2007年茂名市中考数学试题及答案

1、.不用注册，资源免费茂名市2007年初中毕业生学业考试与高中阶段学校招生考试数学试卷考生须知1.       全卷分第一卷选择题，总分值40分，共2页和第二卷非选择题，总分值110分，共8页，全卷总分值150分;考试时间120分钟2.       请认真填写答题卡和第二卷密封线内的有关内容，并在试卷右上角的座位号处填上自己的座位号3.       考试完毕，将第一卷、第二卷和答题卡一并交回温馨提示： 亲爱的同学，请你

2、沉着冷静，充满自信，认真审题，仔细答卷，祝你考出好成绩! 第一卷选择题， 总分值40分，共2页一、精心选一选本大题共10小题，每题4分，共40分每题给出四个答案，其中只有一个是正确的1列计算正确的选项是A  B   C    D2?茂名日报?2007年5月18日报道，刚刚投产半年的茂名百万吨乙烯工程传来喜讯，正在创造全国最好的效益，每月为国家创利30 000万元，这个数用科学记数法表示是A  B  C D         &#

3、160;                  3在一组数据3，4，4，6，8中，以下说法正确的选项是A平均数小于中位数         B平均数等于中位数  C平均数大于中位数   D平均数等于众数4 的角平分线AD交BC于  点D，那么点D到AB的间隔 是A1    &

4、#160;  B2     C3    D45 某商场2006年的销售利润为，预计以后每年比上一年增长b%，那么2020年该商场的销售利润将是   A                     B  C          D 6 在

5、一张由复印机复印出来的纸上，一个多边形图案的一条边由原来的1cm变成2cm，那么这次复印出来的多边形图案面积是原来的A1倍      B2倍      C3倍      D4倍7上午九时，阳光灿烂，小李在地面上同时摆弄两根长度不相等的竹竿，假设它们的影子长度相等，那么这两根竹竿的相对位置可能是    A两根都垂直于地面         &

6、#160;    B两根都倒在地面上    C两根不平行斜竖在地面上  D两根平行斜竖在地面上8右图是由一些一样的小正方体搭成的几何体的三视图，那么搭成这个几何体的小正方体   的个数是   A4 B5  C6 D79 某村今年的荔枝总产量是吨是常数，设该村荔枝的人均产量为y吨，人口总数为x人，那么y与x之间的函数图象是          10 如图是一个圆柱形饮料罐，底面半径是5，高是12，上底

7、   面中心有一个小圆孔，那么一条到达底部的直吸管在罐内部分   的长度罐壁的厚度和小圆孔的大小忽略不计范围是   AB   C D 茂名市2007年初中毕业生学业考试与高中阶段学校招生考试数学试卷第二卷非选择题，共8页，总分值110分二、耐心填一填本大题共5小题，每题4分，共20分请你把答案填在横线的上方11化简：12现有一个测试间隔 为5m的视力表，根据这个  视力表，小华想制作一个测试间隔 为3m的视力表，  那么图中的13假设实数满足，那么14如图是一盏圆锥形灯罩AOB，两母线

8、的夹角，  假设灯炮O离地面的高OO1是2米时，那么光束照射到地面  的面积是米2答案准确到0.115 在数学中，为了简便，记    ，    那么三、细心做一做 本大题共3小题，每题8分，共24分                  16 此题总分值8分如图，阴影部分是由5个小正方形组成的一个直角图形，请用二种方法分别在以下图方格内添涂黑二个小正方

9、形，使它们成为轴对称图形                    17 此题总分值8分正方形和圆的面积均为求正方形的周长和圆的周长用含的代数式表示，并指出它们的大小       18 此题总分值8分一纸箱中放有大小均匀的只白球和只黄球，从箱中随机地取出一只白球的概率是 1试写出与的函数关系式；     

10、;            4分2当时，再往箱中放进20只白球，求随机地取出一只黄球的概率4分       四、沉着冷静，周密考虑本大题共2小题，每题8分，共16分                    19 此题总分值8分某学校为了学生的身

