人教版-2021年营口市中考数学试卷及答案

．．．．年数学试卷考时120分试卷分分注事．试分一分(客观题)和二分(主观)两部答前考务将己的姓、考号写答卡。．答一分时选出小答后笔答卡对题的案号黑如改，橡擦净，选其答标，在试上无。．答二分(主题)时，将案在题对的域，在本卷或题卡定区外效．试束，本卷答卡一交。第一部分(观)一选题(下各的选案，有个正的每题分共30分．下列计算正确的是A2

B2

C．

19

D．2．如右图，是由若干个相同的小立方体搭成的几何体的俯视图和左视图，则小立方的个数有可能是A5或．或7C．或5或6D5或6或．函数y

-

中自变量x的值范围是俯视图左视图第2题图A≥－3B．x

C．x－3或D．x≥－3且eq\o\ac(□,)ABCD中对角线与交于，∠º，∠º，则∠COD是A61ºB．º

C．º

D．º18

人数/A

D

178O

5B

C

0

捐款金额/元50200以上第4题图

第题图．云南鲁甸发生地震后，某社区开展献爱心活动，社区党员积极向灾区捐款，如图该社区部分党员捐款情况的条形统计图，那么本次捐款钱数的众数和中位数分别是A元，100元B元，200元．元，元D．200元元．若关于x的式方程

2x有增根，则的是x3AmB．

C．m

D．或m=3．将弧长为、心角为的扇形围成一个圆锥的侧面，则这个圆锥的高及侧面积分别是A2cm,3B．2cm,3πC．22cm,6π

D．10cm,6

14C2123n-114C2123n-1．如图，△ABE和△CDE是点为位似中心的位似图形，已知点，，点C，，点D，，则点D的对应点B的标是A，2)B．，C，D．(51)y

A

y

BC

B

A

BO

D

x

O

x

NPO

O

M

A第8题图

第题图

第10题图．如图，在平面直角坐标系中A(－3，，以点O为角顶点作等腰直角三角形AOB，双曲线y1

1

在第一象限内的图象经过点设直线AB的解析式为y，当22yy时的值范围是1AB0<x<1或x<CD．0或10如图，点P是∠AOB内意一点，点和N分别是射线OA和线上的动点，△周的最小值是5cm则∠的数AB

C．D．第二部分(观)二填题(每题3分共分)．分解因:

2

．12过度包装既浪费资源又污染环境．据测算，如果全国每年减少十分之一的包装纸用量，那么能减少吨氧化碳的排放量．把数据000用学记数法表示为．13不等式组

x12的有正整数解的和为．x2)14圆内接正六边形的边心距为3cm，则这个正六边形的面积为

cm2

．15图正方形内的阴影部分是由四个直角边长都是和的直角三角形组成的假设以在正方形内部随意取点，那么这个点取在阴影部分的概率为．yC

BB

B

n-1

BC

n-1F

E

…DA

C

CC1

4第15题图

B

第17题图

CA

4

第18题图

AA

．．．．．．123n-112n-111233．．．．．．123n-112n-111233n-1n-11n-1016某服装店购进单价为15元装若干件，销售一段时间后发当销售价为元平均每天能售出件而当销售价每降低元平每天能多售出4件当每件的定为元时，该服装店平均每天的销售利润最大．17定义:只有一组对角是角的四边形叫做损矩形，连接它的两个非直角顶点的线段叫做这个损矩形的直径，即损矩形外接圆的直．如图eq\o\ac(△,，)，∠，以AC为一边向形外作菱形，点D是形ACEF角线的交点，连接，若=60º，∠=15º，=，菱形ACEF的积为．18如图，边长为n的方形OABC的、分在x轴轴正半轴上，、、A…A为OA的n等点B…B为的n等分点接ABB、．AB…A，分别交时，则=

1n

x2)点C、，BCA25三解题小10分，小10分共20分19先化简，再求:

2mm

1

．其中满一元二次方程m

(53tan

)m

．20霾天气严重影响市民的生活质量今寒假期某校八年一班的综合实践小组同学对“雾霾天气的主要成因”随机调查了所在城市部分市民，并对调查结果进行了整理，绘制了如下不完整的统计图表，观察分析并回答下列问题．⑴本次被调查的市民共有多少⑵分别补全条形统计图和扇形统计图，并计算图中区域B所应的扇形圆心角的度数．⑶若该市有100万人口，请估计持有AB两主要成因的市民有多少人？

