2009年广东省肇庆市中考数学试题和答案

肇庆市2009年初中毕业生学业考试数 学 试 题说明：全卷共 4 页，考试时间为 100 分钟，满分 120 分．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4个选项中，只有一项是符合题目要求的．）1．2008 年肇庆市工业总产值突破千亿大关，提前两年完成“十一五”规划预期目标．用科学记数法表示数 1 千亿，正确的是（ ） A ．1000×108 B ．1000×109 C ．1011 D ．1012 2．实数2-，0.3，17，π-中，无理数的个数是（ ）A ．2B ．3C ．4D ．5 3．下列图形中，不是轴对称图形的是（ ）A ．等边三角形B ．平行四边形C ．圆D ．等腰梯形4．如图1是1998年参加国际教育评估的15个国家学生的数学平均成绩的统计图，则平均成绩大于或等于60的国家个数是（ ）A ．4B ．8C ．10D ．125．某几何体的三视图如图2，则该几何体是（ ）A ．球B ．圆柱C ．圆锥D ．长方体 6．函数y =x 的取值范围是（ ）A ．2x >B ．2x <C ．2x ≥D ．2x ≤ 7．若分式33x x -+的值为零，则x 的值是（ ）A ．3B ．3-C ．3±D ．08．如图3，R t A B C △中， 90A C B ∠=°，DE 过点C ，且D E AB ∥，若 55A C D ∠=°，则∠B 的度数是（ ） A ．35° B ．45° C ．55° D ．65°主视图俯视图左视图图2图1成绩A BC DE图3图49．如图 4，⊙O 是正方形 ABCD 的外接圆，点 P 在⊙O 上，则∠APB 等于（ ） A ． 30° B ． 45° C ． 55° D ． 60°10．若1O ⊙与2O ⊙相切，且125O O =，1O ⊙的半径12r =，则2O ⊙的半径2r 是（ ） A ． 3 B ． 5 C ． 7 D ． 3 或7 二、填空题（本大题共 5 小题，每小题 3 分，共15 分．）11．在平面直角坐标系中，点(23)P -，关于原点对称点P '的坐标是 ． 12．某校九年级（2）班（1）组女生的体重（单位：kg ）为：38，40，35，36，65，42， 42，则这组数据的中位数是 ．13．75°的圆心角所对的弧长是2.5π，则此弧所在圆的半径为 ． 14．若正六边形的边长为2，则此正六边形的边心距为 ． 15．观察下列各式：11111323⎛⎫=- ⎪⨯⎝⎭，111135235⎛⎫=- ⎪⨯⎝⎭，111157257⎛⎫=- ⎪⨯⎝⎭，…，根据观察计算：1111133557(21)(21)n n ++++⨯⨯⨯-+＝ ．（n 为正整数）三、解答题（本大题共10小题，共75分．解答应写出文字说明，证明过程或演算步骤．） 16．（本小题满分 6 分）计算：11|sin 452-⎛⎫+-+ ⎪⎝⎭°17．（本小题满分 6 分）2008 年北京奥运会，中国运动员获得金、银、铜牌共 100 枚，金牌数位列世界第一． 其 中金牌比银牌与铜牌之和多 2 枚，银牌比铜牌少 7 枚．问金、银、铜牌各多少枚？18．（本小题满分 6 分）掷一个骰子，观察向上一面的点数，求下列事件的概率： （1）点数为偶数；（2）点数大于 2 且小于5．如图 5，ABCD 是菱形，对角线AC 与BD 相交于O ，306A C D B D ∠==°，． （1）求证：△ABD 是正三角形；（2）求 AC 的长（结果可保留根号）．20．（本小题满分 7 分）已知20082009x y ==，，求代数式22x y xy y x xx ⎛⎫--÷- ⎪⎝⎭的值．21．（本小题满分 7 分）如图 6，ABCD 是正方形．G 是 BC 上的一点，DE ⊥AG 于 E ，BF ⊥AG 于 F ． （1）求证：A B F D A E △≌△； （2）求证：D E EF FB =+．22．（本小题满分 8 分）如图 7，已知一次函数1y x m =+（m 为常数）的图象与反比例函数 2k y x=（k 为常数，0k ≠）的图象相交于点 A （1，3）．（1）求这两个函数的解析式及其图象的另一交点B 的坐标； （2）观察图象，写出使函数值12y y ≥的自变量x 的取值范围．O D CB A图5ADEF CGB图6图7如图 8，在A B C △中，36A B A C A =∠=，°，线段 AB 的垂直平分线交 AB 于 D ，交 AC 于 E ，连接BE ．（1）求证：∠CBE =36°；（2）求证：2AE AC EC = ．24．（本小题满分 10 分）已知一元二次方程2 10x px q +++=的一根为 2． （1）求q 关于p 的关系式；（2）求证：抛物线2 y x px q =++与x 轴有两个交点；（3）设抛物线2y x px q =++的顶点为 M ，且与 x 轴相交于A （1x ，0）、B （2x ，0）两点，求使△AMB 面积最小时的抛物线的解析式．25．（本小题满分 10 分）如图 9，O ⊙的直径2 A B A M =，和B N 是它的两条切线，D E 切O ⊙于E ，交AM 于D ， 交BN 于C ．设AD x BC y ==，． （1）求证：A M B N ∥； （2）求y 关于x 的关系式；（3）求四边形A B C D 的面积S ，并证明：2S ≥．A E CBD图8CB 图9肇庆市2009年初中毕业生学业考试 数学试题参考答案和评分标准一、选择题（本大题共 10 小题，每小题3分，共30 分．） 题号12345678 9 10 答案 C A B D B C A A BD二、填空题（本大题共 5 小题，每小题 3 分，共15 分．）题号 11 12 131415答案(23)-，40621nn +三、解答题（本大题共10小题，共75分．） 16．（本小题满分 6 分）解：原式212=⨯+ ··················································································· （4分） 1= ·········································································································· （6分） 17．