2018-2019学年江苏省苏州市姑苏区八年级(下)期末数学试卷

2018-2019学年江苏省苏州市姑苏区八年级(下)期末数学试卷
2018-2019学年江苏省苏州市姑苏区八年级(下)期末数学试卷

2018-2019学年江苏省苏州市姑苏区八年级(下)期末数学试卷

一、选择题(本大题共10小题,每小题2分,共20分.)

1.(2分)若二次根式有意义,则x的取值范围是()

A.x>1B.x≥1C.x<1D.x≤1

2.(2分)剪纸是潍坊特有的民间艺术,在如图所示的四个剪纸图案中.既是轴对称图形又是中心对称图形的是()

A.B.C.D.

3.(2分)下列二次根式中,可与合并的二次根式是()

A.B.C.D.

4.(2分)完成以下任务,适合用抽样调查的是()

A.调查你班同学的年龄情况

B.为订购校服,了解学生衣服的尺寸

C.对北斗导航卫星上的零部件进行检查

D.考察一批炮弹的杀伤半径.

5.(2分)下列说法:①“掷一枚质地均匀的硬币,朝上一面可能是正面”;②“从一副普通扑克牌中任意抽取一张,点数一定是3”()

A.只有①正确B.只有②正确C.①②都正确D.①②都错误

6.(2分)若P1(x1,y1),P2(x2,y2)是函数y=图象上的两点,当x1>x2>0时,下列结论正确的是()A.0<y1<y2B.0<y2<y1C.y1<y2<0D.y2<y1<0

7.(2分)如图,在平行四边形ABCD中,点E在边DC上,连接AE并延长交BC的延长线于点F,若AD=3CF,那么下列结论中正确的是()

A.FC:FB=1:3B.CE:CD=1:3C.CE:AB=1:4D.AE:AF=1:2.

8.(2分)如图,点A是射线y═(x≥0)上一点,过点A作AB⊥x轴于点B,以AB为边在其右侧作正方形ABCD,过点A的双曲线y=交CD边于点E,则的值为()

A.B.C.D.1

9.(2分)如图,四边形OABC和四边形BDEF都是正方形,反比例函数y=在第一象限的图象经过点E,若两正方形的面积差为12,则k的值为()

A.12B.6C.﹣12D.8

10.(2分)如图,正方形纸片ABCD的边长为4cm,点M、N分别在边AB、CD上.将该纸片沿MN折叠,使点D 落在边BC上,落点为E,MN与DE相交于点Q.随着点M的移动,点Q移动路线长度的最大值是()

A.2 cm B.4 cm C.cm D.1 cm

二.填空题(本大题共8小题,每小题2分,共16分.把答案直接填在答题卡相应位置上.)

11.(2分)抛掷一枚质地均匀的骰子1次,朝上一面的点数不小于3的概率是.

12.(2分)一水塘里有鲤鱼、鲢鱼共10000尾,一渔民通过多次捕捞试验后发现,鲤鱼出现的频率为0.36,则水塘

有鲤鱼尾.

13.(2分)已知=,则=.

14.(2分)当x=1﹣时,x2﹣2x+2028=.

15.(2分)如图,比例规是一种画图工具,使用它可以把线段按一定的比例伸长或缩短,它是由长度相等的两脚AD和BC交叉构成的,如果把比例规的两脚合上,使螺丝钉固定在刻度3的地方(即同时使OA=3OD,OB=3OC),然后张开两脚,使A、B两个尖端分别在线段l的两端上,若CD=2,则AB的长是.

16.(2分)如图,已知在△ABC中,BC边上的高AD与AC边上的高BE交于点F,且∠BAC=45°,BD=6,CD =4,则△ABC的面积为.

17.(2分)如图,在边长相同的小正方形网格中,点A、B、C、D都在这些小正方形的顶点上,AB,CD相交于点P,则△PBD与△P AC的面积比为.