组别雾天气的主要成因百分比A工污染45B汽尾气排放C炉烟气排放％D其(砍滥伐)人数/人908070605040302010

A45%DC15%BABD组别/组图1第题图四解题小12分，小12分共24分

图

21某化妆品专卖店，为了吸引顾客，母亲节当举办了甲、乙两种品牌化妆品有奖酬

宾活动，凡购物满88元，均可得到一次摇奖的机会已知在摇奖机内装有2个球和2个白球除色外其它都相同奖者必须从摇奖机中一次连续摇出两个球据球的颜色决定送礼金券的多(下表:甲种品牌化妆品乙种品牌化妆品

球礼金卷()球礼金卷()

两红两红

一红一白一红一白

两白两白(1)请你用列表法或画树状图求一次连续摇出一红一白两球的概率；(2)如果一个顾客当天在本店购物88元若只考虑获得最多的礼品，请你帮助分析选择购买哪种品牌的化妆品？并说明理由．22如图，我南海某海域处一艘捕鱼船在作业时突遇特大风浪，船长马上向我国渔政搜救中心发出求救信号，此时一艘渔政船正巡航到捕鱼船正西方向的处该渔政船收到渔政求救中心指令后前去救援但两船之间有大片暗礁无直线到达于决定马上调整方向，先向北偏东60方以每小时海的速度航行半小时到达处同时捕鱼船低速航行到点的正北1.5海里处渔政船航行点C时测得点D南偏东53方向上．(1)求点的距离；(2)渔政船决定再次调整航向前去援船航速不变在点处会合的正弦值．

(参考数据:sin

34，cos53，53)5北C东

º

Eº

DB

A第图五解题小题12分，24小题12分共24分)23如图，点是⊙O外点切O点AAB是⊙O的径，连接OP，过点B作∥OP交O于C，连交OP于D．(1求证PC是O的线；

(2若=

163

cmAC=8cm求图中阴影部分的面积；︵(3在)的条件下，若点E是AB的中点，连接，的．EOADCP第23题图24粮油超市平时每天都将一定数量的某些品种的粮食进行包装以便出售每天包装5大黄米的质量是包装江米质量的倍，且每天包装大黄米和江米的质量之和为45千克．4(1)求平均每天包装大黄米和江米质量各是多少千?(2)为迎接今年6月日的端节市定在节日前天增加每天包装大黄米和江米的质量二者的包装质量与天的变化情况如图所示日又恢复到原来每天的包装质量．分别求出在这内每天包装大黄米和江米的质量随天数变化的函数关系式，并写出自变量的取值范围．假该超市每天都会将当天包装后的大黄米和江米全部出售，已知大黄米成本价为每千克元，江米成本价为每千克9.5元二者包装费平均每千克均为0.5元，大黄米售价为每千克10元，江米售价为每千克12，那么在这中有哪几天销售大黄米和江米的利润之和大于元?[利=售价额－成本－包装费]每天包装的质量/克40380

15天数天第24题图

BB六解题(本满14分)25题探】(1)如图1eq\o\ac(△,)ABC中以为向外作等eq\o\ac(△,)和eq\o\ac(△,)ACD，AD=AC∠=∠，接，，试猜想BD与CE大小关系，并说明理由．【深探】(2)如图，四边形ABCD中=7cm=3cm∠ABC=∠ADC=45，求长．(3)如图，在(2)的条件下，eq\o\ac(△,)在段AC左侧时，求BD的．A

A

D

AE

DDB图1图

B

图

C第图

OO七解题(本满14分)26如图，一条抛物线与x轴交于，B两(A在的)，与

y

轴交于点，当x－1和x时y的值相等．直线y

1521x与抛物线有两个交点，其中一个交84的横坐标是6另一个交点是这条抛物线的顶点M(1)求这条抛物线的表达式．(2)动点从原点出发，在线段OB上每秒1个位长度的速度向点B运，同时动点从出发在线段上每秒2个位长度的速度向点运当个点到达终点时，另一个点立即停止运动，设运动时间为

