柳州市2020年中考数学试卷含答案

2020年柳州市初中毕业升学考试试卷

数学

（考试时间共120分钟，全卷满分120分）

第Ⅰ卷（选择题，共36分）

注意事项：

1．答题前，考生务必先将自己的姓名、准考证号用蓝、黑色墨水笔或圆珠笔填写在试卷左边的密封线内．

2．第Ⅰ卷为第1页至第2页．答题时，请用2B 铅笔把各小题正确答案序号填涂在答题卡对应的题号内．如需改动，须用橡皮擦干净后，再填涂其它答案． 在第Ⅰ卷上答题无效．

一、选择题（本大题共12小题，每小题3分，满分36分．在每个小题给出的四个选项中，只有一项是正确的，每小题选对得3分，错选、不选或多选均得零分） 1．5-的相反数是

A

．5 Ｂ．5- Ｃ．55-

Ｄ．55

2．如图1，点A B C 、、是直线l 上的三个点，图中共有线段条数是

A ．1条 Ｂ．2条 Ｃ．3条 Ｄ．4条 3．三条直线a b c 、、，若a c ∥，b c ∥，则a 与b 的位置关系是

A ．a b ⊥ Ｂ．a b ∥ Ｃ．a b a b ⊥或∥ Ｄ．无法确定 4．图2的几何体中，主视图、左视图、俯视图均相同的是

图1

图2

5

．若分式

2

3x

-

有意义，则x的取值范围是

A．3

x≠Ｂ．3

x=Ｃ．3

x<Ｄ．3

x>

6．不等式5

x+≥8的解集在数轴上表示为

A． B. C. D.

7．一个正多边形的一个内角为120度，则这个正多边形的边数为

A．9 Ｂ．8 Ｃ．7 Ｄ．6

8．如图3，Rt ABC

△中，90

C

∠=°，ABC

∠的平分线BD交AC于D，若3cm

CD=，则点D 到AB的距离DE是

A．5cm Ｂ．4cm Ｃ．3cm Ｄ．2cm

图3 图4 图5

9.如图4，在正方形ABCD的外侧作等边ADE

∠的度数为

△，则AEB

A．10°Ｂ．12.5°Ｃ．15°Ｄ．20°

10．上海“世界博览会”某展厅志愿者的年龄分布如图5，这些志愿者年龄的众数是A．19岁Ｂ．20岁Ｃ．21岁Ｄ．22岁

11．抛物线2

=-++上部分点的横坐标x，纵坐标y的对应值如下表：

y x bx c

x…2-1-0 1 2 …

y…0 4 6 6 4 …

从上表可知，下列说法正确的个数是

①抛物线与x轴的一个交点为(20)

-，②抛物线与y轴的交点为(06)，

③抛物线的对称轴是：1

x=④在对称轴左侧y随x 增大而增大

A．1 Ｂ．2 Ｃ．3 Ｄ．4

12．如图6，四边形ABCD是边长为9的正方形纸片，将其沿MN折叠，使点B落在CD边上的B'处，点A对应点为A'，且3

B C'=，则

图6

AM的长是

A．1.5Ｂ．2Ｃ．2.25Ｄ．2.5

2010年柳州市初中毕业升学考试试卷

第Ⅱ卷（非选择题，共84分）

注意事项：

1．答题前，考生务必先将自己的姓名、准考证号用蓝、黑色墨水笔或圆珠笔填写在试卷左边的密封线内．

2．第Ⅱ卷为第3页至第10页．答题时，用蓝黑色墨水笔或圆珠笔直接将答案写在试卷上．二、填空题（本大题共6小题，每小题3分，满分18分．请将答案直接填写在题中横线上的空白处）

