山东省烟台市初中学生九数学学业考试试题(含答案)(word)

2009年烟台市初中学生学业考试数学说明：1．本试题分为Ⅰ卷和Ⅱ卷两部分．第Ⅰ卷为选择题，第Ⅱ卷为非选择题．考试时间为120分钟，满分150分．2．答题前将密封线内的项目填写清楚．3．考试过程中允许考生进行剪、拼、折叠等实验．第Ⅰ卷注意事项：请考生将自己的姓名、准考证号、考试科目涂写在答题卡上．选择题选出答案后，用2B铅笔把答题卡对应题目的答案标号涂黑，不能答在本试题上．如要改动，必须先用橡皮擦干净，再选涂另一个答案．一、选择题（本题共12个小题，每小题4分，满分48分）每小题给出标号为A，B，C，D四个备选答案，其中有且只有一个是正确的．1．|3|的相反数是（）对数视力110.14.0A．3B．3C．D．332．视力表对我们来说并不陌生．如图是视力表的一部分，0.124.1其中开口向上的两个“E”之间的变换是（）．平移．旋转．对称．位似ABCD0.154.2x32x3．学完分式运算后，老师出了一道题“化简：”x2x24（第2题图）(x3)(x2)x2x2x6x2x28小明的做法是：原式；x24x24x24x24小亮的做法是：原式(x3)(x2)(2x)x2x62xx24；x3x2x31x31小芳的做法是：原式1．x2(x2)(x2)x2x2x2其中正确的是（）A．小明B．小亮C．小芳D．没有正确的4．设a，b是方程x2x20090的两个实数根，则a22ab的值为（）A．2006B．2007C．2008D．200945．一个长方体的左视图、俯视图及相关数据如图所示，3则其主视图的面积为（）A．6B．82C．12D．24左视图6．如图，数轴上A，B两点表示的数分别为1和3，俯视图（第5题图）点B关于点A的对称点为C，则点C所表示的数为（）A．23B．13CAOB（第6题图）C．23D．137．某校初一年级有六个班，一次测试后，分别求得各个班级学生成绩的平均数，它们不完全相同，下列说法正确的是（）A．全年级学生的平均成绩一定在这六个平均成绩的最小值与最大值之间B．将六个平均成绩之和除以6，就得到全年级学生的平均成绩C．这六个平均成绩的中位数就是全年级学生的平均成绩D．这六个平均成绩的众数不可能是全年级学生的平均成绩y8．如图，直线ykxb经过点A(1，2)和点B(2，0)，x直线y2x过点A，则不等式2xkxb0的解集为（）BOAA．x2B．2x1C．2x0D．1x0（第8题图）9．现有四种地面砖，它们的形状分别是：正三角形、正方形、正六边形、正八边形，且它们的边长都相等．同时选择其中两种地面砖密铺地面，选择的方式有（）A．2种B．3种C．4种D．5种A10．如图，等边△ABC的边长为3，P为BC上一点，且BP1，D为AC上一点，若APD60°，则CD的长为（）60°D3213BCA．B．C．D．P2324（第10题图）11．二次函数yax2bxc的图象如图所示，则一次函数ybxb24ac与反比例函abc数y在同一坐标系内的图象大致为（）xyyyyyxxxxx1O1OOOOA．B．C．D．（第11题图）12．利用两块长方体木块测量一张桌子的高度．首先按图①方式放置，再交换两木块的位置，按图②方式放置．测量的数据如图，则桌子的高度是（）A．73cmB．74cmC．75cmD．76cm80cm70cm①②（第12题图）第Ⅱ卷二、填空题（本题共6个小题，每小题4分，满分24分）13．若3xm5y2与x3yn的和是单项式，则nm．ab14．设ab0，a2b26ab0，则的值等于．ba15．如图，将两张长为8，宽为2的矩形纸条交叉，使重叠部分是一个菱形，容易知道当两张纸条垂直时，菱形的周长有最小值8，那么（第15题图）菱形周长的最大值是．xa≥216．如果不等式组2的解集是0≤x1，那么ab的值为．2xb317．观察下表，回答问题：序号123…图形…第个图形中“△”的个数是“○”的个数的5倍．18．如图，△ABC与△AEF中，ABAE，BCEF，BE，AB交EF于D．给出下列结论：A①AFCC；②DFCF；③△ADE∽△FDB；④BFDCAF．E其中正确的结论是（填写所有正确结论的序号）．D三、解答题（本大题共8个小题，满分78分）BC19．（本题满分6分）F（第18题图）936化简：18(32)0(12)2．2320．（本题满分8分）将如图所示的牌面数字分别是1，2，3，4的四张扑克牌背面朝上，洗匀后放在桌面上．（1）从中随机抽出一张牌，牌面数字是偶数的概率是；（2）从中随机抽出二张牌，两张牌牌面数字的和是5的概率是；（3）先从中随机抽出一张牌，将牌面数字作为十位上的数字，然后将该牌放回并重新洗匀，再随机抽取一张，将牌面数字作为个位上的数字，请用画树状图或列表的方法求组成的两位数恰好是4的倍数的概率．（第20题图）21．（本题满分8分）某市教育行政部门为了了解初一学生每学期参加综合实践活动的情况，随机抽样调查了某校初一学生一个学期参加综合实践活动的天数，并用得到的数据绘制了下面两幅不完整的统计图（如图）．人数4天3天6015%30%50402天10%305%a207天15%105天6天2天3天4天5天6天7天时间（第21题图）请你根据图中提供的信息，回答下列问题：（1）求出扇形统计图中a的值，并求出该校初一学生总数；（2）分别求出活动时间为5天、7天的学生人数，并补全频数分布直方图；（3）求出扇形统计图中“活动时间为4天”的扇形所对圆心角的度数；（4）在这次抽样调查中，众数和中位数分别是多少？（5）如果该市共有初一学生6000人，请你估计“活动时间不少于4天”的大约有多少人？22．(本题满分8分)腾飞中学在教学楼前新建了一座“腾飞”雕塑（如图①）.为了测量雕塑的高度，小明在二楼找到一点C，利用三角板测得雕塑顶端A点的仰角为30°，底部B点的俯角为45°，小华在五楼找到一点D，利用三角板测得A点的俯角为60°（如图②）.若已知CD为10米，请求出雕塑AB的高度．（结果精确到0.1米，参考数据31.73）．DAC①B②（第22题图）23．(本题满分10分)某商场将进价为2000元的冰箱以2400元售出，平均每天能售出8台，为了配合国家“家电下乡”政策的实施，商场决定采取适当的降价措施.调查表明：这种冰箱的售价每降低50元，平均每天就能多售出4台．（1）假设每台冰箱降价x元，商场每天销售这种冰箱的利润是y元，请写出y与x之间的函数表达式；（不要求写自变量的取值范围）（2）商场要想在这种冰箱销售中每天盈利4800元，同时又要使百姓得到实惠，每台冰箱应降价多少元？（3）每台冰箱降价多少元时，商场每天销售这种冰箱的利润最高？最高利润是多少？24．(本题满分10分)如图，AB，BC分别是⊙O的直径和弦，点D为BC上一点，弦DE交⊙O于点E，交AB于点F，交BC于点G，过点C的切线交ED的延长线于H，且HCHG，连接BH，交⊙O于点M，连接MD，ME．