11、体安康，每天开展体育活动一小时，开设排球、篮球、羽毛球、体操课学生可根据自己的爱好任选其中一项，老师根据学生报名情况进行了统计，并绘制了右边尚未完成的扇形统计图和频数分布直方图，请你结合图中的信息，解答以下问题：1该校学生报名总人数有多少人？2分2从表中可知选羽毛球的学生有多少人？选排球和篮球的人数分别占报名总人数的百分之几？3分3将两个统计图补充完好 3分             20 此题总分值8分函数的图象与轴的两交点的横坐标分别是，且，求c及，的值 

12、;           五、开动脑筋，再接再厉 本大题共3小题，每题10分，共30分21此题总分值10分如图，正方形ABCD的边长是2，E是AB的中点，延长BC到点F使CFAE1假设把绕点旋转一定的角度时，能否与重合？请说明理由5分2现把向左平移，使与重合，得，交于点求证：，并求的长5分           22此题总分值10分 品名厂家批发价元/只商场零售价元/只篮

13、球130160排球100120某体育用品商场采购员要到厂家批发购进篮球和排球共100只，付款总额不得超过11 815元两种球厂家的批发价和商场的零售价如右表，试解答以下问题：1该采购员最多可购进篮球多少只？5分2假设该商场把这100只球全部以零售价售出，为使商场获得的利润不低于2580元，那么采购员至少要购篮球多少只，该商场最多可盈利多少元？5分   23此题总分值10分甲、乙两辆汽车同时、同方向从同一地点A出发行驶1假设甲车的速度是乙车的2倍，甲车走了90千米后立即返回与乙车相遇，相遇时乙车走了1小时求甲、乙两车的速度；5分2假设甲、乙每辆车最多只能带200升汽

14、油，每升汽油可以行驶10千米，途中不能再加油，但两车可以互相借用对方的油，假设两车都必须沿原路返回到出发点A，请你设计一种方案使甲车尽可能地远离出发点A，并求出甲车一共行驶了多少千米？5分       六、充满信心，成功在望本大题共2小题，每题10分，共20分24此题总分值10分如图，点A，B，C，D是直径为AB的O上四个点，C是劣弧的中点，AC交BD于点E， AE2， EC11求证：；          3分2试探究四边形ABCD是否

15、是梯形？假设是，请你给予证明并求出它的面积；假设不是，请说明理由  4分3延长AB到H，使BH OB求证：CH是O的切线                 3分         25  此题总分值10分      如图，平面直角坐标系中，有一矩形纸片OABC，O为坐标原点，轴， B3，现将纸

16、片按如图折叠，AD，DE为折痕，折叠后，点O落在点，点C落在点，并且与在同一直线上1求折痕AD 所在直线的解析式； 3分2求经过三点O，C的抛物线的解析式；3分3假设的半径为，圆心在2的抛物线上运动，与两坐标轴都相切时，求半径的值 4分          茂名市2007年初中毕业生学业考试与高中招生考试数学试题参考答案及评分标准说明：1、 假如考生的解法和本解法不同，可根据试题的主要内容，并参照评分标准制定相应的评分细那么后评卷      2、

17、解答题右端所注的分数，表示考生正确做到这一步应得的累加分数一、选择题本大题共10小题，每题4分，共40分题号12345678910答案DBCBBDCADA二、 填空题本大题共5小题，每题4分，共20分11    12或答 13 1  14   150三、本大题共3小题，每题8分，共24分16解：此题答案不唯一，只要在方格内添的二个正方形使整个图形是对称图形就给分，每答对一个给4分，共8分        17解：设正方形的边长为，圆的半径为R，那么， ·&#

18、183;·················································&#

19、183;··········································· 2分，·····

20、3;·················································

21、3;···························· 4分，·····················

22、······························· 6分， ··················&

23、#183;·················································&

24、#183;·················· 8分 18解：1由题意得 ，  ··························

25、3;············································ 2分即·····

26、··················································

27、·············································· 3分 ···&#

28、183;·················································&#

29、183;·················································&#

30、183;···· 4分2由1知当时，···········································

31、3;··············· 6分获得黄球的概率 ·································

32、;························· 8分四、本大题共2小题，每题8分，共16分19 解：1由两个统计图可知该校报名总人数是人············· 2分2选羽毛球的人数是人  ·

