南京市联合体中考二模数学试题及答案Word版.docx
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南京市联合体中考二模数学试题及答案Word版
2019年中考模拟试卷
(二)
数学
化工园 雨花 栖霞 浦口四区联合体
注意事项:
1.本试卷共6页.全卷满分120分.考试时间为120分钟.考生答题全部答在答题卡上,答在本试卷上无效.
2.请认真核对监考教师在答题卡上所粘贴条形码的姓名、考试证号是否与本人相符合,再将自己的姓名、准考证号用0.5毫米黑色墨水签字笔填写在答题卡及本试卷上.
3.答选择题必须用2B铅笔将答题卡上对应的答案标号涂黑.如需改动,请用橡皮擦干净后,再选涂其他答案.答非选择题必须用0.5毫米黑色墨水签字笔写在答题卡上的指定位置,在其他位置答题一律无效.
4.作图必须用2B铅笔作答,并请加黑加粗,描写清楚.
一、选择题(本大题共6小题,每小题2分,共12分.在每小题所给出的四个选项中,恰有一项是符合题目要求的,请将正确选项前的字母代号填涂在答题卡相应位置上)
1.﹣的相反数是(▲)
A.-2B.2C.-D.
2.下列计算正确的是(▲)
A.a3+a4=a7B.2a3•a4=2a7C.(2a4)3=8a7D.a8÷a2=a4
3.为调查某班学生每天使用零花钱的情况,张华随机调查了20名同学,结果如下表:
每天使用零花钱(单位:
元)
1
2
3
4
5
人数
1
3
6
5
5
则这20名同学每天使用的零花钱的众数和中位数分别是(▲)
A.3,3
B.3,3.5
C.3.5,3.5
D.3.5,3
4.小张同学的座右铭是“态度决定一切”,他将这几个字写在一个正方体纸盒的每个面上,其平面展开图如图所示,那么在该正方体中,和“一”相对的字是(▲)
A.态
B.度
C.决
D.切
F
O
(第4题)
5.如图,⊙O是△ABC的外接圆,∠OBC=42°,则∠A的度数是(▲)
A.42°
B.48°
C.52°
D.58°
6.如图,在矩形ABCD中,AB=3,BC=5,以B为圆心BC为半径画弧交AD于点E,连接CE,作BF⊥CE,垂足为F,则tan∠FBC的值为(▲)
A.
B.
C.
D.
二、填空题(本大题共10小题,每小题2分,共20分,请在答题卡指定区域内作答.)
7.代数式有意义,则x的取值范围是▲.
8.分解因式:
a3-4a=▲.
9.计算-2cos30°-|1-|=▲.
10.反比例函数y=的图象经过点(1,6)和(m,-3),则m=▲.
11.如图,在菱形ABCD中,AC=2,∠ABC=60°,则BD=▲.
(第12题)
12.如图,在⊙O中,AO∥CD,∠1=30°,劣弧AB的长为3300千米,则⊙O的周长用科学计数法表示为▲千米.
(第11题)
13.某商品原价100元,连续两次涨价后,售价为144元,若平均增长率为x,则x=▲.
14.直角坐标系中点A坐标为(5,3),B坐标为(1,0),将点A绕点B逆时针旋转90°得到点C,则点C的坐标为▲.
(第15题)
15.二次函数y=ax2+bx+c(a≠0)的图象如图所示,根据图象可知:
方程ax2+bx+c=k有两个不相等的实数根,则k的取值范围为▲.
(第16题)
16.如图,在半径为2的⊙O中,两个顶点重合的内接正四边形与正六边形,则阴影部分的面积为▲.
三、解答题(本大题共11小题,共88分.请在答题卡指定区域内作答,解答时应写出文字说明、证明过程或演算步骤)
17.(6分)解方程组
18.(6分)化简:
(-x)÷.
19.(8分)为了备战初三物理、化学实验操作考试,某校对初三学生进行了模拟训练.物理、化学各有3个不同的操作实验题目,物理用番号①、②、③代表,化学用字母a、b、c表示.测试时每名学生每科只操作一个实验,实验的题目由学生抽签确定.
(1)小张同学对物理的①、②和化学的b、c实验准备得较好.请用树形图或列表法求他两科都抽到准备得较好的实验题目的概率;
(2)小明同学对物理的①、②、③和化学的a实验准备得较好.他两科都抽到准备得较好的实验题目的概率为▲.
20.(8分)据报道,历经一百天的调查研究,南京PM2.5源解析已经通过专家论证.各种调查显示,机动车成为PM2.5的最大来源,一辆车每行驶20千米平均向大气里排放0.035千克污染物.校环保志愿小分队从环保局了解到南京100天的空气质量等级情况,并制成统计图和表:
2014年南京市100天空气质量等级天数统计图
空气质量等级
优
良
轻度污染
中度污染
重度污染
严重污染
天数(天)
10
a
12
8
25
b
n°
25%
中度
(1)表中a=▲,b=▲,图中严重污染部分对应的圆心角n=▲°.