·= ．

13．计算：23

14．因式分解：29

x-=．

15．写出一个经过点(11)，的一次函数解析式．

16．2010年广州亚运会吉祥物取名“乐羊羊”．图7中各图是按照一定规律排列的羊的组图，图①有1只羊，图②有3只羊，……，则图⑩有只羊．

图7

17．关于x 的一元二次方程(3)(1)0x x +-=的根是 ．

18．如图8，AB 是O ⊙的直径，弦2cm BC =，F 是弦BC 的中点，

60ABC ∠=°．若动点E 以2cm/s 的速度从A 点出发沿着A B A →→方向运动，设运动时间为()(03)t s t <≤，连结EF ，当t

值为 s 时，BEF △是直角三角形．

三、解答题（本大题8分，满分66分．解答应写出必要的文字说明、演算步骤或推理过程） 19．（本题满分6分）

计算：30(2)(20103)tan 45-+--°．

20．（本题满分6分）

如图9，在88?的正方形网格中，ABC △的顶点和线段EF 的端点都在边长为1的小正方形的顶点上．

（1）填空：ABC ∠= ．BC = ；

（2）请你在图中找出一点D ，再连接DE DF 、，使以D E F 、、为顶点的三角形与ABC △全

A

C

B F

O

E 图8

等，并加以证明．

21．（本题满分6分）

桌面上有4张背面相同的卡片，正面分别写着数字“1”、“2”、“3”“4”．先将卡片背面朝上洗匀．

（1）如果让小唐从中任意抽取一张，抽到奇数的概率是；

（2）如果让小唐从中同时抽取两张．游戏规则规定：抽到的两张卡片上的数字之和为奇数，则小唐胜，否则小谢胜．你认为这个游戏公平吗？说出你的理由．

22．（本题满分8分）

如图10，从热气球P 上测得两建筑物A B 、的底部的俯角分别为45°和30°，如果A B 、两建筑物的距离为90m ，P 点在地面上的正投影恰好落在线段AB 上，求热气球P 的高度．（结果精确到0.01m ，参考数据：3 1.732≈，2 1.414≈）

23．（本题满发8分）

目前，“低碳”已成为保护地球环境的热门话题．风能是一种清洁能源，近几年我国风电装机容量迅速增长．图11是我国2003年－2009年部分年份的内力发电装机容量统计图（单位：万千瓦），观察统计图解答下列问题．

45°

30°

图10

（1）2007年，我国风力发电装机容量已达 万千瓦；从2003年到2009年，我国风力发电装机容量平均每年增长．．．．．．

万千瓦； （2）求2007～2009这两年装机容量的年平均增长率．．．．．．

；（参考数据： 5.04 2.24≈，1.26 1.12≈，14 3.74≈）

（3）按（2）的增长率，请你预测2010年我国风力发电装机容量．（结果保留到0.1万千瓦）

图11

24．（本题满分10分）

某住宅小区计划购买并种植甲、乙两种树苗共300株．已知甲种树苗每株60元，乙种树苗每株90元．

（1）若购买树苗共用21000元，问甲、乙两种树苗应各买多少株？

（2）据统计，甲、乙两种树苗每株树苗对空气的净化指数分别为0.2和0.6，问如何购买甲、乙两种树苗才能保证该小区的空气净化指数之和不低于90而且费用最低？

25．（本题满分10分）

如图12，AB 为O ⊙直径，且弦CD AB ⊥于E ，过点B 的切线与AD 的延长线交于点F ． （1）若M 是AD 的中点，连接ME 并延长ME 交BC 于N ．求证：MN BC ⊥．

（2）若4

cos 35

C DF ∠==，，求O ⊙的半径．

图12

26．（本题满分12分）

如图13，过点(43)P -，作x 轴、y 轴的垂线，分别交x 轴、y 轴于A B 、两点，交双曲线

(2)

k

y k x

=

≥于E F 、两点． （1）点E 的坐标是 ，点F 的坐标是 ；（均用含k 的式子表示） （2）判断EF 与AB 的位置关系，并证明你的结论；

（3）记PEF OEF S S S =-△△，S 是否有最小值？若有，求出其最小值；若没有，请说明理由．

图13

2010年柳州市初中毕业升学考试

数学参考答案及评分标准

第Ⅰ卷：一、选择题

题

号

1 2 3 4 5 6 7 8 9 10 11 12

答

案

A C

B D A A D

C C B C B 第Ⅱ卷：二、填空题

题

号

13 14 15 16 17 18

答

案6(3)(3)

x x

+-

如y x=，等等（答案不

唯一，只要正确均可得

分）

55 1

x=或

3

x=-

1或1.75

或2.25

（说明：第17题只写对一个结果给2分，两个结果都写对给3分；第18题每写对一个结果给1分）

三、解答题：

19．本题满分6分．

解：原式=811

-+-····································································································· 3分