33、··················································

34、·· 3分因为选排球的人数是100人，所以， ············································

35、83;··········· 4分因为选篮球的人数是40人，所以，即选排球、篮球的人数占报名的总人数分别是25%和10%·······························

36、;··········· 5分3如图每补充完好一个得1分，共3分···································

37、83;························ 8分 20解：令，即，当方程有两个不相等的实数根时，该函数的图象与x轴有两个交点     1分此时即·············

38、3;·················································

39、3;····························· 2分由，····················

40、;··················································

41、;························ 3分，·························&

42、#183;·················································&

43、#183;···· 4分，舍去············································&

44、#183;························ 6分当时，解得························

45、;···· 7分综上：，为所求············································&

46、#183;···· 8分                   五、本大题共3小题，每题10分，共30分21解：1由正方形ABCD得ADDC2，········· 1分AECF1，·········&#

47、183;·················································&#

48、183;······ 2分，··········································

49、3;···· 3分·············································&

50、#183;····· 4分把绕点D旋转一定的角度时能与重合   5分2由1可知，·····································

51、83;··················· 6分即······························

52、;····························· 7分由得，····················

53、;········································· 8分由AE1，AD2，·······

54、;··················································

55、;··········· 9分，即，·················· 10分注：此题由三角形相似或解直角三角形同样可求AG22解：1设采购员最多可购进篮球只，那么排球是100只，··········

56、3;························· 1分依题意得： ·······················&

57、#183;···································· 3分解得     ········&

58、#183;·················································&

59、#183;····························· 4分是整数，=60··················

60、83;·················································

61、83;························ 5分答：购进篮球和排球共100只时，该采购员最多可购进篮球60只····················

62、·········· 6分2由表中可知篮球的利润大于排球的利润，因此这100只球中，当篮球最多时，商场可盈利最多，即篮球60只，此时排球40只，·····························

63、3;·················································

64、3;······································ 8分商场可盈利元··········&

65、#183;····· 9分即该商场可盈利2600元·········································

66、3;································ 10分23 解：1设甲，乙两车速度分别是x千米/时和y千米/时，············&

67、#183;······················· 1分根据题意得：························

68、3;·················································

69、3;· 3分解之得：···············································

70、83;················································· 4分

71、即甲、乙两车速度分别是120千米/时、60千米/时··············································

72、;········ 5分2方案一：设甲汽车尽可能地远离出发点A行驶了x千米，乙汽车行驶了y千米，那么···································

73、··················································

74、·········· 6分即······························ 7分即甲、乙一起行驶到离A点500千米处，然后甲向乙借油50升，乙不再前进，甲再前进1000千米返回到乙停顿处

75、，再向乙借油50升，最后一同返回到A点，此时，甲车行驶了共3000千米     ··········· 10分方案二：画图法如图     此时，甲车行驶了千米····················&

76、#183;·························· 10分方案三：先把乙车的油均分4份，每份50升当甲乙一同前往，用了50升时，甲向乙借油50升，乙停顿不动，甲继续前行，当用了100升油后返回，到乙停处又用了100升油，此时甲没有油了，再向乙借油50升，一同返回到A点此时，甲车行驶了千米 ···&

77、#183;································ 10分六、本大题共2小题，每题10分，共20分241证明：C是劣弧的中点，···········

78、;······························· 1分而公共，·················

79、3;······················ 3分2证明：连结，由得，·························

80、83;·················································

81、83;······························· 4分由，是O的直径， ， ， ，四边形OBCD是菱形，四边形ABCD是梯形············&

82、#183;··································· 5分法一：过C作CF垂直AB于F，连结OC，那么 ··········

83、;··················································

84、;·········································· 6分，·······&

85、#183;······························· 7分法二：接上证得四边形ABCD是梯形又，连结OC，那么，和的边长均为的等边三角形  6分，·········

86、83;·················································

87、83;·· 7分3证明：连结OC交BD于G由2得四边形OBCD是菱形，且  ·········································&

88、#183;··········································· 8分又OBBH，  ···

89、83;·················································

90、83;············· 9分，CH是O的切线··································

91、3;··········· 10分25 解：1由得，····································

92、83;·················································

93、83;········· 1分设直线AD的解析式为把A，D坐标代入上式得：，解得：，···································

94、83;··················································