(2)请你根据“2019年南京市100天空气质量等级天数统计表”计算100天内重度污染和严重污染出现的频率共是多少?
(3)小明是社区环保志愿者,他和同学们调查了机动车每天的行驶路程,了解到每辆车每天平均出行25千米.已知南京市2019年机动车保有量已突破200万辆,请你通过计算,估计2019年南京市一天中出行的机动车至少要向大气里排放多少千克污染物?
21.(8分)如图,在□ABCD中,E、F、G、H分别为AB、BC、CD、AD的中点,AF与EH交于点M,FG与CH交于点N.
(1)求证:
四边形MFNH为平行四边形;
(2)求证:
△AMH≌△CNF.
N
22.(8分)端午节期间,某食堂根据职工食用习惯,购进甲、乙两种粽子260个,其中甲种粽子花费300圆,乙种粽子花费400元,已知甲种粽子单价比乙种粽子单价高20%,乙种粽子的单价是多少元?
甲、乙两种粽子各购买了多少个?
23.(8分)如图,为了测出某塔CD的高度,在塔前的平地上选择一点A,用测角仪测得塔顶D的仰角为30º,在A、C之间选择一点B(A、B、C三点在同一直线上),用测角仪测得塔顶D的仰角为75º,且AB间距离为40m.
(1)求点B到AD的距离;
(2)求塔高CD(结果用根号表示).
75°
24.(8分)小林家、小华家、图书馆依次在一条直线上.小林、小华两人同时各自从家沿直线匀速步行到图书馆借阅图书,已知小林到达图书馆花了20分钟.设两人出发x(分钟)后,小林离小华家的距离为y(米),y与x的函数关系如图所示.
(1)小林的速度为▲米/分钟,a=▲,小林家离图书馆的距离为▲米;
(2)已知小华的步行速度是40米/分钟,设小华步行时与自己家的距离为y1(米),请在图中画出y1(米)与x(分钟)的函数图象;
(3)小华出发几分钟后两人在途中相遇?
第题
a
25.(8分)施工队要修建一个横断面为抛物线的公路隧道,其高度为6米,宽度OM为12米.现以O点为原点,OM所在直线为x轴建立直角坐标系(如图①所示).
(1)求出这条抛物线的函数表达式,并写出自变量x的取值范围;
(2)隧道下的公路是双向行车道(正中间有一条宽1米的隔离带),其中的一条行车道能否行驶宽2.5米、高5米的特种车辆?
请通过计算说明;
(第25题)
26.(10分)如图,已知△ABC,AB=6、AC=8,点D是BC边上一动点,以AD为直径的⊙O分别交AB、AC于点E、F.
(1)如图①若∠AEF=∠C,求证:
BC与⊙O相切;
(2)如图②,若∠BAC=90°,BD长为多少时,△AEF与△ABC相似.
备用
图②
27.(10分)已知直角△ABC,∠ACB=90°,AC=3,BC=4,D为AB边上一动点,沿EF折叠,点C与点D重合,设BD的长度为m.
(1)如图①,若折痕EF的两个端点E、F在直角边上,则m的范围为▲;
(2)如图②,若m等于2.5,求折痕EF的长度;
(3)如图③,若m等于,求折痕EF的长度.
E
2019中考数学模拟试卷
(二)答案
一、选择题(本大题共6小题,每小题2分,共12分.)
题号
1
2
3
4
5
6
答案
D
B
B
A
B
D
二、填空题(本大题共10小题,每小题2分,共20分.)
7.x>18.a(a-2)(a+2)9.+110.﹣211.2
12.3.96×10413.(﹣2,4)14.0.215.k<216.6-2
三、解答题(本大题共11小题,共88分.)
17.
解:
①×2得:
4x+6y=﹣10③
②×3得:
9x-6y=36④
③+④得:
13x=26
解得:
x=2········································································································3分
把x=2代入①得y=﹣3····················································································5分
所以方程组的解为·················································································6分
18.解原式=[-]÷·············································································1分
=×·····························································································2分
=×··································································································3分
=×···················································································4分
=-x(x-1)··············································································································5分
=﹣x2+x················································································································6分
19.
(1)画图或列表正确·····································································································4分
共有9种等可能结果,期中两科都满意的结果有4种··································································5分
P(两科都满意)=.·········································································································6分
(2)···························································································································8分
20.
(1)25;20;72°······································································································3分
(2)45%···············································································································5分
(3)200×0.035×10000×=87500(千克)···········································································8分
21.
(1)证明:
连接BD,∵E、F、G、H分别为AB、BC、CD、AD的中点,
∴EH为△ABD的中位线,∴EH∥BD.
同理FG∥BD.
