2023年中考押题预测卷02（苏州卷）-数学（参考答案）

2023年中考押题预测卷02【苏州卷】

数学·参考答案

9．【答案】5

10．【答案】

11．【答案】

12．【答案】或

13．【答案】/52度

14．【答案】26

15．【答案】90

16．【答案】27

17．【答案】

【分析】根据乘方运算，特殊角的三角函数值，二次根式化简，然后进行计算即可．

【详解】解：

··································································2分

·······································································4分

．················································································5分

【点睛】本题考查乘方的运算、特殊角的三角函数值、二次根式的化简，关键是熟练掌握乘方运算法则、二次根式的化简，牢记特殊角的三角函数值；易错点是运算过程中的符号问题．

18．【答案】

【分析】先去分母，把分式方程化为整式方程，再解出整式方程，然后检验，即可求解．

【详解】解：方程两边同乘，得：

，········································································2分

解得：，········································································3分

检验：当时，，···························································4分

∴是原方程的解．·································································5分

【点睛】本题主要考查了解分式方程，熟练掌握解分式方程的基本步骤，并注意检验是解题的关键．

19．【答案】7

【分析】根据平方差公式、完全平方公式展开，再合并同类项，化简代数式，再将已知变形整体代入即可求解．

【详解】解：∵，

∴，········································································1分

∴

·····························································2分

·······································································4分

．················································································6分

【点睛】本题考查了整式的混合运算，熟练的应用乘法公式是解决问题的关键．

20．【答案】(1)

(2)

【分析】（1）根据概率公式直接求解即可；

（2）设分别表示红、白、蓝、绿四种颜色，根据列表法求概率即可求解．

【详解】（1）解：共有四种颜色的“冰墩墩”，若花花凭购物小票抽奖一次，她抽到的是红色“冰墩墩”的概率为，

故答案为：．······································································2分

（2）解：设分别表示红、白、蓝、绿四种颜色，列表如下

共有16种等可能结果，其中符合题意的有4种，············································4分

含含和花花抽到相同颜色“冰墩墩”的概率．···········································6分

【点睛】本题考查了概率公式求概率，列表法求概率，熟练掌握求概率的方法是解题的关键．

21．【答案】(1)见解析

(2)4

【分析】（1）由线段的中点得到线段相等的条件，再由平行线的性质得到角相等的条件，即可证得；

（2）由平行线的性质及角平分线定义，导出，得到，据此即可求解．

【详解】（1）证明：∵四边形是平行四边形，

∴，，

∴，，·························································1分

又∵，

∴，

∴；··································································3分

（2）解：∵平分，

∴，

∵，

∴，

∴，

∵，

∴，

∴，

∴故答案为：4．·······································································6分

【点睛】本题考查平行四边形的性质、全等三角形的性质和判定以及等腰三角形的判定和性质，解题的关键是灵活运用所学知识解决问题．

22．【答案】(1)3；，

(2)100

(3)

【分析】（1）根据图象，数出直线上方的人数即可；由图象可得：20名患者的指标y的取值范围是，名非患者的指标y的取值范围是，位置相对比较集中，因此即可求解．

（2）利用样本估计总体，用乘样本中非患者指标x低于所占百分比即可．

（3）数出指标x低于，且指标y低于的人数，而患者有人，求出患病的概率即可求出答案．

【详解】（1）解：①根据图象可得，指标x大于的有3人，

故答案为：3．·········································································2分

②由图象可得：20名患者的指标y的取值范围是，名非患者的指标y的取值范围是，位置相对比较集中，

∴，，

故答案为：，．····································································4分

（2）解：由图象可得，调查的名非患者中，指标x低于的有4人，

∴来该院就诊的500名非患者中，指标x低于的大约（人），

故答案为：．······································································6分

（3）解：由图象可得，指标x低于，且指标y低于的有人，而患者有人，

则发生漏判的概率是：．························································8分

【点睛】本题考查了平均数、方差的意义，利用样本估计总体，以及概率公式，从图中获取有用信息是解题关键．

23．【答案】(1)； ；

(2)存在．点坐标为

【分析】（1）先利用待定系数法求一次函数解析式，再利用一次函数解析式确定M点的坐标，然后利用待定系数法求反比例函数解析式；

（2）先利用两点间的距离公式计算出，，再证明，利用相似比计算出，则，于是可得到点坐标．

【详解】（1）解：∵一次函数的图象经过两点，

∴，

解得，

所以一次函数解析式为；······················································2分

把代入得，

解得，

则点坐标为，

把代入，

得，

所以反比例函数解析式为；·······················································4分

（2）解：存在．5分

∵，

∴， ，

∵，

∴，······································································6分

∵，

∴，

∴，即，··························································7分

∴，

∴，

∴点坐标为．································································8分

【点睛】此题主要考查反比例函数与几何综合，解题的关键是熟知反比例函数的图像与性质、待定系数法求解析式、相似三角形的判定与性质．

24．【答案】(1)等边三角形，见解析

(2)

