2009年莆田市初中毕业、升学考试试卷(WORD版 有答案)

2009年莆田市初中毕业、升学考试试卷

数 学

（满分：150分，考试时间：120分钟）

一、细心填一填（本大题共10小题，每小题4分，共40分．直接把答案填在题中的横线上．）

1． 3的相反数是

2．2009年莆田市参加初中毕业、升学考试的学生总人数约为43000人，将43000用科学记数法表示是___________．

3．在组成单词“Probability”（概率）的所有字母中任意取出一个字母，则取到字母“b”的概率是 ．

4．如图，A、B两处被池塘隔开，为了测量A、B两处的距离，在AB

外选一适当的点C，连接AC、BC，并分别取线段AC、BC的中点(第4题图) E、F，测得EF=20m，则AB=__________m．

5．一罐饮料净重500克，罐上注有“蛋白质含量≥0.4%”，则这罐饮料中蛋白质的含量至少为__________克．

6．如图，菱形ABCD的对角线相交于点O，请你添加一个条件： ，使得该菱形为正方形．

7．甲、乙两位同学参加跳高训练，在相同条件下各跳10次，统计各自成绩的方差得S S，则成绩较稳定的同学是___________．（填“甲”或“乙”）

8．已知⊙O1和⊙O2的半径分别是一元二次方程 x 1 x 2 0的两根，且OO则，12 2

2

甲

2乙

C

（第6题图）

⊙O1和⊙O2的位置关系是

9．出售某种文具盒，若每个获利x元，一天可售出 6 x 个，则当x 元时，一天出售该种文具盒的总利润y最大． 10．如图，在

x轴的正半轴上依次截取

y

OA1 A1A2 A2A3 A3A4 A4A5，过点A1、A2、A3、A4、A5

2

分别作x轴的垂线与反比例函数y x 0 的图象相交于点

x

2 P1、P2、P3、P4、P5

，得直角三角形

OPA并设其面积分别为11、A1P2A2、A2P3A3、A3P4A4、A4P5A5，S1、S2、S3、S4、S，5则S5的值为 ．

（第10题图）

二、精心选一选（本大题共6小题，每小题4分，共24分，每小题给出的四个选项中有且只有一个是正确的，请把正确选项的代号写在题后的括号内，答对的得4分；答错、不答或答案超过一个的一律得0分）． 11

x的取值范围是（ ）

A．x≥0 B．x 0 C．x 0 D．x 0

A．a2 a2 a B．ab2

12．下列各式运算正确的是（ ）

2

a2b4

C．a2·a4=a8 D．5ab 5b a 13．如图是一房子的示意图，则其左视图是（ ）

A． B．

C．

Ｄ．

2、3、3、

3，则这组数据14．某班5位同学参加“改革开放30周年”系列活动的次数依次为1

、的众数和中位数分别是（ ）

2 B

． 2.4、3 C． 3、2 D．3、3 A．2、15．不等式组

2x 4，

的解集在数轴上表示正确的是（ ）

x 1 0

A B C

D16．如图1，在矩形MNPQ中，动点R从点N出发，沿N→P→Q→M方向运动至点M处停止．设点R运动的路程为

A．N处

x，△MNR的面积为y，如果y关于x的函数图象如图2

所示，则当x 9时，点R应运动到（ ）

（图1）

（第16题图）

B．P处 C．Q处 D．M处

三、耐心做一做（本大题共9小题，共86分．解答应写出必要的文字说明、证明过程或演

算步骤．）

1

17．（8

分）计算：3 ．

3

x2 4x 4x 2

18．（8分）先化简，再求值：其中x 1． x，2

x 4x 2

ABCD中，过对角线BD的中点O作直线EF分别交DA的延

长线、AB、DC、BC的延长线于点E、M、N、F．

（1）观察图形并找出一对全等三角形：△________≌△____________，请加以证明；

19．（8分）已知：如图在

（2）在（1）中你所找出的一对全等三角形，其中一个三角形可由另一个三角形经过怎样的

变换得到？

C

（第19题图）

20．（8分）（1）根据下列步骤画图并标明相应的字母:（直接在图１中画图） ．．①以已知线段AB（图1）为直径画半圆O；

②在半圆O上取不同于点A、B的一点C，连接AC、BC； ③过点O画OD∥BC交半圆O于点D． （2）尺规作图：（保留作图痕迹，不要求写作法、证明） ．．已知： AOB（图2）． 求作： AOB的平分线．

