2012年湖北省十堰市中考真题(word版含答案)

2012年十堰市初中毕业生学业考试

数学试题

注意事项：

1．本卷共有4页，共有25小题，满分120分，考试时限120分钟.

2．答题前，考生先将自己的姓名、准考证号填写在试卷和答题卡指定的位置，并认真核对条形码上的准考证号和姓名，在答题卡规定的位置贴好条形码.

3．考生必须保持答题卡的整洁，考试结束后，请将本试卷和答题卡一并上交. 一、选择题：（本题共10个小题，每小题3分，共30分） 下面每小题给出的四个选项中，只有一个是正确的，请把正确选项的字母填涂在答题卡中相应的格子内.

1．有理数1，2，0，3中，最小的一个数是（ ）． （A）1 （B）2 （C）0 （D）3 2．点P2，关于x轴对称点的坐标是（ ）. 3󰀀（A）3，2 （B）2，3 （C）2，3 （D）2，3

3．郧阳汉江大桥是国家南水北调中线工程的补偿替代项目，是南水北调丹江口库区最长的跨江大桥，桥长约2100米，将数字2100用科学记数法表示为（ ）． （A）2.110 （B）2.110 （C）2110 （D）2.110

4．如图是某体育馆内的颁奖台，其主视图是（ ）．

5．如图，直线BD∥EF，AE与BD交于点C，若∠ABC=30°，∠BAC=75°，则∠CEF的大小为（ ）． （A）60° （B）75° （C）90° （D）105° 6．下列运算中，结果正确的是（ ）．

22623（A）xxx （B）xyxy

23224（C）x23x5 （D）822 7．下列说法正确的是（ ）．

（A）要了解全市居民对环境的保护意识，采用全面调查的方式

22（B）若甲组数据的方差S甲0.1，乙组数据的方差S乙0.2，则甲组数据比乙组稳定

（C）随机抛一枚硬币，落地后正面一定朝上 （D）若某彩票“中奖概率为1%”，则购买100张彩票就一定会中奖一次

8．如图，梯形ABCD中，AD∥BC，点M是AD的中点，且MBMC，若AD=4，AB=6，BC=8，则梯形ABCD的周长为（ ）． （A）22 （B）24 （C）26 （D）28

9．一列快车从甲地开往乙地，一列慢车从乙地开往甲地，两车同时出发，两车离乙地的路程S（千米）与行驶时间t（小时）的函数关系如图所示，则下列结论中错误的是（ ）． （A）甲、乙两地的路程是400千米 （B）慢车行驶速度为60千米/小时 （C）相遇时快车行驶了150千米 （D）快车出发后4小时到达乙地

10．如图，O是正△ABC内一点，OA=3，OB=4，OC=5，将线段BO以点B为旋转中心逆时针旋转60°得到线段BO′，连接AO′，下列结论：①△BOA可以由△BOC绕点BOB=150逆时针旋转60°得到；②点O与O′的距离为4；③∠A⑤S△AOCS△AOB= 6+°；④S四边形OBA′633 ；

93．其中正确的结论是（ ）． 4（A）①②③⑤ （B）①②③④ （C）①②③④⑤ （D）①②③ 二、填空题：（本题有6个小题，每小题3分，共18分） 11．在函数y12．计算：x2中，自变量x的取值范围是 ．

031 ．

13．某射击小组有20人，教练根据他们某次射击的数据绘制成如图所示的统计图，则这组数据的众数是 ． 14．如图，矩形ABCD中，AB2，AD4，AC的垂直平分线EF交AD于点E、交BC于点F，则EF= ．

，∠B=30°，AB=12cm，以AC为直径的半圆O15．如图，Rt△ABC中，∠ACB=90°交AB于点D，点E是AB的中点，CE交半圆O于点F，则图中阴影部分的面积为 cm2．

16．如图，直线y6x，yk2x分别与双曲线y在第一象限内交于点A，B，若

x3S△AOB8，则k ．

三、解答题：（本题有9个小题，共72分） 17.（6分）先化简，再求值：11a其中a2. ，2a1a1

18.（6分）如图，在四边形ABCD中，ABAD，CBCD．求证：∠B∠D． 19．（6分）一个不透明的布袋里装有3个大小、质地均相同的乒乓球，分别标有数字1，2，3，小华先从布袋中随机取出一个乒乓球，记下数字后放回，再从袋中随机取出一个乒乓球，记下数字．求两次取出的乒乓球上数字相同的概率．

