2023年湖北省荆州市中考真题数学试题及答案

荆州市2023年初中学业水平考试

数学试题参考答案与评分标准

一、选择题（每小题3分）

1.B  2.A  3.C  4.D  5.B  6.B  7.C  8.A  9. C  10.B

二、填空题（每小题3分）

11.2   12.3   13.300  14.1   15. 13.8  16.（，）

三、解答题（按步骤给分）

17.解：·······················································2分

=··················································3分

==·················································5分

······················································7分

原式=.·················································8分

18.解：（1）依题意得：     ········································3分

（2）当k=1时，原方程变为：，则有：

    ····················································6分

    ····················································8分

19.证明：BD为等边△ABC的中线  ，·································2分

···································3分

BD=DE     ⸫∠E=∠3=···················5分

   ·································7分

CD=CE······························8分

20.解：（1）20；·······2分      6，；·············4分

（2）画树状图为：                               

                                                           ·······························6分

或者列表为：

共有12种等可能结果，其中抽中两名女志愿者的结果有2种

P（抽中两名女志愿者）=.··································8分

21.（1）证明：①四边形ABCD是菱形   AB∥CD

DH⊥AB   ∠CDH=∠DHA=，则CD⊥OD···········1分

又D为⊙O的半径的外端点  CD是⊙O的切线.········2分

②连接HF,则有：∠DEF=∠DHF·················3分

DH为⊙O直径，∠DFH=，而∠DHB =

∠DHF=∠DBA=∠DEF   又∠EDF=∠BDA

△DEF∽△DBA.·····························5分

（2）解：连接AC交BD于G.

菱形ABCD，BD=6，   AC⊥BD，AG=GC，DG=GB=3  

在Rt△AGB中，AG==4    AC=2AG=8

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

在Rt△ADH中，  ············································7分

由△DEF∽△DBA得：  ·······································8分

（注：本题其他方法请参照给分.）

22.解：（1）设A种饰品每件的进价为元，则B种饰品每件的进价为

由题意得：  解得：··········································2分

经检验，是所列方程的根，且符合题意.  ··························3分

A种饰品每件进价为10元，B种饰品每件进价为9元. ··················4分

（2）①根据题意得：    解得：·································5分

购进A种饰品件数的取值范围为： ·······························6分

②设采购A种饰品件时的总利润为元.

当时，   即

，随的增大而减小.   当时，有最大值3480.·························7分

当时，

整理得： 3>0，随的增大而增大.

当时，有最大值3630. ········································8分

，w的最大值为3630，此时.····································9分

即当采购A种饰品210件，B种饰品390件时，商铺获利最大，最大利润为3630元.

······················································10分

23.解：（1）作图 （注：只需作三种，每作对1种给1分.）·················3分

（2）①△PCF是等腰直角三角形. 理由为：·······4分

如图，过点C作CN⊥BE交BE的延长线于N.

由折叠得AC=CM，∠CMP=∠CME=∠A=90o，∠1=∠2

AC=AB，∠A=∠PBD=∠N=90o  四边形ABNC为正方形   CN=AC=CM

又CE=CE  Rt△CME≌Rt△CNE（HL）···········5分

∠3=∠4    而∠1+∠2+∠3+∠4=90o，∠CPF=90o

∠PCF=∠2+∠3=∠CFP=45o    

△PCF是等腰直角三角形.····················6分

②过点F作FQ⊥BE于Q，FR⊥PB交PB的延长线于R，则∠R=∠A=90o.

∠1+∠5=∠5+∠6=90o    ∠1=∠6

由△PCF是等腰直角三角形知：PC=PF △APC≌△RFP（AAS）·········7分

AP=FR，AC=PR，而AC=AB   AP=BR=FR 

在Rt△BRF中，BR2+FR2=BF2，BF=k   

AP=BR=FR=k   PB=2AP=2k   AB=AP+PB=BN=3k ·····················8分

由BR=FR，∠QBR=∠R=∠FQB=90o知：四边形BRFQ为正方形，BQ=QF=k

由FQ⊥BN，CN⊥BN得：FQ//CN 

，而QE=BN-NE-BQ=3k-NE-k=2k-NE

即， 解得：NE= ············································9分

由①知：PM=AP=k，ME=NE=   PE=PM+ME=.······················10分

（注：本题其他方法请参照给分.）

24.解：（1）0或2或. （注：填对1个给1分.）···························3分

（2）①如图，设直线l与BC交于点F. 依题意得：

，  解得： ··································4分

抛物线的解析式为：y=.······················5分

可知P (1，9)，C(0，8).······················6分

由B(4，0)，C(0，8)得直线BC的解析式为

F(1，6），则PF=9-6=3

········································7分

②－存在最大值，理由如下：

如图，设直线交轴于H.

由①得：OB=4，AO=2，AB=6，OC=8，AH=2+m，

PH=   由OD//PH得：，

即

OD=·························································9分

，

=                                        

······················································11分

-3<0，0<m<4   当时，有最大值，最大值为.····························12分

（注：本题其他方法请参照给分.）