18.(2分)如图,正方形ABCD中,AB=30,点E在边CD上,且CD=3DE.将△ADE沿AE对折至△AFE,延长EF交边BC于点G.连接AG、CF.下列结论:①△ABG≌△AFG;②BG=15;③△CFG是正三角形;④△FGC的面积为90.其中正确的是(填所有正确答案的序号).

三、解答题(11题,64分)

19.(4分)计算:|1﹣|﹣+()0

20.(4分)解分式方程:﹣1=.

21.(4分)先化简,再求值:(1+)÷.,其中a=+3.

22.(6分)某校有3000名学生.为了解全校学生的上学方式,该校数学兴趣小组以问卷调查的形式,随机调查了该校部分学生的主要上学方式(参与问卷调查的学生只能从以下六个种类中选择一类),并将调查结果绘制成如下不完整的统计图.

某校部分学生主要上学方式扇形统计图某校部分学生主要上学方式条形统计图

根据以上信息,回答下列问题:

(1)参与本次问卷调查的学生共有人,其中选择B类的人数有人.

(2)在扇形统计图中,求E类对应的扇形圆心角α的度数,并补全条形统计图.

(3)若将A、C、D、E这四类上学方式视为“绿色出行”,请估计该校每天“绿色出行”的学生人数.23.(6分)如图,在平面直角坐标系中,以原点O为位似中心,将△OAB放大到原来的2倍后得到△OA′B′,其中A、B在图中格点上,点A、B的对应点分别为A′、B′.

(1)在第一象限内画出△OA′B′;

(2)若△OAB的面积为3.5,求△OA′B′的面积.

24.(6分)反比例函数y1=(x>0)的图象与一次函数y2=﹣x+b的图象交于A,B两点,其中A(1,2)(1)求这两个函数解析式;

(2)在y轴上求作一点P,使P A+PB的值最小,并直接写出此时点P的坐标.

25.(6分)如图,在Rt△ABC中,∠C=90°,△ACD沿AD折叠,使得点C落在斜边AB上的点E处.(1)求证:△BDE∽△BAC;

(2)已知AC=6,BC=8,求线段AD的长度.

26.(6分)如图,△ABC为锐角三角形,AD是BC边上的高,正方形EFMN的一边MN在边BC上,顶点E、F 分别在AB、AC上,其中BC=24cm,高AD=12cm.

(1)求证:△AEF∽△ABC:

(2)求正方形EFMN的边长.

27.(7分)如图,在矩形ABCD中,AC为对角线,点P为BC边上一动点,连

接AP,过点B作BQ⊥AP,垂足为Q,连接CQ.

(1)证明:△ABP∽△BQP;

(2)当点P为BC的中点时,若∠BAC=37°,求∠CQP的度数;

(3)当点P运动到与点C重合时,延长BQ交CD于点F,若AQ=AD,则=.

28.(7分)(1)某学校“智慧方园”数学社团遇到这样一个题目:

如图1,在△ABC中,点O在线段BC上,∠BAO=30°,∠OAC=75°,AO=,BO:CO=1:3,求AB 的长.

经过社团成员讨论发现,过点B作BD∥AC,交AO的延长线于点D,通过构造△ABD就可以解决问题(如图2).请回答:∠ADB=°,AB=.

(2)请参考以上解决思路,解决问题:

如图3,在四边形ABCD中,对角线AC与BD相交于点O,AC⊥AD,AO=,∠ABC=∠ACB=75°,BO:OD=1:3,求DC的长.

29.(8分)如图1,点O是正方形ABCD的中心,点E是AB边上一动点,在BC上截取CF=BE,连接OE,OF.初步探究:在点E的运动过程中:

(1)猜想线段OE与OF的关系,并说明理由.

深入探究:

(2)如图2,连接EF,过点O作EF的垂线交BC于点G.交AB的延长线于点I.延长OE交CB的延长线于点H.

①直接写出∠EOG的度数.

②若AB=2,请探究BH?BI的值是否为定值,若是,请求出其值;反之,请说明理由

2018-2019学年江苏省苏州市姑苏区八年级(下)期末数学试卷

参考答案与试题解析

一、选择题(本大题共10小题,每小题2分,共20分.)