秒．①若eq\o\ac(△,)BP为角三角形，请求出所有符合条件的

值；②求t

为何值时，四边形AC的积有最小值，最小值是多少？(3)如图，当动点P运到的点时，过点P作PD⊥

轴，交抛物线于点D，连接OD，，eq\o\ac(△,)，沿x轴左平移m个位长度，将平移后的三角形eq\o\ac(△,)重部分的面积记为，S与m的数关系式．y

y

yP

P

PABAOQ

BO

BC

M图1

C

DCMM备用图图

D第26题图

数学试卷参考案及评标准说．答仅参，卷前做答。．果生解与解不，可照评标制相评细。．阅方，解中推步写较详细但许生解过中合省非键的算骤．答端注数表考正确到一应的加数一选题(每题3分共分)．．．4．．．A．．．D．二填题(每题3分共分)11

a)()

．

．6．

2415

13

．22．

12

．三解题小题分20小10分共20分．:

1m

m=m2m

················································································2分=

m

·············································································分=

m

=

1

．·······························································································5化简方程30)o

得m2m，··································································································6分解得···························································································8分1因为当时式无意义，所以舍去；····································································9分1当m，式=2

1．··············································································10分6

DD20解:)200(人．答本次被调查的市民共有人···········································································分2)补全统人数/人908070605040302010

A45%B

％C15%300ABCD19题答1由题意可得，n，图区域B所应扇形圆心角:

组别∕组

第19题答图2360································································································分(3由题意得，

=100=75万(人．答估计持有A、种主要成因的市民有万．····················································10分四解题小题分22小12分共24分21(1)法一:用列表列出所有可能出现的结:第二个第一个红1红2白1白2

红(红红(白红(白红

红2(红，红(白红2)(白，红

白1(红白1)(红白1)(白白1)

白2(红1，白(红2，白(白1，白························································································································分由表格可知，所有可能出现的结果共有12种且每种结果出现的可能性相同其中恰好连续摇出一红一白的结果有8种所以(一红一白)=

8=．····················································51232答一次连续摇出一红一白两球的概率为．································································6分3解法二用树状图列出所有可能的结:

开始红1

红

白

白红白白红1白白红红2白2红红2白1························································································3分由树状图可知，所有可能出现的结果共有12种，且每种结果出现的可能性相同，其中恰

22好连续摇出一红一白的结果有8种所以(一红一白)=

=．·················································分答一次连续摇出一红一白两球的概率为．································································6分(2)若顾客在本店购物满元我认为该顾客应选择购买甲品牌的化妆品．·······················分理由如下由(1)得P(两红)==，P(两白)=．····························································9分62若购买甲品牌化妆品，则获得礼品卷为+6×元)；·······························1若购买乙品牌化妆品，则获得礼品卷为12×=8(元．·································116因为108，所以顾客应选择购买甲品牌的化妆品．··················································12分22解:如图，过点作CG⊥AB于G，DF⊥于，···········································1分则在eq\o\ac(△,t)中，由题意知，∴CG=

BC=

115，···············································································2分22∵∠DAG，∴四边形ADFG是矩形，∴=AD=1.5，∴==7.－，······························在Req\o\ac(△,t)CDF中∠CFD，∵∠，，∴∠=CD

B

北º

东

CºFG第22题答图

H

EDA∴CD

(海里)．·············································································7分cos5335答两距离为海里．····················································································分(2)如图，设渔政船调整方向后t小能与捕渔船相会合，由题意知=30，DEt=3，∠，······················································分过点作⊥于H则∠EHD∠CHE，EH∴sin∠EDH，ED∴=sin53°=tt

，·················································································11分∴在eq\o\ac(△,t)EHC中，sin∠=

tCE30t25

．答∠=

．································································································12分

2∴=－=2∴=－=O五解题小题分24小12分共24分23证明:⑴图，连接OC∵PA切于A．∴∠．·····································································································1分∵OP∥，∴∠=，∠=∠．∵OC=OB，∴∠OBC=，∴∠=．······························································································3又∵OAOC，=，∴△≌()．∴∠∠=90，又∵OC是O的径，∴PC是⊙O的线．·····························································································⑵解法一由(1)得PAPC都为圆的切线，∴=PC，平分∠APC，∠=∠=90，∴∠PAD+∠DAO=∠∠AOD，∴∠=AOD，∴△ADO△PDA······························································································分∴

