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襄阳市樊城区2019-2020学年度下学期期末学业质量调研测试八年级数学试题(word版附答案)

襄阳市樊城区2019-2020学年度下学期期末学业质量调研测试八年级数学试题(word版附答案)
襄阳市樊城区2019-2020学年度下学期期末学业质量调研测试八年级数学试题(word版附答案)

襄阳市樊城区2019-2020学年度下学期期末学业水平测试

八年级数学试题

一、选择题(每小题3分,共30分)

1.二次根式在实数范围内有意义,则x应满足的条件是()

A.x≥1

B.x>1

C.x>-1

D.x≥-1

2.以下列各组数据为边长作三角形,其中能组成直角三角形的是()

A.5,12,13

B.3,5,

C.6,9,14

D.4,10,13

3.下列二次根式,可以与合并的是()

A. B. C. D.

4.某共享单车前a公里1元,超过a公里的,每公里2元,若要使使用该共享单车50%的人只花1元钱,a应该要取什么数()

A.平均数

B.中位数

C.众数

D.方差

5.已知正比例函数y=kx,且y随x的增大而减少,则直线y=2x+k的图象是()

A. B. C. D.

6.矩形具有而菱形不一定具有的性质是()

A.对角线相等

B.四边相等

C.对角线互相垂直

D.对角线互相平分

7.甲、乙两人在一次赛跑中,路程s和t时间的关系如图所示(实线部分为甲,虚线部分为乙),李丽同学得到如下信息,其中错误的是()

A.甲、乙两人的速度不相等

B.甲乙两人同时起跑

C.两人跑的路程相等

D.乙用的时间少

8.如图,将矩形ABCD的四个角向内折叠铺平,恰好拼成一个无缝隙无重叠的矩形EFGH,若EH=5,EF=12.则矩形ABCD的面积是()

A.13

B.

C.60

D.120

第7题图第8题图第9题图第10题图

9.如图,在□ABCD中,分别以A,C为圆心,以大于的长为半径画弧,两弧

相交于M,N两点,直线MN交AD于点E,若△CDE的周长是12,则□ABCD的周长为()

A.22

B.24

C.32

D.44

10.如图所示,梯子AB 斜靠在墙面上,AO ⊥BO ,AO =BO =2米,当梯子的顶点A 沿AO 方向向下滑动以a(0<a <2)米时,梯足B 沿OB 方向滑动b(0<b <2)米,则a 与b 的大小关系是() A.a =b B.a <b C.a >b D.不确定 二、填空题(每小题3分,共18分) 11.已知

,则

__________.

12.若一次函数y =kx+b 的图像不经过第三象限,则k ,b 的取值范围分别为k__________0,b__________0. 13.数据-2,-1,0,1,2的方差是__________.

14.如图,已知菱形ABCD 的周长为16,∠BCD =120°,E 为AB 的中点,若P 为对角线BD 上一动点,则EP+AP 的最小值为________.

15.如图,在平面直角坐标系xOy 中,若直线y 1=-x+a 与直线y 2=bx-4相交于点P(1,-3),则关于x 的不等式-x+a ≤bx-4的解集是__________. 16.如图,一只蚂蚁沿着边长为2的正方体表面从点A 出发,经过3个面爬到点B ,如果它运动的路径是最短的,则此最短路径的长为__________. 三、解答题(共72分) 17.(6分)计算:

18.(8分)八年级甲、乙两班各有学生50人,为了解这两个班学生上网课期间身体素质情

况,进行了抽样调查,过程如下,请补充完整:

收集数据:从甲、乙两个班各随机抽取10名学生进行身体素质测试,测试成绩百分制,规定测试成绩等次如下表:

成绩分段 50≤x <60 60≤x <70 70≤x <80 80≤x <90 90≤x ≤100 等次评定

不合格

合格

良好

优秀

特优

收集成绩如下:

甲班:65,75,75,85,55,50,75,100,95,65 乙班:90,60,85,70,65,70,90,80,60,70 整理描述数据:用下面直方图描述这两组样本数据,

分析数据:两组样本数据的平均数、中位数、众数、方差如上表所示. 结论运用:

(1)整理数据并补全图中的直方图;表中x =__________,y =__________.

(2)请估计乙班50名学生中身体素质为优秀以上的学生约有__________人,甲班50名学生中身体素质为不合格的学生约有__________人;

(3)你认为哪个班的学生身体素质较好?请至少从两个方面说明理由.

班级 平均数 中位数 众数 方差

甲班

74 75 75 234 乙班

74

x

y

119

19.(7分)如图,四边形ABCD中,AB⊥AD,已知AD=3cm,AB=4cm,CD=13cm,BC=12cm,求四边形ABCD的面积.

20.(7分)如图,直线AB与x轴交于点A(1,0),与y轴交于点B(0,-2).

(1)求直线AB的解析式;

(2)若直线AB上的点C在第一象限,且S△BOC=2,求点C的坐标.

21.(7分)如图,点E、F是□ABCD的对角线BD上两点,且BE=DF.求证:四边形AECF是平行四边形.

22.(7分)如图,正方形网格中的每个小正方形的边长都是1,每个小格的顶点叫格点.

(1)在图1中以格点为顶点画一个面积为5的正方形.

(2)如图2所示,点A、B、C是小正方形的顶点,则__________.

23.(8分)如图,△ABC是以BC为底的等腰三角形,AD是边BC上的高,点E、F分别是AB、AC的中点.

(1)求证:四边形AEDF是菱形;

(2)如果四边形AEDF的周长为12,两条对角线的和等于7,求四边形AEDF的面积S.

