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    广东省2009年初中毕业生学业考试数学试题(含答案).pdf

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    广东省2009年初中毕业生学业考试数学试题(含答案).pdf

    - 1 - 2009 年广东省初中毕业生学业考试 数学 说明: 1全卷共 4 页,考试用时100 分钟,满分为120 分 2答卷前,考生务必用黑色字迹的签字笔或钢笔在答题卡填写自己的准考证号、姓名、试 室号、座位号用2B 铅笔把对应该号码的标号涂黑 3选择题每小题选出答案后,用2B 铅笔把答题卡上对应题目选项的答案信息点涂黑,如 需改动,用像皮檫干净后,再选涂其他答案,答案不能答在试题上 4非选择题必须用黑色字迹钢笔或签字笔作答、答案必须写在答题卡各题目指定区域内相 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;不准使用铅笔和涂改液 不按以上要求作答的答案无效 5考生务必保持答题卡的整洁考试结束时,将试卷和答题卡一并交回 一、选择题(本大题5 小题,每小题3 分,共 15 分)在每小题列出的四个选项中,只有一 个是正确的,请把答题卡上对应题目所选的选项涂黑 14 的算术平方根是() A2±B2C2±D2 2计算 ( ) 2 3 a的结果是() A 6 a B 9 a C 5 a D 8 a 3如图所示几何体的主(正)视图是() 4 广东省 2009 年重点建设项目计划(草案) 显示,港珠澳大桥工程估算总投资726 亿元, 用科学记数法表示正确的是() A 10 7.26 10×元B 9 72.6 10×元 C 11 0.726 10×元D 11 7.26 10×元 5如图所示的矩形纸片,先沿虑线按箭头方向向右对折,接着将对折后的纸片沿虑线剪下 一个小圆和一个小三角形,然后将纸片打开是下列图中的哪一个() ABDC CDAB - 2 - 二、填空题(本大题5 小题,每小题4 分,共 20 分)请将下列各题的正确答案填写在答题 卡相应的位置上 6分解因式 3 28xx-= _ 7已知O的直径8AB=cm,C为O上的一点, 30BAC=°,则BC=_cm 8一种商品原价120 元,按八折(即原价的80%)出售, 则现售价应为_元 9在一个不透明的布袋中装有2 个白球和n个黄球, 它们除颜色不同外,其余均相同若从中随机摸出一个球,摸到黄球的概率是 4 5 ,则 n = _ 10用同样规格的黑白两种颜色的正方形瓷砖,按下图的方式铺地板,则第(3)个图形中 有黑色瓷砖_块,第n个图形中需要黑色瓷砖_块(用含n的 代数式表示) 三、解答题(一) (本大题5 小题,每小题6 分,共 30 分) 11 (本题满分6 分)计算 1 2 -+9-sin () 30 3+ 0 ° 12 (本题满分6 分)解方程 2 21 11xx = - - A B C O 第 7 题图 ( 1)(2)(3) 第 10题图 - 3 - 13 (本题满分6 分)如图所示,在平面直角坐标系中,一次函数1ykx=+的图象与反比例 函数 9 y x = 的图象在第一象限相交于点A过点A分别作 x轴、y轴的垂线,垂足为点B、 C如果四边形OBAC是正方形,求一次函数的关系式 14 (本题满分6 分)如图所示, ABC 是等边三角形,D点是 AC的中点,延长BC到E, 使CECD= (1)用尺规作图的方法,过D点作DMBE,垂足是M(不写作法,保留作图痕迹); (2)求证: BMEM= 第 14题图 A B C E D 第 13题图 A C O B x - 4 - 15 (本题满分6 分)如图所示,A、B两城市相距100km现计划在这两座城市间修筑一 条高速公路(即线段 AB) ,经测量,森林保护中心P在A城市的北偏东30° 和B城市的 北偏西 45°的方向上已知森林保护区的范围在以P点为圆心, 50km 为半径的圆形区域 内请问计划修筑的这条高速公路会不会穿越保护区为什么? (参考数据:31.73221.414,) 四、解答题(二) (本大题4 小题,每小题7 分,共 28 分) 16 (本题满分7 分)某种电脑病毒传播非常快,如果一台电脑被感染,经过两轮感染后就 会有 81 台电脑被感染 请你用学过的知识分析,每轮感染中平均一台电脑会感染几台电 脑?若病毒得不到有效控制,3 轮感染后,被感染的电脑会不会超过700 台? A 第 15 题图 B F E P 45° 30° - 5 - 17 (本题满分7 分)某中学学生会为了解该校学生喜欢球类活动的情况,采取抽样调查的 方法,从足球、乒乓球、篮球、排球等四个方面调查了若干名学生的兴趣爱好,并将调 查的结果绘制成如下的两幅不完整的统计图(如图 1,图 2,要求每位同学只能选择一种 自己喜欢的球类;图中用乒乓球、足球、排球、篮球代表喜欢这四种球类的某一种球类 的学生人数) ,请你根据图中提供的信息解答下列问题: ( 1)在这次研究中,一共调查了多少名学生? ( 2)喜欢排球的人数在扇形统计图中所占的圆心角是多少度? ( 3)补全频数分布折线统计图 18 (本题满分7 分)在菱形 ABCD中,对角线AC与BD相交于点O,56ABAC=, 过点D作DEAC交BC的延长线于点E ( 1)求BDE的周长; ( 2)点P为线段BC上的点,连接PO并延长交AD于点Q求证:BPDQ= A QD E B PC O 第 18 题图 图 2 人数 乒乓球 20%足球 排球 篮球 40% 50 40 30 20 10 O 项目 足球乒乓球篮球排球 图 1 第 17 题图 - 6 - 19 (本题满分7 分)如图所示,在矩形ABCD中,1220ABAC=,两条对角线相交 于点O以OB、OC为邻边作第1个平行四边形 1 OBB C;对角线相交于点 1 A;再以 11 AB 、 1 AC为邻边作第 2 个平行四边形 111 ABCC, 对角线相交于点 1 O; 再以 11 O B、 11 OC 为邻边作第3 个平行四边形 1121 O B B C依次类推 ( 1)求矩形ABCD的面积; ( 2)求第 1 个平行四边形 11 OBBC、第 2 个平行四边形 111 ABCC和第 6 个平行四边形的 面积 五、解答题(三) (本大题3 小题,每小题9 分,共 27 分) 20 (本题满分9 分) (1) 如图 1, 圆内接ABC中,ABBCCAOD=,、OE为O的半径,ODBC 于点F,OE AC 于点G, 求证:阴影部分四边形 OFCG的面积是ABC 的面积的 