新疆维吾尔自治区 新疆生产建设兵团
2008年初中毕业生学业考试
数 学 试 题 卷
考生须知：
1．本试卷分为试题卷和答题卷两部分．
2．试题卷共4页，满分150分．考试时间120分钟． 3．答题卷共4页，所有答案均写在答题卷上．．．．．．．．．．．，写在试题卷上的无效．．．．．．．．．
． 4．答题前，考生应先在答题卷密封区内认真填写准考证号、姓名、考场号、座位号、地（州、市、师）、县（市、区、团场）和学校． 5．答题时可以使用科学计算器．．．．．．．．．．
一、精心选择（本大题共10题，每题所给四个选项中，只有一个是正确的．每题5分，共50分．）
1．｜4-｜等于（ ） A ．2- B ．2 C ．4- D ．4 2．2008年5月12日，四川省汶川县发生了里氏8.0级大地震．新
疆各族群众积极捐款捐物，还紧急烤制了2×104个饱含新疆各族人民深情的特色食品——馕（n áng ），运往灾区．每个馕厚度约为2cm ，若将这批馕摞成一摞，其高度大约相当于（ ） A ．160层楼房的高度（每层高约2.5m ） B ．一棵大树的高度 C ．一个足球场的长度 D ．2000m 的高度
3．如图，下列推理不正确．．．的是（ ）
A ．∵A
B ∥CD ∴∠AB
C +∠C =180° B ．∵∠1=∠2 ∴A
D ∥BC C ．∵AD ∥BC ∴∠3=∠4
D ．∵∠A +∠ADC =180° ∴AB ∥CD 4．下列事件属于必然事件的是（ ） A ．打开电视，正在播放新闻 B ．我们班的同学将会有人成为航天员 C ．实数a ＜0，则2a ＜0 D ．新疆的冬天不下雪 5．下列调查方式中，合适的是（ ）
A ．要了解约90万顶救灾帐蓬的质量，采用普查的方式
B ．要了解外地游客对旅游景点“新疆民街”的满意程度，采用抽样调查的方式
C ．要保证“神舟七号”飞船成功发射，对主要零部件的检查采用抽样调查的方式
D ．要了解全疆初中学生的业余爱好，采用普查的方式 6．在函数1y x =
的图象上有三个点的坐标分别为（1，1y ）、（12
，2y ）、（3-，3y ），函数值y 1、y 2、y 3的大小关系是（ ）
A ．y 1＜y 2＜y 3
B ．y 3＜y 2＜y 1
C ．y 2＜y 1＜y 3
D ．y 3＜y 1＜y 2
7．如图，ABC △中BC 边上的高为1h ，DEF △中DE 边上的高为2h ，下列结论正确的是（ ）
A ．12h h >
B ．12h h <
C ．12h h =
D ．无法确定
8．傍晚，小明陪妈妈在路灯下散步，当他们经过路灯时，身体的影长（ ） A ．先由长变短，再由短变长 B ．先由短变长，再由长变短
C ．保持不变
D ．无法确定
9．如图，圆内接四边形ABCD 是由四个全等的等腰梯形组成，AD 是⊙O 的直径，则∠BEC 的度数为（ ） A ．15° B ．30° C ．45° D ．60°
10．古尔邦节，6位朋友均匀地围坐在圆桌旁共度佳节．圆桌半径为60cm ，每人离圆桌的距离均为10cm ，现又来了两名客人，每人向后挪动了相同的距离，再左右调整位置，使8人都坐下，并且8人之间的距离与原来6人之间的距离（即在圆周上两人之间的圆弧的长）相等．设每人向后挪动的距离为x ，根据题意，可列方程（ ）
A ．
2π(6010)2π(6010)
68x +++=
B ．2π(60)2π60
86
x +?=
C ．2π(6010)62π(60)8x +?=+?
D ．2π(60)82π(60)6x x -?=+?
