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2009年上海市中考数学试题及答案 上海市初中试题及答案

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上海市初中试题及答案
2009年上海市初中毕业统一学业考试 数 学 卷 (满分150分,考试时间100分钟) 考生注意: 1.本试卷含三个大题,共25题; 2.答题时,考生务必按答题要求在答题纸规定的位置上作答,在草稿纸、本试卷上答题一律无效. 3.除第一、二大题外,其余各题如无特别说明,都必须在答题纸的相应位置上写出证明或计算的主要步骤. 一、选择题:(本大题共6题,每题4分,满分24分) 【下列各题的四个选项中,有且只有一个选项是正确的,选择正确项的代号并填涂在答题纸的相应位置上.】 1.计算(a)的结果是( ) A.a 532B.a 6C.a 8D.a 92.不等式组x10,的解集是( ) x21B.x3 C.1x3 D.3x1 A.x1 3.用换元法解分式方程x13xx110时,如果设y,将原方程化为关于y的xx1xB.y3y10 D.3yy10 22整式方程,那么这个整式方程是( ) A.yy30 C.3yy10 22 2 4.抛物线y2(xm)n(m,n是常数)的顶点坐标是( ) A.(m,n) B.(m,n) n) C.(m,n) D.(m,5.下列正多边形中,中心角等于内角的是( ) A.正六边形 B.正五边形 C.正四边形 C.正三边形 6.如图1,已知AB∥CD∥EF,那么下列结论正确的是( ) A C E 图1 B D F ADBC DFCECDBCC. EFBEA.BCDF CEADCDADD. EFAFB. 二、填空题:(本大题共12题,每题4分,满分48分) 【请将结果直线填入答题纸的相应位置】 港中数学网 7.分母有理化:1 . 58.方程x11的根是 . 9.如果关于x的方程xxk0(k为常数)有两个相等的实数根,那么k . 21,那么f(3) . 1x211.反比例函数y图像的两支分别在第 象限. x10.已知函数f(x)12.将抛物线yx2向上平移一个单位后,得以新的抛物线,那么新的抛物线的表达式是 . 13.如果从小明等6名学生中任选1名作为“世博会”志愿者,那么小明被选中的概率是 . 14.某商品的原价为100元,如果经过两次降价,且每次降价的百分率都是m,那么该商品现在的价格是 元(结果用含m的代数式表示). A 15.如图2,在△ABC中,AD是边BC上的中线,设向量ABa,2BCb,如果用向量a,b表示向量AD,那么AD= . 16.在圆O中,弦AB的长为6,它所对应的弦心距为4,那么半径OA . 17.在四边形ABCD中,对角线AC与BD互相平分,交点为在不添加任何辅助线的前提下,要使四边形ABCD成为矩形,O.B 还需添加一个条件,这个条件可以是 . 18.在Rt△ABC中,BAC90°,AB3,M为边BC上的点,联结AM(如图3所示).如果将△ABM沿直线AM翻折后,点B恰好落在边AC的中点处,那么点M到AC的距离是 . 三、解答题:(本大题共7题,满分78分) 19.(本题满分10分) B A D 图2 C M 图3 C 2a2a21(a1)2计算:. a1a2a1 20.(本题满分10分) 解方程组: 港中数学网 yx1,2xxy20.2①② 21.(本题满分10分,每小题满分各5分) 如图4,在梯形ABCD中,AD∥BC,ABDC8,B60° ,BC12,联结AC.
(1)求tanACB的值;
(2)若M、N分别是AB、DC的中点,联结MN,求线段MN的长. A D B C 图4 22.(本题满分10分,第
(1)小题满分2分,第
(2)小题满分3分,第
(3)小题满分2分,第
(4)小题满分3分) 为了了解某校初中男生的身体素质状况,在该校六年级至九年级共四个年级的男生中,分别抽取部分学生进行“引体向上”测试.所有被测试者的“引体向上”次数情况如表一所示;各年级的被测试人数占所有被测试人数的百分率如图5所示(其中六年级相关数据未标出). 次数 人数 0 1 1 1 2 2 3 2 4 3 5 4 6 2 7 2 8 2 9 0 10 1 表一 根据上述信息,回答下列问题(直接写出结果): 八年级 九年级 25%
(1)六年级的被测试人数占所有被测试人数的百分率30% 是 ;
(2)在所有被测试者中,九年级的人数是 ; 七年级 25% 六年级
(3)在所有被测试者中,“引体向上”次数不小于6的人数所占的百分率是 ; 图5
(4)在所有被测试者的“引体向上”次数中,众数是 . 23.(本题满分12分,每小题满分各6分) D A 已知线段AC与BD相交于点O,联结AB、DC,E为OBO 的中点,F为OC的中点,联结EF(如图6所示).
