2023年3月山东省济南市高新区九年级二模检测数学卷

这是一份2023年3月山东省济南市高新区九年级二模检测数学卷，文件包含202303高新二模-数学-试题docx、202303高新二模-数学-评分标准docx等2份试卷配套教学资源，其中试卷共12页， 欢迎下载使用。
2023年高新区学考模拟测试数学试题参考答案及评分标准2023.03一、选择题题号12345678910答案ACABAACABA二、填空题:（本大题共6个小题，每小题4分，共24分．）11．（m+2）（m﹣2）．  12．．  13．2． 14．2023．  15．y＝2x+2．  16．②③．三、解答题:（本大题共10个小题，共86分．解答应写出文字说明、证明过程或演算步骤．）17．（本题6分）解：原式＝12+21··························································································6分18．（本题6分）解：解第一个不等式得：x0········································································2分解第二个不等式得：x≤1········································································4分∴不等式组的正整数解是：0x···························································5分则整数解是：1······················································································6分19．（本题6分）证明：∵四边形ABCD为菱形，∴AD＝CD＝AB＝BC，∠A＝∠C···························································2分∵BM＝BN，∴AB﹣BM＝BC﹣BN，即AM＝CN·······················································································4分在△AMD和△CND中，，∴△AMD≌△CND（SAS）···································································5分∴DM＝DN·······················································································6分20．（本题8分）解：（1）50，18··························································································2分（2）5，6·····························································································4分（3）（8×4+5×18+6×20+7×4）＝5.4（篇）·············································6分答：本次抽查的学生平均每人阅读的篇数为5.4篇；（4）抽查学生中阅读4篇的有8人，占抽查学生的16%，所以1000×16%＝160（人）··································································8分答：估计该校学生在这一周内文章阅读的篇数为4篇的人数有160人．21．（本题8分）解：（1）∵斜坡的坡度为1：3，∴·····················································1分∵BD＝CD﹣CB＝2.2（米）·····································································2分 在Rt△ABD中，AB＝3BD＝6.6（米）························································3分 故AD7.04（米）···············································4分答：斜面AD的长度应约为7.04米．（2）过C作CE⊥AD，垂足为E，∴∠DCE+∠CDE＝90°，∵∠BAD+∠ADB＝90°，∴∠DCE＝∠BAD，∴tan∠BAD＝tan∠DCE·······························································5分设DE＝x米，则EC＝3x米，在Rt△CDE中，3.22＝x2+（3x）2····························································6分解得：x≈1.011则3x＝3.033·························································································7分∵3.033＞2.8，∴货车能进入地下停车场········································································8分22．（本题8分）（1）证明：连接OB，如图，∵AD是⊙O的直径，∴∠ABD＝90°······················································································1分∴∠A+∠ADB＝90°，∵BC为切线，∴OB⊥BC····························································································2分∴∠OBC＝90°，∴∠OBA+∠CBP＝90°，而OA＝OB，∴∠A＝∠OBA······················································································3分∴∠CBP＝∠ADB··················································································4分（2）解：∵OP⊥AD，∴∠POA＝90°，∴∠P+∠A＝90°，∴∠P＝∠D·························································································5分∴△AOP∽△ABD··················································································6分∴，即·············································································7分∴BP＝14·····························································································8分23．（本题10分）解：（1）设每个足球的进价为x元，则每个排球的进价为（x+15）元···················1分根据题意得····································································································3分解得x＝75···················································································································4分经检验x＝75是原分式方程的解·······················································································5分∴x+15＝75+15＝90（元）．∴篮球的进价为75元，排球的进价为90元．答：足球的单价为75元，排球的单价为90元······································································6分（2）设该学校可以购进排球a个，则购进足球（100﹣a）个·················································7分根据题意，得90a+75（100﹣a）≤8000···············································································8分解得·················································································································9分∵a是整数，∴a＝33，答：最多可以购进排球33个··························································································10分24．（本题10分）解：（1）将A（1，a）和B（b，2）代入y1＝-2x+8，解得点A（1，6），B（3，2）··························································································2分将点A（1，6）代入y2，解得m＝6，即y2··································································3分（2）作B点关于y轴的对称点B'，连接AB'交y轴于点P，连接PB，∴PB＝PB'，∴PB+PA+AB＝PB'+AP+AB≥AB'+AB，当A、P、B'三点共线时，△PAB的周长最小，∵B（3，2），∴B'（﹣3，2）·············································································································4分设直线AB'的解析式为y＝k'x+b'，∴，解得，∴y＝x+5·····················································································································6分∴P（0，5）················································································································7分（3）D点坐标为（4，3）或（﹣2，9）或（2，1）····························································10分25．（本题12分）解：（1）1，··························································································2分（2）仍然存在证明：∵AB＝AC，2AD=AB，2AE=AC，∴AD=AE∵∠DAE＝∠BAC，∴∠DAE﹣∠BAE＝∠BAC﹣∠BAE，即∠BAD＝∠CAE·········································································································3分在△ABD和△ACE中，，∴△ABD≌△ACE（SAS）·······························································································4分∴BD＝CE，即BD：CE=1······························································································5分∵△ABC是等腰直角三角形，AN⊥BC，∴ABAN由旋转的性质知，∠DAB＝∠MAN＝α，∵2AD=AB，2AE=AC，∴△ADE∽△ABC·········································································································6分∴，∴△AMN∽△ADB········································································································7分∴，∴，即BDMN······························································································8分（3）FB的长为或·······························································································12分26．（本题12分）解：（1）由题意得：····························································2分解得．抛物线y1所对应的函数解析式为···········································3分（2）当x＝﹣1时，，∴D（﹣1，1）·······················································4分设直线AD的解析式为y＝kx+b，∴，解得，∴直线AD的解析式为······················································································5分如答图1，当M点在x轴上方时，∵∠M1CB＝∠DAC，∴DA∥CM1，设直线CM1的解析式为···················································································6分∵直线经过点C，∴，解得：··························································································7分∴直线CM1的解析式为，∴，解得：，（舍去），∴··················8分（3）P点坐标为或···············································································12分