2008年沈阳市中等学校招生统一考试

数学试卷

*考试时间120分钟 试卷满分150分

一、选择题（下列各题的备选答案中，只有一个答案是正确的，将正确答案的序号填在题后的括号内，每小题3分，共24分）

1．沈阳市计划从2008年到2012年新增林地面积253万亩，253万亩用科学记数法表示正确的是（ ） A．25.310亩

5B．2.5310亩

6C．25310亩

4D．2.5310亩

72．如图所示的几何体的左视图是（ ）

正面

第2题图

A． B． C． D．

3．下列各点中，在反比例函数y2图象上的是（ ） x1) A．(2，

B．，3

231) C．(2，，2) D．(14．下列事件中必然发生的是（ ）

A．抛两枚均匀的硬币，硬币落地后，都是正面朝上 B．掷一枚质地均匀的骰子，朝上一面的点数是3 C．通常情况下，抛出的篮球会下落 D．阴天就一定会下雨

5．一次函数ykxb的图象如图所示，当y0时，x的取 值范围是（ ） A．x0 B．x0

y 3 O x

2

C．x2



D．x2

第5题图

6．若等腰三角形中有一个角等于50，则这个等腰三角形的顶角的度数为（ ） A．50

B．80

2C．65或50



D．50或80

7．二次函数y2(x1)3的图象的顶点坐标是（ ）

A

D

F E

C

第8题图

，3) A．(1

，3) B．(1

，3) C．(1，3) D．(18．如图所示，正方形ABCD中，点E是CD边上一点，连接AE， 交对角线BD于点F，连接CF，则图中全等三角形共有（ ）

B 2008年沈阳市中等学校招生统一考试 1/12

A．1对 B．2对 C．3对 二、填空题（每小题3分，共24分）

D．4对

9．已知A与B互余，若A70，则B的度数为 ． 10．分解因式：2m8m ．

11．已知△ABC中，A60，ABC，ACB的平分线交于点O，

3A D

O 则BOC的度数为 ．

C 12．如图所示，菱形ABCD中，对角线AC，BD相交于点O，若再补 B 第12题图 充一个条件能使菱形ABCD成为正方形，则这个条件是 （只

填一个条件即可）． C B 13．不等式2xx6的解集为 ．

14．如图所示，某河堤的横断面是梯形ABCD，BC∥AD，迎水坡

A E 12AB长13米，且tanBAE，则河堤的高BE为 米．

第14题图 515．观察下列图形的构成规律，根据此规律，第8个图形中有 个圆．

D

„„

第1个 第2个 第3个 第15题图

第4个

，，点B的坐标为(111)，，点C到直线AB的16．在平面直角坐标系中，点A的坐标为(11)距离为4，且△ABC是直角三角形，则满足条件的点C有 个．

三、（第17小题6分，第18，19小题各8分，第20小题10分，共32分）

117．计算：(1)052723．

2

18．解分式方程：

11x2． x33x

19．先化简，再求值：

1y(xy)(xy)2x22y2，其中x，y3．

3

2008年沈阳市中等学校招生统一考试 2/12

20．如图所示，在66的方格纸中，每个小方格都是边长为1的正方形，我们称每个小正方形的顶点为格点，以格点为顶点的图形称为格点图形，如图①中的三角形是格点三角形． （1）请你在图①中画一条直线将格点三角形分割成两部分，将这两部分重新拼成两个不同的格点四边形，并将这两个格点四边形分别画在图②，图③中； （2）直接写出这两个格点四边形的周长．

图① 图② 图③

第20题图

四、（每小题10分，共20分）

21．如图所示，AB是O的一条弦，ODAB，垂足为C，交O于点D，点E在O上．

（1）若AOD52，求DEB的度数；

E O （2）若OC3，OA5，求AB的长．

B A C D 第21题图

22．小刚和小明两位同学玩一种游戏．游戏规则为：两人各执“象、虎、鼠”三张牌，同时各出一张牌定胜负，其中象胜虎、虎胜鼠、鼠胜象，若两人所出牌相同，则为平局．例如，小刚出象牌，小明出虎牌，则小刚胜；又如，两人同时出象牌，则两人平局． （1）一次出牌小刚出“象”牌的概率是多少？

