2007年长沙市

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2007年长沙市

 

 2007 年长沙市初中毕业学业考试试卷 数

 学 考生注意:

 本试卷共 26 道小题, 时量 120 分钟, 满分 120 分.

 一、 填空题(本题共 8 个小题, 每小题 3 分, 满分 24 分)

 1. 如图, 已知直线 ab∥,135=∠, 则2∠ 的度数是

  .

 2. 请写出一对互为相反数的数:

  和

 .

 3. 计算xyxyxy−=−−

  .

 4.5. 投掷一枚质地均匀的普通骰子, 朝上的一面为 6 点的概率是

 .

 ABC△中, DE,分别是 ABAC,的中点, 当10cmBC =时, DE =

  cm.

 6. 计算:188−=

  .

 7. 单独使用正三角形、 正方形、 正六边形、 正八边形四种地砖, 不能镶嵌(密铺)

 地面的是

 .

 8. 如图, 点 AB,在数轴上对应的实数分别为 m含mn,的式子表示)

  n,, 则 AB,间的距离是

 .(用 二、 选择题(本题共 8 个小题, 每小题 3 分, 满分 24 分)

 请将你认为正确的选择支的代号填在下面的表格里:

 题号 9 10 11 12 13 14 15 16 答案

  9. 在平面直角坐标系中, 点 (34)−,所在的象限是(

 )

 A. 第一象限

 10. 下列说法正确的是(

 )

 A. 有两个角为直角的四边形是矩形

 C. 等腰梯形的对角线相等

 11. 某校社会实践小组八位成员上街卖报, 一天的卖报数如下表:

 A

 B

 B. 第二象限

 C. 第三象限

 D. 第四象限 B. 矩形的对角线互相垂直 D. 对角线互相垂直的四边形是菱形

  成员 C

 D

 E

 F

 G

 H

 卖报数(份)

 25 28 29 28 27 28 32 25 则卖报数的众数是(

 )

 A. 25

 B. 26

 12. 经过任意三点中的两点共可以画出的直线条数是(

 )

 A. 一条或三条

 B. 三条

 13. 星期天, 小王去朋友家借书, 下图是他离家的距离 y (千米)

 与时间 x(分钟)

 的函数C. 27

 D. 28 C. 两条

 D. 一条 图象, 根据图象信息, 下列说法正确的是(

 )

 A. 小王去时的速度大于回家的速度 B. 小王在朋友家停留了 10 分钟 C. 小王去时所花的时间少于回家所花的时间 abc

 1 2 AmBn

 0x

 y (千米)0x (分钟)

 220 30 40

 D. 小王去时走上坡路, 回家时走下坡路 14. 把抛物线22yx= −向上平移1个单位, 得到的抛物线是(

 )

 A.22(1)yx= −+

 B.22(1)yx= −− C.221yx= −+

 D.221yx= −−

 15. 圆锥侧面展开图可能是下列图中的(

 )

  16. 在密码学中, 直接可以看到内容为明码, 对明码进行某种处理后得到的内容为密码. 有一种密码, 将英文 26 个字母abc,,, …, z (不论大小写)

 依次对应 1, 2, 3, …, 26这 26 个自然数(见表格). 当明码对应的序号 x 为奇数时, 密码对应的序号12xy+=; 当明码对应的序号 x 为偶数时, 密码对应的序号132xy =+.

 字母 a

 b

 c

 d

 e

 f

 g

 h

 i

 j

 k

 l

 m

 序号 1 n

 2 o

 3 p

 4 q

 5 r

 6 s

 7 t 8 u

 9 v

 10 w

 11 x

 12 y

 13 z

 字母 序号 按上述规定, 将明码“love” 译成密码是(

 )

 A. gawq

 B. shxc

 14 15 16 17 18 19 20 21 22 23 24 25 26 C. sdri

 D. love 三、 解答题(本题共 6 个小题, 每小题 6 分, 满分 36 分)

 17. 计算:211( 3)−22−− −+.

