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数学(徐州卷)-2024年中考数学考前押题卷
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这是一份数学(徐州卷)-2024年中考数学考前押题卷,文件包含数学徐州卷全解全析docx、数学徐州卷参考答案及评分标准docx等2份试卷配套教学资源,其中试卷共40页, 欢迎下载使用。
第Ⅰ卷
一、选择题(本大题共8个小题,每小题3分,共24分.在每个小题给出的四个选项中,只有一项符合题目要求,请选出并在答题卡上将该项涂黑)
第Ⅱ卷
二、填空题(本大题共10小题,每小题3分,共30分)
9.3.84×10510.811.7或9或1112.26°
13.1214.24°15.40°或75°或130°16.10
17.m<1818.2-2或1或2
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)
19.(10分)
解:(1)原式=1+2﹣4+14
=-34;·······································································5分
(2)原式=2xx+1-2(x+3)(x+1)(x-1)•(x-1)2x+3
=2xx+1-2(x-1)x+1
=2x+1.·······································································10分
20.(10分)
解:(1)3x+2y=13①2x+3y=-8②,
①×3﹣②×2,得:5x=55,
解得x=11,
将x=11代入①,得:33+2y=13,
解得:y=﹣10.
∴方程组的解为x=11y=-10;····························································5分
(2)5y-23-1>3y-522(y-3)≤0,
解第一个不等式得x>﹣5,
解第二个不等式得y≤3.
故不等式组的解集为﹣5<y≤3.·······················································10分
21.(6分)
解:(1)90÷30%=300,
故答案为:300;······································································2分
(2)合唱人数:300×10%=30(人),
舞蹈人数:300﹣120﹣90﹣30=60(人),
补全条形统计图如图所示:
····················································4分
(3)800×30+60300=240(人),
答:该校喜欢合唱和舞蹈社团的学生共有240人.··········································6分
22.(8分)
解:(1)第一次摸出一个球,球上的数字是偶数的概率是13,
故答案为:13;··········································································4分
(2)画树状图如下:
························································6分
共有9种等可能的结果,其中两次摸出球上的数字的积为奇数的结果有4种,
∴两次摸出球上的数字的积为奇数的概率为49.·············································8分
23.(7分)
解:设B品牌的呼吸机每台的进价是x万元,则A品牌的呼吸机每台的进价是(x+0.2)万元,
依题意,得:20x+0.2=18x,·····························································4分
解得:x=1.8,
经检验:x=1.8是原方程的解,且符合题意,
∴x+0.2=2.·········································································7分
答:A品牌的呼吸机每台的进价是2万元,B品牌的呼吸机每台的进价是1.8万元.
24.(8分)
(1)证明:连接OD,
∵OA=OD,
∴∠OAD=∠ODA,
∵AD平分∠BAC,
∴∠OAD=∠BAD,
∴∠ODA=∠BAD,
∴OD∥AB,
∴∠ODC=∠B=90°,
∴半径OD⊥BC于点D,
∴BC是⊙O的切线;··································································3分
(2)解:连接 OF,DE,
∵∠B=90°,tan∠ADB=3,
∴∠ADB=60°,∠BAD=30°,
∵BD=5,
∴AD=2BD=10,
∵AE是⊙O的直径,
∴∠ADE=90°,
∵AD平分∠BAC,
∴∠DAE=∠BAD=30°,
在 Rt△ADE 中,AD=10,
∵cs∠DAE=ADAE=32,
∴AE=2033,
∴OA=12AE=1033,
∵AD平分∠BAC,
∴∠BAC=2∠BAD=60°,
∵OA=OF,
∴△AOF 是等边三角形,
∴∠AOF=60°,
∵OD∥AB,
∴S△ADF=S△AOF,
∴S阴影=S扇形OAF=60π×(1033)2360=50π9.················································8分
25.(7分)
解:延长AC交EF于点G,
由题意得:AC=BD=1.8米,AB=CD=FG=0.8米,
设CG=x米,则AG=AC+CG=(x+1.8)米,
在Rt△AEG中,∠EAG=45°,
∴EG=AG•tan45°=(x+1.8)米,·······················································3分
在Rt△ECG中,∠ECG=53°,
∴EG=CG•tan53°≈1.33x(米),
∴x+1.8=1.33x,
解得:x=6011,
∴EF=EG+FG=1.33×6011+0.8≈8.1(米),
∴电池板离地面的高度EF的长约为8.1米.···············································7分
26.(10分)
解:(1)如图,连接OB′、AB′,
∵四边形OABC为矩形,A(4,0),C(0,m),
∴B(4,m),OA=4,OC=m,
∵将矩形OABC绕点O逆时针旋转90°得到矩形OA′B′C′,
∴A′(0,4),C′(﹣m,0),B′(﹣m,4),
∴B′C⊥x轴,B′C=4,
∴S△OAB′=12×4×4=8,
∴△OAB′的面积是8.··································································3分
(2)如图①,连接A′A,∵∠A′OA=90°,OA′=OA=4,
∴AA′=OA'2+OA2=42+42=42,
∵四边形OA′B′C′是矩形,OB′、A′C′交于点D,
∴C′D=A′D,
∵AD⊥A′C′,
∴AC′=AA′=42,
∴OC′=AC′﹣OA=42-4,
∴OC=OC′=42-4,
∴C(0,42-4),
∴m的值是42-4.··································································6分
(3)如图②,∵四边形OCEF是平行四边形,
∴EF∥OC,EF=OC=m,
∴∠EFB=∠A′CB=90°,
∴EF⊥BC,
∵E是A′B的中点,
∴CE=BE=A′E=12A′B,
∴CF=BF,
∴A′C=2EF=2m,
∵A′C+OC=4,
∴2m+m=4,
解得m=43,
∴m的值是43.·······································································10分
27.(10分)
解:(1)∵抛物线y=x2+bx+c经过点B(1,0),C(0,﹣3),
∴1+b+c=0c=-3,
解得:b=2c=-3,
∴抛物线的函数表达式为:y=x2+2x﹣3.················································2分
(2)①当x=0时,y=x2+2x﹣3=﹣3,
∴点C(0,﹣3).