24.(10分)某运动品商场欲购进篮球和足球共100个,两种球进价和售价如下表所示.设

25.(12分)如图,点P是正方形ABCD对角线AC上一动点,点E在射线BC上,连接PD,O 为AC中点.

(1)如图1,当点P在线段AO上时,若PD=PE,试猜想PE与PD的位置关系,并说明理由;

(2)如图2,当点P在线段OC上时,若PD⊥PE,试猜想PE与PD的数量关系,并说明理由;

(3)如图2,在(2)的条件下,若∠PDC=30°,,直接写出CE的值.

襄阳市樊城区2019-2020学年度下学期期末学业水平测试

八年级数学试题参考答案

1 2 3 4 5 6 7 8 9 10

A A D

B D A B D B C

11.;12.>,≤;13.2;2;15.x≥1;16.

三、解答题

17.解:原式=

===

18.解:(1)由收集的数据得知:乙班成绩的:60,60,65,70,70,70,80,85,90,90,乙班成绩不合格人数为0人,合格人数3人,良好人数3人,优秀人数2人,特优人数2人,补全图中的直方图如图所示:

70,70;

(2)估计乙班50名学生中身体素质为优秀的学生有=20(人),

甲班50名学生中身体素质为不合格的学生约有=10(人),

故答案为2010;

(3)可以推断出乙班学生的身体素质更好一些,理由一:两班级学生的平均数相同,而乙班的方差小于甲班的方差,说明乙班学生的身体素质更好一些;

理由二:从条形统计图上可以看出乙班的不合格人数为0,而甲班的不合格人数2人,良好,特优人数相同,其余的乙班人数都高于甲班,说明乙班学生的身体素质更好一些.

19.解:连接BD,如图,∵AD=4cm,AB=3cm,AB⊥AD,

∴BD==5(cm)

∴S

△ABD

=AB?AD=6(cm2).

在△BDC中,∵52+122=132,即BD2+BC2=CD2,

∴△BDC为直角三角形,即∠DBC=90°,∴S

△DBC

=BD?BC=30(cm2),

∴S

四边形ABCD =S

△BDC

-S

△ABD

=30-6=24(cm2).

答:四边形ABCD的面积为24cm2.

20.解:(1)令直线AB解析式为y=kx+b,把A(1,0),B(0,-2)两点坐标代入上式得,,解之,,∴y=2x-2;

=2得,②,

(2)令点C(m,n)可得,n=2m-2①,由S

△BOC

由①,②得,m=2,n=2,∴C(2,2).

21.证明:连接AC,交BD于点O,如图所示:

∵四边形ABCD是平行四边形,∴OA=OC,OB=OD,

∵BE=DF,∴OB-BE=OD-DF,即OE=OF,

∵OA=OC,∴四边形AECF是平行四边形.

22.解:(1)如图1所示:

(2)由图2,可知:AB=,BC=,

∴.故答案为.

23.(1)证明:在等腰△ABC中,AD⊥BC,∴D为BC中点

∵E是AB中点,∴DE//AC,且

∵F是AC中点,∴,DE//AF,且DE=AF,

∴四边形AEDF是平行四边形.

由,,AB=AC,∴AE=AF,∴□AEDF是菱形;

(2)解:如图,连接EF交AD于O,

∵菱形AEDF的周长为12,∴AE=3,

设EF=x,AD=y,则x+y=7,∴x2+2xy+y2=49,①

∵AD⊥EF于O,∴Rt△AOE中,AO2+EO2=AE2,

∴(y)2+(x)2=32,即x2+y2=36,②

把②代入①,可得2xy=13,∴xy=,

∴菱形AEDF的面积S=xy=.

24.解:(1)W=(76-62)x+(60-54)(100-x)=8x+600(0<x≤100,x为整数); (2)由5800≤62x+54(100-x)≤6000得50≤x≤75

W=(76-a-62)x+(60-54)(100-x)=(8-a)x+600(0<a<14)

①当8-a<0,即8<a<14时,W随x的增大而减小,x=50时,利润最大,

∴此时购进篮球50个,足球50个;

②当8-a=0,即a=8时,W=600,

利润不变,

∴此时购进篮球和足球共100个即可;

③当8-a>0,即0<a<8时,W随x的增大而增大,x=75时,利润最大,

∴此时购进篮球75个,足球25个.

答:当0<a<8时,篮球购进75个,足球购进25个获利最大;当a=8时,无论怎样进货,利润是定值;当a>8时,篮球购进50个,足球购进50个获利最大.

25.解:(1)PE⊥PD.

理由:过P分别作PM⊥BC于M,PN⊥CD于N.

在正方形ABCD中,AC平分∠BCD,∴PM=PN

在Rt△PDN与Rt△PEM中,

,Rt△PDN≌Rt△PEM(HL)

∴∠1=∠2

在四边形PMCN中,

∠MPN=360°-90°-90°-90°=90°

∴∠2+∠EPN=90°

∴∠1+∠EPN=90°

∴PE⊥PD.

(2)PE=PD.

理由:过P分别作PM⊥BC于M,PN⊥CD于N.

由(1)可得,PN=PM,∠MPN=90°∴∠2+∠EPM=90°

∵PE⊥PD,∴∠1+∠2=90°

∴∠1=∠EPM

在Rt△PEM与Rt△PDN中,

∴Rt△PDN≌Rt△PEM(ASA)

∴PE=PD.

(3),理由如下:

连接DE,如图:

∵四边形ABCD是正方形

∴∠PCN=45°

在Rt△PNC中

∴PN=NC=

在Rt△PND中

∵∠PDC=30°

∴PD=2PN=2

∴DN=

∴DC=DN+NC=

∵PD⊥PE,PD=PE

∴DE=

在Rt△DCE中

CE==(负值舍去).

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