20．（8分）一辆汽车开往距离出发地180千米的目的地，按原计划的速度匀速行驶60千米后，再以原来速度的1.5倍匀速行驶，结果比原计划提前40分钟到达目的地，求原计划的行驶速度．

21．（8分）如图，为了测量某山AB的高度，小明先在山脚下C点测得山顶A的仰角为45°，然后沿坡角为30°的斜坡走100米到达D点，在D点测得山顶A的仰角为30°，求山AB的高度．（参考数据：3≈1.73） 22．（6分）阅读材料： 例：说明代数式x1解：x122x324的几何意义，并求它的最小值．

2x324x0122x32如图，建立平面直角坐标系，22，点Px，0是x轴上一点，则x012可以看成点P与点A01，的距离，

x3222可以看成点P与点B3，2的距离，所以原代数式的值可以看成线段PA与

PB长度之和，它的最小值就是PAPB的最小值．

′设点A关于x轴的对称点为A′，则PAPA，因此，求PAPB的最小值，只需求

PA′PB的最小值，PB的最小值为线段A′B而点A′、B间的直线段距离最短，所以PA′′CB，因为AC=3，CB=3，所以A′B32，即原式的长度．为此，构造直角三角形A的最小值为32．

根据以上阅读材料，解答下列问题： （1）代数式

x121x229的值可以看成平面直角坐标系中点Px，0与点

（填写点B的坐标） A11，、点B___________的距离之和．（2）代数式x249x212x37的最小值为_____________．

23．（10分）某工厂计划生产A、B两种产品共50件，需购买甲、乙两种材料．生产一件A产品需甲种材料30千克、乙种材料10千克；生产一件B产品需甲、乙两种材料各20千克．经测算，购买甲、乙两种材料各1千克共需资金40元，购买甲种材料2千克和乙种材料3千克共需资金105元．

（1）甲、乙两种材料每千克分别是多少元？

（2）现工厂用于购买甲、乙两种材料的资金不超过38000元，且生产B产品不少于28件，问符合条件的生产方案有哪几种？

（3）在（2）的条件下，若生产一件A产品需加工费200元，生产一件B产品需加工费300元，应选择哪种生产方案，使生产这50件产品的成本最低？（成本=材料费+加工费）

AB是直径，OD∥AC，BD=BA∠C24．（10分）如图1，O是△ABC的外接圆，且∠C，

OD交O于点E．

（1）求证：BD是O的切线；

（2）若点E为线段OD的中点，证明：以O、A、C、E为顶点的四边形是菱形；

（3）作CFAB于点F，连接AD交CF于点G（如图2），求

FG的值. FC25．（12分）抛物线yx2bxc经过点A、B、C，已知A13． ，0，C0，（1）求抛物线的解析式；

C的（2）如图1，P为线段BC上一点，过点P作y轴平行线，交抛物线于点D，当△BD面积最大时，求点P的坐标；

（3）如图2，抛物线顶点为E，EFx轴于F点，Mm，0是x轴上一动点，N是线段

EF上一点，若∠MNC=90°，请指出实数m的变化范围，并说明理由．

2012年十堰市初中毕业生学业考试

数学试题参及评分说明

一、选择题：（本题共10个小题，每小题3分，共30分）

1.B 2.C 3.A 4.A 5.D 6.D 7.B 8.B 9.C 10.A 二、填空题：（本题有6个小题，每小题3分，共18分）

11.x≥2 12.3 13.7 14.5 15.3三、解答题：（本题有9个小题，共72分）

3 16.6 a211a1 17．解：原式= ··············································································· 2分 2a1aa2a1a ························································· 4分 .

a1a1aa1当a2时，原式的值为2.································································································ 6分

18.证明：连接AC， ·································································································· 1分 ∵ABAD，CBCD，ACAC， ······································································· 4分 ∴△ABC≌△ADC， ································································································ 5分 ∴∠B∠D ·············································································································· 6分 19.解：列表如下：

1 2 3 1 （1，1） （2，1） （3，1） 2 （1，2） （2，2） （3，2） 3 （1，3） （2，3） （3，3）

由以上表格可知：有9种可能结果，两个数字相同的只有3种······························· 4分 所以，P两个数字相同31. ······················································································ 6分 9320.解：设原计划的行驶速度为x千米/时， ······································································ 1分

