山东省济南市中考数学试题word版含答案.docx
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山东省济南市中考数学试题word版含答案.docx
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山东省济南市中考数学试题word版含答案
2020年山东省济南市中考数学试题(word版)(含答案)
济南市2018年初三年级学业水平考试
温室气体排放约28400吨.将28400吨用科学记数法表示为
6.以下各选项的运算结果正确的选项是
7.在一次体育课上,体育老师对九年级一班的
x
A.
y
x1
x7
B.
y1
C.
y3
236
A.(2x)8x
623
C.xxx
2
B.5ab
2a2b
22
D.(ab)a
3
b2
40名同学进行了
跳远项目的测试,测试所得分数及相应的人数如下图,那么这
试的平均分为
x3D.
y1
人数〔人〕
20
15
10
立定
次测
6分8分10分分数
第7题图
5.二元一次方程组
11.观看以下图形及图形所对应的算式,依照你发觉的规律运算1+8+16+24+⋯⋯+8n〔n是正整数〕的结果为
济南市2018年初三年级学业水平考试
数学试题
本卷须知:
1.第二卷共6页.用蓝、黑色钢笔或圆珠笔直截了当答在考试卷上.
2.答卷前将密封线内的项目填写清晰.
第二卷〔非选择题共72分〕
二、填空题〔本大题共5个小题,每题3分,共15分.把答案填在题
中的横线上.〕
2
13.分解因式:
x
14.如下图,△DEF
2x1=.
是△ABC沿水平方向向右平
移后的对应
图形,假设∠B=31°,∠C=79°,那么∠D的度数是度.
15.解方程23的结果是.
x12x3
1
16.如下图,点A是双曲线y1在第二象限的分支上的任意一点,点B、C、D分不是点A关于x轴、
原点、y轴的对称点,那么四边形ABCD的面积是.
17.如下图,△ABC的三个顶点的坐标分不为A(-1,3)、B(-2,-2)、C(4,-2),那么
△ABC外接圆半径的长度为.
2x≤4
⑵如下图,在梯形
求证:
BM=CM.
ABCD中,BC∥AD,AB=DC,点M是AD的中点.
(本小题总分值
3)0
7分)
⑵如下图,△ABC中,∠C=90°,∠B=30°,AD是△ABC的角平分线,假设AC=3.
求线段AD的长.
如下图,有一个能够自由转动的圆形转盘,被平均分成四个扇形,四个扇形内分不标有数字
(本小题总分值8分)
1、2、-3、
a、b〔假设指针恰好指
-4.假设将转盘转动两次,每一次停止转动后,指针指向的扇形内的数字分不记为
在分界线上,那么该次不计,重新转动一次,直至指针落在扇形内〕.
请你用列表法或树状图求a与b的乘积等于2的概率.
16米的墙,打算用
形草坪ABCD.求该矩形草坪BC边的长.
32米长的围栏靠墙围成一个面积为120平方米的矩
16米
草坪
B
第21题图
如下图,菱形ABCD的顶点A、B在x轴上,点A在点B的左侧,点D在y轴的正半轴上,∠BAD=60°,
点A的坐标为(-2,0).
⑴求线段AD所在直线的函数表达式.
A→D→C→B→A的顺序在菱形的边上匀速
1为半径的圆与对角线AC相切?
⑵动点P从点A动身,以每秒1个单位长度的速度,按照
运动一周,设运动时刻为t秒.求t为何值时,以点P为圆心、以
得分评卷人
23.(本小题总分值9分)
⑴如图1所示,点
⑵如图2所示,点
点,你认为∠MP1N+∠
M、P、N分不是边AB、BC、CA的中点.求证:
∠MPN=∠A.
AM1AN1
M、N分不在边AB、AC上,且AM1,AN1,点P1、P2是边BC的三等分
AB3AC3
MP2N=∠A是否正确?
请讲明你的理由.
△ABC是任意三角形.
⑶如图3所示,点M、N分不在边AB、AC上,且
是边BC的2018等分点,那么∠MP1N+∠MP2N+⋯⋯+
〔请直截了当将该小咨询的答案写在横线上.〕
AA
BMNM
PCBP1P2
第23题图1第23题图2
,,点P1、P2、⋯⋯、P2018
AB2010AC2010
∠MP2018N=.
A
MN
NCBP1P2⋯⋯⋯⋯P2018C
第23题图3
24.(本小题总分值9分)
如下图,抛物线yx22x3与x轴交于
物线的对称轴l与直线BD交于点C、与x轴交于点
⑴求A、B、C三个点的坐标.
⑵点P为线段AB上的一个动点〔与点A、点
A、B两点,直线BD的函数表达式为y3x33,抛
E.
B不重合〕,以点A为圆心、以AP为半径的圆弧与线段
AC交于点M,以点B为圆心、以
①求证:
AN=BM.
②在点P运动的过程中,四边形
BP为半径的圆弧与线段BC交于点N,分不连接AN、BM、MN.
AMNB的面积有最大值依旧有最小值?
