Question

07和09年济南数学中考试题及答案

最好是Word的，很急。

Answer 1

教务教研室有题库啊

Answer 2

去中学校长网下,我试过

Answer 3

一一一一、、、、 选择题选择题选择题选择题（（（（本大题共本大题共本大题共本大题共12个小题个小题个小题个小题，，，，每小题每小题每小题每小题4分分分分，，，，共共共共48分分分分．）．）．）．） 1．3−的相反数是（ ） A．3 B．3− C．13 D．13− 2．图中几何体的主视图是（ ） 3．如图，ABCD∥，直线EF与AB、CD分别相交于G、H．60AGE=°∠，则EHD∠的度数是（ ） A．30° B．60° C．120° D．150° 4．估计20的算术平方根的大小在（ ） A．2与3之间 B．3与4之间 C．4与5之间 D．5与6之间 5．2009年10月11日，第十一届全运会将在美丽的泉城济南召开．奥体中心由体育场，体育馆、游泳馆、网球馆，综合服务楼三组建筑组成，呈“三足鼎立”、“东荷西柳”布局．建筑面积约为359800平方米，请用科学记数法表示建筑面积是（保留三个有效数字）（ ） A．535.910×平方米 B．53.6010×平方米 C．53.5910×平方米 D．435.910×平方米 6．若12xx，是一元二次方程2560xx−+=的两个根，则12xx+的值是（ ） A．1 B．5 C．5− D．6 7．“只要人人都献出一点爱，世界将变成美好的人间”．在今年的慈善一日捐活动中，济南市某中学八年级三班50名学生自发组织献爱心捐款活动．班长将捐款情况进行了统计，并绘制成了统计图．根据右图提供的信息，捐款金额．．的众数和中位数分别是（ ） A．20、20 B．30、20 C．30、30 D．20、30 8．不等式组213351xx+>−≤的解集在数轴上表示正确的是（ ） A C E B F D H G （第3题图） 正面 （第2题图）A．B．C．D．1 2 0 A． B． 1 2 0 C． 1 2 0 D． 1 2 0 捐款人数 金额（元）0 5 10 15 20 61320 8 3 20 30 50 100 （第7题图） 10 13．分解因式：29x−= ． 14．如图，O⊙的半径5cmOA=，弦8cmAB=，点P为弦AB上一动点，则点P到圆心O的最短距离是 cm． 15．如图，AOB∠是放置在正方形网格中的一个角，则cosAOB∠的值是 ． 16．“五一”期间，我市某街道办事处举行了“迎全运，促和谐”中青年篮球友谊赛．获得男子篮球冠军球队的五名主力队员的身高如下表：（单位：厘米） 号码 4 7 9 10 23 身高 178 180 182 181 179 则该队主力队员身高的方差是 厘米2． 17．九年级三班小亮同学学习了“测量物体高度”一节课后，他为了测得右图所放风筝的高度，进行了如下操作： （1）在放风筝的点A处安置测倾器，测得风筝C的仰角60CBD=°∠； （2）根据手中剩余线的长度出风筝线BC的长度为70米； （3）量出测倾器的高度1.5AB=米． 根据测量数据，计算出风筝的高度CE约为 米．（精确到0.1米，31.73≈） 三三三三、、、、解答题解答题解答题解答题（（（（本大题共本大题共本大题共本大题共7个小题个小题个小题个小题，，，，共共共共57分分分分.解答应写出文字说明解答应写出文字说明解答应写出文字说明解答应写出文字说明、、、、证明过程或演算步骤证明过程或演算步骤证明过程或演算步骤证明过程或演算步骤）））） 18．（本小题满分7分） （1）计算：()()2121xx++− （2）解分式方程：2131xx=−−． 19．（本小题满分7分） （1）已知，如图①，在ABCD▱中，E、F是对角线BD上的两点，且BFDE=．求证：AECF=． （2）已知，如图②，AB是O⊙的直径，CA与O⊙相切于点A．连接CO交O⊙于点D，CO的延长线交O⊙于点E．连接BE、BD，30ABD=°∠，求EBO∠和C∠的度数． 20．（本小题满分8分） 有3张不透明的卡片，除正面写有不同的数字外，其它均相同．将这三张卡片背面朝上洗匀后，第一次从中随机抽取一张，并把这张卡片标有的数字记作一次函数表达式中的k，第二次从余下．．的两张卡片中再随机抽取一张，上面标有的数字记作一次函数表达式中的b． （1）写出k为负数的概率； （2）求一次函数ykxb=+的图象经过二、三、四象限的概率．（用树状图或列表法求解） 21．（本小题满分8分） 自2008年爆发全球金融危机以来，部分企业受到了不同程度的影响，为落实“促民生、促经济”政策，济南市某玻璃制品销售公司今年1月份调整了职工的月工资分配方案，调整后月工资由基本保障工资和计件奖励工资两部分组成（计件奖励工资=销售每件的奖励金额×销售的件数）．下表是甲、乙两位职工今年五月份的工资情况信息： 职工 甲 乙 月销售件数（件） 200 180 月工资（元） 1800 1700 （1）试求工资分配方案调整后职工的月基本保障工资和销售每件产品的奖励金额各多少元？ （2）若职工丙今年六月份的工资不低于2000元，那么丙该月至少应销售多少件产品？ 22．（本小题满分9分） 已知：如图，正比例函数yax=的图象与反比例函数kyx=的图象交于点()32A，． （1）试确定上述正比例函数和反比例函数的表达式； （2）根据图象回答，在第一象限内，当x取何值时，反比例函数的值大于正比例函数的值？ （3）()Mmn，是反比例函数图象上的一动点，其中03m<<，过点M作直线MNx∥轴，交y轴于点B；过点A作直线ACy∥轴交x轴于点C，交直线MB于点D．当四边形OADM的面积为6时，请判断线段BM与DM的大小关系，并说明理由． 23．（本小题满分9分） 1−2− 3− 正面 背面 （第22题图）y x OA D M C B 如图，在梯形ABCD中，354245ADBCADDCABB====°∥，，，，∠．动点M从B点出发沿线段BC以每秒2个单位长度的速度向终点C运动；动点N同时从C点出发沿线段CD以每秒1个单位长度的速度向终点D运动．设运动的时间为t秒． （1）求BC的长． （2）当MNAB∥时，求t的值． （3）试探究：t为何值时，MNC△为等腰三角形． 24．（本小题满分9分） 已知：抛物线()20yaxbxca=++≠的对称轴为1x=−，与x轴交于AB，两点，与y轴交于点C，其中()30A−，、()02C−，． （1）求这条抛物线的函数表达式． （2）已知在对称轴上存在一点P，使得PBC△的周长最小．请求出点P的坐标． （3）若点D是线段OC上的一个动点（不与点O、点C重合）．过点D作DEPC∥交x轴于点E．连接PD、PE．设CD的长为m，PDE△的面积为S．求S与m之间的函数关系式．试说明S是否存在最大值，若存在，请求出最大值；若不存在，请说明理由． 济南市济南市济南市济南市2009年高中阶段学校招生考试年高中阶段学校招生考试年高中阶段学校招生考试年高中阶段学校招生考试 数学试题参考答案及评分标准数学试题参考答案及评分标准数学试题参考答案及评分标准数学试题参考答案及评分标准 一一一一、、、、选择题选择题选择题选择题（（（（本大题共本大题共本大题共本大题共12个小题个小题个小题个小题，，，，每小题每小题每小题每小题4分分分分，，，，共共共共48分分分分）））） 题号 1 2 3 4 5 6 7 8 9 10 11 12 答案 A B C C B B C C C D B B 二二二二、、、、填空题填空题填空题填空题（（（（本大题共本大题共本大题共本大题共5个小题个小题个小题个小题，，，，每小题每小题每小题每小题3分分分分，，，，共共共共15分分分分）））） 13． ()()33xx+− 14．3 15．22 16．2 17．62.1 三三三三、、、、解答题解答题解答题解答题（（（（本大题共本大题共本大题共本大题共7个小题个小题个小题个小题，，，，共共共共57分分分分）））） 18．