二、合理填空（本大题共4题，每题5分，共20分）
11．根据下列图形的排列规律，第2008个图形是福娃 （填写福娃名称即可）．
12．如图，在平面直角坐标系中，线段11A B 是由线段AB 平移得到的，已知A B ，两点的坐
标分别为(23)A -，
，(31)B -，，若1A 的坐标为(34)，，则1B 的坐标为 ．
13．已知一元二次方程有一个根是2，那么这个方程可以是 （填上一个符合条件的
方程即可）．
14．如图，一束光线从y 轴上点A （0，1）发出，经过x 轴上点C 反射后，经过点B （6，2），则光线从A 点到B 点经过的路线的长度为 ．（精确到0.01）
三、准确解答（本大题共10题，共80分）
15．（6分）计算：201
2(π6)4
-÷--．
16（6分）化简分式2211
211
x x x x x -+-++-，并从2-、1-、0、1、2中选一个能使分式有意义
的数代入求值． 17．（6分）城区某中学要从自愿报名的张、王、李、赵4名老师中选派2人下乡支教，请用画树状图（或列表）的方法求出张、王两位老师同时被选中的概率． 18．（8分）如图，⊙O 的半径10cm OC =，直线l ⊥CO ，垂足为H ，交⊙O 于A 、B 两点，16cm AB =，直线l 平移多少厘米时能与⊙O 相切？
（1）请你根据以上数据填写下表：
（2）补全折线统计图．
（3）请你从以下两个不同的方面对这两种水果在去年3月份至8月份的销售情况进行分析：
①根据平均数和方差分析；
②根据折线图上两种水果销售量的趋势分析．
20．（8分）如图，某市区南北走向的北京路与东西走向的喀什路相交于点O处．甲沿着喀什路以4m/s的速度由西向东走，乙沿着北京路以3m/s的速度由南向北走．当乙走到O点以北50m处时，甲恰好到点O处．若两人继续向前行走，求两个人相距85m时各自的位置．
21．（8分）如图，在△ABC中，∠C=2∠B，AD是△ABC的角平分线，∠1=∠B．
求证：AB=AC+CD．
22．（9分）某社区计划购买甲、乙两种树苗共600棵，甲、乙两种树苗单价及成活率见下表：
（1
（2）若希望这批树苗成活率不低于90%，并使购买树苗的费用最低，应如何选购树苗？购买树苗的最低费用为多少？
23．（10分）（1）请用两种不同的方法，用尺规在所给的两个矩形中各作一个不为正方形的菱形，且菱形的四个顶点都在矩形的边上．（保留作图痕迹）
（2）写出你的作法．
24．（10分）某工厂要赶制一批抗震救灾用的大型活动板房．如图，板房一面的形状是由矩形和抛物线的一部分组成，矩形长为12m，抛物线拱高为5.6m．
（1）在如图所示的平面直角坐标系中，求抛物线的表达式．
（2）现需在抛物线AOB的区域内安装几扇窗户，窗户的底边在AB上，每扇窗户宽1.5m，高1.6m，相邻窗户之间的间距均为0.8m，左右两边窗户的窗角所在的点到抛物线的水平距离至少为0.8m．请计算最多可安装几扇这样的窗户？
新疆维吾尔自治区 新疆生产建设兵团
2008年初中毕业生学业考试
数学试卷参考答案及评分标准
（满分150分）
说明：本参考答案供阅卷教师评卷时使用．阅卷中，考生如有其它解法，只要正确、合理，均可得相应分值．
11．欢欢 12．（2，2） 13．2
4x =（答案不惟一） 14．6.71 三、解答题（本大题共10题，共80分）
15．（6分）解：原式11
144
=
÷+ ··············································································· 4分
=······························································································· 6分 16．（6分）解：原式2
(1)(1)1
(1)1
x x x x x -++=
-+- ······································································· 1分 11
11
x x x x -+=
-+- ························································································································ 2分 22
(1)(1)(1)(1)
x x x x --+=
-+ ··············································································································· 3分 241x
x -=
- ·································································································································· 5分 把0x =代入 原式0= ·································································································································· 6分 或把2x =代入
原式2428
213-?==-- ················································································································ 6分 或把2x =-代入 原式2
4(2)8
(2)13
-?-=
=--． ················································ 6分17．（6分）解：方法1：画树状图
·············································································· 4分
张、王两位老师同时被选中的概率是16
． ············································································ 6分 方法2：列表
张、王两位老师同时被选中的概率是
1
6
． ············································································ 6分
18．（8分）解法1：如图，连结OA ，延长CO 交⊙O 于D ， ∵l ⊥OC ，
∴OC 平分AB ． ∴AH =8． ······················································································ 3分
在Rt △AHO 中，6OH ， ·
······ 6分 ∴4cm 16cm CH DH ==，．
答：直线AB 向左移4cm ，或向右平移16cm 时与圆相切． ················································ 8分 解法2：设直线AB 平移cm x 时能与圆相切，
222(10)810x -+=
················································································································ 3分 116x =
24x =
∴4cm 16cm CH DH ==，． ····························································································· 8分
答：略．
（只答一个方向的平移扣2分） 19．（9分） 解：（1）
（2）
·························································· （7分）
（3）①由于平均数相同，22
S S <大枣葡萄
，所以大枣的销售情况相对比较稳定． ················ 8分 ②从图上看，葡萄的月销售量呈上升趋势． ········································································· 9分 （答案不惟一，合理均可得分） 20．（8分）解法1：设经过x 秒时两人相距85m ································································· 1分 根据题意得：222(4)(503)85x x ++= ················································································ 4分 化简得：2
121890x x +-=
解得：12921x x ==-，（不符合实际情况，舍去） ·························································· 6分
当9x =时，43650377x x =+=，
∴当两人相距85m 时，甲在O 点以东36m 处，乙在O 点以北77m 处． ·························· 8分 解法2：设甲与O 处的距离为x m 时，两人相距85m
则乙与O 处的距离为350m 4x ??
