资阳市2009年中考数学试题及答案

数 学

全卷分为第Ⅰ卷（选择题）和第Ⅱ卷（非选择题）两部分．第Ⅰ卷1至2页，第Ⅱ卷3至8页．全卷满分120分，考试时间共120分钟．

答题前，请考生务必在答题卡上正确填涂自己的姓名、考号和考试科目，并将试卷密封线内的项目填写清楚；考试结束，将试卷和答题卡一并交回．

第Ⅰ卷（选择题 共30分）

注意事项：每小题选出的答案不能答在试卷上，须用2B铅笔在答题卡上把对应题目的答案标号．．．．涂黑．如需改动，用橡皮擦擦净后，再选涂其它答案．

一、选择题：（本大题共10个小题，每小题3分，共30分）在每小题给出的四个选项中，只有一个选项符合题意．

1. –3的绝对值是（ ） A. 3

B. –3

C. ±3

D. 9

2. 下列计算正确的是（ ） A. a+2a2=3a3

B. a2·a3=a6

C. (a3)2=a9

D. a3÷a4=a1（a≠0）

3. 吴某打算用同一大小的正多边形地板砖铺设家中的地面，则该地板砖的形状不能是（ ） A. 正三角形

B. 正方形

C. 正六边形

D. 正八边形

4. 若一次函数y=kx+b(k≠0)的函数值y随x的增大而增大，则（ ） A. k<0

B. k>0

C. b＜0

D. b>0

5. 化简4x的结果是（ ） A. 2x

B. ±2x

C. 2x

D. ±2x 1x1,6. 在数轴上表示不等式组2的解集，正确的是（ ）

x1

7. 如图1，在矩形ABCD中，若AC=2AB，则∠AOB的大小是（ ） A. 30°

B. 45° C. 60°

1

D.90°

图 1

8. 按图2中第一、二两行图形的平移、轴对称及旋转等变换规律，填入第三行“？”处的图形应是（ ）

9. 用a、b、c、d四把钥匙去开X、Y两把锁，其中仅有a钥匙能够打开X锁，仅有b钥匙能打开Y锁．在求“任意取出一把钥匙能够一次打开其中一把锁”的概率时，以下分析正确的是（ ）

图 2

A. 分析1、分析2、分析3 C. 分析1

B. 分析1、分析2 D. 分析2

10. 如图3，已知Rt△ABC的直角边AC=24，斜边AB=25，一个以点P为圆心、半径为1的圆在△ABC内部沿顺时针方向滚动，且运动过程中⊙P一直保持与△ABC的边相切，当点P第一次回到它的初始位置时所经过路径的长度是（ ）

A. C.

56 3

B. 25 D. 56

112 3图 3

资阳市2009年高中阶段学校招生统一考试 数 学

第Ⅱ卷（非选择题 共90分）

题号 得分 二 三 17 18 19 20 21 22 23 24 25 总 分 总分人 注意事项：本卷共6页，用黑色或蓝色钢笔或圆珠笔直接答在试卷上．请注意准确理解题意、

明确题目要求，规范地表达、工整地书写解题过程或结果．

2

二、填空题：（本大题共6个小题,每小题3分,共18分）把答案直接填在题中横线上．

11. 甲、乙两人进行跳远训练时，在相同条件下各跳10次的平均成绩相同，若

甲的方差为0.3，乙的方差为0.4，则甲、乙两人跳远成绩较为稳定的是_________(填“甲”或“乙”）．

2xy5,12. 方程组的解是_____________．

xy413. 若两个互补的角的度数之比为1∶2，则这两个角中较小角的度数是_____________． ．．14. 如图4，已知直线AD、BC交于点E，且AE=BE，欲证明△AEC≌△BED，需增加的条件可以是__________________(只填一个即可)．

15. 若点A(–2，a)、B(–1，b)、C(1，c)都在反比例函数y=则用“<”连接a、b、c的大小关系为___________________．

16. 若n为整数，且n≤xk(k<0)的图象上，x1图4

和

11119801980198030个111120092009200930个的值，可以确定x=

11111119801981198220082009的整数部分是______．

三、解答题：（本大题共9个小题，共72分）解答应写出必要的文字说明、证明过程或演算步骤．

17．（本小题满分7分） 解方程：

3

x210． x318．（本小题满分7分）

如图5，已知□ABCD的对角线AC、BD相交于点O，AC =12，BD=18，且△AOB的周长l=23，求AB的长．

19．（本小题满分8分）

已知Z市某种生活必需品的年需求量y1(万件)、供应量y2(万件)与价格x(元/件)在一定范围内分别近似满足下列函数关系式：y1= –4x＋190，y2=5x–170．当y1=y2时，

