数是()

A

.-

1

2

B.

1

2

C.-2 D.2

2. 下列图形中,是轴对称图形,但不是中心对称图形的是()

A.B.C.D.

3. 某桑蚕丝的直径约为0.000016米,将0.000016用科学计数法表示是()

A.4

1.610-

?B.5

1.610-

?C.7

1.610-

?D.4

1610-

?

4. 在下列的计算中,正确的是()

A.m3+m2=m5B.m5÷m2=m3C.(2m)3=6m3D.(m+1)2=m2+1

5.关于2、6、1、10、6的这组数据,下列说法正确的是( )

A.这组数据的众数是6 B.这组数据的中位数是1

C.这组数据的平均数是6 D.这组数据的方差是10

6.如图,△ABC内接于⊙O,若∠A=α,则∠OBC等于()

A.180°-2α B.2α C.90°+α D.90°-α

7.将函数y=x2的图象用下列方法平移后,所得的图象不经过点A(1,4)的方法是()

A.向左平移1个单位B.向右平移3个单位

C.向上平移3个单位D.向下平移1个单位

8. 如图在矩形ABCD中,AB=5,AD=3,动点P满足

1

3

PAB ABCD

S S

?

=

四边形

,则点P到

A、B两点距离之和P A+PB的最小值为()

A.29B.34C.52D.41

二、填空题(本大题有10小题,每题3分,满分30分,将答案填在答题纸上)

(第6题)

(第8题)

9. 二次根式2a 中字母a 的取值范围是 .

10. 如图,转盘中6个扇形的面积都相等.任意转动转盘一次,当转盘停止转动时,指针指向奇数的概率

是 .

11. 如图,在△ABC 中,∠ABC 和∠ACB 的角平分线相交于点O ,若∠A =50°,则∠BOC = . 12. 已知反比例函数y =2

x

,当x <-1时,y 的取值范围为___________.

学科网ZXXK]

13. 如图,直线a ∥b ∥c ,直线l 1,l 2与这三条平行线分别交于点A 、B 、C 和点D 、E 、F .若AB ︰BC =1︰2,

DE =3,则EF 的长为 . 14. 已知a 2+a =1,则代数式3-a 2-a 的值为

15. 如图所示的正六边形 ABCDEF ,连结 FD ,则∠FDC 的大小为 .

16. 如图,AC 是⊙O 的切线,BC 是⊙O 的直径,AB 交⊙O 于点D ,连接OD ,若∠A =50°,则∠COD 的度数为 .

17. 在矩形ABCD 中,AB =2,BC =3,若点E 为边CD 的中点,连接AE ,过点B 作BF ⊥AE 于点F ,则BF

长为 .

(第10题)

32

1321

(第11题) O

A

(第13题)

c

b a F E D

C

B A l 2

l 1(第15题) (第16题)

D

O

A

(第17题)

F

E A

B

18. 某广场用同一种如图所示的地砖拼图案,第一次拼成的图形如图1所示的图案,第二次拼成图形如图2所示的图案,第三次拼成的图形如图3所示的图案,第四次拼成的图形如图4所示的图案......按照这样的规律进行下去,第n次

拼成的图形共用地砖块.

三、解答题(本大题共10小题,共86分.解答应写出文字说明、证明过程或演算步骤.)

19.(本题10分)

(1)(-2018)°-(

3

1

)-1+9;(2)

4

4

4

2

2

-

+

-

a

a

a

÷

a

a

a

2

2

2+

-

-3.

20. (本题10分)

(1)解方程:

1

x-2

+2=

1-x

2-x

;(2)解不等式组:

2157

10

2

3

()

x x

x

x

?+>-

?

?+

>

??

.

