泰安市东平明湖中学2022-2023学年度高中段学校招生考试试题和答案（一）

这是一份泰安市东平明湖中学2022-2023学年度高中段学校招生考试试题和答案（一），共11页。试卷主要包含了本试题分第Ⅰ卷和第Ⅱ卷两部分，分式方程的解是，如图，在中，的度数为是上一点，，在同一直角坐标系中，函数和等内容，欢迎下载使用。
试卷类型：A泰安市东平明湖中学2022-2023学年度高中段学校招生考试数  学  试  题（一） 注意事项：1．本试题分第Ⅰ卷和第Ⅱ卷两部分．第Ⅰ卷3页为选择题，36分；第Ⅱ卷8页为非选择题，84分；共120分．考试时间120分钟．2．答第Ⅰ卷前务必将自己的姓名、考号、考试科目涂写在答题卡上．考试结束，试题和答题卡一并收回．3．第Ⅰ卷每题选出答案后，都必须用2B铅笔把答题卡上对应题目的答案标号（ABCD）涂黑，如需改动，必须先用橡皮擦干净，再改涂其他答案，不能答在试卷上． 第Ⅰ卷（选择题  共36分） 一、选择题（本大题共12小题，在每小题给出的四个选项中，只有一个是正确的，请把正确的选项选出来，每小题选对得3分，选错、不选或选出的答案超过一个，均记零分）1．的相反数是（    ）A．  B．  C．  D．2．如图是由相同小正方体组成的立体图形，它的左视图为（    ）3．下列运算正确的是（    ）A．   B．     C．    D．4．如图，下列条件之一能使是菱形的为（    ）①  ②  ③  ④A．①③  B．②③  C．③④  D．①②③5．分式方程的解是（    ）A．  B．  C．  D．6．如图，在中，的度数为是上一点，是上不同的两点（不与两点重合），则的度数为（    ）A．  B．  C．  D．7．在0，1，2三个数中任取两个，组成两位数，则在组成的两位数中是奇数的概率为（    ）A．  B．  C．  D．8．直角三角形纸片的两直角边长分别为6，8，现将如图那样折叠，使点与点重合，折痕为，则的值是（    ）A．  B．  C．  D．9．函数的图象如图所示，下列对该函数性质的论断不可能正确的是（    ）A．该函数的图象是中心对称图形  B．当时，该函数在时取得最小值2C．在每个象限内，的值随值的增大而减小D．的值不可能为1 10．在同一直角坐标系中，函数和（是常数，且）的图象可能是（    ）       11．如图，圆锥的侧面积恰好等于其底面积的2倍，则该圆锥侧面展开图所对应扇形圆心角的度数为（    ）A．  B．C．  D．12．如图所示是二次函数的图象在轴上方的一部分，对于这段图象与轴所围成的阴影部分的面积，你认为与其最接近的值是（    ）A．4  B．  C．  D．  泰安市东平明湖中学2022-2023学年度高中段学校招生考试数  学  试  题（一） 第Ⅱ卷（非选择题  共84分） 注意事项：1．答卷前将密封线内的项目填写清楚．2．第Ⅱ卷共4页，用蓝黑钢笔或圆珠笔直接答在试卷上．二、填空题（本大题共7小题，满分21分．只要求填写结果，每小题填对得3分）13．计算的结果是        ．14．将分解因式的结果是        ．15．在如图所示的单位正方形网格中，将向右平移3个单位后得到（其中的对应点分别为），则的度数是　　　　　．16．不等式组的解集为         ． 17．若等腰梯形的上、下底之和为4，并且两条对角线所夹锐角为，则该等腰梯形的面积为           （结果保留根号的形式）．18．四边形的对角线的长分别为，可以证明当时（如图1），四边形的面积，那么当所夹的锐角为时（如图2），四边形的面积         ．（用含的式子表示）            19．如图，将边长为1的正三角形沿轴正方向连续翻转2008次，点依次落在点的位置，则点的横坐标为        ．        三、解答题（本大题共7小题，满分63分．解答要写出必要的文字说明、证明过程或推演步骤）20．（本小题满分8分）（1）先化简，再求值：，其中．（2）用配方法解方程：．   21．（本小题满分7分）为了解某品牌A，B两种型号冰箱的销售状况，王明对其专卖店开业以来连续七个月的销售情况进行了统计，并将得到的数据制成如下的统计表：月份一月二月三月四月五月六月七月A型销售量（单位：台）10141716131414B型销售量（单位：台）6101415161720 （1）完成下表（结果精确到0.1）：  平均数中位数方差A型销售量 14 B型销售量14 18.6（2）请你根据七个月的销售情况在图中绘制成折线统计图，并依据折线图的变化趋势，对专卖店今后的进货情况提出建议（字数控制在20~50字）．        22．（本小题满分9分）两个大小不同的等腰直角三角形三角板如图1所示放置，图2是由它抽象出的几何图形，在同一条直线上，连结．（1）请找出图2中的全等三角形，并给予证明（说明：结论中不得含有未标识的字母）；（2）证明：．     23．（本小题满分9分）某厂工人小王某月工作的部分信息如下：信息一：工作时间：每天上午8∶20~12∶00，下午14∶00~16∶00，每月25元；信息二：生产甲、乙两种产品，并且按规定每月生产甲产品的件数不少于60件．生产产品件数与所用时间之间的关系见下表：生产甲产品件数（件）生产乙产品件数（件）所用总时间（分）10103503020850信息三：按件计酬，每生产一件甲产品可得1.50元，每生产一件乙产品可得2.80元．根据以上信息，回答下列问题：（1）小王每生产一件甲种产品，每生产一件乙种产品分别需要多少分？（2）小王该月最多能得多少元？此时生产甲、乙两种产品分别多少件？   24．（本小题满分10分）如图所示，是直角三角形，，以为直径的交于点，点是边的中点，连结．（1）求证：与相切；（2）若的半径为，，求．  25．（本小题满分10分）某市种植某种绿色蔬菜，全部用来出口．为了扩大出口规模，该市决定对这种蔬菜的种植实行政府补贴，规定每种植一亩这种蔬菜一次性补贴菜农若干元．经调查，种植亩数（亩）与补贴数额（元）之间大致满足如图1所示的一次函数关系．随着补贴数额的不断增大，出口量也不断增加，但每亩蔬菜的收益（元）会相应降低，且与之间也大致满足如图2所示的一次函数关系．       （1）在政府未出台补贴措施前，该市种植这种蔬菜的总收益额为多少？（2）分别求出政府补贴政策实施后，种植亩数和每亩蔬菜的收益与政府补贴数额之间的函数关系式；（3）要使全市这种蔬菜的总收益（元）最大，政府应将每亩补贴数额定为多少？并求出总收益的最大值．     26．（本小题满分10分）在等边中，点为上一点，连结，直线与分别相交于点，且． （1）如图1，写出图中所有与相似的三角形，并选择其中一对给予证明；（2）若直线向右平移到图2、图3的位置时（其它条件不变），（1）中的结论是否仍然成立？若成立，请写出来（不证明），若不成立，请说明理由；（3）探究：如图1，当满足什么条件时（其它条件不变），？请写出探究结果，并说明理由．（说明：结论中不得含有未标识的字母） 泰安市2022-2023学年度高中段学校招生考试数学试题（A）参考答案及评分标准  一、选择题（每小题3分，共36分）题号123456789101112答案CADAABACCDDB 二、填空题（本大题共7小题，每小题3分，共21分）13．