称该商品的价格为稳定价格，需求量为稳定需求量；当y1y2时，称该商品的供求关系为供不应求．

(1) (4分) 求该商品的稳定价格和稳定需求量；

(2) (4分) 当价格为45(元/件)时，该商品的供求关系如何？为什么？

4

图5

20．（小题满分8分）

根据W市统计局公布的数据，可以得到下列统计图表．请利用其中提供的信息回答下列问题：

W市近几年生产总值(GDP)(亿元) 467.6 500409.5300.1400 300263.4200 1000 2005年2006年2007年2008年 (1) (3分) 从2006年到2008年，W市的GDP 哪一年比上一年的增长量最大？

(2) (3分) 2008年W市GDP分布在第三产业 的约是多少亿元？(精确到0.1亿元)

(3) (2分) 2008年W市的人口总数约为多少万人？(精确到0.1万人)

21．（本小题满分8分）

某市在举行“5.12汶川大地震”周年纪念活动时，根据地形搭建了一个台面为梯形(如图6所示)的舞台，且台面铺设每平方米售价为a元的木板．已知AB=12米，AD=16米，∠B=60°，∠C=45°，计算购买铺设台面的木板所用资金是多少元．(不计

铺设损耗，结果不取近似值)

5

W市2008年GDP结构分布第一产业第二产业28%第三产业26%46%W市近3年人均GDP(元) 年 份 人均GDP 2006年 2007年 2008年 7900 10600 12000 图6

22．（本小题满分8分）

已知关于x的一元二次方程x2+kx–3=0，

(1) (4分) 求证：不论k为何实数，方程总有两个不相等的实数根；

(2) (4分) 当k=2时，用配方法解此一元二次方程．

23．（本小题满分8分）

如图7，已知四边形ABCD、AEFG均为正方形，∠BAG=α (0°<α<180°)． (1) (6分) 求证： BE=DG，且 BE⊥DG；

(2) (2分) 设正方形ABCD、AEFG的边长分别是3和2，线段BD、DE、EG、GB所围成封闭图形的面积为S．当α变化时，指出S的最大值及相应的α值．(直接写出结果，不必说明理由)

6

图7

24．（本小题满分9分）

如图8-1，已知O是锐角∠XAY的边AX上的动点，以点O为圆心、R为半径的圆与射线AY切于点B，交射线OX于点C．连结BC，作CD⊥BC，交AY于点D．

(1) (3分) 求证：△ABC∽△ACD；

(2) (6分) 若P是AY上一点，AP=4，且sinA=35，

① 如图8-2，当点D与点P重合时，求R的值；

② 当点D与点P不重合时，试求PD的长(用R表示)．

7

图8-1

图8-2

25．（本小题满分9分）

12

x–2x＋1的顶点为P，A为抛物线与y轴的交点，过A2与y轴垂直的直线与抛物线的另一交点为B，与抛物线对称轴交于点O′，过点B和P的直线l交y轴于点C，连结O′C，将△ACO′沿O′C翻折后，点A落在点D的位置．

如图9，已知抛物线y=

(1) (3分) 求直线l的函数解析式； (2) (3分) 求点D的坐标；

(3) (3分) 抛物线上是否存在点Q，使得S△DQC= S△DPB? 若存在，求出所有符合条件的点Q的坐标；若不存在，请说明理由．

图9

8

资阳市2009年高中阶段学校招生统一考试 数学试题参考答案及评分意见

说 明：

1. 解答题中各步骤所标记分数为考生解答到这一步应得的累计分数．

2. 参考答案一般只给出该题的一种解法，如果考生的解法和参考答案所给解法不同，请参照本答案及评分意见给分．

3. 考生的解答可以根据具体问题合理省略非关键步骤．

4. 评卷时要坚持每题评阅到底，当考生的解答在某一步出现错误、影响了后继部分时，如果该步以后的解答未改变问题的内容和难度，可视影响程度决定后面部分的给分，但不得超过后继部分应给分数的一半；如果这一步后面的解答有较严重的错误，就不给分；若是几个相对独立的得分点，其中一处错误不影响其他得分点的得分．

