2023年江苏省无锡市宜兴市中考数学一模试卷及答案

  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

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