1.【解答】解:∵二次根式有意义,

∴x﹣1≥0,

∴x≥1.

故选:B.

2.【解答】解:A、此图形沿一条直线对折后不能够完全重合,∴此图形不是轴对称图形,不是中心对称图形,故此选项错误;

B、此图形沿一条直线对折后能够完全重合,∴此图形是轴对称图形,不是中心对称图形,故此选项错误.

C、此图形沿一条直线对折后能够完全重合,∴此图形是轴对称图形,旋转180°能与原图形重合,是中心对称

图形,故此选项正确;

D、此图形沿一条直线对折后不能够完全重合,∴此图形不是轴对称图形,是中心对称图形,故此选项错误.

故选:C.

3.【解答】解:A、原式==,符合题意;

B、原式==,不符合题意;

C、与不是同类二次根式,不符合题意;

D、原式=3,不符合题意,

故选:A.

4.【解答】解:A、人数不多,容易调查,宜采用全面调查;

B、为订购校服,了解学生衣服的尺寸是要求精确度高的调查,适合全面调查;

C、对北斗导航卫星上的零部件进行检查,因为调查的对象比较重要,应采用全面调查;

D、考察一批炮弹的杀伤半径适合抽样调查;

故选:D.

5.【解答】解:掷一枚质地均匀的硬币,朝上一面可能是正面,可能是反面,所以①正确;

从一副普通扑克牌中任意抽取一张,点数不一定是3,所以②错误,

故选:A.

6.【解答】解:把点P1(x1,y1)、P2(x2,y2)代入y=得y1=,y2=,

则y1﹣y2=﹣=,

∵x1>x2>0,

∴x1x2>0,x2﹣x1<0,

∴y1﹣y2=<0,

即y1<y2.

故选:A.

7.【解答】解:∵在平行四边形ABCD中,

∴AD∥BC,

∴△ECF∽△ADE,

∵AD=3CF,

A、FC:FB=1:4,错误;

B、CE:CD=1:4,错误;

C、CE:AB=1:4,正确;

D、AE:AF=3:4.错误;

故选:C.

8.【解答】解:设点A的横坐标为m(m>0),则点B的坐标为(m,0),把x=m代入y=x得:y=m,

则点A的坐标为:(m,m),线段AB的长度为m,点D的纵坐标为m,∵点A在反比例函数y=上,

∴k=m2,

即反比例函数的解析式为:y=,

∵四边形ABCD为正方形,

∴四边形的边长为m,

点C,点D和点E的横坐标为m+m=m,

把x=m代入y=得:

y=m,

即点E的纵坐标为m,

则EC=m,DE=m﹣m=m,

=,

故选:A.

9.【解答】解:设正方形OABC、BDEF的边长分别为a和b,则D(a,a﹣b),F(a+b,a),所以E(a+b,),

所以=a﹣b,

∴(a+b)(a﹣b)=k,

∴a2﹣b2=k,

∵两正方形的面积差为12,

∴k=12.

故选:A.

10.【解答】解:如图,取AB、CD中点K、G,连接KG、BD交于点O.

由题意可知点Q运动的路线就是线段OG,

∵DO=OB,DG=GC,

∴OG=BC=×4=2.

∴点Q移动路线长度的最大值是2.

故选:A.

二.填空题(本大题共8小题,每小题2分,共16分.把答案直接填在答题卡相应位置上.)

11.【解答】解:∵抛掷一枚质地均匀的骰子1次共有6种等可能结果,朝上一面的点数不小于3的有4种结果,所以朝上一面的点数不小于3的概率是=,

故答案为:.

12.【解答】解:水塘约有鲤鱼10000×0.36=3600(尾),

故答案为:3600.

13.【解答】解:∵=,

∴=,

∴﹣=,

=.

故答案为:.

14.【解答】解:当x=1﹣时,

x2﹣2x+2028=(x﹣1)2+2027

=(1﹣﹣1)2+2027

=(﹣)2+2027,

=3+2027

=2030,

故答案为:2030.