A

A

图1

B

图2

B

（第20题图）

21．（8分）某校课题研究小组对本校九年级全体同学体育测试情况进行调查，他们随机抽

查部分同学体育测试成绩（由高到低分A、B、C、D四个等级），根据调查的数据绘

制成如下的条形统计图和扇形统计图．

等级

（第21题图）

请根据以上不完整的统计图提供的信息，解答下列问题：

（1）该课题研究小组共抽查了__________名同学的体育测试成绩，扇形统计图中B级所占的百分比b=___________； （2）补全条形统计图；

（3）若该校九年级共有400名同学，请估计该校九年级同学体育测试达标（测试成绩C级以上，含C级）约有___________名．

22．（10分）已知，如图，BC是以线段AB为直径的⊙O的切线，AC交⊙O于点D，过点D作弦DE AB，垂足为点F，连接BD、BE．． （1）仔细观察图形并写出四个不同的正确结论：①________，②________ ，③________，④____________（不添加其它字母和辅助线，不必证明）； （2） A=30°，CD

=

，求⊙O的半径r． 3

C

（第22题图）

23．（10分）面对全球金融危机的挑战，我国政府毅然启动内需，改善民生．国务院决定从2009年2月1日起，“家电下乡”在全国范围内实施，农民购买人选产品，政府按原价购买．．总额的给予补贴返还．某村委会组织部分农民到商场购买人选的同一型号的冰箱、电．．．13%．．．

视机两种家电，已知购买冰箱的数量是电视机的2倍，且按原价购买冰箱总额为40000元、电视机总额为15000元．根据“家电下乡”优惠政策，每台冰箱补贴返还的金额比每台电视机补贴返还的金额多65元，求冰箱、电视机各购买多少台？ （1）设购买电视机

台，依题意填充下列表格：

（2）列出方程（组）并解答．

24.（12分）已知：等边△ABC的边长为a．

探究（1）：如图１，过等边△ABC的顶点A、B、C依次作AB、BC、CA的垂线围成

△MNG，求证：△MNG是等边三角形且

．MN ；

探究（2）：在等边△ABC内取一点O，过点O分别作OD AB、OE BC、OF CA，垂足分别为点D、E、F．

①如图2，若点O是△ABC的重心，我们可利用三角形面积公式及等边三角形性质得到两个正确结论（不必证明）：结论1

．OD OE OF 2．AD BE CF

a；结论2

3

a； 2

2是否仍然成立？如果成立，②如图3，若点O是等边△ABC内任意一点，则上述结论1、

请给予证明；如果不成立，请说明理由．

B

C （图1）

G

E （图2） F E （图3）

E （图4）

N

（第24题图）

1 作平行于x轴的直线l，抛物线y 25．（14分）已知，如图1，过点E 0，

12

x上4

的两点A、B的横坐标分别为 1和4，直线AB交y轴于点F，过点A、B分别作直线l的垂线，垂足分别为点C、D，连接CF、DF． （1）求点A、B、F的坐标； （2）求证：CF DF； （3）点P是抛物线y

12

x对称轴右侧图象上的一动点，过点P作PQ⊥PO交x轴4

于点Q，是否存在点P使得△OPQ与△CDF相似？若存在，请求出所有符合条件的点P的坐标；若不存在，请说明理由．

备用图

（第25题图）

（图1）

2009年莆田市初中毕业、升学考试试卷

数学试卷参考答案及评分标准

说明：

（一）考生的解法与“参考答案”不同时，可参照“答案的评分标准”的精神进行评分 （二）如解答的某一步计算出现错误，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骤应得的累计分数． （四）评分的最小单位是１分，得分或扣分都不能出现小数． 一、细心填一填（本大题共10小题，每小题4分，共40分．）

2

4．40 5．2 11

1

6．AB BC或AC BD或AO BO等 7．甲 8．相交 9．3 10．

5

1．3 2．4.3 104（不必考虑有效数字） 3． 二、精心选一选（本大题共6小题，每小题4分，共24分．） 11．A 12．B 13．C 14．D 15．A 16．C 三、耐心做一做（本题共9小题，共86分）