H求证：（1）DEAB；（2）HMDMHEMEH．DMCGABOFE（第24题图）25．(本题满分14分)如图，直角梯形ABCD中，AD∥BC，BCD90°，且CD2AD，tanABC2，过点D作DE∥AB，交BCD的平分线于点E，连接BE．（1）求证：BCCD；（2）将△BCE绕点C，顺时针旋转90°得到△DCG，连接EG.．求证：CD垂直平分EG.（3）延长BE交CD于点P．AD求证：P是CD的中点．EGBC（第25题图）26．(本题满分14分)如图，抛物线yax2bx3与x轴交于A，B两点，与y轴交于C点，且经过点(2，3a)，对称轴是直线x1，顶点是M．（1）求抛物线对应的函数表达式；（2）经过C,M两点作直线与x轴交于点N，在抛物线上是否存在这样的点P，使以点P，A，C，N为顶点的四边形为平行四边形？若存在，请求出点P的坐标；若不存在，请说明理由；（3）设直线yx3与y轴的交点是D，在线段BD上任取一点E（不与B，D重合），经过A，B，E三点的圆交直线BC于点F，试判断△AEF的形状，并说明理由；（4）当E是直线yx3上任意一点时，（3）中的结论是否成立？（请直接写出结论）．yxAO1B3CM（第26题图）2009年烟台市初中学生学业考试数学试题参考答案及评分意见本试题答案及评分意见，供阅卷评分使用．考生若写出其它正确答案，可参照评分意见相应评分．一、选择题（本题共12个小题，每小题4分，满分48分）题号123456789101112答案BDCCBAABBBDC二、填空题（本题共6个小题，每小题4分，满分24分）113．14．215．1716．117．2018．①，③，④4三、解答题（本题共8个小题，满分78分）19．（本题满分6分）936解：18(32)0(12)2233322(12)1|12|．····························································2分2332212121．·································································4分2321····································································································6分220．（本题满分8分）1解：（1）···································································································1分21（2）········································································································3分3（3）根据题意，画树状图：·············································································6分开始第一次1234第二次1234123412341234（第20题图）由树状图可知，共有16种等可能的结果：11，12，13，14，21，22，23，24，31，32，33，34，41，42，43，44．其中恰好是4的倍数的共有4种：12，24，32，44．41所以，P（4的倍数）．·····································································8分164或根据题意，画表格：····················································································6分第一次1234第二次111121314221222324331323334441424344由表格可知，共有16种等可能的结果，其中是4的倍数的有4种，所以，41P（4的倍数）．··············································································8分16421．（本题满分8分）解：（1）a1(10%15%30%15%5%)25%．····································1分初一学生总数：2010%200（人）．·····························································2分（2）活动时间为5天的学生数：20025%50（人）．活动时间为7天的学生数：2005%10（人）．················································3分频数分布直方图（如图）人数6050403020102天3天4天5天6天7天时间（第21题图）····················4分（3）活动时间为4天的扇形所对的圆心角是360°30%108°．···························5分（4）众数是4天，中位数是4天．····································································7分（5）该市活动时间不少于4天的人数约是6000(30%25%15%5%)4500（人）．·················································8分22．（本题满分8分）D解：过点C作CE⊥AB于E．D90°6030°，ACD90°30°60°，CAD90°．1CD10，ACCD5．·························3分2A在Rt△ACE中，BC5AEACsinACE5sin30°，···············4分25BCEACcosACE5cos30°3，·············5分2（第22题图）在Rt△BCE中，5BCE45°，BECEtan45°3，····················································6分2555ABAEBE3(31)≈6.8（米）．222所以，雕塑AB的高度约为6.8米．···································································8分23．