(1)添加条件AD,OEFOFE, F E 求证:ABDC. B C 图6
(2)分别将“AD”记为①,“OEFOFE”记为②,“ABDC”记为③,添加条件①、③,以②为结论构成命题1,添加条件②、③,以①为结论构成命题2.命题1是 命题,命题2是 命题(选择“真”或“假”填入空格). 港中数学网 24.(本题满分12分,每小题满分各4分) 在直角坐标平面内,O为原点,点A的坐标为(14),直线CM∥x,0),点C的坐标为(0,轴(如图7所示).点B与点A关于原点对称,直线yxb(b为常数)经过点B,且与直线CM相交于点D,联结OD. yxb
(1)求b的值和点D的坐标; y
(2)设点P在x轴的正半轴上,若△POD是等腰三M D 角形,求点P的坐标; 4 C
(3)在
(2)的条件下,如果以PD为半径的圆P与3 圆O外切,求圆O的半径. 2 1 A B x 1 O 1 图7 25.(本题满分14分,第
(1)小题满分4分,第
(2)小题满分5分,第
(3)小题满分5分) 已知ABC90°为线段BD上的动点,点Q在射线,AB2,BC3,AD∥BC,PAB上,且满足PQAD(如图8所示). PCAB
(1)当AD2,且点Q与点B重合时(如图9所示),求线段PC的长;
(2)在图8中,联结AP.当AD为x,3,且点Q在线段AB上时,设点B、Q之间的距离2S△APQS△PBCy,其中S△APQ表示△APQ的面积,S△PBC表示△PBC的面积,求y关于x的函数解析式,并写出函数定义域;
(3)当ADAB,且点Q在线段AB的延长线上时(如图10所示),求QPC的大小. A D A P P Q B 图8 C (Q) B C 图9 Q B 图10 D A D P C 港中数学网 2009年上海市初中毕业统一学业考试 数学卷答案要点与评分标准 说明: 1. 解答只列出试题的一种或几种解法.如果考生的解法与所列解法不同,可参照解答中评分标准相应评分; 2. 第一、二大题若无特别说明,每题评分只有满分或零分; 3. 第三大题中各题右端所注分数,表示考生正确做对这一步应得分数; 4. 评阅试卷,要坚持每题评阅到底,不能因考生解答中出现错误而中断对本题的评阅.如果考生的解答在某一步出现错误,影响后继部分而未改变本题的内容和难度,视影响的程度决定后继部分的给分,但原则上不超过后继部分应得分数的一半; 5. 评分时,给分或扣分均以1分为基本单位. 一.选择题:(本大题共6题,满分24分) 1. B; 2.C; 3.A; 4.B; 5.C; 6.A. 二.填空题:(本大题共12题,满分48分) 5; 8.x2; 9.1; 10.1; 11.一、三; 425122112.yx1; 13.; 14.100(1m); 15.ab; 6216.5; 17.ACBD(或ABC90等); 18. 2. 7.三.解答题:(本大题共7题,满分78分) 2(a1)1(a1)(a1) ··········································· (7分) 2a1a1(a1)2a1 = ······································································· (1分) a1a11a = ·············································································· (1分) a1 =1. ················································································ (1分) 20.解:由方程①得yx1, ③ ························································ (1分) 19.解:原式=将③代入②,得2xx(x1)20, ·········································· (1分) 整理,得xx20, ······························································ (2分) 解得x12,x21, ·································································· (3分) 分别将x12,x21代入③,得y13,y20, ·························· (2分) 22x21, ····································· (1分) y0.221.解:
(1) 过点A作AEBC,垂足为E. ··········································· (1分) 在Rt△ABE中,∵B60,AB8, ∴BEABcosB8cos604, ·············································· (1 分) 所以,原方程组的解为x12, y3;1AEABsinB8sin6043. ·················································· (1分) ∵BC12,∴EC8. ······························································· (1 分) 在Rt△AEC中,tanACBAE433. ··································· (1分) 港中数学网
(2) 在梯形ABCD中,∵ABDC,B60, ∴DCBB60. ········································································ (1分) 过点D作DFBC,垂足为F,∵DFCAEC90,∴AE//DF. ∵AD//BC,∴四边形AEFD是平行四边形.∴ADEF. ···················· (1分) 在Rt△DCF中, FCDCcosDCF8cos604, ···················· (1分) ∴EFECFC4.∴AD4. ∵M、N分别是AB、DC的中点,∴MNADBC412······ (2分) 8. ·22 22.