ADDO，PDAD∴AD

，16∵，PD，31∴=，OD，AO，················································································分2由题意知OD为△的位线，∴OD=6，．·························································································分25

答阴影部分的面积为

482

cm．··········································································分解法二∵AB是O的径，OP∥，∴∠=ACB=90º．∵∠PCO=，∴∠+=∠∠，

E即∠=．又∵∠=∠，∴∠=，

A

D

O

M

B∴△∽ACB，································分∴AC16又∵AC=8，PD3

．，

P

C第23题答图

22∴=，PC=2．·······································································分320∴，∴，AB，·································································································8分∴=S=

2

2

答阴影部分的面积为

482

cm

．··········································································分(3)如图，连接AE，过点作BM⊥点M．·····················································10分∴∠=EMB=∠，︵又∵点是的中点，∴∠ECB=∠CBM=∠=，=MB2，BE=ABº=，···························11分∴EMBM∴=+EM

=42，答CE的为cm．·························································································24解:解法一:设平均每天包装大黄米和江米的质量分别为a千和千，········································分则

，解得

．················································································分答平均每天包装大黄米和江米的质量分别为千和20千．····································解法二设平均每天包装大黄米的质量为a克，那么包装江米的质(a)千克，·····················则(45)，得a，所以45．····················································答平均每天包装大黄米和江米的质量分别为千和20千．····································观察图象，可设平均每天包装大黄米的质量与天数的关系式为yx，平均每天包装江米的质量与天数的关系式为．22①当x由题意x的象过点1

分分分分则可列方程组为

25k1，得1b11

，∴y25．······································································································分1由题意y的图象过点22

111112则可列方程组为

2k2

，解得

,∴20．··································································································65②当由题意yx的象过点则可列方程组为

k1，解得k

k1851

，∴y．···································································································7分由题意yx的图象过点2则可列方程组为

20k38

，解得

k，∴y

185

x92．································································································分∴

x

x，y18

．························分设第①当0.51.6yy12x．由题意4x120，x，∴x的值范围为由题意知．···················································································10分②当0.5

181.6yy1.685x320．

由题意x，x

503

，∴x的值范围为x．由题意知．··································································································11分答由①、②可知在第11，，1415，16中销售大黄米和江米的总利润大于元．··························································································12分六解题(本满14分)

222222222225．(1):BD．································································································理由:∵∠BAE∠，∴∠+∠BAC=∠CAD+∠，即∠EAC∠BAD，···················································又∵AE=AB，∴△EAC≌△(SAS)，∴=．··········································································································(2)解如图△ABC外部点A为角顶点作等腰直角三角形∠BAE，=AB连接、、EC··················································································∵ACDADC45

分分分分∴ACAD，CAD∴∠=

E∴∠+∠BAC=∠CAD+∠，即∠EAC=∠，

A∴△EAC≌△(SAS)，·····················7分∴=．∵AE=，

D∴BE7+7=7，∠AEC∠AEB．

B

第

C题答图又∵∠ABC，∴∠ABC+∠ABE，···············································································分∴=BC=

=107，∴BD答BD是107cm．····························································································分(3)如图，在线段AC的右侧过点作AE⊥于，交的长线于点，·················10分∴∠，又∵∠ABC，∴∠E∠ABC，∴AE=，7=7．··········································································11分又∵∠=∠=45，∴∠=∠DAC，∴∠BAEBAC即∠EAC=∠，∴△EAC≌△(SAS)，

D

A∴=．································分∵，

B

C

E第25题答2∴==2

3(cm)．答BD是

72

3)cm．······················································································14分

12=12=七解题(本满14分)26解:∵x，的值相等，抛物线的对称轴为直线x，x和别代入

中得顶点(1,)，一个交点坐标(，，···············分则可设抛物线的表达式为(x，(，代入其中，解得，82733∴抛物线的表达式为y(，2x．········································3分8843(2)如图，当时xx得xx．由题意知，(4B，0)

，0)所以OA=4当x时以点C由股定理知BC=5，OP=1×=t，2t①∵∠PB是角，

．·····················································································分∴有∠P或∠Q两情况:当∠PB=90º时，可eq\o\ac(△,)PB∽△，BQPBt4∴，，BO∴t；·············································