=8-··············································································································· 6分

20．本题满分6分．

（1）135

ABC

∠=°，22

BC=，··················································2分

（2）（说明：D 的位置有四处，分别是图中的1234D D D D 、、、．此处画出D 在1D 处的位置及证明，D 在其余位置的画法及证明参照此法给分） 解：EFD △的位置如图所示． ················································ 3分

证明：

FD BC ===Q ························································ 4分

9045135EFD ABC ∠=∠==°+?° ·············································································· 5分

2EF AB ==

EFD ABC ∴△≌△ ······································································································· 6分 （说明：其他证法参照此法给分） 21．本题满分6分．

解：（1）1

2

················································································································· 2分

（2）（方法一）

这个游戏不公平． ······································································································· 3分 理由如下：任意抽取两个数，共有6种不同的抽法，其中和为奇数的抽法共有4种．

P ∴（和为奇数）=

42

63

= ··························································································· 4分 P （和为偶数）=1

3

····································································································· 5分 （方法二）设2008年的风力发电装机容量为a 万千瓦．

5002520500a a

a

--=

········································································································ 4分 21260000a = ···········································································································

0a >Q

1122a ∴≈ ·············································································································· 5分 经检验，1122a ≈是所列方程的根．

则2007到2009这两年装机容量的年增长率为

1122500

1.24124%500

-=≈ ····· 6分

答：2007到2009这两年装机容量的年平均增长率约为124%．

（3）(1 1.24)25205644.8+?=Q ················································································ 7分

∴2010年我国风力发电装机容量约为5644.8万千瓦． ······································· 8分

24．本题满分10分．

解：（1）设甲种树苗买x 株，则乙种树苗买(300)x -株． ································ 1分

6090(300)21000x x +-= ···························································································· 3分

200x =

······························································································ 4分 300200100-=

····························································································· 5分 答：甲种树苗买200株，乙种树苗买100株．

（2）设买x 株甲种树苗，(300)x -株乙种树苗时该小区的空气净化指数之和不低于90．

0.20.6(300)90x x +-≥ ······························································································· 6分

0.21800.690x x +-≥

0.490x --≥

225x ≤························································································· 7分

此时费用6090(300)y x x =+-

3027000y x =-+

·························································································· 8分 y Q 是x 的一次函数，y 随x 的增大而减少

∴当225x =最大时，302252700020250y =-?+=最小（元） ································ 9分

即应买225株甲种树苗，75株乙种树苗时该小区的空气净化指数之和不低于90，费用最小为20250元． ············································································································ 10分 （说明：其他解法参照此法给分） 25．本题满分10分 （1）（方法一） 连接AC ．

AB Q 为O ⊙的直径，且AB CD ⊥于E ，

由垂径定理得：点E 是CD 的中点．························· 1分 又Q M 是AD 的中点

ME ∴是DAC △的中位线 ············································ 2分 MN AC ∴∥

···································································· 3分 AB Q 为O ⊙直径，90ACB ∴∠=°，·························· 4分 90MNB ∴∠=°即MN BC ⊥ ········································· 5分 （方法二）

AB CD ⊥Q ，90AED BEC ∴∠=∠=°

························ 1分 M 是AD 的中点，ME AM ∴=，即有MEA A ∠=∠

·············································· 2分 又MEA BEN ∠=∠Q ，由A ∠与C ∠同对?BD

知C A ∠=∠ C BEN ∴∠=∠ ·············································································································· 3分 又90C CBE ∠+∠=Q °

90CBE BEN ∴∠+∠=°

································································································ 4分 90BNE ∴∠=°，即MN BC ⊥． ················································································ 5分 （方法三）

AB CD ⊥Q ，90AED ∴∠=° ······················································································ 1分 由于M 是AD 的中点，ME MD ∴=，即有MED EDM ∠=∠

又CBE ∠Q 与EDA ∠同对?