（本小题满分 6 分）解：设金、银牌分别为x 枚、y 枚，则铜牌为(7)y +枚，·····································（1 分）依题意，得(7)100(7) 2.x y y x y y +++=⎧⎨=+++⎩，········································································ （3分）解以上方程组，得5121x y ==，， ······································································（5 分） 所以 721728y +=+=．答：金、银、铜牌分别为51枚、21枚、28 枚． ··················································· （6分）18．（本小题满分 6 分）解：掷一个骰子，向上一面的点数可能为 1，2，3，4，5，6，共6 种. 这些点数出现的可能性相等． （1）点数为偶数有3种可能，即点数为 2，4，6， ∴P （点数为偶数）3162==； ··············································································（3 分）（2）点数大于2 且小于5有2种可能，即点数为 3，4， ∴P （点数大于2且小于5）2163=．·····································································（6 分）19．（本小题满分 7 分）（1）证明：∵AC 是菱形ABCD 的对角线， ∴AC 平分∠BCD ．又∠ACD=30°，∴∠BCD =60°． ····································（1 分）∵∠BAD 与∠BCD 是菱形的一组对角，∴∠BAD =∠BCD =60°． ···················································（2 分） ∵AB 、AD 是菱形的两条边，∴AB AD =．··························································（3 分）ODCB A图5∴△ABD 是正三角形． ··························································································（4 分） （2）解：∵O 为菱形对角线的交点， ∴123902A C O C O DB DC OD ===∠=，，°． ··············································· （5分） 在R t C O D △中，tan tan 30O D O C D O C=∠=°，∴tan 303OD OC ===°， ··········································································· （6分）∴2AC O C ==A C的长为 ························································· （7分） 20．（本小题满分 7 分） 解：22x y xy y x xx ⎛⎫--÷- ⎪⎝⎭222x y x xy y x x --+=÷ ············································ （2分） 2()x y x x x y -=⨯- ····················································· （4分）1x y=- ····································································· （5分）∵20082009x y ==，，∴原式11120082009x y===---． ··························· （7分） 21．（本小题满分 7 分）证明：（1）∵DE ⊥AG ，BF ⊥AG ，∴∠AED =∠AFB =90°． ·············································· （1 分） ∵ABCD 是正方形，DE ⊥AG ，∴∠BAF +∠DAE =90°，∠ADE +∠DAE =90°，∴∠BAF =∠ADE ． ··················································· （2 分） 又在正方形ABCD 中，AB =AD ． ··································· （3 分） 在△ABF 与△DAE 中，∠AFB =∠DEA =90°， ∠BAF =∠ADE ，AB =DA ，∴△ABF ≌△DAE ． ······················································ （5 分）（2）∵△ABF ≌△DAE ，∴AE =BF ，DE =AF ． ··················································· （6 分） 又 AF=AE+EF ，∴AF=EF+FB ，∴DE=EF+FB ． ················································· （7 分） 22．（本小题满分 8 分）解：（1）由题意，得31m =+， ··········································································（1 分） 解得2m =，所以一次函数的解析式为12y x =+．···············································（2 分） 由题意，得31k =， ·······························································································（3 分）ADE F CGB图6解得3k =，所以反比例函数的解析式为23y x=． ·············（4 分）由题意，得32x x+=，解得1213x x ==-，．