5. 给分和扣分都以1分为基本单位．

6. 正式阅卷前应进行试评，在试评中须认真研究参考答案和评分意见，不能随意拔高或降低给分标准，统一标准后须对全部试评的试卷予以复查，以免阅卷前后期评分标准宽严不同．

一、选择题（每小题3分，共10个小题，满分30分）： 1-5. ADDBC ；6-10. DCBAC．

二、填空题（每小题3分，共6个小题，满分18分）：

x3,11．甲；12．13．60°；14．∠A=∠B或∠C=∠D或CE=DE；15．cy1;三、解答题（共9个小题，满分72分）：

17．原方程可变形为：3(x–2)–x=0， ··················································································· 3分

整理，得 2x=6， ·········································································································· 5分 解得 x=3． ··················································································································· 6分 经检验，x=3是原方程的解． ······················································································ 7分 18．∵□ABCD的对角线AC、BD相交于点O，AC =12，BD=18， ································· 1分

1∴ AO=AC=6， ··········································································································· 3分

21BO=BD=9． ·············································································································· 5分

2又∵△AOB的周长l=23，∴ AB=l–(AO+BO)=23–(6+9)=8． ······································· 7分 19．(1) 由y1=y2，得：–4x＋190=5x–170， ········································································ 2分

解得 x=40．·················································································································· 3分 此时的需求量为 y1= –4×40+190=30．········································································· 4分 因此，该商品的稳定价格为40元/件，稳定需求量为30万件． (2) 当x=45时，y1= – 4×45＋190=10， ······································································· 5分 y2= 5×45–170=55， ······································································································· 6分 ∴ y19

20．(1) 观察条形统计图可知，W市的GDP2007年比上一年的增长量最大． ················ 3分

(2) 2008年W市GDP分布在第三产业的约是： 467.6×26%≈121.6(亿元)． ····························································································· 6分 (3) 2008年W市人口总数约为：467.6×104÷12000≈389.7 (万人)． ···························· 8分 21．作AE⊥BC于点E，DF⊥BC于点F，易知ADFE为矩形． ······································· 1分

在Rt△ABE中，AB=12米，∠B=60°，∴ BE =12×cos60°=6(米)， ······························ 2分

AE =12×sin60°=63(米) ． ························································································· 3分 在矩形ADFE中，AD=16米，

∴ EF=AD=16米，DF=AE=63米． ··········································································· 4分 在Rt△CDF中，∠C=45°，∴ CF =DF=63 (米) ． ················································· 5分 ∴ BC=BE+EF+CF=(22+63) (米)， ············································································ 6分 ∴ S梯形ABCD =(AD+BC)·AE=[16+(22+63)]×63=(54+1143) (米2)， ············· 7分 ∴购买木板所用的资金为(54+1143)a 元． ······························································ 8分

22

22. (1) 方程的判别式为 Δ=k –4×1×(–3)= k +12， ···························································· 2分

不论k为何实数，k2≥0，k2 +12>0，即Δ>0， ····························································· 3分 因此，不论k为何实数，方程总有两个不相等的实数根．········································ 4分 (2) 当k=2时，原一元二次方程即 x2+2x–3=0， ∴ x2+2x+1=4， ·············································································································· 5分

2

∴ (x+1)=4， ················································································································· 6分 ∴ x+1=2或x+1= –2， ··································································································· 7分 ∴ 此时方程的根为 x1=1，x2= –3． ············································································· 8分 23. (1) 证法一：∵四边形ABCD、AEFG均为正方形，

∴ ∠DAB=∠GAE=90°，AD=AB，AG=AE． ······························································ 2分 ∴ 将AD、AG分别绕点A按顺时针方向旋转90°，它们恰好分别与AB、AE重合，即点D与点B重合，点G与点E重合， ································································································ 3分

∴ DG绕点A顺时针旋转90°与BE重合， ································································· 5分 ∴ BE=DG，且BE⊥DG． ··························································································· 6分

证法二：∵四边形ABCD、AEFG均为正方形， ∴ ∠DAB=∠GAE=90°，AD=AB，AG=AE． ······························································ 2分 ∴ ∠DAB+α=∠GAE+α，∴ ∠DAG=∠BAE． ① 当α≠90°时，由前知 △DAG≌△BAE (S.A.S.)， ················································· 2分 ∴ BE=DG， ················································································································· 3分 且∠ADG=∠ABE． ······································································································ 4分