+
???
·
······················································································ 1分 2
2
2350854x x ??
++= ???
·
········································································································· 4分 解得：123684x x ==-，（不符合实际情况，舍去 ） ······················································ 6分 当33650774
x x =+=， ········································································································· 7分 答：当两人相距85米时，甲在O 点以东36米处，乙在O 点以北77米处． ···················· 8分 21．（8分）证明： ∵∠1=∠B
∴∠AED =2∠B ，DE =BE ········································································································ 2分 ∴∠C =∠AED ························································································································· 3分 在△ACD 和△AED 中
CAD EAD AD AD
C AE
D ∠=∠??
=??∠=∠?
∴△ACD ≌△AED ··················································································································· 5分
∴AC =AE ，CD =DE ，∴CD =BE ． ·························································································· 6分 ∴AB =AE +EB =AC +CD ． ········································································································· 8分 22．（9分）解：（1）设最多可购买乙树苗x 棵，则购买甲树苗（600 x -）棵 ················ 1分
60(600)8044000x x -+≤ ································································································· 3分 400x ≤．
答：最多可购买乙树苗400棵． ···························································································· 5分 （2）设购买树苗的费用为y 则60(600)80y x x =-+
2036000y x =+ ················································································································ 6分
根据题意 0.88(600)0.960.9600x x -+?≥
150x ≥
∴当150x =时，y 取最小值． ······························································································ 8分
min 2015036000y =?+
39000=．
答：当购买乙树苗150棵时费用最低，最低费用为39000元． ·········································· 9分 （本题不答不扣分） 23．（10分）解：（1）所作菱形如图①、②所示．
说明：作法相同的图形视为同一种．例如类似图③、图④的图形视为与图②是同一种．
（作出一个图形得3分） （2）图①的作法：
作矩形A 1B 1C 1D 1四条边的中点E 1、F 1、G 1、H 1； 连接H 1E 1、E 1F 1、G 1F 1、G 1H 1． 四边形E 1F 1G 1H 1即为菱形． 图②的作法：
在B 2C 2上取一点E 2，使E 2C 2＞A 2E 2且E 2不与B 2重合； 以A 2为圆心，A 2E 2为半径画弧，交A 2D 2于H 2；
以E 2为圆心，A 2E 2为半径画弧，交B 2C 2于F 2； 连接H 2F 2，则四边形A 2E 2F 2H 2为菱形． （写对一个作法得2分）
（此题答案不惟一，只要画法及作法合理、正确，均可酌情得分．）
24．（10分）解：（1）设抛物线的表达式为2y ax = ····· 1分
点(6 5.6)B -，
在抛物线的图象上． ∴ 5.636a -=
7
45
a =-
·········································································· 3分 ∴抛物线的表达式为2
745y x =- ·························································································· 4分 （2）设窗户上边所在直线交抛物线于C 、D 两点，D 点坐标为（k ，t ）
已知窗户高1.6m ，∴ 5.6( 1.6)4t =---=- ······································································· 5分
2
7445
k --=
125.07 5.07k k -≈，≈（舍去） ·
························································································ 6分 ∴ 5.07210.14CD =?≈（m ） ···························································································· 7分
又设最多可安装n 扇窗户
∴1.50.8(1)10.14n n ++≤ ··································································································· 9分
4.06n ≤．
答：最多可安装4扇窗户． ································································································· 10分 （本题不要求学生画出4个表示窗户的小矩形）