2009年重庆市中考数学试题及答案

这是一份2009年重庆市中考数学试题及答案，共7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。
2009年重庆市中招考试数学试卷（A卷）一、选择题：（本大题10个小题，每小题4分，共40分）1．的相反数是（    ）A．5  B．  C．  D．2．计算的结果是（    ）A．  B．  C．  D．3．函数的自变量的取值范围是（    ）A．  B．  C．  D．4．如图，直线相交于点，．若，则等于（    ）A．70°  B．80°  C．90°  D．100°5．下列调查中，适宜采用全面调查（普查）方式的是（    ）A．调查一批新型节能灯泡的使用寿命B．调查长江流域的水污染情况C．调查重庆市初中学生的视力情况D．为保证“神舟7号”的成功发射，对其零部件进行检查6．如图，是的外接圆，是直径．若，则等于（    ）A．60°  B．50°  C．40°  D．30°7．由四个大小相同的正方体组成的几何体如图所示，那么它的左视图是（    ）    A．       B．       C．         D．8．观察下列图形，则第个图形中三角形的个数是（    ）     A．  B．  C．  D．9．如图，在矩形中，，，动点P从点B出发，沿路线作匀速运动，那么的面积S与点P运动的路程之间的函数图象大致是（    ）                          10．如图，在等腰中，，F是AB边上的中点，点D、E分别在AC、BC边上运动，且保持．连接DE、DF、EF．在此运动变化的过程中，下列结论：①是等腰直角三角形；②四边形CDFE不可能为正方形，③DE长度的最小值为4；④四边形CDFE的面积保持不变；⑤△CDE面积的最大值为8．其中正确的结论是（    ）A．①②③  B．①④⑤  C．①③④  D．③④⑤二、填空题：（本大题6个小题，每小题4分，共24分）11．据重庆市统计局公布的数据，今年一季度全市实现国民生产总值约为7840000万元．那么7840000万元用科学记数法表示为          万元．12．分式方程的解为         ．13．已知与相似且面积比为4∶25，则与的相似比为        ．14．已知的半径为3cm，的半径为4cm，两圆的圆心距为7cm，则与的位置关系是       ．15．在平面直角坐标系中，直线与两坐标轴围成一个．现将背面完全相同，正面分别标有数1、2、3、、的5张卡片洗匀后，背面朝上，从中任取一张，将该卡片上的数作为点P的横坐标，将该数的倒数作为点P的纵坐标，则点P落在内的概率为        ．16．某公司销售A、B、C三种产品，在去年的销售中，高新产品C的销售金额占总销售金额的40%．由于受国际金融危机的影响，今年A、B两种产品的销售金额都将比去年减少20%，因而高新产品C是今年销售的重点．若要使今年的总销售金额与去年持平，那么今年高新产品C的销售金额应比去年增加         %．三、解答题：（本大题4个小题，每小题6分，共24分）17．计算：．      18．解不等式组：     19．作图，请你在下图中作出一个以线段AB为一边的等边．（要求：用尺规作图，并写出已知、求作，保留作图痕迹，不写作法和结论） 已知： 求作：          20．为了建设“森林重庆”，绿化环境，某中学七年级一班同学都积极参加了植树活动，今年4月该班同学的植树情况的部分统计如下图所示：           （1）请你根据以上统计图中的信息，填写下表：该班人数植树株数的中位数植树株数的众数   （2）请你将该条形统计图补充完整．           四、解答题：（本大题4个小题，每小题10分，共40分）21．先化简，再求值：，其中．              22．已知：如图，在平面直角坐标系中，直线AB分别与轴交于点B、A，与反比例函数的图象分别交于点C、D，轴于点E，．（1）求该反比例函数的解析式；（2）求直线AB的解析式．                    23．有一个可自由转动的转盘，被分成了4个相同的扇形，分别标有数1、2、3、4（如图所示），另有一个不透明的口袋装有分别标有数0、1、3的三个小球（除数不同外，其余都相同），小亮转动一次转盘，停止后指针指向某一扇形，扇形内的数是小亮的幸运数，小红任意摸出一个小球，小球上的数是小红的吉祥数，然后计算这两个数的积．（1）请你用画树状图或列表的方法，求这两个数的积为0的概率；（2）小亮与小红做游戏，规则是：若这两个数的积为奇数，小亮赢；否则，小红赢．你认为该游戏公平吗？为什么？如果不公平，请你修改该游戏规则，使游戏公平．                                     24．已知：如图，在直角梯形ABCD中，AD∥BC，∠ABC=90°，DE⊥AC于点F，交BC于点G，交AB的延长线于点E，且．（1）求证：；（2）若，求AB的长．                                        五、解答题：（本大题2个小题，第25小题10分，第26小题12分，共22分）25．某电视机生产厂家去年销往农村的某品牌电视机每台的售价y（元）与月份x之间满足函数关系，去年的月销售量p（万台）与月份x之间成一次函数关系，其中两个月的销售情况如下表：月份1月5月销售量3.9万台4.3万台（1）求该品牌电视机在去年哪个月销往农村的销售金额最大？最大是多少？（2）由于受国际金融危机的影响，今年1、2月份该品牌电视机销往农村的售价都比去年12月份下降了，且每月的销售量都比去年12月份下降了1.5m%．国家实施“家电下乡”政策，即对农村家庭购买新的家电产品，国家按该产品售价的13%给予财政补贴．受此政策的影响，今年3至5月份，该厂家销往农村的这种电视机在保持今年2月份的售价不变的情况下，平均每月的销售量比今年2月份增加了1.5万台．若今年3至5月份国家对这种电视机的销售共给予了财政补贴936万元，求的值（保留一位小数）．（参考数据：，，，）                            26．已知：如图，在平面直角坐标系中，矩形OABC的边OA在y轴的正半轴上，OC在x轴的正半轴上，OA=2，OC=3．过原点O作∠AOC的平分线交AB于点D，连接DC，过点D作DE⊥DC，交OA于点E．（1）求过点E、D、C的抛物线的解析式；（2）将∠EDC绕点D按顺时针方向旋转后，角的一边与y轴的正半轴交于点F，另一边与线段OC交于点G．如果DF与（1）中的抛物线交于另一点M，点M的横坐标为，那么EF=2GO是否成立？若成立，请给予证明；若不成立，请说明理由；（3）对于（2）中的点G，在位于第一象限内的该抛物线上是否存在点Q，使得直线GQ与AB的交点P与点C、G构成的△PCG是等腰三角形？若存在，请求出点Q的坐标；若不存在，请说明理由．                      重庆市2009年初中毕业暨高中招生考试数学试题参考答案及评分意见一、选择题1．A  2．B  3．C  4．B  5．D  6．C  7．A  8．D  9．B  10．B二、填空题11．      12．    13．    14．外切     15．     16．30三、解答题17．解：原式·························································（5分）            ．····························································（6分）18．解：由①，得．····················································（2分）由②，得．····················································（4分）所以，原不等式组的解集为．······································（6分）19．解：已知：线段．··················································（1分）求作：等边．·························································（2分）作图如下：（注：每段弧各1分，连接线段各1分）    ········································（6分） 20．