（2）如果用A，B，C分别表示小刚的象、虎、鼠三张牌，用A1，B1，C1分别表示小明的象、虎、鼠三张牌，那么一次出牌小刚胜小明的概率是多少？用列表法或画树状图（树形图）法加以说明．

小刚 小明

A B C A1 B1 C1

第22题图

2008年沈阳市中等学校招生统一考试 3/12

五、（本题12分）

23．在学校组织的“喜迎奥运，知荣明耻，文明出行”的知识竞赛中，每班参加比赛的人数相同，成绩分为A，B，C，D四个等级，其中相应等级的得分依次记为100分，90分，80分，70分，学校将某年级的一班和二班的成绩整理并绘制成如下的统计图：

一班竞赛成绩统计图 二班竞赛成绩统计图 人数 12 D级 12 16% 10 A级 8 6 C级 44% 5 6 36% 4 2 2 0 A B C D B级4% 等级

第23题图 请你根据以上提供的信息解答下列问题：

（1）此次竞赛中二班成绩在C级以上（包括C级）的人数为 ； （2）请你将表格补充完整：

一班 二班 平均数（分） 中位数（分） 众数（分） 87.6 87.6 90 100 （3）请从下列不同角度对这次竞赛成绩的结果进行分析：

①从平均数和中位数的角度来比较一班和二班的成绩； ②从平均数和众数的角度来比较一班和二班的成绩；

③从B级以上（包括B级）的人数的角度来比较一班和二班的成绩． 六、（本题12分）

24．一辆经营长途运输的货车在高速公路的A处加满油后，以每小时80千米的速度匀速行驶，前往与A处相距636千米的B地，下表记录的是货车一次加满油后油箱内余油量y（升）与行驶时间x（时）之间的关系： 行驶时间x（时） 余油量y（升） 0 100 1 80 2 60 2.5 50 （1）请你认真分析上表中所给的数据，用你学过的一次函数、反比例函数和二次函数中的一种来表示y与x之间的变化规律，说明选择这种函数的理由，并求出它的函数表达式；（不要求写出自变量的取值范围）

（2）按照（1）中的变化规律，货车从A处出发行驶4.2小时到达C处，求此时油箱内余油多少升？

（3）在（2）的前提下，C处前方18千米的D处有一加油站，根据实际经验此货车在行驶中油箱内至少保证有10升油，如果货车的速度和每小时的耗油量不变，那么在D处至少加多少升油，才能使货车到达B地．（货车在D处加油过程中的时间和路程忽略不计）

2008年沈阳市中等学校招生统一考试 4/12

七、（本题12分）

BACDAE，25．已知：如图①所示，在△ABC和△ADE中，ABAC，ADAE，

且点B，A，D在一条直线上，连接BE，CD，M，N分别为BE，CD的中点． （1）求证：①BECD；②△AMN是等腰三角形．

（2）在图①的基础上，将△ADE绕点A按顺时针方向旋转180，其他条件不变，得到图②所示的图形．请直接写出（1）中的两个结论是否仍然成立；

△PBD∽△AMN．（3）在（2）的条件下，请你在图②中延长ED交线段BC于点P．求证：

C

C N

N E D A B M M D B A E

图② 图①

第25题图 八、（本题14分） 26．如图所示，在平面直角坐标系中，矩形ABOC的边BO在x轴的负半轴上，边OC在y轴的正半轴上，且AB1，OB3，矩形ABOC绕点O按顺时针方向旋转60后得到

矩形EFOD．点A的对应点为点E，点B的对应点为点F，点C的对应点为点D，抛物线yaxbxc过点A，E，D． （1）判断点E是否在y轴上，并说明理由； （2）求抛物线的函数表达式；

