2009年广东省清远市中考数学试题含答案

年清远市初中毕业生学业考试数学科试题说明：1．全卷共4页，考试时间为100分钟，满分120分．2．选择题每小题选出答案后，用2B铅笔把答题卡上对应题的答案标号涂黑；如需改动，用橡皮擦干净后，再选涂其他答案标号；不能答在试卷上．3．非选择题必须用黑色字迹的钢笔或签字笔作答，涉及作图的题目，用2B铅笔画图，再用黑色字迹的钢笔或签字笔描黑．答案必须写在答题卡各题指字区域内的相应位置上；如需改动，先划掉原来的答案，然后再写上新的答案；改动的答案也不能超出指定的区域．不准使用铅笔、圆珠笔和涂改液．不按以上要求作答的答案无效．4．考生必须保持答题卡的清洁，考试结束后，将本试卷和答题卡一半交回．一、选择题（本大题共10小题，每小题3分，共30分）在每小题给出的四个选项中，只有一个是正确的，请将所选选项的字母涂在相应题号的答题卡上．1．5等于（）A．5B．5C．15D．152．不等式20x≤的解集在数轴上表示正确的是（）A．B．C．D．3．今年我国参加高考人数约为10200000，将10200000用科学记数法表示为（）A．710.210B．71.0210C．70.10210D．7102104．某物体的三视图如图1所示，那么该物体形状可能是（）A．圆柱B．球C．正方体D．长方体5．小明记录某社区七次参加“防甲型H1N1流感活动”的人数分别如下：33，32，32，31，32，28，26．这组数据的众数是（）A．28B．31C．32D．336．方程216x的解是（）A．4xB．4xC．4xD．16x7．已知O⊙的半径r，圆心O到直线l的距离为d，当dr时，直线l与O⊙的位置关系是（）A．相交B．相切C．相离D．以上都不对8．计算：23ab（）A．22abB．23abC．26abD．6ab3210123321012332101233210123图1主视图左视图俯视图．如图2，ABCD∥，EFAB于EEF，交CD于F，已知160°，则2（）A．20°B．60°C．30°D．45°10．如图3，AB是O⊙的直径，弦CDAB于点E，连结OC，若5OC，8CD，则tanCOE=（）A．35B．45C．34D．43二、填空题（本大题共6小题，每小题3分，共18分）请把下列各题的正确答案填写在相应题号的答题卡上．11．计算：3(2)=．12．当x时，分式12x无意义．13．已知反比例函数kyx的图象经过点(23)，，则此函数的关系式是．14．如果a与5互为相反数，那么a．15．如图4所示，转盘平面被等分成四个扇形，并分别填上红、黄两种颜色，自由转动这个转盘，当它停止转动时，指针停在黄色区域的概率为．16．如图5，若111ABCABC△≌△，且11040AB°，°，则1C=．三、解答题（本大题共5小题，每小题5分，共25分）17．在“我喜欢的体育项目”调查活动中，小明调查了本班30人，记录结果如下：（其中喜欢打羽毛球的记为A，喜欢打乒乓球的记为B，喜欢踢足球的记为C，喜欢跑步的记为D）ＡＡＣＢＡＤＣＣＢＣＡＤＤＣＣＢＢＢＢＣＢＤＢＤＢＡＢＣＡＢ求Ａ的频率．18．计算：201(1)π34．图4红红红黄ABCC1A1B1图5CDBAEF12图2ACDBEO图3．已知图形B是一个正方形，图形A由三个图形B构成，如右图所示，请用图形A与B合拼成一个轴对称图形，并把它画在答题卡的表格中．20．解分式方程：132xx21．如图6，某飞机于空中A处探测到地平面目标B，此时从飞机上看目标B的俯角为，若测得飞机到目标B的距离AB约为2400米，已知sin0.52，求飞机飞行的高度AC约为多少米？四、解答题（本大题共4小题，每小题6分，共24分）22．化简：222692693xxxxxx23．如图7，已知正方形ABCD，点E是AB上的一点，连结CE，以CE为一边，在CE的上方作正方形CEFG，连结DG．求证：CBECDG△≌△24．在一个不透明的口袋中装有红球2个、黑球2个，它们只有颜色不同，若从口袋中一次摸出两个球，求摸到两个都是红球的概率．（要求画出树状图）ABBCA图6EBCGDFA图7．已知二次函数2yaxbxc中的xy，满足下表：x…21012…y…40220…求这个二次函数关系式．五、解答题（本大题共3小题，第26小题7分，第27、28小题各8分，共23分）26．如图8，已知AB是O⊙的直径，过点O作弦BC的平行线，交过点A的切线AP于点P，连结AC．（1）求证：ABCPOA△∽△；（2）若2OB，72OP，求BC的长．27．某饮料厂为了开发新产品，用A种果汁原料和B种果汁原料试制新型甲、乙两种饮料共50千克，设甲种饮料需配制x千克，两种饮料的成本总额为y元．（1）已知甲种饮料成本每千克4元，乙种饮料成本每千克3元，请你写出y与x之间的函数关系式．（2）若用19千克A种果汁原料和17.2千克B种果汁原料试制甲、乙两种新型饮料，下表是试验的相关数据；每千克饮料果汁含量果汁甲乙A0.5千克0.2千克B0.3千克0.4千克请你列出关于x且满足题意的不等式组，求出它的解集，并由此分析如何配制这两种饮料，可使y值最小，最小值是多少？28．如图9，已知一个三角形纸片ABC，BC边的长为8，BC边上的高为6，B和C都为锐角，M为AB一动点（点M与点AB、不重合），过点M作MNBC∥，交AC于点N，在AMN△中，设MN的长为x，MN上的高为h．（1）请你用含x的代数式表示h．