（本小题满分7分） （1）解：()()2121xx++− =22122xxx+++− ···························································································· 2分 =23x+··················································································································· 3分 （2）解：去分母得：()213xx−=− ·············································································· 1分 A D C B M N （第23题图）A C x y解得1x=− ········································································································ 2分 检验1x=−是原方程的解 ················································································· 3分 所以，原方程的解为1x=−·············································································· 4分 19．（本小题满分7分） （1）证明：∵四边形ABCD是平行四边形， ∴ADBCADBC=，∥． ∴ADEFBC=∠∠ ······················································································ 1分 在ADE△和CBF△中， ∵ADBCADEFBCDEBF===，∠∠， ∴ADECBF△≌△ ···················································································· 2分 ∴AECF= ····································································································· 3分 （2）解：∵DE是O⊙的直径 ∴90DBE=°∠ ······························································································ 1分 ∵30ABD=°∠ ∴903060EBODBEABD=−=°−°=°∠∠∠ ········································· 2分 ∵AC是O⊙的切线 ∴90CAO=°∠ ······························································································ 3分 又260AOCABD==°∠∠ ∴180180609030CAOCCAO=°−−=°−°−°=°∠∠∠ ······················· 4分 20．（本小题满分8分） 解：（1）k为负数的概率是23 ····································································································· 3分 （2）画树状图 或用列表法： 第二次 第一次 1− 2− 3 1− （1−，2−） （1−，3） 2− （2−，1−） （2−，3） 3 （3，1−） （3，2−） ·········································································· 5分 共有6种情况，其中满足一次函数ykxb=+经过第二、三、四象限， 即00kb<<，的情况有2种 ··························································································· 6分 所以一次函数ykxb=+经过第二、三、四象限的概率为2163= ··································· 8分 21．（本小题满分8分） 解：（1）设职工的月基本保障工资为x元，销售每件产品的奖励金额为y元 ··················· 1分由题意得20018001801700xyxy+=+= ·························································································· 3分 解这个方程组得8005xy== ······························································································ 4分 答：职工月基本保障工资为800元，销售每件产品的奖励金额5元. ································· 5分 （2）设该公司职工丙六月份生产z件产品 ··········································································· 6分 由题意得80052000z+≥ ·························································································· 7分 解这个不等式得240z≥ 答：该公司职工丙六月至少生产240件产品 ········································································· 8分 22．解：（1）将()32A，分别代入kyyaxx==，中，得2323ka==， ∴263ka==， ······································································································ 2分 ∴反比例函数的表达式为：6yx= ········································································· 3分 正比例函数的表达式为23yx= ··········································································· 4分 （2）观察图象，得在第一象限内， 当03x<<时，反比例函数的值大 于正比例函数的值． ··························· 6分 （3）BMDM= ···································································································· 7分 理由：∵132OMBOACSSk==×=△△ ∴33612OMBOACOBDCOADMSSSS=++=++=△△矩形四边形 即12OCOB=i ∵3OC= ∴4OB= ················································································································ 8分 即4n= ∴632mn== ∴3333222MBMD==−=， ∴MBMD= ··········································································································· 9分 23.