180601806040················································································· 5分 x1.5x60解得x60，

经检验：x60是原方程的解，且符合题意，所以x60．

则：

答：原计划的行驶速度为60千米/时. ······································································ 8分 21.解：过D作DEBC于E，作DFAB于F，设AB=x，

，CD=100， 在Rt△DEC中，∠DCE=30°∴DE50，CE503 ·················································································· 2分 在Rt△ABC中，∠ACB=45°，∴BCx

则AFABBFABDEx50

···································································· 4分 DFBEBCCEx+503 ·

，tan30°=在Rt△AFD中，∠ADF=30°∴AF ， FDx503， ··························································································· 6分 3x503∴x503+3≈236.5（米），

答：山AB的高度约为236.5米. ················································································ 8分

22.（1）（2，3） ·········································································································· 3分 （2）10 ·························································································································· 6分 23.解：（1）设甲材料每千克x元，乙材料每千克y元，则xy40 ················ 2分，2x3y105

∴x15所以甲材料每千克15元，乙材料每千克25元 ············································ 3分 ，y25（2）设生产A产品m件，则生产这50件产品的材料费为：

1530m+2510m+152050m+252050m=40000100m，

由题得：40000100m≤38000 ··················································································· 4分 m≥20， ·························································································································· 5分 又50m≥28m≤22，∴20≤m≤22，∴m20， 21，22， ······················ 6分 生产方案如下表：

A（件） B（件） 20 30 21 29 22 28 ·································································································································· 7分 （3）设总生产成本为W元，加工费为：200m30050m，

则W=40000100m200m30050m=200m55000， ······························ 8分 因为W随m的增大而减小，又m20，21，22，

所以当m22时，总成本最低，此时W50600元． ············································· 10分 24. （1）证明：∵AB是O的直径，

∴∠BCA=90°，

∴∠ABC+∠BAC=90°， ······························································································· 1分 又∠CBD=∠BAC，

∴∠ABC+∠CBD=90°， ∴∠ABD=90°， ·············································································································· 2分 又点B在O上，∴BD为O的切线 ·········································································· 3分

（2）连接CE、OC，∵OEED，∴OD2OB，又∵∠OBD90°，

，∠BOE60° ·∴∠ODB30°··················································································· 4分

又AC∥OD，∴∠OAC60°，又OAOC，∴ACOAOE ∴AC∥OE且ACOE，

∴四边形OACE是平行四边形， ····················································································· 5分 又OAOE，∴四边形OACE是菱形 ············································································ 6分 （3）∵CFAB，∴∠AFC∠OBD=90°，又AC∥OD， ∴∠CAF∠DOB，∴△AFC∽△OBD，∴又∠FAG∠BAD，∠AFG∠ABD90°， ∴∠AFG∽△ABD，∴∴

FCAF ······································ 8分 BDOBFGAF ············································································· 9分 BDABFGOB1． ········································································································ 10分 FCAB21bc0b225.解：（1）由题，解得：，

c3c3所以抛物线解析式为yx22x3 ············································································· 3分 （2）令x2x30，∴x11，x23 即B3，0

设直线BC的解析式为ykxb′，

2∴3b′k1 ∴ 033kb′b′故直线BC的解析式为yx3， ················································································ 4分

a22a3，设Pa， 3a，则Da，PDa22a33aa23a ······································································ 5分

S△BDCS△PDCS△PDB

111PDaPD3aPD3 2223a23a ················································································································· 6分 23327a

228当a2333时，△BDC的面积最大，此时P，. ······················································· 7分 222

2xx23x1（3）由（1），y42，

所以OF1，EF4，OC3，过C作CHEF于H点，则CHEH1.

当M在EF左侧时，因为∠MNC90°，

MFFN， NHHC1mn， 设FNn，则NH3n，∴································································· 8分 3n1则△MNF∽△NCH，得

2即n3nm10，关于n的方程有解，34m1≥0，

254当M在EF右侧时，Rt△CHE中，CHEH1，∠CEH=45°，即∠CEF=45°， 作EMCE交x轴于点M，则∠FEM45°， ∵FMEF4，∴OM5，

即N为点E时，OM5，∴m≤5. ·········································································· 11分

5综上，m的变化范围为：≤m≤5.

4

得m≥. ······················································································································· 9分