2023年江苏省无锡市宜兴市中考数学一模试卷

这是一份2023年江苏省无锡市宜兴市中考数学一模试卷，文件包含2023宜兴市初三中考适应性练习九年级数学试卷pdf、适应性卷答案及评分标准20224docx等2份试卷配套教学资源，其中试卷共9页， 欢迎下载使用。
  2023年春季初三数学中考适应性测试参考答案与评分标准一、选择题（本大题共10小题，,每小题3分，共30分．）1．D 2．D  3．D 4．B  5．C 6．A 7．C 8．B  9．B 10．A二、填空题（本大题共8小题，每小题3分，共24分．）11 ．    12．    13．（略）14．(不唯一) 15．145       16．      17．     18．90，或三、解答题（本大题共10小题，共84分．） 19. （本题满分8分）（1）解：原式=········································3分                                      =+1·················································4分（2）原式=····························································3分                =2····························································4分20. （本题满分8分）解：（1）由①得：由②得：···············································2分∴不等式组的解集为·················································4分（2）              1分···························································3分∴              4分21. （本题满分10分）证明：（1）□ABCD中，AB∥CD，AB=CD∴∠B+∠BCD=180°···················2分∵AB=AE ∴AE=CD，∠B=∠AEB=··········································4分∵∠AEB+∠AEC=180°∴∠AEC=∠BCD  ····································5分 ∵EC=CE  ∴△AEC≌△DCE  ∴AC=DE······································6分(2)由（1）得△AEC≌△DCE∴∠OEC=∠OCE ································8分∴OE=OC····························································10分22.（本题满分10分）解：（1）由图知，抽取的这1200名学生每周参加家庭劳动时间的中位数为第600个和第601个数据的平均数，308+295=603，故中位数落在第二组；              4分（2）（人，答（略）····················································7分（3）由统计图可知，该地区中小学生每周参加家庭劳动时间大多数都小于h，建议学校多开展劳动教育，养成劳动的好习惯．（答案不唯一）．              10分23. （本题满分10分）解：（1）0.5 ··························································3分（2）列表或树状图如下：················································8分  12451 121415221 242544112 455515254 出现的等可能性结果有12种，符合条件共有8种·································9分∴P（两位数能被3整除）=．··············································10分24. （本题满分10分）（1）证明：连接OE，OC∵ AE平分∠CAB，∴ ∠CAE=∠BAE .......................................1分∴ ，∴ ∠COE=∠BOE..................................................2分∵OC=OB∴ OE⊥BC....................................................3分∵ BC∥EF，∴ OE⊥EF.................................................4分∵ OE是⊙O的半径   ∴ EF是⊙O的切线；....................................5分（2）解：如图，设⊙O的半径为x，则OE=OB=x，OF=x+5，在Rt△OEF中，由勾股定理，得 OE2+EF2=OF2，∴ x2+122=（x+9）2，解得：x=3.5，⊙O的半径为3.5............................6分∵ AB是⊙O的直径，∴ ∠AEB=90°，∵∠OEF=90° ∴∠BEF=∠AEO∵OA=OE  ∴∠BAE=∠AEO  ∴∠BEF=∠BAE∵∠F=∠F，∴ △EBF∽△AEF， ..........................................7分∴ ，∴ AE=BE，在Rt△ABE中， ，即，解得 ，∴ AE=5.6，.....................................................9分∵ BC∥EF，∴ ，即，∴ AD=...........................................10分25. （本题满分10分）解：（1）设乙种图书进价每本x元，则甲种图书进价为每本1.4x元由题意得：   解得：x＝20·················································3分经检验，x＝20是原方程的解···············································4分∴甲种图书进价为每本元答：甲种图书进价每本28元，乙种图书进价每本20元；···························5分（2）设甲种图书进货a本，总利润ｗ元，则              6分∵              7分解得              8分∵2>0,w随a的增大而增大∴当a最大时w最大····································9分∴当本时，w最大=13000 . 此时，乙种图书进货本数为（本）答（略）····························································10分26. （本题满分10分）解：（1）过点A作MN的垂线，垂足为P．·····································2分在MN上截取QP=PA．····················································3分过点Q作MN的垂线，与AQ的垂直平分线的交点为圆心O··························5分以O为圆心，OA（或OQ）为半径，作⊙O．···································6分（2），·····························································10分27. （本题满分10分）解：（1）当M落在CD上时，AP的长度达到最大，∵四边形是矩形，∴AB=CD=5，BC=AD=4，∠A=∠C=∠D=90°，∵△ABP沿直线翻折，∴∠PMB=∠A=90°，BM=AB=5， ·························1分∴  DM=5-3=2··························································2分∴∠PMD+∠BMC=90°，∠PMD+∠MPD=90°，∴∠BMC=∠MPD ∴△PDM∽△MCB ·······································3分 ∴，  ∴PD=，AP= ·····················································5分∴AP的取值范围是······················································6分（2）如图，由折叠性质得：∠ABP=∠MBP， ∴∠ABM =2∠ABP∵∠ABM =2∠ADG，∴∠ABP =∠ADG，∵∠A=∠A∴△ADG∽△ABP，∴， 设AP=5x，AG=4x····································7分过M作MH⊥AD于H，由折叠性质得：=5x，，∴=∠ADG，∴，∴DH=AH=2，HP=2-5x∵∠BAD=∠MHA=90°∴MN∥AG，∴为的中位线，则MN=AG=2x，············································8分在Rt△PHM中，，······················································9分∴（舍去），·························································10分     28. （本题满分10分）解：（1）将A（－1，0）、B（3，0）代入y＝ax2＋bx－2，得 表达式为·························································2分（2）由题意得，   ，····················································4分设BD与y轴交于E，过点C作CP⊥BE点P，    ∵BE=5，CE=2，OB=3    ∴CP=··········································6分∵BC=，∴sin∠CBD=. ···················································7分（3）点的坐标为：，，·················································10分