·················· （5分）当23x =-时，121y y ==-，所以交点(31)B --，．·········（6 分） （2）由图象可知，当30x -<≤或1x ≥时，函数值12y y ≥．·······························································（8 分） 23．（本小题满分8分）证明：（1）∵DE 是A B 的垂直平分线，∴EA EB =，∴36EBA A ∠=∠=°． ······················································ （1 分） ∵36A B A C A =∠=，°，∴72A B C C ∠=∠=°． ··········· （2 分） ∴36C B E A B C E B A ∠=∠-∠=°．··································· （3 分） （2）由（1）得，在△BCE 中，7236C C B E ∠=∠=°，°， ∴72B E C C ∠=∠=°，∴B C B E A E ==．····················· （4 分） 在△ABC 与△BEC 中，C B E A ∠=∠，C C ∠=∠，∴A B C B E C △∽△． ······················································· （6 分） ∴A CBC B CE C=，即2 BC AC EC = ． ···································································· （7分）故2 AE AC EC = ． ····························································································· （8分） 24．（本小题满分 10 分）（1）解：由题意，得22210p q +++=，即(25)q p =-+． ······························（2 分） （2）证明：∵一元二次方程20x px q ++=的判别式24p q ∆=-，由（1）得2224(25)820(4)40p p p p p ∆=++=++=++>，························（3 分） ∴一元二次方程20x px q ++=有两个不相等的实根． ·········································（4 分） ∴抛物线2y x px q =++与x 轴有两个交点． ·······················································（5 分） （3）解：抛物线顶点的坐标为2424p q p M ⎛⎫-- ⎪⎝⎭，，············································· （6分） ∵12x x ，是方程20x px q ++=的两个根，∴1212.x x p x x q +=-⎧⎨=⎩，∴12||||AB x x =-==··········································· （7分）∴22141||(4248AM B q p S AB p q -==-△， ········································ （8分）A E CBD 图8要使AM B S △最小，只须使24p q -最小．而由（2）得224(4)4p q p -=++，所以当4p =-时，有最小值4，此时AM B S △13q ==，．······································ （9分） 故抛物线的解析式为243y x x =-+． ·································································（10分） 25．（本小题满分 10 分）（1）证明：∵AB 是直径，AM 、BN 是切线， ∴A M A B B N A B ⊥，⊥，∴A M B N ∥． ················· （2 分） 解：（2）过点D 作 D F BC ⊥于F ，则AB D F ∥．由（1）A M B N ∥，∴四边形A B F D 为矩形. ∴2D F AB ==，BF AD x ==．································ （3 分） ∵DE 、DA ，CE 、CB 都是切线， ∴根据切线长定理，得D E D A x ==，CE CB y ==． ··································· （4 分）在R t D FC △中，2DF DC DE CE x y CF BC BF y x ==+=+=-=-，，，∴222()2()x y y x +=+-，··················································································（5 分） 化简，得1(0)y x x=>． ······················································································· （6分）（3）由（1）、（2）得，四边形的面积111()222S AB AD BC x x ⎛⎫=+=⨯⨯+ ⎪⎝⎭， 即1(0)S x x x=+>． ··························································································· （8分）∵211220x x x x ⎛⎛⎫+-=-+= ⎪⎝⎭⎝≥，当且仅当1x =时，等号成立．∴12x x+≥，即2S ≥． ···················································································（10分）CB图9F。