2009年浙江省宁波市中考数学试题含参考答案及评分标准(word版)

2009年浙江省宁波市中考数学试题及答案一、选择题（每小题3分，共36分，在每小题给出的四个选项中，只有一项符合要求） 1．下列四个数中，比0小的数是 （ ）A ．23B C ．π D ．1-2．等腰直角三角形的一个底角的度数是 （ ） A ．030 B ．045 C ．060 D ．0903．一个不透明的布袋装有4个只有颜色的球，其中2个红色，1个白色，1个黑色，搅匀后从布袋里摸出1个球，摸到红球的概率是 （ ） A ．12B ．13C ．14D ．164．据《宁波市休闲基地和商务会议基地建设五年行动计划》，预计到2012年，宁波市接待游客容量将达到4640万人，其中4640万用科学记数法可表示为 （ ） A ．90.46410⨯ B ．84.6410⨯ C ．74.6410⨯ D ．746.410⨯5x 的取值范围是 （ ）A ．B ．C ．D ．A ．2x ≠B ．2x >C ．2x ≤D ．2x ≥6．如图是由４来个立方块组成的立体图形，它的俯视图是 （ ）7．下列调查适合作普查的是 （ ） A ．了解在校大学生的主要娱乐方式． B ．了解宁波市居民对废电池的处理情况． C ．日光灯管厂要检测一批灯管的使用寿命．D ．对甲型H1N1流感患者的同一车厢乘客进行医学检查． 8．以方程组21y x y x =-+⎧⎨=-⎩的解为坐标的点(,)x y 在平面直角坐标系中的位置是（ ）A ．第一象限B ．第二象限C ．第三象限D ．第四象限9．如图，1∠、2∠、3∠、4∠是五边形ABCD 的外角，且0123470∠=∠=∠=∠=， 则A E D ∠的度数是 （ ）A ．0110 B ．0108 C ．0105 D ．010010、反比例函数k y x=在第一象限的图象如图所示，则k 的值可能是（ ）A ．1B ．2C ．3D ．411．如图，菱形ABCD 中，对角线AC 、BD 相交于点O ，M 、N 分别是边AB 、AD 的中点，连结OM 、ON 、MN ，则下列叙述正确的是 （ ）A ．△AOM 和△AON 都是等边三角形B ．四边形MBON 和四边形MODN 都是菱形C ．四边形AMON 和四边形ABCD 都是位似图形 D ．四边形MBCO 和四边形NDCO 都是等腰梯形12．如图，点A 、B 、C 、D 在一次函数2y x m =-+的图象上，它们的横坐标依次为-1、1、2，分别过这些点作x 轴与y 轴的垂线，则图中阴影部分的面积这和是 （ ）A ．1B ．3C ．3(1)m -D ．3(2)2m -二、填空题（每小题3分，共18分） 13．实数8的立方根是 ．14．不等式组6020x x -<⎧⎨->⎩的解是 ．15．甲、乙、丙三名射击手的20次测试的平均成绩都是8环，方差分别是220.4()s =甲环223.2()s =已环，221.6()s =丙环，则成绩比较稳定的是 ．（填甲，乙，丙中的一个）16．如图，在坡屋顶的设计图中，AB=AC ，屋顶的宽度l 为10米，坡角α这35°，则坡屋顶的高度h 为 米．（结果精确到0．1米）17．如图，梯形ABCD 中，A D ∥BC ，∠B=70°，∠C=40°，作DE ∥AB 交BC 于点E ，若AD=3，BC=10，则CD 的长是 18．如图，⊙A 、⊙B 的圆心A 、B 在直线l 上，两圆半径都为1cm ，开始时圆心距AB=4cm ，现⊙A 、⊙B 同时沿直线l 以每秒2cm 的速度相向移动，则当两圆相切时，⊙A 运动的时间为 秒三、解答题（第19～21题各题6分，第22题20分，第23～24题各题8分，第25题10分，第26题12分，共66分）19．先化简，再求值：(2)(2)(2)a a a a-+--，其中1a=-20．如图，点A，B在数轴上，它们所对应的数分别是-4，2235xx+-，且点A、B到原点的距离相等，求x的值21．（1）如图1，把等边三角形的各边三等分，分别以居中那条线段为一边向外作等边三角形，并去掉居中的那条线段，得到一个六角星，则这个六角星的边数是（2）如图2 ，在55⨯的网格中有一个正方形，把正方形的各边三等分，分别以居中那条线段为一边向外作正方形，去掉居中的那条线段，请把得到的图画在图3中，并写出这个图形的边数（3）现有一个正五边形，把正五边形的各边三等分，分别以居中的那条线段为边向外作正五边形，并去掉居中的那条线段，得到的图的边数是多少？22．2009年宁波市初中毕业生升学体育集中测试项目包括体能（耐力）类项目和速度（跳跃、力量、技能）类项目．体能类项目从游泳和中长跑中任选一项，速度类项目从立定跳远、50米跑等6项中任选一项．某校九年级共有200名女生在速度类项目中选择了立定跳远，现从这200名女生中随机抽取10名女生进行测试，下面是她们测试结果的条形图．（另附：九年级女生立定跳远的计分标准）(1)求这10名女生在本次测试中，立定跳远距离．．的极差，立定跳远得．．分的众数和平均数．(2)请你估计该校选择立定跳远的200名女生得满分的人数．23．如图抛物线254y ax x a =-+与ｘ轴相交于点Ａ、Ｂ，且过点Ｃ（５，４）． (1)求a 的值和该抛物线顶点P 的坐标． (2)请你设计一种．．平移的方法，使平移后抛物线的顶点落要第二象限，并写出平移后抛物线的解析式．24．已知：如图，⊙O 的直径AB 与弦CD 相交于Ｅ，弧BC ＝弧BD ，⊙O 的切线BF 与弦AD 的延长线相交于点F ． (1)求证:C D ∥BF ．(2)连结BC,若⊙O 的半径为4,co s ∠BCD=34,求线段AD 、CD 的长．25．2009年4月7日，国务院公布了《医药卫生体制改革近期重点实施方案（2009～2011）》，某市政府决定2009年投入6000万元用于改善医疗卫生服务，比例2008年增加了1250万元．投入资金的服务对象包括“需方”（患者等）和“供方”（医疗卫生机构等），预计2009年投入“需方”的资金将比2008年提高30%，投入“供方”的资金将比2008年提高20%．（1）该市政府2008年投入改善医疗卫生服务的资金是多少万元？ （2）该市政府2009年投入“需方”和“供方”的资金是多少万元？（3）该市政府预计2011年将有7260万元投入改善医疗卫生服务，若从2009～2011年每年的资金投入按相同的增长率递增，求2009～2011年的年增长率．26．如图1，在平面直角坐标系中，O 为坐标原点，点A 的坐标为（－8，0），直线BC 经过点B （－8，6），将四边形OABC 绕点O 按顺时针方向旋转α度得到四边形OA ′B ′C ′，此时声母OA ′、直线B ′C ′分别与直线BC 相交于P 、Q ． （1）四边形的形状是 ，当α=90°时，BP PQ的值是 ．