 18. 解分式方程:233xx=−.

 19. 如图是某设计师在方格纸中设计图案的一部分, 请你帮他完成余下的工作:

 (1)

 作出关于直线 AB 的轴对称图形;

 (2)

 将你画出的部分连同原图形绕点O 逆时针旋转90;

 A.B.C.D.

 (3)

 发挥你的想象, 给得到的图案适当涂上阴影, 让图案变得更加美丽.

 20. 为了改进银行的服务质量, 随机抽查了 30 名顾客在窗口办理业务所用的时间(单位:分钟). 下图是这次调查得到的统计图. 请你根据图中的信息回答下列问题:

 (1)

 办理业务所用的时间为 11 分钟的人数是

  ;

 (2)

 补全条形统计图;

 (3)

 这 30 名顾客办理业务所用时间的平均数是

 分钟.

  人数

  21. 先化简, 再求值:22 (a a)()bab+−+, 其中2008a =,2007b =.

  22. 如图所示, 某超市在一楼至二楼之间安装有电梯, 天花板与地面平行, 请你根据图中数据计算回答:

 小敏身高 1.78 米, 她乘电梯会有碰头危险吗? 姚明身高 2.29 米, 他乘电梯会有碰头危险吗?

 (可能用到的参考数值:

 sin270.45=, cos270.89=, tan270.51=)

  AOB89 10 11 12 13时间246810二楼一楼4m A4m4m B 27° C

 四、 解答题(本题共 2 个小题, 每小题 8 分, 满分 16 分)

 23. (本题满分 8 分)

 小华准备将平时的零用钱节约一些储存起来, 他已存有 62 元, 从现在起每个月存 12 元; 小华的同学小丽以前没有存过零用钱, 听到小华在存零用钱, 表示从现在起每个月存 20 元,争取超过小华.

 (1)试写出小华的存款总数1y 与从现在开始的月数 x 之间的函数关系式以及小丽存款数2y与月数 x 之间的函数关系式;

 (2)

 从第几个月开始小丽的存款数可以超过小华?

 24. (本题满分 8 分)

 如图, RtABC△中,90C =∠, O 为直角边BC 上一点, 以O 为圆心, OC 为半径的圆恰好与斜边 AB 相切于点 D , 与 BC 交于另一点 E .

 (1)

 求证:AOCAOD△≌△(2)

 若1BE = ,3BD =, 求 ;

 O的半径及图中阴影部分的面积 S .

 五、 解答题(本题共 2 个小题, 每小题 10 分, 满分 20 分)

 25. (本题满分 10 分)

 某班到毕业时共结余经费 1800 元, 班委会决定拿出不少于 270 元但不超过 300 元的资金为老师购买纪念品, 其余资金用于在毕业晚会上给 50 位同学每人购买一件文化衫或一本相册作为纪念品. 已知每件文化衫比每本相册贵 9 元, 用 200 元恰好可以买到 2 件文化衫和 5本相册.

 (1)

 求每件文化衫和每本相册的价格分别为多少元?

 (2)

 有几购买文化衫和相册的方案? 哪种方案用于购买老师纪念品的资金更充足?

 26. (本题满分 10 分)

 如图,ABCD中,4AB =,3BC =,120BAD =∠, E 为 BC 上一动点(不与 B 重合), 作 EFAB⊥于 F , FE, DC 的延长线交于点G , 设 BEx=,DEF△的面积为 S .

 ACBDEO

 (1)

 求证:(2)

 求用 x 表示S 的函数表达式, 并写出 x 的取值范围;

 (3)

 当 E 运动到何处时, S 有最大值, 最大值为多少?

