2008年从化市九年级数学一模参考答案

从化市2008年初中毕业生综合测试数学试题参考答案第1页共5页2008年从化市初中毕业生综合测试数学试题参考答案说明：1．参考答案与评分标准指出了每道题要考查的主要知识和能力，并给出了一种或几种解法供参考，如果考生的解法与参考答案不同，可根据试题主要考查的知识点和能力比照评分标准给以相应的分数．2．对解答题中的计算题，当考生的解答在某一步出现错误时，如果后继部分的解答未改变该题的内容和难度，可视影响的程度决定后继部分的得分，但所给分数不得超过该部分正确解答应得分数的一半；如果后继部分的解答有较严重的错误，就不再给分．3．解答右端所注分数，表示考生正确做到这一步应得的累加分数．4．只给整数分数，选择题和填空题不给中间分．一、选择题：本题考查基本知识和基本运算，每小题3分，满分30分．题号12345678910答案DBADCBADCA二、填空题：本题考查基本知识和基本运算，每小题3分，满分18分．题号111213141516答案71082.36x45°3123,121xx三、解答题：本题考查基本知识和基本运算，及数学能力，满分102分．17、（本题满分9分）本小题主要考查分式方程等基础知识，考查运算求解能力解：移项得：2321xx……………………………………1分方程两边同乘以)2)(2(xx，得)2(32xx……4分解这个方程，得4x………………………………………7分检验：将4x代入原方程，得左边6321右边…………8分所以，4x是原方程的根………………………………9分18、（本题满分9分）本小题主要考查平移和正比例函数等基础知识，考查运算求解能力解：（1）如图所示图正确……………………………………4分（2）由已知设直线OP的函数解析式为：y=kx………6分因为点P的坐标为（－2，3），代入，得3＝－2k……7分32k…………………………………………8分即直线OP的函数解析式为：32yx………………9分xy1-11PMNOM′N′O′P′从化市2008年初中毕业生综合测试数学试题参考答案第2页共5页19、（本题满分12分）本小题主要考查统计等有关知识，考查看图表与识图表能力解：（1）1－28%－34%=38%················································································2分（没有计算过程扣1分）（2）A=1－0.35－0.32－0.08=0.25······························································4分768÷0.32=2400·················································································5分B=2400－768－600－192=840································································7分∴A的值为0.25，B的值840·································································8分（没有计算过程扣2分）（3）408341200％··········································································10分240012002·············································································11分∴该校学生平均每人读2本课外书．························································12分（没有计算过程扣1分）20、（本题满分10分）本小题主要考查概率等基础知识，考查运算求解能力解：由已知得：共组成4组边，即2,3,5；3,3,5；3,4,5；3,5,5，……………………2分（1）依题意，3,3,5；3,4,5；3,5,5，有3组能构成三角形,····························4分∴43)(构成三角形P·········································································6分（2）依题意,3,3,5和5,3,5两组能构成等腰三角形·······································8分∴2142)(构成等腰三角形P·······························································10分21、（本题满分12分）本小题主要考查直角三角形、全等三角形和近似计算等基础知识，考查综合运用知识分析问题和解决问题的能力解：(1)∵AB切⊙O于点B,∴0B⊥AB,即∠B=90°···········································································1分∵CD⊥OA,OA交⊙O于C点,∴CD是⊙O的切线,∠ODC=90°····························································3分在Rt△COD与Rt△B0D中,∵OD=OD,OB=OC∴Rt△COD≌Rt△B0D············································································6分(2)∵∠A=32°,AD=8,ADCDAsin∴832sinCD,则CD=832sin···························································8分∵∠A=32°,∴∠ADC=58°则∠CDB=122°由(1)知Rt△COD≌Rt△B0D,∴∠CDO=∠BDO=21×122°=61°·····················10分∴tan61°=CDOC,∴OC=tan61°×CD=8×32sin×tan61°≈7.65㎝则⊙O的半径约为7.65㎝…………………………………………………………12分从化市2008年初中毕业生综合测试数学试题参考答案第3页共5页22、（本题满分12分）本小题主要考查二元一次方程组的应用等基础知识，考查数学推理论证能力、运算求解能力解：（1）∵100×13＝13001392…………………………………………………………3分∴乙团的人数不少于50人，不超过100人…………………………………………4分（2）设甲、乙两旅行团分别有x人、y人…………………………………………5分则1080)(913921113yxyx……………………………………………………………………9分解得：8436yx…………………………………………………………………………11分答:所以甲、乙两旅行团分别有36人、84人。……………………………………12分23、（本题满分12分）本小题主要考查二次函数的图像与性质等基础知识，考查数形结合的数学思想方法，以及推理论证能力、运算求解能力解：（1）根据图象得y随x的增大而增大的自变量x的取值范围是x＜2····················2分（2）根据图象知抛物线经过点（1，0）和（3，0）两点·····································3分∴03901cbcb……………………………………………………………………5分解得34cb························································································7分（3）方法1:由图象知抛物线经顶点为（2，2），···············································8分结合图象知直线ky与抛物线cbxxy2有交点的条件是k≤1;···········12分方法2：由（2）知3,4cb,即一元二次方程kxx342有实数根,化简得:0342kxx·············8分∴0)3(1442k···································································10分解得k≤1;··························································································12分24、（本题满分12分）本小题主要考查直角三角形和一次函数等基础知识，考查数形结合的数学思想方法，以及推理论证能力、运算求解能力解：（1）在Rt△B′OC中，因为tan∠OB′C＝34，OC＝6，∴43'6OB,…………………………………………………………………………2分解得OB′＝8，即点B′坐标为（8，0）.·····················································4分从化市2008年初中毕业生综合测试数学试题参考答案第4页共5页（2）因为将纸翻折后，使点B恰好落在x轴上，记为B′，折痕为CE，∴△CBE≌△CB′E，∴BE＝B′E，CB′＝CB＝OA，········································5分由勾股定理，得CB′＝22OBOC＝10，··················································6分设AE＝n，则EB′＝EB＝6－n，AB′＝AO－OB′＝2，由勾股定理，得n2+22＝(6－n)2，解得n＝83.∴点E坐标为（10，83），点C坐标为（0，6）.···········································8分设直线CE的解析式y＝kx+b，根据题意得·····················································9分6,810.3bkb解得613bk··························································11分∴即CE所在直线的解析式：y＝－13x+6.···················································12分25、（本题满分14分）本小题主要考查三角形相似等基础知识，考查分析问题和解决问题的能力、以及创新意识解：(1)证明：在矩形ABCD中，∠A=∠D=90º在RtΔABP中，∠1+∠2=90º∵∠BPC=90º，∴∠2+∠3=90º∴∠1=∠3∴△ABP∽△DPC…………………………4分(2)有（1）得：△ABP∽△DPC∴APABDCDP即AB2=AP·DP………………………………………5分设AP=x，则DP=13-x∴62=x(13-x)解得x=4或x=9即AP=4或9……………………………………………………………7分∵∠PBC=∠2tan∠PBC=tan∠2=ABAP∴当AP=4时，tan∠BPC=6342；当AP=9时，tan∠BPC=6293…………………9分(3)解：过点P作PG⊥BC于点G，PG交MN于点H…………………………………10分在矩形ABGP中，PG=AB=6在矩形MEGH中，HG=ME在矩