（本题满分10分）x解：（1）根据题意，得y(24002000x)84，502即yx224x3200．··········································································2分252（2）由题意，得x224x32004800．25整理，得x2300x200000．·····································································4分解这个方程，得x100，x200．································································5分12要使百姓得到实惠，取x200．所以，每台冰箱应降价200元．····························6分2（3）对于yx224x3200，2524当x150时，··········································································8分2225150y(24002000150)84250205000．最大值50所以，每台冰箱的售价降价150元时，商场的利润最大，最大利润是5000元．·········10分24．（本题满分10分）H（1）证明：连接OC，HCHG，HCGHGC．·························1分HC切⊙O于C点，1HCG90°，···········2分DOBOC，12，分M······································3CGHGC3，2390°．·······················4分ABOFBFG90°，即DE⊥AB．····························5分（2）连接BE．由（1）知DE⊥AB．EAB是⊙O的直径，（第24题图）BDBE．·······························································································6分BEDBME．·····················································································7分四边形BMDE内接于⊙O，HMDBED．···········································8分HMDBME．BME是△HEM的外角，BMEMHEMEH．······························9分HMDMHEMEH．····································································10分25．（本题满分14分）证明：（1）延长DE交BC于F．AD∥BC，AB∥DF，AADBF，ABCDFC．····························1分D在Rt△DCF中，tanDFCtanABC2，CDP2，即CD2CF．EGCFCD2AD2BF，BFCF．······················3分BC11FBCBFCFCDCDCD，22（第25题图）即BCCD．·······························································································4分（2）CE平分BCD，BCEDCE．由（1）知BCCD，CECE，△BCE≌△DCE，BEDE．·················6分由图形旋转的性质知CECG，BEDG，DEDG．·····································8分C，D都在EG的垂直平分线上，CD垂直平分EG．·····································9分（3）连接BD．由（2）知BEDE，12．AB∥DE．32．13．·······················································11分AD∥BC，4DBC．由（1）知BCCD．DBCBDC，4BDP．····························12分又BDBD，△BAD≌△BPD，DPAD．·······································13分11ADCD，DPCD．P是CD的中点．········································14分2228．（本题满分14分）3a4a2b3，解：（1）根据题意，得b··············2分1.y2aDa1，E解得b2.NAOx抛物线对应的函数表达式为yx22x3．········3分1N（2）存在．FP在yx22x3中，令x0，得y3．CM令y0，得x22x30，x1，x3．12（第26题图）A(1，0)，B(3，0)，C(0，3)．又y(x1)24，顶点M(1，4)．······························································5分容易求得直线CM的表达式是yx3．在yx3中，令y0，得x3．N(3，0)，AN2．···············································································6分在yx22x3中，令y3，得x0，x2．12CP2，ANCP．AN∥CP，四边形ANCP为平行四边形，此时P(2，3)．·····························8分（3）△AEF是等腰直角三角形．理由：在yx3中，令x0，得y3，令y0，得x3．直线yx3与坐标轴的交点是D(0，3)，B(3，0)．ODOB，OBD45°．······································································9分又点C(0，3)，OBOC．OBC45°．···········································10分由图知AEFABF45°，AFEABE45°．···································11分EAF90°，且AEAF．△AEF是等腰直角三角形．····························12分（4）当点E是直线yx3上任意一点时，（3）中的结论成立．·······················14分