21. (本题7分)中华文明,源远流长,中华汉字,寓意深广。为传承中华优秀传统文化,某校团委组织了一

次全校3000名学生参加的“汉字听写”大赛.为了解本次大赛的成绩,校团委随机抽取了其中200名学生的成绩(成绩x取整数,部分100分)作为样本进行统计,制成如下不完整的统计图表:

根据所给信息,解答下列问题:

(1)m=,n=;

(2)补全频数分布直方图;

(3)这200名学生成绩的中位数会落在分数段;

(4)若成绩在90分以上(包括90分)为“优”等,请你估计该校参加本次比赛的3000名学生中成绩是“优”等

成绩x(分)频数(人)频率

50≤x<60 10 0.05

60≤x<70 30 0.15

70≤x<80 40 n

80≤x<90 m0.35

90≤x<100 50 0.25

第21题图

频数(人)

频数分布直方图

成绩(分)

的约为多少人?

22. (本题7分) 为弘扬中华传统文化,某校举办了学生“国学经典大赛”.比赛项目为:A.唐诗;B.宋词;C.论语;D.三字经.比赛形式分“单人组”和“双人组”.

(1)小丽参加“单人组”,她从中随机抽取一个比赛项目,恰好抽中“三字经”的概率是多少?

(2)小红和小明组成一个小组参加“双人组”比赛,比赛规则是:同一小组的两名队员的比赛项目不能相同,且每人只能随机抽取一次,则小红和小明都没有抽到“论语”的概率是多少?请用画树状图或列表的方法进行说明.

23. (本题8分)如图,在平行四边形ABCD中,边AB的垂直平分线交AD于点E,交CB的延长线于点F,

连接AF,BE.

(1)求证:△AGE≌△BGF;

(2)试判断四边形AFBE的形状,并说明理由.

24. (本题8分) 某内陆城市为了落实国家“一带一路”战略,促进经济发展,增强对外贸易的竞争力,把距离港口420 km的普通公路升级成了同等长度的高速公路,结果汽车行驶的平均速度比原来提高了50%,行驶时间缩短了2h.求汽车原来的平均速度.

25. (本题8分)如图,一座钢结构桥梁的框架是△ABC,水平横梁BC长18米,中柱AD高6米,其中D 是BC的中点,且AD⊥B C.

(1)求sinB的值;

(2)现需要加装支架DE、EF,其中点E在AB上,BE=2AE,且EF⊥BC,垂足为点F,求支架DE的长.

26.(本题9分)定义:数学活动课上,李老师给出如下定义:如果一个三角形有一边上的中线等于这条边的一半,那么称三角形为“智慧三角形”.

图1 图2 图3

理解:

⑴如图1,已知A、B是⊙O上两点,请在圆上找出满足条件的点C,使△ABC为“智慧三角形”(画出点C的位置,保留作图痕迹);

⑵如图2,在正方形ABCD中,E是BC的中点,F是CD上一点,且CF=1

4

CD,试判断△AEF是否为“智慧

三角形”,并说明理由;

运用:

⑶如图3,在平面直角坐标系xOy中,⊙O的半径为1,点Q是直线y=3上的一点,若在⊙O上存在一点P,使得△OPQ为“智慧三角形”,当其面积取得最小值时,直接写出此时点P的坐标.

27.(本题9分)如图,△ABC是边长为4cm的等边三角形,边AB在射线OM上,且OA=6cm,点D从点O 出发,沿OM的方向以1cm/s的速度运动,当D不与点A重合时,将△ACD绕点C逆时针方向旋转60°得到△BCE,连接DE.

(1)求证:△CDE 是等边三角形;

(2)当6<t <10时,的△BDE 周长是否存在最小值?若存在,求出△BDE 的最小周长; 若不存在,请说明理由.

(3)当点D 在射线OM 上运动时,是否存在以D 、B 、E 为顶点的三角形是直角三角形? 若存在,求出此时t 的值;若不存在,请说明理由.