（2）①如图2,当四边形OA ′B ′C ′的顶点B ′落在y 轴正半轴上时,求BP PQ的值;②如图3,当四边形OA ′B ′C ′的顶点B ′落在直线BC 上时,求ΔOPB ′的面积． （3）在四边形OA B C 旋转过程中,当000180α<≤时，是否存在这样的点P 和点Q ，使BP=12B Q ？若存在，请直接写出点P 的坐标；基不存在，请说明理由．宁波市2009年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36分）二、填空题（每小题3分，共18分）三、解答题（共66分）注：1．阅卷时应按步计分，每步只设整分；2．如有其它解法，只要正确，都可参照评分标准，各步相应给分．19．解：原式2242a a a =--+ ··················································································· 2分24a =-． ······························································································ 4分当1a =-时， 原式2(1)4=⨯--6=- ····················································································································· 6分20．解：由题意得，22435x x +=-， ··············································································································· 3分 解得115x =． ··············································································································· 5分经检验，115x =是原方程的解．∴x 的值为115．····························································6分21．（1）12． ·······························································1分 （2）这个图形的边数是20． ·············4分（其中画图2分） （3）得到的图形的边数是30． ·····································6分22．（1）立定跳远距离的极差20517431(cm )=-=． ··············································· 2分 立定跳远距离的中位数199197198(cm )2+==． ························································· 4分根据计分标准，这10名女生的跳远距离得分分值分别是：7，9，10，10，10，8，10，10，9． 所以立定跳远得分的众数是10（分）， ········································································ 6分 立定跳远得分的平均数是9.3（分）． ··········································································· 8分 （2）因为10名女生中有6名得满分，所以估计200名女生中得满分的人数是620012010⨯=（人）．······················································································································10分 23．解：（1）把点(54)C ，代入抛物线254y ax ax a =-+得，252544a a a -+=， ·································································································· 1分解得1a =． ················································································································· 2分∴该二次函数的解析式为254y x x =-+．22595424y x x x ⎛⎫=-+=-- ⎪⎝⎭∴顶点坐标为5924P ⎛⎫- ⎪⎝⎭，．·························································································· 4分 （2）（答案不唯一，合理即正确）如先向左平移3个单位，再向上平移4个单位， ··························································· 6分 得到的二次函数解析式为225917342424y x x ⎛⎫⎛⎫=-+-+=++ ⎪ ⎪⎝⎭⎝⎭，即22y x x =++． ······································································································· 8分24．解：（1） 直径A B 平分 CD ， ∴A B C D ⊥． ············································································································ 1分 BF 与O ⊙相切，A B 是O ⊙的直径，AB BF ∴⊥． ············································································································· 2分 C D B F ∴∥． ············································································································· 3分（2）连结B D ，A B 是O ⊙的直径， 90AD B ∴∠=°， 在R t AD B △中，3cos cos 4A C ∠=∠=，428A B =⨯=．