（本小题满分9分） 解：（1）如图①，过A、D分别作AKBC⊥于K，DHBC⊥于H，则四边形ADHK是矩形 ∴3KHAD==． ······································································································ 1分 在RtABK△中，2sin454242AKAB=°==i． 2cos454242BKAB=°==ii ·····························································在RtCDH△中，由勾股定理得，22543HC=−= ∴43310BCBKKHHC=++=++= ······························································ 3分 （2）如图②，过D作DGAB∥交BC于点，则四边形ADGB是平行四边形 ∵MNAB∥ ∴MNDG∥ ∴3BGAD== ∴1037GC=−= ·································································································· 4分 由题意知，当M、N运动到t秒时，102CNtCMt==−，． ∵DGMN∥ ∴NMCDGC=∠∠ 又CC=∠∠ ∴MNCGDC△∽△ ∴CNCMCDCG= ·········································································································· 5分 即10257tt−= 解得，5017t= ·········································································································· 6分 （3）分三种情况讨论： ①当NCMC=时，如图③，即102tt=− ∴103t= ·················································································································· 7分 ②当MNNC=时，如图④，过N作NEMC⊥于E 解法一： 由等腰三角形三线合一性质得()11102522ECMCtt==−=− 在RtCEN△中，5cosECtcNCt−== 又在RtDHC△中，3cos5CHcCD== ∴535tt−= （第23题图①）A D C B K H （第23题图②）A D C B G M N A D C B M N （第23题图③）（第23题图④）A D C B M N H E 解得258t= ·············································································································· 8分 解法二： ∵90CCDHCNEC=∠=∠=°∠∠， ∴NECDHC△∽△ ∴NCECDCHC= 即553tt−= ∴258t= ·················································································································· 8分 ③当MNMC=时，如图⑤，过M作MFCN⊥于F点.1122FCNCt== 解法一：（方法同②中解法一） 132cos1025tFCCMCt===− 解得6017t= 解法二： ∵90CCMFCDHC=∠=∠=°∠∠， ∴MFCDHC△∽△ ∴FCMCHCDC= 即1102235tt−= ∴6017t= 综上所述，当103t=、258t=或6017t=时，MNC△为等腰三角形 ···················· 9分 24.（本小题满分9分） 解：（1）由题意得129302baabcc=−+==− ··············································································· 2分 解得23432abc===− ∴此抛物线的解析式为224233yxx=+− ······························································ 3分 （2）连结AC、BC.因为BC的长度一定，所以PBC△周长最小，就是使PCPB+最小.B点关于对称轴的对称点是A点，AC与对称轴1x=−的交点即为所求的点P.设直线AC的表达式为ykxb=+ 则302kbb−+==−， ···························································· 4分 解得232kb=−=− ∴此直线的表达式为223yx=−−． ········································································ 5分 把1x=−代入得43y=− ∴P点的坐标为413−−， ····················································································· 6分 （3）S存在最大值 ································································································· 7分 理由：∵DEPC∥，即DEAC∥． ∴OEDOAC△∽△． ∴ODOEOCOA=，即223mOE−=． ∴333322OEmAEOEm=−==，， 方法一： 连结OP OEDPOEPODOEDPDOESSSSSS=−=+−△△△△四边形 =()()13411332132223222mmmm×−×+×−×−×−×− =23342mm−+ ····································································································· 8分 ∵304−< ∴当1m=时，333424S=−+=最大 ··································································· 9分 方法二： OACOEDAEPPCDSSSSS=−−−△△△△ =()1131341323212222232mmmm××−×−×−−××−×× =()22333314244mmm−+=−−+ ····································································· 8分 ∵304−< ∴当1m=时，34S=最大 ···················································································· 9分