中考数学综合练习题44

这是一份中考数学综合练习题44，共15页。试卷主要包含了选择题，填空题，解答题等内容，欢迎下载使用。
中考数学综合练习题44第Ⅰ卷（机读卷  共32分）一、选择题（共8道小题，每小题4分，共32分）下列各题均有四个选项，其中只有一个是符合题意的．用铅笔把“机读答题卡”上对应题目答案的相应字母处涂黑．1．的绝对值等于（    ）A．  B．  C．  D．2．截止到2008年5月19日，已有21 600名中外记者成为北京奥运会的注册记者，创历届奥运会之最．将21 600用科学记数法表示应为（    ）A．  B．  C．  D．3．若两圆的半径分别是1cm和5cm，圆心距为6cm，则这两圆的位置关系是（    ）A．内切  B．相交  C．外切  D．外离4．众志成城，抗震救灾．某小组7名同学积极捐出自己的零花钱支援灾区，他们捐款的数额分别是（单位：元）：50，20，50，30，50，25，135．这组数据的众数和中位数分别是（    ）A．50，20  B．50，30  C．50，50  D．135，505．若一个多边形的内角和等于，则这个多边形的边数是（    ）A．5  B．6  C．7  D．86．如图，有5张形状、大小、质地均相同的卡片，正面分别印有北京奥运会的会徽、吉祥物（福娃）、火炬和奖牌等四种不同的图案，背面完全相同．现将这5张卡片洗匀后正面向下放在桌子上，从中随机抽取一张，抽出的卡片正面图案恰好是吉祥物（福娃）的概率是（    ）A．  B．  C．  D．7．若，则的值为（    ）A．  B．  C．  D．8．已知为圆锥的顶点，为圆锥底面上一点，点在上．一只蜗牛从点出发，绕圆锥侧面爬行，回到点时所爬过的最短路线的痕迹如右图所示．若沿将圆锥侧面剪开并展开，所得侧面展开图是（    ）       第Ⅱ卷（非机读卷  共88分）二、填空题（共4道小题，每小题4分，共16分）9．在函数中，自变量的取值范围是          ．10．分解因式：         ．11．如图，在中，分别是的中点，若，则      cm．12．一组按规律排列的式子：，，，，…（），其中第7个式子是     ，第个式子是        （为正整数）．三、解答题（共5道小题，共25分）13．（本小题满分5分）计算：．解：      14．（本小题满分5分）解不等式，并把它的解集在数轴上表示出来．解：    15．（本小题满分5分）已知：如图，为上一点，点分别在两侧．，，．求证：．证明：         16．（本小题满分5分）如图，已知直线经过点，求此直线与轴，轴的交点坐标．解：     17．（本小题满分5分）已知，求的值．解：    四、解答题（共2道小题，共10分）18．（本小题满分5分）如图，在梯形中，，，，，，求的长．解：    19．（本小题满分5分）已知：如图，在中，，点在上，以为圆心，长为半径的圆与分别交于点，且．（1）判断直线与的位置关系，并证明你的结论；（2）若，，求的长．解：（1）   （2）    五、解答题（本题满分6分）20．为减少环境污染，自2008年6月1日起，全国的商品零售场所开始实行“塑料购物袋有偿使用制度”（以下简称“限塑令”）．某班同学于6月上旬的一天，在某超市门口采用问卷调查的方式，随机调查了“限塑令”实施前后，顾客在该超市用购物袋的情况，以下是根据100位顾客的100份有效答卷画出的统计图表的一部分：             “限塑令”实施后，塑料购物袋使用后的处理方式统计表处理方式直接丢弃直接做垃圾袋再次购物使用其它选该项的人数占总人数的百分比5%35%49%11%请你根据以上信息解答下列问题：（1）补全图1，“限塑令”实施前，如果每天约有2 000人次到该超市购物．根据这100位顾客平均一次购物使用塑料购物袋的平均数，估计这个超市每天需要为顾客提供多少个塑料购物袋？（2）补全图2，并根据统计图和统计表说明，购物时怎样选用购物袋，塑料购物袋使用后怎样处理，能对环境保护带来积极的影响．解：（1）  （2）  六、解答题（共2道小题，共9分）21．（本小题满分5分）列方程或方程组解应用题：京津城际铁路将于2008年8月1日开通运营，预计高速列车在北京、天津间单程直达运行时间为半小时．某次试车时，试验列车由北京到天津的行驶时间比预计时间多用了6分钟，由天津返回北京的行驶时间与预计时间相同．如果这次试车时，由天津返回北京比去天津时平均每小时多行驶40千米，那么这次试车时由北京到天津的平均速度是每小时多少千米？解：    22．（本小题满分4分）已知等边三角形纸片的边长为，为边上的点，过点作交于点．于点，过点作于点，把三角形纸片分别沿按图1所示方式折叠，点分别落在点，，处．若点，，在矩形内或其边上，且互不重合，此时我们称（即图中阴影部分）为“重叠三角形”．        （1）若把三角形纸片放在等边三角形网格中（图中每个小三角形都是边长为1的等边三角形），点恰好落在网格图中的格点上．如图2所示，请直接写出此时重叠三角形的面积；（2）实验探究：设的长为，若重叠三角形存在．试用含的代数式表示重叠三角形的面积，并写出的取值范围（直接写出结果，备用图供实验，探究使用）．       解：（1）重叠三角形的面积为         ；（2）用含的代数式表示重叠三角形的面积为      ；的取值范围为     ．   七、解答题（本题满分7分）23．已知：关于的一元二次方程．（1）求证：方程有两个不相等的实数根；（2）设方程的两个实数根分别为，（其中）．若是关于的函数，且，求这个函数的解析式；（3）在（2）的条件下，结合函数的图象回答：当自变量的取值范围满足什么条件时，．（1）证明：  （2）解：   （3）解：      八、解答题（本题满分7分）24．在平面直角坐标系中，抛物线与轴交于两点（点在点的左侧），与轴交于点，点的坐标为，将直线沿轴向上平移3个单位长度后恰好经过两点．（1）求直线及抛物线的解析式；（2）设抛物线的顶点为，点在抛物线的对称轴上，且，求点的坐标；（3）连结，求与两角和的度数．解：（1）     （2）   （3）   九、解答题（本题满分8分）25．请阅读下列材料：问题：如图1，在菱形和菱形中，点在同一条直线上，是线段的中点，连结．若，探究与的位置关系及的值．小聪同学的思路是：延长交于点，构造全等三角形，经过推理使问题得到解决．        请你参考小聪同学的思路，探究并解决下列问题：（1）写出上面问题中线段与的位置关系及的值；（2）将图1中的菱形绕点顺时针旋转，使菱形的对角线恰好与菱形的边在同一条直线上，原问题中的其他条件不变（如图2）．你在（1）中得到的两个结论是否发生变化？写出你的猜想并加以证明．（3）若图1中，将菱形绕点顺时针旋转任意角度，原问题中的其他条件不变，请你直接写出的值（用含的式子表示）．解：（1）线段与的位置关系是          ；         ．（2）                       数学试卷答案第Ⅰ卷  （机读卷  共32分） 一、选择题（共8道小题，每小题4分，共32分）题号12345678答案ADCCBBBD 第Ⅱ卷  （非机读卷  共88分）二、填空题（共4道小题，每小题4分，共16分）题号9101112答案4三、解答题（共5道小题，共25分）13．（本小题满分5分）解：····································································4分．···································································5分14．（本小题满分5分）解：去括号，得．······················································1分移项，得．····························································2分合并，得．····························································3分系数化为1，得．·······················································4分不等式的解集在数轴上表示如下：  ····································································5分15．（本小题满分5分）证明：，．···································································2分在和中，．···································································4分．···································································5分16．（本小题满分5分）解：由图象可知，点在直线上，············································1分．