3cos 864A D AB A ∴=∠=⨯= ． ················································································ 5分A B C D ⊥于E ，在R t AED △3cos cos 4A C ∠=∠=，sin 4A ∠=．sin 64D E AD A ∴=∠=⨯=····································································· 7分直径A B 平分 CD ，2C D D E ∴== ································································································ 8分25．解：（1）该市政府2008年投入改善医疗服务的资金是：600012504750-=（万元） ······················································································ 2分 （2）设市政府2008年投入“需方”x 万元，投入“供方”y 万元，由题意得4750(130%)(120%)6000.x y x y +=⎧⎨+++=⎩，解得30001750.x y =⎧⎨=⎩， ··········································································································· 4分∴2009年投入“需方”资金为(130%) 1.330003900x +=⨯=（万元），2009年投入“供方”资金为(120%) 1.217502100y +=⨯=（万元）．答：该市政府2009年投入“需方”3900万元，投入“供方”2100万元．····················· 6分（3）设年增长率为x ，由题意得26000(1)7260x +=， ································································································ 8分 解得10.1x =，2 1.1x =-（不合实际，舍去）答：从2009~2011年的年增长率是10%． ····································································10分 26．解：（1）矩形（长方形）； ·················································································· 1分 47BP BQ=． ··················································································································· 3分（2）① P O C B O A ''∠=∠，P C O O A B ''∠=∠90=°，C O P A O B ''∴△∽△． C P O C A B O A ∴='''，即668C P =，92C P ∴=，72B P BC C P =-=． ·············································································· 4分同理B CQ B C O '''△∽△， C Q B C C QB C '∴='''，即10668C Q -=，3CQ ∴=，11BQ BC CQ =+=． ············································································· 5分 722BP BQ∴=． ·············································································································· 6分②在O C P △和B A P ''△中， 90O PC B PA O C P A O C B A ''∠=∠⎧⎪'∠=∠=⎨⎪''=⎩，°，， (AAS)OCP B A P ''∴△≌△． ····················································································· 7分 O P B P '∴=．设B P x '=，在R t O C P △中， 222(8)6x x -+=，解得254x =．·················································· 8分125756244O PB S '∴=⨯⨯=△． ······················································································· 9分（3）存在这样的点P 和点Q ，使12B P B Q =． ·························································10分 点P的坐标是19P ⎛⎫--⎪⎝⎭，2764P ⎛⎫- ⎪⎝⎭，． ··························································12分 对于第（3）题，我们提供如下详细解答，对学生无此要求． 过点Q 画QH OA '⊥于H ，连结O Q ，则QH OC OC '==，12P O Q S P Q O C =△，12P O Q S O P Q H =△，PQ OP ∴=．设B P x =，12BP BQ =，2BQ x ∴=，① 如图1，当点P 在点B 左侧时，3OP PQ BQ BP x ==+=，在R t PC O △中，222(8)6(3)x x ++=，解得11x =+，21x =-9P C B C B P ∴=+=+，19P ⎛⎫∴--⎪⎝⎭．②如图2，当点P 在点B 右侧时，OP PQ BQ BP x ∴==-=，8P C x =-．在R t PC O △中，222(8)6x x -+=，解得254x =．PC BC BP ∴=-257844=-=，2764P ⎛⎫∴- ⎪⎝⎭，．综上可知，存在点19P ⎛⎫--⎪⎝⎭，2764P⎛⎫-⎪⎝⎭，，使12B P B Q=．。