∴EH∥FG·······················································································································2分
在□ABCD中
∴AD∥=BC,
∵H为AD的中点AH=AD,
∵F为BC的中点FC=BC,
∴AH∥=FC
∴四边形AFCH为平行四边形,
∴AF∥CH·······················································································································4分
又∵EH∥FG
∴四边形MFNH为平行四边形···························································································5分
(2)∵四边形AFCH为平行四边形
∴∠FAD=∠HCB···········································································································6分
∵EH∥FG,∴∠AMH=∠AFN
∵AF∥CH
∴∠AFN=∠CNF
∴∠AMH=∠CNF············································································································7分
又∵AH=CF
∴△AMH≌△CNF·············································································································8分
22.解:
设乙种粽子的单价是x元,则甲种粽子的单价为(1+20%)x元,
由题意得,+=260,···················································································4分
解得:
x=2.5,·················································································5分
经检验:
x=2.5是原分式方程的解,························································································6分
(1+20%)x=3,
则买甲粽子为:
=100个,乙粽子为:
=160个.················································7分
答:
乙种粽子的单价是2.5元,甲、乙两种粽子各购买100个、160个.········································8分
23.
(1)作BE⊥AD,垂足为E,
E
在Rt△AEB中,sinA=,
=,BE=20················3分
(2)∠DBC是△ABD的外角
∠ADB=∠DBC-∠A=45°,···············4分
在Rt△DEB中,tan∠EDB=,1=,
ED=20·············································5分
在Rt△AEB中,cos∠EAB=,EA=20······························6分
AD=ED+EA=20+20························································································7分
在Rt△ACD中,sin∠DAC=,EA=10+10·····················································8分
24.
(1)60;960;1200;····························3分
(2)如图略(以(0,0)、(24,960)为端点的线段),····························5分
(3)解法一:
由题意得60x-240=40x,x=12,小华出发12分钟后两人在途中相遇.························8分
解法二:
设小林在4~20分钟的函数表达式为y=kx+b,
则,∴k=60,b=-240,下同解法一··········8分
25.解:
(1)设抛物线的函数表达式为y=a(x-6)2+6,∵图像过点(0,0)∴a=-,…………………2分
∴y=-(x-6)2+6=-x2+2x,…………………3分
0≤x≤12.…………………4分
(2)当x=3时,y=-×9+2×3=4.5.…………………6分
∵4.5<5,∴不能通过.…………………8分
26.
(1)证明:
连接DF,在⊙O中∠AEF=∠ADF····························1分
又∵∠AEF=∠C∴∠ADF=∠C····························2分
∵AD为直径,∴∠AFD=90°∴∠CFD=90°∴∠C+∠CDF=90°
∴∠ADF+∠CDF=90°∴∠ADC=90°····························3分
又∵AD为直径∴BC与⊙O相切.····························4分
(2)情况一:
若△AEF∽△ACB,则∠AEF=∠C,由
(1)知BC与⊙O相切.∴BD=3.6···············7分
情况二:
若△AEF∽△ABC∴∠AEF=∠B,∴EF∥BC,
∵∠EAF为直角,∴EF为直径,∴△AEO∽△ABD,
∴===,∴BD=2EO=EF
∵EF∥BC∴△AEF∽△ABC∴==,即BD=2EO=EF=BC=5……………………10分
27.解:
(1)2≤m≤4;…………………2分
(2)方法一、∵∠ACB=90°,AC=3,BC=4,∴AB=5,∵BD=2.5,∴AD=DB=CD=2.5,
∵点C与点D关于对称,∴DE=CE,CF=DF,∴∠CAD=∠ECD=∠EDC,
∴△ACD∽△CDE,
∴=,即=,
∴CE=;同理CF=;∴EF=.…………………6分
方法二、作DG⊥BC,垂足为G,连接DF,△BGD∽△BCA,∴==
∴DG=,CG=GB=2
在Rt△FDG中,FG2+DG2=DF2,(2-DF)2+1.52=DF2,解得DF=,CF=DF=…………………4分
∵∠CEF+∠ECD=90°,∠DCF+∠ECD=90°,∴∠CEF=∠DCF,又∵∠ECF=∠CGD=90°
∴△ECF∽△CGD∴=∴EF=.…………………6分
(3)作DG⊥BC,垂足为G,作EH⊥BC,垂足为H,连接DF,△BGD∽△BCA,∴==
∴DG=,GB=∴CG=
在Rt△FDG中,FG2+DG2=DF2,(-DF)2+()2=DF2,解得DF=,CF=DF=……………8分
易证∠HEF=∠DCG,又∵∠EHF=∠DGC=90°
∴△EHF∽△CGD∴=∴==,设FH=x,则EH=3x,
∵EH∥AC,∴△EHB∽△ACB∴=∴=解得x=,
∴EF=FH=…………10分
G