期中模拟卷02（江苏）（苏科版九上第1~4章&第6章：一元二次方程、二次函数、圆、概率统计）2023-2024学年九年级数学上学期期中模拟考试试题及答案（含答题卡）

2023-2024学年上学期期中模拟考试

九年级数学

9．x1＝x2＝﹣2     10．1：4        11．

12．               13．（20﹣x）（200+8x）＝8450  14．1

15．2       16．

17．（8分）

解：（1）方程整理得：x2x，

配方得：x2x，即（x）2，

开方得：x±，

解得：x1＝3，x2；······························································4分

（2）方程整理得：（x﹣5）2+2（x﹣5）＝0，

分解因式得：（x﹣5）（x﹣5+2）＝0，

解得：x1＝5，x2＝3．······························································8分

18．（8分）

解：（1）根据题意得：m＝1+2+7+6+4﹣1﹣3﹣5﹣3＝8，

甲队成绩的平均分为（1×6+2×7+7×8+9×6+10×4）＝8.5，························2分

甲队成绩的中位数为8.5，·······················································4分

乙队成绩的众数为8，

乙队成绩的中位数为8，······················································6分

故答案为：8.5；8.5；8；8；

（2）王老师很有可能选择甲队代表学校参加市里比赛，

理由如下：甲队的平均分大于乙队的平均分；乙的方差与甲队的方差相差不大，甲队的中位数高于乙队的中位数．·······································································8分

19．（10分）

解：（1）此次抽查的学生为：60÷20%＝300（人），

故答案为：300；···································································2分

（2）C组的人数为：300×40%＝120（人），

A组的人数为：300﹣100﹣120﹣60＝20（人），

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

（3）此次抽查的学生有300人，随机询问一人，有300种等可能结果，其中活动时间低于1小时的有100+20＝120种结果，

∴P（活动时间低于1小时），

则该生当天在校体育活动时间低于1小时的概率是．····································7分

（4）（人）

答：估计在当天达到国家规定体育活动时间的学生约有960人．···························10分

20．（8分）

（1）证明：Δ＝（k+1）2﹣4×2（k﹣2）··············································1分

＝k2﹣6k+17·······································································2分

＝（k﹣3）2+8，

∵（k﹣3）2≥0，

∴（k﹣3）2+8≥8，

∴不论k取何值，此方程必有两个不相等的实数根；······································4分

（2）解：当b＝a＝4或c＝a＝4时，

把x＝4代入方程x2﹣（k+1）x+2（k﹣2）＝0得16﹣4（k+1）+2（k﹣2）＝0，··············5分

解得k＝4，··········································································6分