28. (本题10分)如图,抛物线y=-x 2+bx+c 与直线AB 交于A (-4,-4),B (0,4)两点,直线AC :y =-1

2

x -6交y 轴与点C 。点E 是直线AB 上的动点,过点E 作EF ⊥x 轴交AC 于点F ,交抛物线于点G 。 (1)求抛物线y=-x 2+bx+c 的表达式;

(2)连接GB 、EO ,当四边形GEOB 是平行四边形时,求点G 的坐标;

(3)①在y 轴上存在一点H ,连接EH 、HF ,当点E 运动到什么位置时,以A 、E 、F 、H 为顶点的四边形是矩形?求出此时点E 、H 的坐标;

②在①的前提下,以点E 为圆心,EH 长为半径作圆,点M 为⊙E 上一动点,求

1

2

AM+CM 的最小值。 E

A

C

B

O

·

D

M

A C

B

O

·

D E

M

D

A C

B O

·

E

参考答案

一、选择题

题号 1 2 3 4 5 6 7 8 答案

A

A

B

B

A

D

D

D

二、填空题 9.a ≥2 10.

2

3

11. 115° 12. -2<y <0 13. 6 14. 2 15. 90° 16. 80° 17. 3

105

18. 2n 2+2n 三、解答题

19. (1)解:(-2018)°-(

3

1)-1

+9=1-3+3=1. (2)解:44422-+-a a a ÷a

a a 222+--3=)2)(2()2(2+--a a a ·2)

2(-+a a a -3=a -3.

20.(1)解:方程两边同乘x -2,得1+2(x -2)=x -1,解得x =2,经检验,x =2是增根,原方程无解.

(2)解:2157

10

23

()①②

x x x x ?+>-?

?+>??

由①得:x <3, 由②得:x <2,

∴不等式的解集为x <2. 21.解:(1)m =70, n =0.2;

(2)频数分布直方图如图所示,

(3) 80≤x <90;

(4)该校参加本次比赛的3000名学生中成绩“优”等的约有:3000×0.25=750(人). 22.解:(1)因为四个比赛项目被抽中的机会均等,所以小丽恰好抽中“三字经”的概率为

4

1

; 频数(人)

频数分布直方图

成绩(分)

(2)列表法列举可能情况如下:

共有12种可能,但两人都没有抽中“论语C ”的有6种可能即:AB ,AD ,BD ,BA ,DA ,DB ,故小红和小明都没有抽到“论语”的概率是

2

1

。或用画树状图法列举如下: .

23.解:(1)证明:∵EF 是AB 的垂直平分线,∴AG =BG .∵四边形ABCD 是平行四边形,∴AD ∥CF ,∴∠AEG =∠BFG ,∠EAG =∠FBG ,在△AGE 和△BGF 中,∠AEG =∠BFG ,∠EAG =∠FBG ,AG =BG ,∴△AGE ≌△BGF (AAS ).

(2)四边形AFBE 是菱形.理由如下:∵△AGE ≌△BGF ,∴AE =BF ,又AD ∥CF ,∴四边形AFBE 是平行四边形,又AB ⊥EF ,∴四边形AFBE 是菱形.

24.解:设汽车原来的平均速度为x km/h ,根据题意,得.

420x -420(1+50%)x

=2. 解这个方程,得 x =70. 经检验x =70是方程的解. 答:汽车原来的平均速度为70km/h.

25.解:(1)在Rt △ABD 中,∵BD =DC =9,AD =6, ∴AB =

=3

∴sinB =.

(2)∵EF ∥AD ,BE =2AE ,

A B C D A AB AC AD B BA BC BD C CA CB CD D

DA

DB

DC

∴,

∴,

∴EF=4,BF=6,

∴DF=3,

在Rt△DEF中,DE==5.

26.解:(1)如图所示:

(说明:画对一个给1分,无画图痕迹不给分) (2)△AEF是“智慧三角形”,理由如下:

∵四边形ABCD是正方形,

∴∠B=∠C=90°,AB=BC=CD.

∴BE=EC=1

2

BC=

1

2

AB,CF=

1

4

AB.

1

==

2 CF EC

BE AB

.

