2023年3月山东省济南市高新区九年级二模检测数学卷

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

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+2

1··························································································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