（2）将AMN△沿MN折叠，使AMN△落在四边形BCNM所在平面，设点A落在平面的点为1A，1AMN△与四边形BCNM重叠部分的面积为y，当x为何值时，y最大，最大值为多少？POACBBCNMA图9年清远市初中毕业生学业考试数学试题参考答案及评分标准一、选择题：（每小题3分，共30分）1．A2．B3．B4．A5．C6．A7．B8．C9．C10．D二、填空题：（每小题3分，共18分）11．6；12．2；13．6yx；14．5；15．14；16．30°三、解答题：（每小题5分，共25分）17．解：A的频率=61305···············································································5分18．解：原式=11123···············································································4分=13··············································································································5分19．解：拼成正确图形之一的给5分，例如20．解：去分母，得36xx·········································································2分解得：3x··································································································3分检验：把3x代入原方程得：左边=右边····························································4分所以3x是原方程的解···················································································5分21．解：由题意得：90BC，°sinsin0.52B≈···················································································2分sinACBABsin24000.521248ACABB·（米）····························5分答：飞机飞行的高度约为1248米．四、解答题：（每小题6分，共24分）22．解：原式=2(3)(3)(3)(3)2(3)xxxxxx······························································4分=(3)(3)22xxxx·······················································································6分23．证明：四边形ABCD和四边形CEFG都是正方形90CBCDCECGBCDECG，，°················································3分90BCEDCE°-90DCGDCE°-BCEDCG························································································5分CBECDG△≌△·······················································································6分．解：画出树状图为：·············································4分摸到两个都是红球的概率P=21126···································································6分25．解：把点(02)，代入2yaxbxc得2c·············································2分再把点(10)(20)，，，分别代入22yaxbx204220abab····························································································4分解得11ab这个二次函数的关系式为：22yxx························································6分（说明：其它解法可参照上述给分）五、解答题：（本大题共3小题，26题7分，27、28题各8分，共23分）26．（1）证明：BCOP∥AOPB·············································1分AB是直径90C°·················································2分PA是O⊙的切线，切点为A90OAP°COAP