1 3 (2)如图 2,若 DOE 保持120°角度不变,求证:当 DOE 绕着O点旋转时,由两 条半径和ABC的两条边围成的图形(图中阴影部分)面积始终是ABC的面积的 1 3 第 20 题图 C E D B O F G A 图 1 D B O C E A 图 2 第 19 题图 A1 A2 B2 C2 C1B1 O1 DA B C O - 7 - 21 (本题满分9 分)小明用下面的方法求出方程 230 x- = 的解,请你仿照他的方法求 出下面另外两个方程的解,并把你的解答过程填写在下面的表格中 方程 换元法得新方 程 解新方程检验求原方程的解 230x -= 令 xt= , 则230t-= 3 2 t = 3 0 2 t = 3 2 x =, 所以 9 4 x= 230xx+-= 240 xx +-= - 8 - 22 (本题满分9 分)正方形 ABCD边长为 4,M 、N分别是 BC、CD上的两个动点, 当M点在BC上运动时,保持AM和MN垂直, (1)证明: RtRtABMMCN ; (2)设BMx=,梯形ABCN的面积为y,求y与x之间的函数关系式;当M点运 动到什么位置时,四边形 ABCN面积最大,并求出最大面积; (3)当M点运动到什么位置时RtRtABMAMN,求此时x的值 D M A B C 第 22 题图 N - 9 - 参考答案与评分建议 一、选择题(本大题5 小题,每小题3 分,共 15 分) 1B2A3B4A5C 二、填空题(本大题5 小题,每小题4 分,共 20 分) 6 ()() 222x xx+- 748969810 10,3 1n+ 三、解答题(一) (本大题5 小题,每小题6 分,共 30 分) 11解:原式 = 11 31 22 +-+ ···························································································4分 =4 ·········································································································6 分 12解:方程两边同时乘以 ()() 11xx+-,··································································2 分 2= () 1x-+ ,················································································································4 分 3x= -,·······················································································································5分 经检验: 3x = - 是方程的解·····················································································6 分 13解:(1)作图见答案13 题图, ·····················································2分 (2) ABCQ 是等边三角形, D是AC的中点 BD平分ABC(三线合一) , 2ABCDBE= ,··································································································4分 CECD=Q, CEDCDE= Q, 又 ACBCEDCDE= + Q 2ACBE= ,·······································································································5分 又 ABCACB= Q , 22DBCE =, DBCE= , BDDE= , 又DMBEQ, BMEM=············································································································6 分 A B E D C M 答案 13 题图 - 10- 14解:依题意可得:9xyOB OC=········································································2 分 又四边形 ABCD为正方形,所以3OCOB= 所以有 ()3 3A, ,···········································································································3 分 直线1ykx=+过点A,所以得 331k=+, 所以 2 3 k =····················································································································4 分 故有直线 2 1 3 yx=+ ······································································································6分 15解:过点P作PC ABC, 是垂足, 则 3045APCBPC=°,° , ·················································································2分 ACPC=·tan30BCPE=°,· tan45°, ACBCAB+=Q ,······································································································4分 PC·tan30PC+°·tan45°=100, 3 1100 3 PC ? += ? ? ? ,································································································5分 () () 50 335031.73263.450PC=-×- 答:森林保护区的中心与直线AB的距离大于保护区的半径,所以计划修筑的这条高速公路 不会穿越保护区·········································································································6 分 四、解答题(二) (本大题4 小题,每小题7 分,共 28 分) 16 解:设每轮感染中平均每一台电脑会感染x台电脑, ············································1 分 依题意得: 1+ ()181 xx x+= , () 2 181x+=,··············································································································3分 19x+= 或 19x+ = - , 1 8 x = 或 2 10x = - (舍去),·························································································5 分 ()() 33 118729700x+=+=··················································································6 分 答:每轮感染中平均每一台电脑会感染8 台电脑, 3 轮感染后, 被感染的电脑会超过700 台 P 答案 15 题图 F BCA E - 11- ·····································································································································7 分 17解:(1)20 20%÷ =100(人) ·············································································1分 (2) 3 100%30% 100 ×= ,··························································································2 分 120%40%30%10%-=, 36010%36×=°°······································································································3分 (3)喜欢篮球的人数:40% 10040×= (人) ,·····························································4分 喜欢排球的人数:10%10010×=(人) ······································································5 分 ·····················································7分 18 (1)因为四边形ABCD为菱形, 所以BE ADACDE, , 故四边形 ABCD为平行四边形, 则有 5ABADBCCE= ,所以 10BEBCCE=+= ,·······································1 分 6ACDE=,···········································································································2 分 又 6 11 35 22 OAACABOA ? ? = ? ? ? ? ,垂直于OB, 所以在RtABC中有 222 ABOBOA=+, 所以 1 48 2 OBBDBD=, ,···················································································3分 故三角形BDE的周长为 861024BDDEBE+=+= ··················································································4 分 (2)因为四边形ABCD为菱形, 所以OBODBEAD=,则DBC=DOQ 又 BOPDOQ= , 所以BOP全等于DOQ ·························································································6分 故有BPDQ=··············································································································7 分 5OF =时,CD与O相切于F点, O 20 30 40 10 50 答案 17 题图 足球 乒乓球 篮球 排球 项目 人数 - 12- 即 310 3 5 23 mm=,···························································································6 分 当 10 3 3 m = 时,CD与 O 相切 ····································································7分 19解:(1)在RtABC中, 2222 201216BCACAB=-=-= , 12 16192 ABCD SAB BC=×= 矩形 ·············································································2分 (2)Q矩形 