(1) 20%; ················································································· (2分)
(2) 6; ··················································································· (3分)
(3) 35%; ················································································ (2分)
(4) 5. ······················································································ (3分) 23.
(1) 证明:OEFOFE, ∴OEOF. ··································································· (1分) ∵E为OB的中点,F为OC的中点, ∴OB2OE,OC2OF. ············································· (1分) ∴OBOC. ··································································· (1分) ∵AD,AOBDOC, ∴△AOB≌△DOC. ························································ (2分) ABDC. ··································································· (1分)
(2) 真; ························································································ (3分) 假. ··························································································· (3分) 24.解:
(1) ∵点A的坐标为(1,0),点B与点A关于原点对称, ∴点B的坐标为(1································································ (1分) ,0). ·∵直线yxb经过点B,∴1b0,得b1. ··························· (1分) ∵点C的坐标为(0,······· (1分) 4),直线CM//x轴,∴设点D的坐标为(x,4). ∵直线yx1与直线CM相交于点D,∴x3.∴D的坐标为(3, 4).…(1分)4),∴OD5. ·
(2) ∵D的坐标为(3,·············································· (1分) 当PDOD5 时,点P的坐标为(6,···································· (1分) 0); ·当POOD5 时,点P的坐标为(5,···································· (1分) 0), ·0)(x0), 当POPD 时,设点P的坐标为(x,2525,∴点P的坐标为(,·········· (1分) 0). ·6625综上所述,所求点P的坐标是(6,0). 0)、(5,0)或(,6
(3) 当以PD为半径的圆P与圆O外切时, 0),则圆P的半径PD5,圆心距PO6, 若点P的坐标为(6,∴圆O的半径r1. ····································································· (2分) 0),则圆P的半径PD25,圆心距PO5, 若点P的坐标为(5,∴x(x3)242,得x∴圆O的半径r525. ·························································· (2分) 综上所述,所求圆O的半径等于1或525. 港中数学网 25.解:
(1) ∵AD//BC, ∴ADBDBC. ∵ADAB2,∴ABDADB.∴DBCABD. ∵ABC90.∴PBC45. ················································ (1分) PQAD,ADAB,点Q与点B重合,∴PBPQPC. PCAB∴PCBPBC45. ······························································ (1分) ∴BPC90. ········································································· (1分) 32在Rt△BPC中,PCBCcosC3cos45. ···················· (1分) 2
(2) 过点P作PEBC,PFAB,垂足分别为E、F. ···················· (1分) ∴PFBFBEBEP90.∴四边形FBEP是矩形. ∴PF//BC,PEBF. PFAD∵AD//BC,∴PF//AD.∴. BFAB3PF3∵AD,AB2,∴··············································· (1分) . ·2PE42x3∵AQABQB2x,BC3,∴S△APQPF,S△PBCPE. 22SAPQ2x2x∴,即y . ················································· (2分) 4SPBC47函数的定义域是0≤x≤. ··························································· (1分) 8
(3) 过点P作PMBC,PNAB,垂足分别为M、N. 易得四边形PNBM为矩形,∴PN//BC,PMBN,MPN90. PNADPNAD∵AD//BC,∴PN//AD.∴.∴. ·············· (1分) BNABPMABPQADPNPQ∵,∴. ······················································ (1分) PCABPMPC又∵PMCPNQ90,∴Rt△PCM∽Rt△PQN. ··············· (1分) ∴CPMQPN. ··································································· (1分) ∵MPN90,∴CPMQPMQPNQPMMPN90, 即QPC90. ········································································· (1分) ∵ 港中数学网 上海市初中试题及答案。
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