并求出该最大值或最小值
y
MCNx
AOEPB
济南市2018年初三年级学业水平考试
二、填空题
13.(x1)2
三、解答题
18.
(1)解:
数学试题参考答案及评分标准
题号
1
2
3
4
5
6
7
8
9
10
11
12
答案
C
D
C
B
D
A
B
B
B
C
A
C
14.7015.x916.417.13
x2>x①
2x≤4②
解不等式①,得x>1,·····················································1分
解不等式②,得x≥-2,·····················································2分
∴不等式组的解集为x>1.······················································3分
(2)证明:
∵BC∥AD,AB=DC,
∴∠BAM=∠CDM,·····················································1分
∵点M是AD的中点,
∴AM=DM,·······························································2分
∴△ABM≌△DCM,···············································
·····3分
(2)解:
∵△ABC中,∠C=90o,∠B=30o,
∴∠BAC=60o,
AD是△ABC的角平分线,
CAD=30o,·····························································1分
AC
Rt△ADC中,ADAC·······································2分
cos30
=3×······································3分
=2.·············································4分
20.解:
a与b的乘积的所有可能显现的结果如下表所示:
ab
1
2
-3
-4
1
1
2
-3
-4
2
2
4
-6
-8
-3
-3
-6
9
12
-4
-4
-8
12
16
··························································································6分总共有16种结果,每种结果显现的可能性相同,其中ab=2的结果有2种,
··································································································7分∴a与b的乘积等于2的概率是1.··············································8分
8
21.解:
设BC边的长为x米,依照题意得········································1分
x32x120,································································4分
2
解得:
x112,x220,···························································6分
20>16,
x220不合题意,舍去,····················································7分
答:
该矩形草坪BC边的长为12米.·····································8分
22.解:
⑴∵点A的坐标为(-2,0),∠BAD=60°,∠AOD=90°,
∴OD=OA·tan60°=23,
∴点D的坐标为〔0,23〕,················································1分
设直线AD的函数表达式为ykxb,
k3
b23
∴直线AD的函数表达式为y3x23.·······························3分
⑵∵四边形ABCD是菱形,
∴∠DCB=∠BAD=60°,
∴∠1=∠2=∠3=∠4=30°,
AD=DC=CB=BA=4,··························································5分如下图:
①点P在AD上与AC相切时,
AP1=2r=2,
∴t1=2.·············································································6分
②点P在DC上与AC相切时,
CP2=2r=2,
∴AD+DP2=6,
∴t2=6.······························7分
③点P在BC上与AC相切时,
CP3=2r=2,
∴AD+DC+CP3=10,
∴t3=10.······························8分
④点P在AB上与AC相切时,
AP4=2r=2,
AD+DC+CB+BP4=14,
t4=14,
t=2、6、10、14时,以点P为圆心、以1为半径的圆与对角线AC相切.
···············································9分
23.⑴证明:
∵点M、P、N分不是AB、BC、CA的中点,
∴线段MP、PN是△ABC的中位线,
∴MP∥AN,PN∥AM,···············1分
∴四边形AMPN是平行四边形,····2分
∴∠MPN=∠A.······················3分
⑵∠MP1N+∠MP2N=∠A正确.···············4分
如下图,连接MN,·························5分
1
,∠A=∠A,
3
AMN∽△ABC,
MN∥BC,MN=1BC,··················6分
3
P1、P2是边BC的三等分点,
MN与BP1平行且相等,MN与P1P2平行且相等,MN与P2C平行且相等,
MBP1N、MP1P2N、MP2CN差不多上平行四边形,
MB∥NP1,MP1∥NP2,MP2∥AC,
·······················································7分
MP1N=∠1,∠MP2N=∠2,∠BMP2=∠A,
MP1N+∠MP2N=∠1+∠2=∠BMP2=∠A.
·····················································8分
⑶∠A.········································9分
2
24.解:
⑴令x2x30,
解得:
x11,x23,
∴A(-1,0),B(3,0)························2分22
∵yx22x3=(x1)24,
∴抛物线的对称轴为直线x=1,
将x=1代入y3x33,得y=23,
∴C〔1,23〕.····························3分
⑵①在Rt△ACE中,tan∠CAE=CE3,
AE
∴∠CAE=60o,
抛物线的对称性可知l是线段AB的垂直平分线,
∴AC=BC,
∴△ABC为等边三角形,·····················································4分
∴AB=BC=AC=4,∠ABC=∠ACB=60o,
又∵AM=AP,BN=BP,
∴BN=CM,
ABN≌△BCM,
AN=BM.·········································································5分
6分
②四边形AMNB的面积有最小值.
设AP=m,四边形AMNB的面积为S,
AB=BC=4,BN=CM=BP,
S△ABC=3×42=43,
4
∴CM=BN=BP=4-m,CN=m,
过M作MF⊥BC,垂足为F,
那么MF=MC?
sin60o=3(4m),
2
113
∴S△CMN=CNMF=m?
(4m)=
222
3m23m,····················
4
7分
S=S△ABC-S△CMN
=43-〔m2
4
=3(m2)233
4
3m〕
8分
m=2时,S取得最小值33.
9分
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