AC ，CBE EDA ∴∠=∠ ················································· 2分 又MED NEC ∠=∠Q

NEC CBE ∴∠=∠ ········································································································· 3分 又90C CBE ∠+∠=Q °

90NEC C ∴∠+∠=° ·····································································································

4分

即有90CNE ∠=°，MN BC ∴⊥ ················································································ 5分 （2）连接BD

BCD ∠Q 与BAF ∠同对?BD

，C A ∴∠=∠ 4

cos cos 5

A C ∴∠=∠=

············································· 6分 BF Q 为O ⊙的切线，90ABF ∴∠=° 在Rt ABF △中，4

cos 5

AB A AF ∠=

= 设4AB x =，则5AF x =，由勾股定理得：3BF x = ························································································ 7分 又AB Q 为O ⊙直径，BD AD ∴⊥

ABF BDF ∴△∽△ BF DF

AF BF ∴=

··················································································································· 8分 即3353x x x

= 5

3

x = ···················································································································· 9分

∴直径520

4433AB x ==?=

则O ⊙的半径为10

3 ····································································································· 10分

（说明：其他解法参照此法给分） 26．本题满分12分．

解：（1）44k E ??-- ???，，33k F ?? ???

， ·············································································· 3分 （说明：只写对一个点的坐标给2分，写对两个点的坐标给3分）

（2）（证法一）结论：EF AB ∥ ··········································································· 4分

证明：(43)P -Q ，

44k E ??∴-- ???，，33k F ??

???

，， 即得：3443

k k

PE PF =+=+， ·················································································· 5分

312412

12123443

PA PB k k PE k PF k ====

++++Q

， APB EPF ∠=∠

PAB PEF ∴△∽△

PAB PEF ∴∠=∠

·········································································································· 6分 EF AB ∴∥ ···················································································································· 7分 （证法二）结论：EF AB ∥ ······················································································ 4分

证明：(43)P -Q ，44k E ??∴-- ???，，33k F ??

???

，， 即得：3443

k k

PE PF =+=+， ·················································································· 5分

在Rt PAB △中，4

tan 3

PB PAB PA ∠==

在Rt PEF △中，4

4

3tan 3

34

k PF PEF k PE +∠===+

tan tan PAB PEF ∴∠=∠

PAB PEF ∴∠=∠

·········································································································· 6分 EF AB ∴∥ ···················································································································· 7分 （3）（方法一）

S 有最小值 ···················································································································· 8分

EAO FBO PAOB PEOF S S S =+Q △△矩形四边形+S

=12k +

12EOF PEF PEF PEOF S S S S k ∴=-=-+△△△四边形 ····························································· 9分

由（2）知，11342243PEF k k S PE PF ????

==++ ???????

△·· 212PEF OEF PEF S S S S k ∴=---△△△= ······································································· 10分

221

(6)31212

k k k =+=+- ···················································································· 11分

又2k Q ≥，此时S 的值随k 值增大而增大，

∴当2k =时，73

S =

最小 S ∴的最小值是7

3

······································································································ 12分

（方法二）

S 有最小值 ···················································································································· 8分 分别过点E F 、作PF PE 、的平行线，交点为P '

由（2）知，34k k P ??

'- ???， Q 四边形PEP F '为矩形

P EF PEF S S '∴=△△ PEF OEF S S S ∴=-△△

=P EF OEF S S '-△△

=OME ONF OMP N S S S '++△△矩形 ················································································· 9分

=22122

k k k

++ ········································································································ 10分 =2

12

k k + =

21

(6)312

k +- ··································································································· 11分 又2k Q ≥，此时S 的值随k 值增大而增大，

∴当2k =时，73

S =

最小 S ∴的最小值是7

3

． ·································································································· 12分

（说明：其他解法参照此法给分）