3 14．或  15．  16．17．或  18．  19．2008三、解答题（本大题共7小题，满分63分）20．（本小题满分8分）（1）解：原式····································································2分····································································3分当时，原式····································································4分（2）解：原式两边都除以6，移项得········································5分配方，得，，···································································7分即或所以，·······························································8分21．（本小题满分7分）（1）型销售量平均数14；型销售量中位数15；型销售量方差4.3．··················3分（2）           ····································································5分建议如下，从折线图来看，型冰箱的月销售量呈上升趋势，若考虑增长势头，进货时可多进型冰箱．·······················7分22．（本小题满分9分）（1）解：图2中························································1分证明如下：与均为等腰直角三角形，，·································································3分即···································································4分····································································6分（2）证明：由（1）知····································································7分又····································································9分23．（本小题满分9分）（1）解：设生产一件甲种产品需分，生产一件乙种产品需分，由题意得：····································································2分即解这个方程组得：生产一件甲产品需要15分，生产一件乙产品需要20分．···························4分（2）解：设生产甲种产品用分，则生产乙种产品用分，则生产甲种产品件，生产乙种产品件．·······················5分····································································7分又，得由一次函数的增减性，当时取得最大值，此时（元）此时甲有（件），乙有：（件）············································9分24．（本小题满分10分）（1）证明：连结是直径····································································1分是的中点····································································2分又即···································································4分但····································································5分是的切线·····························································6分（2）····································································9分···································································10分25．（本小题满分10分）解：（1）政府没出台补贴政策前，这种蔬菜的收益额为（元）·······························································2分（2）由题意可设与的函数关系为将代入上式得得所以种植亩数与政府补贴的函数关系为·······································4分同理可得每亩蔬菜的收益与政府补贴的函数关系为······························6分（3）由题意···························································7分····································································9分所以当，即政府每亩补贴450元时，全市的总收益额最大，最大为7260000元．···································································10分26．（本小题满分10分）（1）与······························································2分以为例，证明如下：····································································4分（2）均成立，均为，····················································6分（3）平分时，．·······················································7分证明：平分····································································8分又···································································10分注：所有其它解法均酌情赋分．