又∵∠B=∠C=90°,

∴△ECF∽△ABE,

∴∠CEF=∠BAE.

∵∠BAE+∠AEB=90°,

∴∠CEF+∠AEB=90°,

∴∠AEF=90°,

∴△AEF是直角三角形.

∵Rt△AEF斜边AF上的中线等于AF的一半,∴△AEF是“智慧三角形”.

(3)1221()33P -,,2221

()33

P ,

27.解:(1)证明:∵△BCE 是由△ACD 绕点C 逆时针方向旋转60°所得,∴∠DCE =60°,DC =EC ,∴△CDE 是等边三角形.

(2)存在,当6<t <10时,由旋转可知, ∴BE =AD .

C △DBE =BE +DB +DE =AB +DE =4+DE ,又由(1)可知,△CDE 是等边三角形. ∴DE =C

D ,

∴C △DBE =CD +4,由垂线段最短可知,当CD ⊥AB 时,△BDE 的周长最小, 此时,CD =32cm ,∴△DBE 的最小周长C △DBE =CD +4=32+4(cm ). (3)存在,

①∵当点D 与点A 重合时,D 、E 、B 不能构成三角形;当点D 与点B 重合时,显然不合题意. ∴t ≠6s ,t ≠10s ,

②当0≤t <6s 时,由旋转可知∠ABE =60°,∠BDE <60°,从而∠BED =90°,由(1)可知△CDE 是等边三角形, ∴∠DEC =60°,∴∠CEB =30°,

∴∠CDA =30°,∵∠CAB =60°,∴∠ACD =∠ADC =30°, ∴DA =CA =4,∴OD =OA -DA =6-4=2,∴t =2÷1=2s , ③当6<t <10s 时,由∠DBE =120°>90°,∴此时不存在;

④当t >10s 时,由旋转可知∠DEB =60°,又由(1)知∠CDE =60°, ∴∠BDE =∠CDE +∠BDC =60°+∠BDC ,而∠BDC >0°, ∴∠BDE >60°,

∴只能∠BDE =90°,从而∠BDC =30°, ∴BD =BC =4,∴OD =14cm ,∴t =14÷1=14s 。

综合①②③④得,当t =2s 或t =14s 时,以D 、E 、B 为顶点的三角形是直角三角形.

28.解:(1)y =-x 2-2x+4

(2)设直线AB 的表达式为y=kx+b

∵直线AB 过点A (-4,-4),B (0,4),∴444k b b -=-+??=?,解得2

4k b =??=?

,∴y=2x+4

设E (m ,2m+4),则G (m ,-m 2-2m+4)

∵四边形GEOB是平行四边形,∴GE=OB=4,∴-m2-2m+4-2m-4=4,解得m=-2 ∴G(-2,4)

(3)①设E(m,2m+4),则F(m,-1

2

m-6)

过A作AN⊥EG,过H作HQ⊥EG

四边形AFHE是矩形,∴△PFN≌△HEQ,∴AN=QH,∴m+4=-m,解得m=-2,E(-2,0)

EQ=FN=-4+1

2

m+6=1

∴H(0,-1)

②由题意可得,E(-2,0),H(0,-1),∴EH=5,即⊙E的半径为5,∵M点在⊙E上,∴EM=5

∵A(-4,-4),E(-2,0),∴AE=25

在AE上截取EP=1

2

EM,则EP=

1

5

2

,连接PM,

在ΔEPM与ΔEMA中,∵EP

EM

=

5

2

5

=

1

2

=

5

25

=

EM

EA

,∠PEM=∠MEA,∴ΔEPM∽ΔEMA∴PM=

1

2

AM

∴线段PC的长即为1

2

AM+CM的最小值

由EP=1

2

EM=

1

4

AE=

1

4

×25=

1

5

2

,AP=AE-PE=

3

5

2

, AC=25∴PC=

55

即1

2

AM+CM的最小值为

55

【最新】徐州市中考英语模拟试卷含答案

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