解：（1）填表如下：该班人数植树株数的中位数植树株树的众数5032·······························（4分）（2）补图如下：         ···················（6分）四、解答题：21．解：原式·························································（4分）···································································（6分）．··································································（8分）当时，原式．·························································（10分）22．解：（1），．轴于点．，．································································（1分）点的坐标为．·························································（2分）设反比例函数的解析式为．将点的坐标代入，得，··················································（3分）．··································································（4分）该反比例函数的解析式为．···············································（5分）（2），．····························································（6分），，．································································（7分）设直线的解析式为．将点的坐标分别代入，得················································（8分）解得································································（9分）直线的解析式为．······················································（10分）23．解：（1）画树状图如下：  ················（4分）   或列表如下：        幸运数积吉祥数12340000011234336912···································································（4分）由图（表）知，所有等可能的结果有12种，其中积为0的有4种，所以，积为0的概率为．·················································（6分）（2）不公平．························································（7分）因为由图（表）知，积为奇数的有4种，积为偶数的有8种．所以，积为奇数的概率为，···············································（8分）积为偶数的概率为．····················································（9分）因为，所以，该游戏不公平．游戏规则可修改为：若这两个数的积为0，则小亮赢；积为奇数，则小红赢．·························（10分）（只要正确即可）24．（1）证明：于点，．······································（1分），········································（2分）．······································（3分）连接，···································（4分），．······································（5分）．······································（6分）（2）解：，．··································································（7分）．，··································································（8分）．··································································（9分）．··································································（10分）五、解答题：25．解：（1）设与的函数关系为，根据题意，得···································································（1分）解得所以，．·························································（2分）设月销售金额为万元，则．···············································（3分）化简，得，所以，．当时，取得最大值，最大值为10125．答：该品牌电视机在去年7月份销往农村的销售金额最大，最大是10125万元．········（4分）（2）去年12月份每台的售价为（元），去年12月份的销售量为（万台），·········································（5分）根据题意，得．·······················································（8分）令，原方程可化为．．，（舍去）答：的值约为52.8．····················································（10分）26．解：（1）由已知，得，，，．．··································································（1分）设过点的抛物线的解析式为．将点的坐标代入，得．将和点的坐标分别代入，得···································································（2分）解这个方程组，得故抛物线的解析式为．··················································（3分）（2）成立．··························································（4分）点在该抛物线上，且它的横坐标为，点的纵坐标为．·······················································（5分）设的解析式为，将点的坐标分别代入，得   解得的解析式为．·························································（6分），．································································（7分）过点作于点，则．，．又，．．．··································································（8分）．（3）点在上，，，则设．，，．①若，则，解得．，此时点与点重合．．··································································（9分）②若，则，解得 ，，此时轴．与该抛物线在第一象限内的交点的横坐标为1，点的纵坐标为．．··································································（10分）③若，则，解得，，此时，是等腰直角三角形．过点作轴于点，则，设，．．解得（舍去）．．·····································（12分）综上所述，存在三个满足条件的点，即或或．