当y=0时,x2+2x﹣3=0,
解得:x1=﹣3,x2=1,
∴A(﹣3,0),
设直线AC的解析式为y=kx+n,
把A(﹣3,0),C(0,﹣3)代入,
得:-3k+n=0n=-3,解得:k=-1n=-3,
∴直线AC的解析式为:y=﹣x﹣3.
∵OA=OC=3,
∴∠OAC=∠OCA=45°.
过点E作EK⊥y轴于点K,
∵EG⊥AC,
∴∠KEG=∠KGE=45°,
∴EG=EKsin45°=2EK=2OD,
设P(m,m2+2m﹣3),则E(m,﹣m﹣3),
∴PE=﹣m﹣3﹣(m2+2m﹣3)=﹣m2﹣3m,
∴PE+2EG=PE+2OD=﹣m2﹣3m﹣2m=﹣m2﹣5m=﹣(m+52)2+254,
由题意有﹣3<m<0,且﹣3<-52<0,﹣1<0,
当m=-52时,PE+2EG取最大值,PE+2EG的最大值为254;······························5分
②作EK⊥y轴于K,FM⊥y轴于M,记直线EG与x轴交于点N.
∵EK⊥y轴,PD⊥x轴,∠KEG=45°,
∴∠DEG=∠DNE=45°,
∴DE=DN.
∵∠KGE=∠ONG=45°,
∴OG=ON.
∵y=x2+2x﹣3的对称轴为直线x=﹣1,
∴MF=1,
∵∠KGF=45°,
∴GF=MFsin45°=2MF=2.
∵∠FDG=45°,
∴∠FDN=∠DEG.
又∵∠FDG=∠DEG,
∴△DGF∽△EGD,
∴DGFG=EGDG,
∴DG2=FG•EG=2×2(﹣m)=﹣2m,
在Rt△ONG中,OG=ON=|OD﹣DN|=|OD﹣DE|=|﹣m﹣(m+3)|=|﹣2m﹣3|,OD=﹣m,
在Rt△ODG中,
∵DG2=OD2+OG2=m2+(2m+3)2=5m2+12m+9,
∴5m2+12m+9=﹣2m,
解得m1=﹣1,m2=-95.·······························································10分
28.(10分)
(1)解:∵四边形ABCD是正方形,
∴AB=CD=AD,∠BAD=∠C=∠D=90°,
由旋转的性质得:△ABE≌△ADM,
∴BE=DM,∠ABE=∠D=90°,AE=AM,∠BAE=∠DAM,
∴∠BAE+∠BAM=∠DAM+∠BAM=∠BAD=90°,
即∠EAM=90°,
∵∠MAN=45°,
∴∠EAN=90°﹣45°=45°,
∴∠MAN=∠EAN,
在△AMN和△AEN中,
AM=AE∠MAN=∠EANAN=AN,
∴△AMN≌△AEN(SAS),
∴MN=EN,
∵EN=BE+BN=DM+BN,
∴MN=BN+DM,
在Rt△CMN中,由勾股定理得:MN=CN2+CM2=62+82=10,
则BN+DM=10,
设正方形ABCD的边长为x,则BN=BC﹣CN=x﹣6,DM=CD﹣CM=x﹣8,
∴x﹣6+x﹣8=10,
解得:x=12,
即正方形ABCD的边长是12;
故答案为:12;·······································································3分
(2)证明:设BN=m,DM=n,
由(1)可知,MN=BN+DM=m+n,
∵∠B=90°,tan∠BAN=13,
∴tan∠BAN=BNAB=13,
∴AB=3BN=3m,
∴CN=BC﹣BN=2m,CM=CD﹣DM=3m﹣n,
在Rt△CMN中,由勾股定理得:(2m)2+(3m﹣n)2=(m+n)2,
整理得:3m=2n,
∴CM=2n﹣n=n,
∴DM=CM,
即M是CD的中点;··································································6分
(3)解:延长AB至P,使BP=BN=4,过P作BC的平行线交DC的延长线于Q,延长AN交PQ于E,连接EM,如图③所示:
则四边形APQD是正方形,
∴PQ=DQ=AP=AB+BP=12+4=16,
设DM=a,则MQ=16﹣a,
∵PQ∥BC,
∴△ABN∽△APE,
∴BNPE=ABAP=1216=34,
∴PE=43BN=163,
∴EQ=PQ﹣PE=16-163=323,
由(1)得:EM=PE+DM=163+a,
在Rt△QEM中,由勾股定理得:(323)2+(16﹣a)2=(163+a)2,
解得:a=8,
即DM的长是8;
故答案为:8.·········································································10分1
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