此时方程化为x2﹣5x+4＝0，解得x1＝4，x2＝1，

∴等腰三角形的三边分别为4、4、1，符合题意；

∵Δ＞0，

∴b≠c，

综上所述，k的值为4．······························································8分

21．（8分）

解：（1）∵∠ACB＝90°，

∴∠BCE+∠GCA＝90°，····························································1分

∵CG⊥BD，

∴∠CEB＝90°，

∴∠CBE+∠BCE＝90°，

∴∠CBE＝∠GCA，·································································2分

又∵∠DCB＝∠GAC＝90°，

∴△BCD∽△CAG，·································································3分

∴，

∴，

∴．·········································································4分

（2）∵∠GAC+∠BCA＝180°，

∴GA∥BC，

∴，

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

∴，

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

又∵3，

∴S△AFC．···································································8分

22．（8分）

（1）解：CE与⊙O相切．························································1分

证明如下：∵AB为⊙O的直径，

∴∠ACB＝90°，

∵EC＝ED，

∴∠DCE＝∠EDC，

在Rt△DCF中，∠DCE+∠ECF＝90°，

∴∠CDE+∠ECF＝90°，

∵∠CDE+∠F＝90°，

∴∠ECF＝∠F，···································································2分

 连接OC，

∵OF⊥AB，

∴∠DOA＝90°，

∴∠A+∠ADO＝90°，

∵OA＝OC，

∴∠A＝∠OCA，

∴∠OCA+∠ADO＝90°，

∵∠ADO＝∠CDE，

∴∠OCA+∠CDE＝90°，

∵∠CDE＝∠DCE，

∴∠OCA+∠DCE＝90°，

∴EC⊥OC，

∴EC是⊙O的切线；·································································4分

（2）解：∵EF＝3，ED＝EF，

∴EC＝DE＝3，

∴OE5，·····················································5分

∴OD＝OE﹣DE＝2，

在Rt△OAD中，AD2 ，····································6分

在Rt△AOD和Rt△ACB中，

∵∠A＝∠A，∠ACB＝∠AOD，

∴Rt△AOD∽Rt△ACB，······························································7分

∴，

即，

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

23．（8分）

解：（1）如图，点C即为所求；

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

（2）如图，连接OD交AB于点F．

∵∠PCA＝∠PCB，

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

∴OD⊥AB，

∵DE⊥OA，

∴∠OFA＝∠OED＝90°，····················································6分

∵∠FOA＝∠EOD，OA＝OD，

∴△OFA≌△OED（AAS），····················································7分

∴OE＝OF＝4，

∵OD⊥AB，

∴BF＝AF，

∵OC＝OA，

∴BC＝2OF＝8．·································································8分

24．（8分）

解：如图，延长DF交MG于Q，则DQ⊥MG，DQ＝PG＝23.6，·······················2分

∵BC⊥AP，MG⊥AP，

∴BC∥MG，

∴△ABC∽△ANG，··························································3分

∴，即，

∴NG＝12，·································································4分

同理得：△DEF∽△DMQ，

∴，

∵EF＝0.1米，DF＝0.2米，

∴DF＝2EF，································································6分

∴MQDQ23.6＝11.8（米），

∴MN＝MQ+QG﹣GN＝11.8+1.5﹣12＝1.3（米）．

答：旗帜的宽度MN是1.3米．·················································8分

25．（10分）

解：（1）销售量：100+20100+160＝260，·······································2分

利润：（100+160）（6﹣4﹣0.8）＝312，···············································4分

则每天的销售量为260千克、销售利润为312元；

故答案为：260，312；

（2）将这种水果每千克降低x元，则每天的销售量是10020＝100+200x（千克）；

故答案为：（100+200x）；································································6分

（3）设这种水果每千克降价x元，

根据题意得：（6﹣4﹣x）（100+200x）＝300，···············································7分

2x2﹣3x+1＝0，

解得：x＝0.5或x＝1，··································································8分

当x＝0.5时，销售量是100+200×0.5＝200＜240；

当x＝1时，销售量是100+200＝300＞240．

∵每天至少售出240千克，

∴x＝1．

6﹣1＝5，

答：张阿姨应将每千克的销售价降至5元．···············································10分

26．（12分）

解：（1）连接BB1，如图①所示：

则MN是BB1的垂直平分线，·······················································1分

作NN1⊥AB于N1，则NN1＝BC＝AD＝3，

∴∠ABB1+∠N1MN＝90°，

∵∠N1NM+∠N1MN＝90°，

∴∠ABB1＝∠N1NM，

∵∠A＝∠NN1M＝90°，

∴△ABB1∽△N1NM，······························································2分

∴，

∵MN1＝BM﹣CN，

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

（2）AC5，作BH⊥AC于H，

①当点P在CH上时，CP＝CQ，作QE⊥AC交AC延长线于E，如图②﹣1所示：

则∠PQE+∠QPE＝90°，

∵∠BPH+∠PBH＝90°，∠QPE+∠BPH＝180°﹣∠BPQ＝180°﹣90°＝90°，

∴∠PBH＝∠QPE，

∵∠BHP＝∠PEQ＝90°，

∴△BHP∽△PEQ，······························································4分

∴，

∵∠QEC＝∠D＝90°，∠QCE＝∠ACD，

∴△CEQ∽△CDA，······························································5分

∴，

∴，

同理可得：，

∴CECQCP，QECQCP，

∵S矩形ABCD＝AB•AD＝2AC•BH，

即4×3＝5BH，

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

∵∠ABC＝∠BHC＝90°，∠BCA＝∠HCB，

∴△CBH∽△CAB，

∴，即，

∴CH，

∵，

∴，

解得：CP＝1，

∴AP＝AC﹣CP＝5﹣1＝4；··························································8分

②当点P在AH上时，PQ＝CQ，过点Q作QE⊥AC于E，如图②﹣2所示：

则PE＝CE，

同理可得：CP，

∴AP＝AC﹣CP＝5；

综上所述，如果△CPQ是等腰三角形，AP的值为4或；·································10分

（3）延长ON交CB于P，延长NO交AD于Q，作OH⊥AD于H，如图③所示：

∵四边形ABCD是矩形，

∴OD＝OB，AD∥BC，

∴∠QDO＝∠PBO，

在△ODQ和△OBP中，，

∴△ODQ≌△OBP（ASA），

∴DQ＝PB，

设DQ＝PB＝x，则PE＝x﹣2，

∵AD∥BC，

∴△AMD∽△BME，△AMQ∽△BMP，

∴，

∴，

即，

解得：x＝6，

则PE＝6﹣2＝4，

在Rt△OHQ中，OHAB＝2，HQAD+DQ3+6，

由勾股定理得：OQ，

AQ＝AD+DQ＝3+6＝9，

∵AD∥BC，

∴△PNE∽△QNA，

∴，

设PN＝4a，则NQ＝9a，PQ＝13a，OP＝OQa，ON＝OP﹣PNa，

由a，

解得：a，

∴ONa，

故答案为：．·································································12分