17．（1）解：原式

=34 1 ························································································6分

=····································································································8分

1 注：3 32分）

4（2分）， =1（2分）

3

x 2 x 2 x

18.解：原式= ···········································································6分

x 2x 2x 2

=1 x ·······································································································7分 当x 1时原式=1 1 0 ······················································································8分

2

注：x 4x 4 x 2 、x 4 x 2 x 2 、？

2

2

2

x 2x 2

? （各2分） x 2x 2

A 19． （1）①△DOE≌△BOF； ························· 2分

证明：∵四边形ABCD是平行四边形

∴AD∥BC ·········································· 3分 F

C

∴ EDO FBO， E F······························ 4分

（第19题图）

又∵OD OB

∴△DOE≌△BOF AAS ·············································································5分

②△BOM≌△DON ······················································································2分

证明：∵四边形ABCD是平行四边形

∴AB∥CD ······································································································3分

∴ MBO NDO， BMO DNO ···························································4分 又∵BO DO

∴△BOM≌△DON AAS ···········································································5分

③△ABD≌△CDB； ·····················································································2分

证明：∵四边形ABCD是平行四边形

∴AD CB，AB CD ····················································································3分 又∵BD DB···································································································4分

∴△ABD≌△CDB SSS ···············································································5分 （2）绕点O旋转180°后得到或以点O为中心作对称变换得到．·······························8分

20．（1）正确完成步骤①、②、③，各得1分，字母标注完整得1分，满分4分．

（2）说明：①以点O为圆心，以适当长为半径作弧交OA、OB于两点C、D ·············5分

②分别以点C、D为圆心，以大于

1

CD长为半径作弧， 2

两弧相交于点E ····················································································7分

③作射线OE ···························································································8分

A B

（第20题图）

21．（1）80 ················································································ 2分 40% ··························································································· 4分

（2）补全条形图（如右图） ······················································ 6分 （3）380························································································ 8分 DF FE，BD BE，△BDF≌△BEF，22．（1）BC AB，AD BD， △BDF∽△BAD， BDF BEF， A E，DE∥BC等 等级

（每写出一个正确结论得１分，满分４分．）

（第21题图） （2）解:又

O

图1

D

图2

AB是⊙O的直径 ADB 90° ································ 5分

E 30°

A 30° ·············································································· 6分

1

BD AB r ····································································· 7分

2

又BC是⊙O的切线 CBA 90° ············································································· 8分 C 60

C

（第22题图）

在Rt△

BCD中，CD

3

BDr

····································································································9分 tan60° ·

DC······························································································································10分 r 2 ·

23．（1）每个空格填对得1分，满分5分．

（2）解:依题意得

-··························································7分 65·

2xx

解得x 10··························································································································8分

经检验x 10是原分式方程的解 ··························································································9分

2x 20．

答:冰箱、电视机分别购买20台、10台 ····································10分 24．证明：如图1，△ABC为等边三角形

A

ABC 60°

G BC MN，BA MG

B ∴ CBM BAM 90°

C

ABM 90°- ABC 30 ················································ 1分 M 90 - ABM 60 ···················································· 2分 （图1） N 同理： N G 60 △MNG为等边三角形． ·····················

·····························

·················································3分

在Rt△ABM中，BM

ABa

sinMsin60 在Rt△

BCN中，BN

BCa a································································4分 tanNtan60 MN BM BN ························································································································

··········· 5分

（2）②：结论1成立.

证明；方法一：如图2，连接AO、BO、CO 由S△ABC S△AOB S△BOC S△AOC=作AH BC，垂足为H，

D

1

a OD OE OF ·············· 7分 2

C

则AH ACsin ACB a sin60 2

E H （图2）

S△ABC

11BC·AH a· 222

11 a

OD OE OF a

22 OD OE OF

··································································································8分 ·方法二：如图3，过点O作GH∥BC，分别交AB、AC于点G、H，过点

H作HM⊥BC于点M，

DGO B 60°， OHF C 60° △AGH是等边三角形 GH AH ···································································· 6分 OE⊥BC OE∥HM

四边形OEMH是矩形 HM OE······································································· 7分