ABCD,对角线相交于点O, 4 ABCDOBC SS= ,·····································································································3 分 Q四边形 1 OBBC是平行四边形, 11 OBCBOCBB, , 11 OBCBCBOCBBBC=, 又 BCCB=Q , 1 OBCB CB , 1 1 296 2 OBB COBCABCD SSS= ,·············································································5分 同理, 1111 111 48 222 A B C COBB CABCD SSS=× ×= , ···························································6 分 第 6 个平行四边形的面积为 6 1 3 2 ABCD S=···································································7 分 五、解答题(三) (本大题3 小题,每小题9 分,共 27 分) 20证明:( 1)如图 1,连结 OA、OC, 因为点O是等边三角形 ABC的外心, 所以Rt RtRtOFCOGCOGA ······················2 分 2 OFCGOFCOAC SSS= , 因为 1 3 OACABC SS= , 所以 1 3 OFCGABC SS= ·································································································4分 (2)解法一: 连结OA、OB和OC,则 12AOCCOBBOA = , ,··························5分 不妨设OD交BC于点 FOE, 交AC于点G, 3412054120AOCDOE= + = + =°,° , 答案 20 题图( 1) A E O G F BC D - 13- 35 = ,················································································································7分 在 OAG 和 OCF 中 12 35 OAOC = ? ? = ? ? = ? OAGOCF ,··················································8分 1 3 OFCGAOCABC SSS= ·······································9分 解法二: 不妨设OD交BC于点F,OE交AC于点G, 作OH BCOKAC, , 垂足分别为点H、K,·················································5分 在四边形 HOKC中, 9060OHCOKCC= =°,°, 360909060120HOK=° -° -° -° =°,················6分 即 12120 + =° , 又23120GOF= + =Q°, 13 = ,················································································································7 分 ACBC=Q, OHOK=, OGKOFH ,·································································································8分 1 3 OFCGOHCKABC SSS= ·······················································································9 分 21解: 方程 换元法得新方 程 解新方程检验求原方程的解 230xx+-= 令xt = ,则 2 230tt+-= 1 分 1213tt= -, 2 分 1 10 t = , 2 30t = - , 220 t = - (舍去) 8 分 21x-= ,所以 213xx-=, 9 分 22解:(1)在正方形ABCD中, 490ABBCCDBC= = =,° , AMMNQ, 90AMN=°, 答案 20 题图( 2) A E O G F B C D 1 2 3 4 5 答案 20 题图( 3) A E O G F B C D 1 3 2 H K N D A C D B M 答案 22 题图 - 14- 90CMNAMB+ =°, 在Rt ABM 中, 90MABAMB+ =° , CMNMAB= , RtRtABMMCN , ···········································2 分 (2)RtRtABMMCNQ, 4 4 ABBMx MCCNxCN = - , , 2 4 4 xx CN -+ =,·····································································································4 分 () 2 2 21411 4428210 2422 ABCN xx ySxxx ?-+ =+= -+= -+ ? ? 梯形 · , 当2x =时,y取最大值,最大值为10·····································································10分 (3)90BAMN= =Q°, 要使ABMAMN,必须有 AMAB MNBM = ,······················································7分 由( 1)知 AMAB MNMC =, BMMC=, 当点M 运动到BC的中点时, ABMAMN ,此时 2 x = ····························9分 (其它正确的解法,参照评分建议按步给分)

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