2009年广东省清远市中考数学试题含答案

2009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 ．5 B ．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 ．28 B ．31 C ．32 D ．33 6．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d r =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 ．22a b B ．23a bC ．26a b D ．6ab3-3-3-3-图1主视图 左视图 俯视图9．如图2，AB CD ∥，E F A B ⊥于E EF ，交CD 于F ，已知160∠=°，则2∠=（ ） A ．20° B ．60° C ．30° D ．45°10．如图3，A B 是O ⊙的直径，弦CD AB ⊥于点E ，连结OC ，若5OC =，8CD =，则tan COE ∠=（ ） 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k y x=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111A B C ABC △≌△，且11040A B ∠=∠=°，°，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 ）Ａ Ａ Ｃ Ｂ Ａ ＤＣ Ｃ Ｂ Ｃ Ａ Ｄ Ｄ Ｃ ＣＢ Ｂ Ｂ Ｂ Ｃ Ｂ Ｄ Ｂ Ｄ Ｂ Ａ Ｂ Ｃ Ａ Ｂ求Ａ的频率．18．计算：21(1)π3--++-图4 A B C C 1 A 1 B 1 图5C D B A EF1 2 图2 A B 图319．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：132x x=-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α，若测得飞机到目标B 的距离A B 约为2400米，已知sin 0.52α=，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：222692693x x x xx x-+-÷-+23．如图7，已知正方形ABCD ，点E 是A B 上的一点，连结C E ，以C E 为一边，在C E 的上方作正方形CEFG ，连结DG ．求证：CBE CDG △≌△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）A BB C A α图6E BC GDFA 图725．已知二次函数2y ax bx c =++中的x y ，满足下表：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A B 是O ⊙的直径，过点O 作弦BC 的平行线，交过点A 的切线A P 于点P ，连结AC ．（1）求证：ABC POA △∽△； （2）若2OB =，72O P =，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；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A B 一动点（点M 与点A B 、不重合），过点M 作MN BC ∥，交AC 于点N ，在A M N △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A M N △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 ，1A M N△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？BCNM A 图92009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．A 2．B 3．B 4．A 5．C 6．A 7．B 8．C 9．C 10．D 二、填空题：（每小题3分，共18分） 11．6-； 12．2； 13．6y x=； 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36x x =- ····························································································· 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：90B C α∠=∠∠=，°sin sin 0.52B α∴=≈ ········································································································· 2分 sin AC B AB=s i n 24000.521A C AB B ∴==⨯=·（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)x x x x x x -+⨯+-- ··············································································· 4分=(3)(3)22x x x x --⨯=-- ·············································································································· 6分23．证明： 四边形ABCD 和四边形CEFG 都是正方形90CB CD CE CG BCD ECG ∴==∠=∠=，，° ····························································· 3分 90BCE DCE ∴∠=∠°- 90DCG DCE ∠=∠°-BCE DCG ∴∠=∠ ··············································································································· 5分 CBE CDG ∴△≌△ ·············································································································· 6分24．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2y ax bx c =++得2c =- ························································· 2分再把点(10)(20)-，，，分别代入22y ax bx =+- 204220a b a b --=⎧⎨+-=⎩···················································································································· 4分 解得11a b =⎧⎨=-⎩∴这个二次函数的关系式为：22y x x =-- ······································································· 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：BC OP ∥AOP B ∴∠=∠ ·························································· 1分 A B 是直径90C ∴∠=° ······························································· 2分 P A 是O ⊙的切线，切点为A90OAP ∴∠=°C OAP ∠=∠ ····························································· 3分 ABC POA ∴△∽△ ·················································· 4分 （2）ABC POA △∽△B C A B O AP O∴= ·························································································································· 5分722O B PO == ，24OA AB ∴==，4722B C ∴=······························································································································ 6分716827BC BC ∴==， ················································ (7)分开始 红 红 红 红红红红红黑黑黑 黑 黑 黑 黑 黑27．解：（1）依题意得：43(50)150y x x x =+-=+ ······················································· 3分（2）依题意得：0.50.2(50)19(1)0.30.4(50)17.2(2)x x x x +-⎧⎨+-⎩≤…………≤……… ···················································· 5分解不等式（1）得：30x ≤解不等式（2）得：28x ≥∴不等式组的解集为2830x ≤≤ ······················································································· 7分150y x =+ ，y 是随x 的增大而增大，且2830x ≤≤ ∴当甲种饮料取28千克，乙种饮料取22千克时，成本总额y 最小，28150178y =+=最小（元） ································································· 8分 28．解：（1）MN BC ∥ AMN ABC ∴△∽△68h x ∴=34x h ∴=····································································· 3分（2）1AM N A M N △≌△1A M N ∴△的边MN 上的高为h ，①当点1A 落在四边形BCNM 内或BC 边上时， 1A M N y S =△=211332248M N h x x x ==··（04x <≤） ························································ 4分②当1A 落在四边形BCNM 外时，如下图(48)x <<，设1A EF △的边E F 上的高为1h ， 则132662h h x =-=-11EF M NA EF A M N ∴ ∥△∽△11A M N ABC A EF ABC ∴ △∽△△∽△1216A E F S h S ⎛⎫= ⎪⎝⎭△△ABC168242A B C S =⨯⨯= △ 22363224122462E Fx S x x ⎛⎫- ⎪∴==⨯=-+ ⎪⎪⎝⎭1△A BCNM AMNCBEFAA 11122233912241224828A M N A EF y S S x x x x x ⎛⎫=-=--+=-+- ⎪⎝⎭△△ 所以 291224(48)8y x x x =-+-<< ············································································ 6分 综上所述：当04x <≤时，238y x =，取4x =，6y =最大当48x <<时，2912248y x x =-+-，取163x =，8y =最大86>∴当163x =时，y 最大，8y =最大 ······················································································ 8分。