15.【解答】解:根据题意,可知:△ABO∽△DCO,

∴=,即=3,

∴AB=6.

故答案为:6.

16.【解答】解:∵AD⊥BC,BE⊥AC,∴∠AEF=∠BEC=∠BDF=90°,∵∠BAC=45°,

∴AE=EB,

∵∠EAF+∠C=90°,∠CBE+∠C=90°,

∴∠EAF=∠CBE,

∴△AEF≌△BEC,

∴AF=BC=10,设DF=x.

∵△ADC∽△BDF,

∴=,

∴=,

整理得x2+10x﹣24=0,

解得x=2或﹣12(舍弃),

∴AD=AF+DF=12,

∴S△ABC=?BC?AD=×10×12=60.

故答案为60.

17.【解答】解:∵BD∥AC,BD=1,AC=3,

∴△DBP∽△CAP,

∴=()2=,

故答案为.

18.【解答】解:①在正方形ABCD中,AD=AB,∠D=∠B=∠C=90°,又∵△ADE沿AE对折至△AFE,延长EF交边BC于点G

∴∠AFG=∠AFE=∠D=90°,AF=AD,

即有∠B=∠AFG=90°,AB=AF,AG=AG,

在直角△ABG和直角△AFG中,

∴△ABG≌△AFG;正确.

②∵AB=6,点E在边CD上,且CD=3DE,

∴DE=FE=10,CE=20,

不妨设BG=FG=x,(x>0),

则CG=30﹣x,EG=10+x,

在Rt△CEG中,(10+x)2=202+(30﹣x)2

解得x=15,于是BG=GC=15;正确.

③∵BG=GF=CG,

∴△CFG是等腰三角形,

∵BG=AB,

∴∠AGB≠60°,

则∠FGC≠60°,

∴△CFG不是正三角形.错误.

④∵=,

∴=,

∴S△FGC=S△EGC=××20×15=90.正确.正确的结论有①②④.

故答案为:①②④.

三、解答题(11题,64分)

19.【解答】解:|1﹣|﹣+()0

=﹣1﹣2+1

=﹣

20.【解答】解:两边都乘以3(x﹣1),得:3x﹣3(x﹣1)=2x,

解得:x=1.5,

检验:x=1.5时,3(x﹣1)=1.5≠0,

所以分式方程的解为x=1.5.

21.【解答】解:原式=?

当a=+3时,

原式=

22.【解答】解:(1)参与本次问卷调查的学生共有162÷36%=450人,其中选择B类的人数有450×14%=63人,故答案为:450、63;

(2)E类对应的扇形圆心角α的度数360°×(1﹣36%﹣14%﹣20%﹣16%﹣4%)=36°,

C方式的人数为450×20%=90人、D方式人数为450×16%=72人、E方式的人数为450×10%=45人,F方式的人数为450×4%=18人,

补全条形图如下:

(3)估计该校每天“绿色出行”的学生人数为3000×(1﹣14%﹣4%)=2460人.

23.【解答】解:(1)△OA′B′如图所示:

(2)∵将△OAB放大到原来的2倍后得到△OA′B′,

∴S△OAB:S△OA′B′=1:4,

∴S△OA′B′=4×S△OAB=4×3.5=14.

24.【解答】解:(1)将点A(1,2)代入y1=,得:k=2,

则y1=;

将点A(1,2)代入y2=﹣x+b,得:﹣1+b=2,

解得:b=3,

则y2=﹣x+3;

(2)作点A关于y轴的对称点A′(﹣1,2),连接A′B,交y轴于点P,即为所求,如图所示:

由得:或,

∴B(2,1),

设A′B所在直线解析式为y=mx+n,

根据题意,得:,

解得:,

则A′B所在直线解析式为y=﹣x+,

当x=0时,y=,

所以点P(0,).