解得．·······························································2分直线的解析式为．······················································3分令，可得．直线与轴的交点坐标为．·················································4分令，可得．直线与轴的交点坐标为．·················································5分17．（本小题满分5分）解：····································································2分．···································································3分当时，．·····························································4分原式．·······························································5分四、解答题（共2道小题，共10分）18．（本小题满分5分）解法一：如图1，分别过点作于点，于点．···································1分．又，四边形是矩形．．······································2分，，，．．，····································································4分在中，，．···································································5分解法二：如图2，过点作，分别交于点．······································1分，．，．在中，，，，····································································2分在中，，，，．．···································································4分在中，，．···································································5分19． （本小题满分5分）解：（1）直线与相切．··················································1分证明：如图1，连结．，．，        ．又，．．直线与相切．··························································2分（2）解法一：如图1，连结．是的直径，       ．，．···································································3分，，．···································································4分，       ．······························································5分解法二：如图2，过点作于点．      ．，．·····························3分，，．········································4分，．···································································5分五、解答题（本题满分6分）解：（1）补全图1见下图．···············································1分         （个）．这100位顾客平均一次购物使用塑料购物袋的平均数为3个．·······················3分．估计这个超市每天需要为顾客提供6000个塑料购物袋．···························4分（2）图2中，使用收费塑料购物袋的人数所占百分比为．·························5分根据图表回答正确给1分，例如：由图2和统计表可知，购物时应尽量使用自备袋和押金式环保袋，少用塑料购物袋；塑料购物袋应尽量循环使用，以便减少塑料购物袋的使用量，为环保做贡献．·······················6分六、解答题（共2道小题，共9分）21．解：设这次试车时，由北京到天津的平均速度是每小时千米，则由天津返回北京的平均速度是每小时千米．·······················1分依题意，得．··························································3分解得．·······························································4分答：这次试车时，由北京到天津的平均速度是每小时200千米．····················5分22．解：（1）重叠三角形的面积为．···········································1分（2）用含的代数式表示重叠三角形的面积为；·································2分的取值范围为．························································4分七、解答题（本题满分7分）23．（1）证明：是关于的一元二次方程，．当时，，即．方程有两个不相等的实数根．··············································2分（2）解：由求根公式，得．或．·································································3分，．，，．·································································4分．即为所求．····························5分（3）解：在同一平面直角坐标系中分别画出与的图象．·····································6分由图象可得，当时，．····················7分八、解答题（本题满分7分）24．解：（1）沿轴向上平移3个单位长度后经过轴上的点，．设直线的解析式为．在直线上，．解得．直线的解析式为．······················································1分抛物线过点，解得抛物线的解析式为．·····················································2分（2）由．可得．，，，．可得是等腰直角三角形．，．如图1，设抛物线对称轴与轴交于点，．过点作于点．．可得，．在与中，，，．，．解得．点在抛物线的对称轴上，点的坐标为或．························································5分（3）解法一：如图2，作点关于轴的对称点，则．连结，可得，．由勾股定理可得，．又，．是等腰直角三角形，，．．．即与两角和的度数为．···················································7分解法二：如图3，连结．同解法一可得，．在中，，，．在和中，，，．．．．，．即与两角和的度数为．···················································7分九、解答题（本题满分8分）25．解：（1）线段与的位置关系是；．···································································2分（2）猜想：（1）中的结论没有发生变化．证明：如图，延长交于点，连结．是线段的中点，      ．由题意可知．．，      ．，．四边形是菱形，，．由，且菱形的对角线恰好与菱形的边在同一条直线上，可得．       ．四边形是菱形，．        ．．，．．即．，，，．．···································································6分（3）．······························································8分