【分析】（1）如图：连接，先说明是的直径，则，即；根据是的切线可得，即；再根据结合直角三角形的性质和对顶角的性质可得，进而得到即可；

（2）根据直角三角形的性质可得，再根据勾股定理求得，根据是等边三角形和可得，然后解直角三角形可得，最后根据即可解答．

【详解】（1）解：是等边三角形，理由如下：·······································1分

如图：连接，

∵，，是的外接圆，

∴是的直径，

∴,

∴，······························································2分

∵是的切线，

∴，

∴，

∵，，

∴，

∴，

∴，

∴是等边三角形；·······························································4分

（2）解：∵，，，

∴，

∴，····························································5分

∵是等边三角形，，

∴，

∴，······································································6分

∵，，

∴，···········································7分

∴．··················································8分

【点睛】本题主要考查了圆的内接三角形、圆周角定理、切线的性质、等腰三角形的判定、解直角三角形等知识点，灵活运用相关性质、定理是解答本题的关键．

25．【答案】(1)种笔记本每本12元，种笔记本每本15元

(2)20

【分析】（1）设种笔记本每本元，则种笔记本每本元，由题意得，，计算可得的值，进而可得的值；

（2）设第二次购进种笔记本本，则购进种笔记本本，由题意得，，可得，设获得的利润为元，由题意得，，由一次函数的性质可知，当时，的值最大，最大值为，令，求解满足要求的解即可．

【详解】（1）解：设种笔记本每本元，则种笔记本每本元，

由题意得，，···················································2分

解得，，

∴，

∴种笔记本每本12元，种笔记本每本15元；········································4分

（2）解：设第二次购进种笔记本本，则购进种笔记本本，

由题意得，，····················································5分

解得，，

∴，·····································································6分

设获得的利润为元，由题意得，

，··································································7分

，

随的增大而减小，······························································8分

当时，的值最大，最大值为，

由题意得，

解得，，·····································································9分

为正整数，

的最小值为20．·································································10分

【点睛】本题考查了一元一次方程的应用，一元一次不等式的应用，一次函数的应用等知识．解题的关键在于根据题意正确的列等式和不等式．

26．【答案】(1)，

(2)

(3)

(4)

【分析】（1）把和点代入求出b和c的值，即可得出函数表达式，将其化为顶点式，即可求出点D的坐标；

（2）先求出点C的坐标，再根据两点之间的距离公式，求出，根据勾股定理逆定理，得出，最后根据直角三角形的外心与斜边中点重合，即可求解；

（3）过点P作于点M，作关于x轴的对称线段，

则，点M关于x轴的对称点在，，通过证明，得出，则

当点三点共线时，取最小值，即为的长度，用等面积法求出的长度即可；

（4）连接，先求出点，根据，，可设，，再根据两点之间的距离公式得出，，，，然后根据勾股定理可得：，即可得出n关于m的表达式，将其化为顶点式后可得当时，n随m的增大而减小，当时，n随m的增大而增大，再求出当时，点N经过的路程为，以及当时，点N经过的路程为，即可求解．

【详解】（1）解：把和点代入得：

，解得：，

∴该二次函数的表达式为：，···········································1分

∵，

∴点D的坐标为；······························································2分

（2）解：把代入得，

∴，

∵，，，

∴，

∴，

∴，·····································································3分

∴外接圆半径；·················································4分

（3）解：过点P作于点M，作关于x轴的对称线段，

则，点M关于x轴的对称点在上，，

，

，  

，···························································5分

，

当点三点共线且时，取最小值，即为的长度，

，

，即的最小值为．···········································6分

（4）解：连接，

把代入得，

解得：，  

∴，

∵，，

∴设，，

 ∴，，，，··········7分

根据勾股定理可得：，

∴，

整理得：，

∴，····················································8分

∴当时，n随m的增大而减小，当时，n随m的增大而增大，

∵动点M从点C出发，直线b与直线a重合时运动停止，，

∴，

∵当时，，

当时，，

当时，，

∴当时，点N经过的路程为：，

当时，点N经过的路程为：，··········································9分

∴点N经过的总路程为：．··············································10分

【点睛】本题主要考查了二次函数综合，解题的关键是掌握用待定系数法求解函数表达式，直角三角形外接圆圆心为斜边中点，胡不归问题的解决方法，以及勾股定理和二次函数图象上点的坐标特征和勾股定理．

27．【答案】(1)

(2)，理由见解析

(3)

(4)

【分析】（1）证明，即可得出结论；

（2）证明，即可得出结论；

（3）过点作于点，先证明，得到，再证明，得到，推出，即可得出结论；

（4）连接交于点，过点作于点，由（3）推出，设，则，求出，根据，得到，进而求出，利用二次函数的性质，求出最值即可得出结果．

【详解】（1）∵四边形是菱形，，

∴，

∴和都是等边三角形，

∴，

∵，

∴，

∴是等边三角形，，

∴，

在和中，

∴，

∴；

故答案为：，································································2分

（2）解：，理由如下：

∵四边形是菱形，，

∴，

∴，

∵，

∴，

∴

∴，

∴

∴；········································································4分

（3）如图3，过点作于点，

∵四边形是菱形，，

∴，

∵，

∴，

∴，

∴，········································································5分

又∵，

∴，

∴，········································································6分

∵，

∴，

∴，

∴··························································7分

（4）如图4，连接交于点，过点作于点，

由（3）可得：， ， ，

∴，

∴，

∴，

设，则，

∵四边形是菱形，

∴，

∴

∴，

∵，

∴，，

∴

∴ ·············································8分

∵，

∴，

∴ ，

∴，

即

∵，······································································9分

∴当时， 有最大值：．

故答案为：．··································································10分

【点睛】本题考查菱形的性质，全等三角形的判定和性质，等边三角形的判定和性质，等腰三角形的判定和性质，相似三角形的判定和性质，解直角三角形以及利用二次函数的性质求最值．本题的综合性强，难度大，属于中考压轴题．熟练掌握菱形的性质，证明三角形全等和相似，是解题的关键．