中考数学试卷精选合辑(补充)52之13-初中毕业生学业考试题及答案

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（náng），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．∵AB∥CD ∴∠ABC+∠C=180° B．∵∠1=∠2 ∴AD∥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y?11的图象上有三个点的坐标分别为（1，y1）、（，y2）、（?3，y3），函数x2

D．y3＜y1＜y2

值y1、y2、y3的大小关系是（ ）

A．y1＜y2＜y3 B．y3＜y2＜y1C．y2＜y1＜y3

7．如图，△ABC中BC边上的高为h1，△DEF中DE边上的高为h2，下列结论正确的是（ ） A．h1?h2 C．h1?h2

B．h1?h2 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（ ）

2π(60?10)2π(60?10?x) ?682π(60?x)2π?60B． ?86A．

C．2π(60?10)?6?2π(60?x)?8 D．2π(60?x)?8?2π(60?x)?6

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A1B1是由线段AB平移得到的，已知A，B两点的坐标分别为A(?2，3)，B(?31)4)，则B1的坐标为 ． ，，若A1的坐标为(3，

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：2

?2??1?18?(π?6)0． 4x2?1x?116（6分）化简分式2，并从?2、?1、0、1、2中选一个能使分式有意义?x?2x?1x?1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OC?10cm，直线l⊥CO，垂足为H，交⊙O于A、B两点，AB?16cm，直线l平移多少厘米时能与⊙O相切？

19．（9分）某水果销售公司去年3至8月销售吐鲁番葡萄、哈密大枣的情况见下表： 吐鲁番葡萄（吨） 哈密大枣（吨） 3月 4 8 4月 8 7 5月 5 9 6月 8 7 7月 10 10 8月 13 7 （1）请你根据以上数据填写下表：

吐鲁番葡萄 哈密大枣

（2）补全折线统计图．

平均数 8 方差 9

（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：

种类 甲 乙 单价（元） 60 80 成活率 88% 96% （1）若购买树苗资金不超过44000元，则最多可购买乙树苗多少棵？

（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．

一、选择题（本大题共10题，每题5分，共50分） 题号 选项 1 D 2 A 3 C 4 C 5 B 6 D 7 C 8 A 9 B 10 A 二、填空题（本大题共4题，每题5分，共20分） 11．欢欢 12．（2，2） 13．x?4（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?211·············································································· 4分 ??32?1 ·

44?32 ································································································ 6分

16．（6分）解：原式?(x?1)(x?1)x?1 ······································································· 1分 ?2(x?1)x?1?x?1x?1 ························································································································ 2分 ?x?1x?1(x?1)2?(x?1)2? ··············································································································· 3分

(x?1)(x?1)?4x ·································································································································· 5分 x2?1把x?0代入 原式?0 ·································································································································· 6分 或把x?2代入

?4?28原式?2··············································································································· 6分 ?? ·

2?13或把x??2代入 ?原式??4?(?2)8················································ 6分17．（6分）解：方法1：画树状图 ?． 2(?2)?13

·············································································· 4分

张、王两位老师同时被选中的概率是方法2：列表 张 王 李 赵 张 王张 李张 赵张 王 张王 李王 赵王 李 张李 王李 赵李 赵 张赵 王赵 李赵 1． ············································································ 6分 6 ······························································· 4分 张、王两位老师同时被选中的概率是

1． ············································································ 6分 618．（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，OH?AO2?AH2?102?82?6， ······· 6分

∴CH?4cm，DH?16cm．

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xcm时能与圆相切，

(10?x)2?82?102 ················································································································ 3分 x1?16

x2?4

∴CH?4cm，DH?16cm． ····························································································· 8分 答：略．

（只答一个方向的平移扣2分） 19．（9分） 解：（1）

吐鲁番葡萄 哈密大枣 平均数 8 8 方差 9 4 3??（4分）

（2）

·························································· （7分）

（3）①由于平均数相同，S大枣?S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：(4x)?(50?3x)?85 ················································································ 4分 化简得：x?12x?189?0

解得：x1?9，x2??21（不符合实际情况，舍去） ·························································· 6分 当x?9时，4x?36，50?3x?77

∴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xm时，两人相距85m 则乙与O处的距离为?2222222?3?······················································································ 1分 x?50?m ·

4???3?x??x?50??852 ·········································································································· 4分

?4?2解得：x1?36，x2??84（不符合实际情况，舍去 ） ······················································ 6分 当x?36，x?50?77 ········································································································· 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

34??CAD??EAD? ?AD?AD??C??AED?∴△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 ································································································ 3分 60(600?x)?80x≤44000 ·

x≤400．

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y?60(600?x)?80x

y?20 ················································································································ 6分 x?36000根据题意 0.88(600?x)?0.96x≥0.9?600

x≥150

∴当x?150时，y取最小值． ······························································································ 8分 ymin?20?150?36000

?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A1B1C1D1四条边的中点E1、F1、G1、H1； 连接H1E1、E1F1、G1F1、G1H1． 四边形E1F1G1H1即为菱形． 图②的作法：

在B2C2上取一点E2，使E2C2＞A2E2且E2不与B2重合； 以A2为圆心，A2E2为半径画弧，交A2D2于H2；

以E2为圆心，A2E2为半径画弧，交B2C2于F2； 连接H2F2，则四边形A2E2F2H2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y?ax ····· 1分 点B(6，?5.6)在抛物线的图象上． ∴?5.6?36a

27 ·········································································· 3分 4572∴抛物线的表达式为y??························································································· 4分 x ·

45a??（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，∴t??5.6?(?1.6)??4 ······································································· 5分

?4??72k 45k1≈5.07，k2≈?5.07（舍去） ························································································· 6分

∴CD?5.07?2≈10.14（m） ···························································································· 7分 又设最多可安装n扇窗户

∴1.5n?0.8(n?1)≤10.14 ··································································································· 9分

n≤4.06．

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）

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