（3）在x轴的上方是否存在点P，点Q，使以点O，B，P，Q为顶点的平行四边形的面积是矩形ABOC面积的2倍，且点P在抛物线上，若存在，请求出点P，点Q的坐标；若不存在，请说明理由．

y E A B F C O 第26题图

D x

2

2008年沈阳市中等学校招生统一考试 5/12

2008年沈阳市中等学校招生统一考试

数学试题参考答案及评分标准

一、选择题（每小题3分，共24分） 1．B 2．A 3．D 4．C 5．C 二、填空题（每小题3分，共24分） 9．20

6．D

7．A 8．C

10．2m(m2)(m2)

11．120

13．x4

14．12

12．BAD90（或ADAB，ACBD等）

15．65 16．8 三、（第17小题6分，第18，19小题各8分，第20小题10分，共32分）

17．解：原式1(2)27523 ············································································ 4分 ········································································································ 5分 1233523 ·

································································································································ 6分 36 ·

18．解：12(x3)x ······································································································· 2分

12x6x x7 ······································································································································· 5分

1检验：将x7代入原方程，左边右边 ······································································· 7分

4所以x7是原方程的根 ········································································································ 8分 （将x7代入最简公分母检验同样给分）

19．解：原式xyyx2xyyx2y····························································· 4分

22222xy ····································································································································· 6分

1当x，y3时，

3原式31 ············································································································· 8分 20．解：（1）答案不唯一，如分割线为三角形的三条中位线中任意一条所在的直线等．

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

拼接的图形不唯一，例如下面给出的三种情况：

图① 图② 图③ 图④

2008年沈阳市中等学校招生统一考试 6/12

13

图⑤ 图⑥ 图⑦

图⑧ 图⑨

图①~图④，图⑤~图⑦，图⑧~图⑨，画出其中一组图中的两个图形． ···························· 6分 （2）对应（1）中所给图①~图④的周长分别为425，8，425，425； 图⑤~图⑦的周长分别为10，825，825；

图⑧~图⑨的周长分别为245，445．结果正确． ·············································· 10分 四、（每小题10分，共20分）

 ·21．解：（1）ODAB，·········································································· 3分 ADDB11AOD5226 ·················································································· 5分 22（2）ODAB，ACBC，△AOC为直角三角形， OC3，OA5， DEB由勾股定理可得ACOA2OC252324··························································· 8分

AB2AC8 ················································································································· 10分

122．解：（1）P(一次出牌小刚出“象”牌) ································································· 4分

3（2）树状图（树形图）：

2008年沈阳市中等学校招生统一考试 7/12

小刚

A

小明

A1

B1 C1 A1

开始 B B1 C1 A1

C B1 C1

··································································································· 8分

或列表

小刚 小明 A1 (A，A1) (B，A1) (C，A1) B1 (A，B1) (B，B1) (C，B1) C1 (A，C1) (B，C1) (C，C1) A B C ············································································ 8分

由树状图（树形图）或列表可知，可能出现的结果有9种，而且每种结果出现的可能性相同，其中小刚胜小明的结果有3种． ···························································································· 9分

P(一次出牌小刚胜小明)1． ························································································ 10分 3五、（本题12分） 23．解：（1）21 ······················································································································ 2分 （2）一班众数为90，二班中位数为80 ················································································ 6分 （3）①从平均数的角度看两班成绩一样，从中位数的角度看一班比二班的成绩好，所以一班成绩好； ······························································································································ 8分 ②从平均数的角度看两班成绩一样，从众数的角度看二班比一班的成绩好，所以二班成绩好； ··············································································································································· 10分 ③从B级以上（包括B级）的人数的角度看，一班人数是18人，二班人数是12人，所以一班成绩好． ························································································································ 12分 六、（本题12分） 24．解：（1）设y与x之间的关系为一次函数，其函数表达式为ykxb ···················· 1分

100)，(180)，代入上式得， 将(0，b100k20 解得 kb80b1002008年沈阳市中等学校招生统一考试 8/12

y20x100 ·················································································································· 4分