25.【解答】证明:(1)∵∠C=90°,△ACD沿AD折叠,∴∠C=∠AED=90°,

∴∠DEB=∠C=90°,

又∵∠B=∠B,

∴△BDE∽△BAC;

(2)由勾股定理得,AB=10.

由折叠的性质知,AE=AC=6,DE=CD,∠AED=∠C=90°.∴BE=AB﹣AE=10﹣6=4,

在Rt△BDE中,由勾股定理得,

DE2+BE2=BD2,

即CD2+42=(8﹣CD)2,

解得:CD=3,

在Rt△ACD中,由勾股定理得AC2+CD2=AD2,

即32+62=AD2,

解得:AD=.

26.【解答】(1)证明:∵四边形EFMN是正方形,

∴EF∥BC,

∴∠AEF=∠B,∠AFE=∠C,

∴△AEF∽△ABC.

(2)解:设正方形EFMN的边长为x.

∵△AEF∽△ABC,AD⊥BC,

∴=,

∴=,

∴x=8,

∴正方形的边长为8cm.

27.【解答】(1)证明:∵四边形ABCD是矩形,∴∠ABP=90°,

∵BQ⊥AP,

∴∠BQP=∠ABP=90°,

∵∠BPQ=∠APB,

∴△ABP∽△BQP.

(2)解:∵△ABP∽△BQP,

∴=,

∴PB2=PQ?P A,

∵PB=PC,

∴PC2=PQ?P A,

∴=,

∵∠CPQ=∠APC,

∴△CPQ∽△APC,

∴∠PQC=∠ACP,

∵∠BAC=37°,

∴∠ACB=90°﹣37°=53°,

∴∠CQP=53°.

(3)解:连接AF.

∵∠D=∠AQF=90°,AF=AF,AD=AQ,

∴Rt△ADF≌Rt△AQF(HL),

∴DF=QF,设AD=AQ=BC=m,DF=FQ=x,FC=y,CQ=a,∵∠BCF=∠CQB=∠CQF=90°,

∴∠BCQ+∠FCQ=90°,∠∠CBQ=90°,

∴∠FCQ=∠CBQ,

∴△BCQ∽△CFQ,

∴=,

∴=,

∴=,

∵CF∥AB,

∴=,

∴=,

∴=,

∴x2+xy﹣y2=0,

∴x=y或y(舍弃),

∴=,

∴=.

故答案为.

28.【解答】解:(1)∵BD∥AC,

∴∠ADB=∠OAC=75°.

∵∠BOD=∠COA,

∴△BOD∽△COA,

∴==.

又∵AO=,

∴OD=AO=,

∴AD=AO+OD=4.

∵∠BAD=30°,∠ADB=75°,

∴∠ABD=180°﹣∠BAD﹣∠ADB=75°=∠ADB,∴AB=AD=4.

故答案为:75;4.

(2)过点B作BE∥AD交AC于点E,如图所示.∵AC⊥AD,BE∥AD,

∴∠DAC=∠BEA=90°.

∵∠AOD=∠EOB,

∴△AOD∽△EOB,

∴==.

∵BO:OD=1:3,

∴==.

∵AO=3,

∴EO=,

∴AE=4.

∵∠ABC=∠ACB=75°,

∴∠BAC=30°,AB=AC,

∴AB=2BE.

在Rt△AEB中,BE2+AE2=AB2,即(4)2+BE2=(2BE)2,解得:BE=4,

∴AB=AC=8,AD=12.

在Rt△CAD中,AC2+AD2=CD2,即82+122=CD2,

解得:CD=4.

29.【解答】解:(1)OE=OF,OE⊥OF,

连接AC,BD,

∵点O是正方形ABCD的中心

∴点O是AC,BD的交点

∴BO=CO,∠ABO=∠ACB=45°,∠BOC=90°

∵CF=BE,BO=CO,∠ABO=∠ACB

∴△BEO≌△CFO(SAS)

∴OE=OF,∠BOE=∠COF

∵∠COF+∠BOF=90°,

∴∠BOE+∠BOF=90°

∴∠EOF=90°,

∴EO⊥FO,

(2)

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