在Rt△

ODG中，OD OG·sin DGO OG·sin60

A

在Rt△

OFH中，OF OH·sin OHF OH·sin60

在Rt△

HMC中，HM HC·sinC HC·

sin60

HC O

OD OE OF OD HM OF

HC 222

E M （图3）

（2）②:结论2成立．

Aa ···························8分

GH HC 222

E

E

证明：方法一：如图４，过顶点A、B、C依次作边

AB、B、CCA△MNG，NG为等边

的垂线围成由（1）得△M

三角形且MN ··············································· 9分

G

D MN于D ,OE NG于NG于点过点O分别作OD MGE ，OF于点F

由结论1得：

N

（图4）

OD OE OF

又

3

··································································10分 MN a ·

22

OD AB，AB MG，OF MG

ADO DAF OF A 90 四边形ADOF 为矩形 OF AD

同理：OD BE，OE CF····························································································11分

3

······························································································ 12分 AD BE CF OD OE OF a ·2

方法二：（同结论1方法二的辅助线） 在Rt△

OFH中，FH

OF

tan OHFA

HM在Rt△

HMC中，HC ························· 9分

·

sinC CF HC FH

33

O

E M （图3）

同理：AD

···············································10分 ，BE OE ·

3333

AD BE CF

OD OE OF ········································································································11分 由结论1

得：OD OE OF

a AD BE CF 3 a···················································································12分 22

A D

F

方法三：如图5，连接OA、OB、OC，根据勾股定理得：

BE2 OE2 OB2 BD2 OD2① CF OF OC CE OE②

2

2

2

2

2

E

（图5）

22222

AD OD AO AF OF③···················································································9分

①+②+③得：

············································································10分 BE2 CF2 AD2 BD2 CE2 AF2 ·

BE2 CF2 AD2 a AD a BE a CF

········································11分 a2 2ADa AD2 a2 2BEa BE2 a2 2CFa CF2 ·整理得：2a AD BE CF 3a

2

222

AD BE CF

3

·····································································································12分 a ·2

1 4

25．（1）解:方法一，如图1，当x 1时,y 当x 4时,y 4

∴A 1············································································· 1分 ·

1 4

（图1）

B 4，4 ······················································································ 2分

设直线AB的解析式为y kx b ·············································· 3分

13

k b k 则 4 解得 4 4k b 4 b 1

∴直线AB的解析式为y 当x 0时,y 1

3

x 1 ············································· 4分 4

F 01，···························································································································5分 ·

方法二:求A、B两点坐标同方法一,如图2,作FG BD,AH BD,垂足分别为G、H,交y轴于点N,则四边形FOMG和四边形

NOMH均为矩形,设FO x···················································· 3分 △BGF∽△BHA

BGFG （图2） BHAH4 x4

·······················································································································4分 ·

54 4解得x 1

F 0，1 ·························································································································5分

(2)证明：方法一：在Rt△CEF中，CE 1,EF 2

CF2 CE2 EF2 12 22

5

························································································································6分 CF ·

在Rt△DEF中，DE 4，EF 2

DF2 DE2 EF2 42 22

20

DF 由（1）得C 1， 1 ，D 4， 1

CD 5 CD2 52 25

·································································································· 7分 CF2 DF2 CD2 ·

CFD 90° ··············································································································· 8分 CF⊥DF·

55方法二：由 （1

）知AF ，AC

44 AF AC ················································································································· 6分

同理：BF BD ACF AFC AC∥EF

ACF CFO AFC CFO ······································································································ 7分 同理： BFD OFD

CFD OFC OFD 90° 即CF⊥DF ················································································································ 8分

（3）存在.

解：如图3，作PM⊥x轴，垂足为点M ··········· 9分 又

PQ⊥OP

Rt△OPM∽Rt△OQP

PMOM

PQOP

PQPM

···················································· 10分

OPOM

图3

设P xx2 x 0 ，则PM

14

12

x，OM x 4

①当Rt△QPO∽Rt△CFD时，

PQCF1

·······························································································11分 ·

OPDF2

12x

PM1 OMx2解得x 2

···················································································································12分 P， ·1 21②当Rt△OPQ∽Rt△CFD时，

PQDF·······························································································13分 2 ·

OPCF12

x

PM 2 OMx解得x 8

P2 816，

，， 使得△OPQ与△CDF相似. ·综上，存在点P··································14分 、P2 8161 21

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