  BEFCEG△∽△;

  2007 年长沙市初中毕业学业考试试卷 数学参考答案及评分标准 一、 填空题 1. 35

 2. 1,1− (答案不唯一)

  3. 1

 4. 5

 5.16

 6.2

  7. 正八边形

 8. nm− 二、 选择题 题号 9 10 11 12 13 14 15 16 答案 D C D A B C D B 三、 解答题 17. 原式11922= −+ ············································································································· 3 分

  9= ··························································································································· 6 分 18. 去分母, 得23(3)xx=− ······························································································· 2 分 去括号, 移项, 合并, 得x =9x = ····························································································· 5 分 是原方程的根.

 ····························································································· 6 分 19. 图略. 三步各计 2 分, 共 6 分.

 20. (1)

 5;

 ···························································································································· 2 分 (2)

 图略;

 ···························································································································· 4 分 (3)

 10.

 ································································································································ 6 分 检验, 得921. 原式22222(2)aabaabb=+−++ ··············································································· 2 分

  222222aabaabb=+−−− ·················································································· 3 分

  22ab=− ················································································································· 4 分 ACBD

 EFG

 当2008a =,2007b =时,

 原式22( 2008)( 2007)200820071=−=−=

 ································································ 6 分 22. 作CDAC⊥交 AB 于 D , 则27CAB =∠,

 ····························································· 1 分 在Rt

 4 0.51ACD△中,tanCDACCAB2.04=∠= ············································································· 3 分 (米)

 ·································································· 4 分 = ×所以小敏不会有碰头危险, 姚明则会有碰头危险.

 ····························································· 6 分 四、 解答题 23. (1)162 12+yx=,220yx= ······················································································ 4 分 (2)

 由 20所以从第 8 个月开始小丽的存款数可以超过小华.

 ····························································· 8 分 24. (1)AB切O于 D ,ODAB∴⊥ ········································································· 1 分 62 12+xx>得7.75x >,

 ·················································································· 7 分 在RtAOC△和 RtAOD△中,OCODAOAO==, ······································································ 3 分 RtRt(HL)AOCAOD∴△≌△ ·························································································· 4 分 (2)

 设半径为 r , 在RtODB△中,2223(1)rr+=+, 解得4r = ································· 6 分 由(1)

 有 ACAD=,2229(3)ACAC∴+=+, 解得12AC = ······································ 7 分 22111112 9× −454 82222SAC BCr∴=−π=×π×=− π.

 ············································· 8 分 五、 解答题 25. (1)

 设文化衫和相册的价格分别为 x 元和 y 元, 则 ······················································ 1 分 925200xyxy−=+= ····················································································································· 3 分 解得3526xy== 答:

 文化衫和相册的价格分别为 35 元和 26 元.

 ································································· 5 分 (2)

 设购买文化衫t件, 则购买相册(50) t−本, 则 15003526(50)1530tt+−≤≤ ·························································································· 7 分 解得20023099t≤ ≤ t 为正整数,第一种方案:

 购文化衫 23 件, 相册 27 本, 此时余下资金 293 元;

 第二种方案:

 购文化衫 24 件, 相册 26 本, 此时余下资金 284 元;

 23t∴ =, 24 , 25 , 即有三种方案.

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

 第三种方案:

 购文化衫 25 件, 相册 25 本, 此时余下资金 275 元;

 ··································· 9 分 所以第一种方案用于购买教师纪念品的资金更充足.

 ······················································· 10 分 26. (1)

 证明略; ··················································································································· 3 分 (2)

 由(1)

 DG 为DEF△中 EF 边上的高,

 在RtBFE△中,60B =∠,3sin2EFBEBx==,

 ··················································· 4 分 在RtCEG△中,3CEx= −,3(3)cos602xCGx−=−=,

 112xDGDCCG−∴=+=,

 ······························································································ 5 分 21311 3288SEF DGxx∴== −+,

 ·············································································· 6 分 其中 03x<≤.

 ···················································································································· 7 分 (3)308a = −<, 对称轴112x =, ∴ 当 03x<≤时, S 随 x 的增大而增大,

 ∴ 当3x =, 即 E 与C 重合时, S 有最大值.

 ····································································· 9 分 3 3S=最大.

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