设直线DG分别与直线BA、BE交于点M、N，又∵∠AMD=∠BMN，∠ADG+∠AMD=90°， ∴∠ABE+∠BMN=90°， ······························································································· 5分 ∴∠BND=90°，∴BE⊥DG． ······················································································· 6分 ② 当α=90°时，点E、点G分别在BA、DA的延长线上，显然BE=DG，且BE⊥DG． (说明：未考虑α=90°的情形不扣分)

25(2) S的最大值为， ·································································································· 7分

2

10

1212当S取得最大值时，α=90°． ······················································································· 8分 24．(1) 由已知，CD⊥BC，∴ ∠ADC=90°–∠CBD， ·························································· 1分

又∵ ⊙O切AY于点B，∴ OB⊥AB，∴∠OBC=90°–∠CBD， ··································· 2分 ∴ ∠ADC=∠OBC．又在⊙O中，OB=OC=R，∴∠OBC=∠ACB，∴∠ACB=∠ADC． 又∠A=∠A，∴△ABC∽△ACD ． ··············································································· 3分

3(2) 由已知，sinA=，又OB=OC=R，OB⊥AB，

5∴ 在Rt△AOB中，AO=

5OBR54==R，AB=(R)2R2=R，

33sinA33558∴ AC=R+R=R ． ··································································································· 4分

33ACAD由(1)已证，△ABC∽△ACD，∴ ， ··························································· 5分 ABAC8RAD163∴，因此 AD=R． ·················································································· 6分 483RR33163① 当点D与点P重合时，AD=AP=4，∴R=4，∴R=． ···································· 7分

34② 当点D与点P不重合时，有以下两种可能：

316i) 若点D在线段AP上(即043316ii) 若点D在射线PY上(即R>)，PD=AD–AP=R–4． ······································· 9分

433163综上，当点D在线段AP上(即0)时，

43416316PD=R–4．又当点D与点P重合(即R=)时，PD=0，故在题设条件下，总有PD=|R–4|(R>0)．

343125．(1) 配方,得y=(x–2)2 –1，∴抛物线的对称轴为直线x=2，顶点为P(2，–1) ． ····· 1分

21取x=0代入y=x2 –2x＋1，得y=1，∴点A的坐标是(0，1)．由抛物线的对称性知，点A(0，1)

2与点B关于直线x=2对称，∴点B的坐标是(4，1)． ························································ 2分

设直线l的解析式为y=kx＋b(k≠0)，将B、P的坐标代入，有 14kb,k1,解得∴直线l的解析式为y=x–3． ············································· 3分 12kb,b3.(2) 连结AD交O′C于点E，∵ 点D由点A沿O′C翻折后得到，∴ O′C垂直平分AD． 由(1)知，点C的坐标为(0，–3)，∴ 在Rt△AO′C中，O′A=2，AC=4，∴ O′C=25．

1148据面积关系，有 ×O′C×AE=×O′A×CA，∴ AE=5，AD=2AE=5．

2255 11

作DF⊥AB于F，易证Rt△ADF∽Rt△CO′A，∴∴ AF=

AFDFAD， ACOAOCAD16AD8·AC=，DF=·O′A=，5分 OC5OC583163又 ∵OA=1，∴点D的纵坐标为1–= –，∴ 点D的坐标为(，–)． ············ 6分

5555(3) 显然，O′P∥AC，且O′为AB的中点， ∴ 点P是线段BC的中点，∴ S△DPC= S△DPB ． 故要使S△DQC= S△DPB，只需S△DQC=S△DPC ． ······················································································7分

过P作直线m与CD平行，则直线m上的任意一点与CD构成的三角形的面积都等于S△DPC ，故m与抛物线的交点即符合条件的Q点．

1633容易求得过点C(0，–3)、D(，–)的直线的解析式为y=x–3，

54535据直线m的作法，可以求得直线m的解析式为y=x–．

421357351令x2–2x＋1=x–，解得 x1=2，x2=，代入y=x–，得y1= –1，y2=， 242242871因此，抛物线上存在两点Q1(2，–1)(即点P)和Q2(，)，使得S△DQC= S△DPB．9分

28(仅求出一个符合条件的点Q的坐标，扣1分)

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