验证：当x2时，y20210060，符合一次函数； 当x2.5时，y202.510050，也符合一次函数．

可用一次函数y20x100表示其变化规律，

而不用反比例函数、二次函数表示其变化规律． ································································· 5分

y与x之间的关系是一次函数，其函数表达式为y20x100 ··································· 6分

（2）当x4.2时，由y20x100可得y16

即货车行驶到C处时油箱内余油16升． ·············································································· 8分 （3）方法不唯一，如：

方法一：由（1）得，货车行驶中每小时耗油20升， ························································· 9分 设在D处至少加油a升，货车才能到达B地．

636804.22010a16， ································································· 11分

80解得，a69（升） ············································································································ 12分

依题意得，

方法二：由（1）得，货车行驶中每小时耗油20升， ························································· 9分 汽车行驶18千米的耗油量：

18204.5（升） 80D，B之间路程为：636804.218282（千米）

2822070.5（升） ······································································································· 11分 80汽车行驶282千米的耗油量：

70.510(164.5)69（升） ······················································································· 12分

方法三：由（1）得，货车行驶中每小时耗油20升， ························································· 9分 设在D处加油a升，货车才能到达B地．

636804.22010≤a16，

80解得，a≥69 ······················································································································ 11分

······························································ 12分 在D处至少加油69升，货车才能到达B地． ·

依题意得，

七、（本题12分）

25．证明：（1）①BACDAE BAECAD

ABAC，ADAE △ABE≌△ACD BECD ···························································································································· 3分 ②由△ABE≌△ACD得ABEACD，BECD M，N分别是BE，CD的中点，BMCN ······························································ 4分 又ABAC

2008年沈阳市中等学校招生统一考试 9/12

△ABM≌△ACN

AMAN，即△AMN为等腰三角形 ············································································ 6分

（2）（1）中的两个结论仍然成立． ······················································································ 8分 （3）在图②中正确画出线段PD

由（1）同理可证△ABM≌△ACN CANBAM BACMAN 又BACDAE

MANDAEBAC

△AMN，△ADE和△ABC都是顶角相等的等腰三角形 ··········································· 10分 PBDAMN，PDBADEANM △PBD∽△AMN ··········································································································· 12分 八、（本题14分） 26．解：（1）点E在y轴上 ··································································································· 1分 理由如下：

连接AO，如图所示，在Rt△ABO中，AB1，BO3，AO2

sinAOB1，AOB30 2由题意可知：AOE60

BOEAOBAOE306090

·················································································· 3分 点B在x轴上，点E在y轴上． ·（2）过点D作DMx轴于点M

OD1，DOM30

在Rt△DOM中，DM点D在第一象限， 点D的坐标为13，OM 2231 ····································································································· 5分 2，2由（1）知EOAO2，点E在y轴的正半轴上

2) 点E的坐标为(0，······································································································ 6分 点A的坐标为(31)， ·

抛物线yax2bxc经过点E，

2008年沈阳市中等学校招生统一考试 10/12

c2

由题意，将A(31)，，D31，代入yax2bx2中得 2283a3b21a9 解得 331b2ab534229853······························································· 9分 x2 ·所求抛物线表达式为：yx299（3）存在符合条件的点P，点Q． ··················································································· 10分 理由如下：矩形ABOC的面积ABBO3 以O，B，P，Q为顶点的平行四边形面积为23．

由题意可知OB为此平行四边形一边， 又OB3

OB边上的高为2 ··············································································································· 11分

2) 依题意设点P的坐标为(m，853x2上 点P在抛物线yx299853m2m22

99解得，m10，m253 853P2，P(0，2)218， 以O，B，P，Q为顶点的四边形是平行四边形，

PQ∥OB，PQOB3， 2)时， 当点P1的坐标为(0，点Q的坐标分别为Q1(3，2)，Q2(3，2)；

A B 2008年沈阳市中等学校招生统一考试 11/12

y E F C D O M x 当点P2的坐标为5328，时，

13333，22点Q的坐标分别为Q3······················································ 14分

，Q4，． ·88（以上答案仅供参考，如有其它做法，可参照给分）

2008年沈阳市中等学校招生统一考试

12/12