吉林省中考数学试题含答案word版.docx
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吉林省中考数学试题含答案word版
吉林省2009年初中毕业生学业考试
数学试卷
一、填空题(每小题2分,共20分
1.数轴上A、B两点所表示的有理数的和是.
2.计算25
(3aa·=.
3.为鼓励大学生创业,某市为在开发区创业的每位大学生提供货款150000元,这个数据用科学记数法表示为元.
4.不等式23xx>-的解集为.
5.如图,点A关于y轴的对称点的坐标是.
6.方程
312
x=-的解是.
7.若a5,2,0,babab==->+=且则.
8.将一个含有60°角的三角板,按图所示的方式摆放在半圆形纸片上,O为圆心,则ACO∠=度.
9.如图,OAB△的顶点B的坐标为(4,0,把OAB△沿x轴向右平移得到CDE△,如果1,CB=那么OE
的长为.
10.将一张矩形纸片折叠成如图所示的形状,则∠ABC=度.
(第5题
4-0123
A
B
(第1题
3-2-1-
(第8题
B
(第9题
C34°
B
A
(第10题
二、单项选择题(每小题3分,共18分11.化简2
244
xyyxx--+的结果是(
A.
2
xx+B.2
xx-C.
2
yx+D.
2
yx-
12.下列图案既是轴对称图形,又是中心对称图形的是(
A.
B.
C.
D.13.下列几何体中,同一个几何体的主视图与俯视图不同的是(
A.
B.
C.
D.
14.A种饮料B种饮料单价少1元,小峰买了2瓶A种饮料和3瓶B种饮料,一共花了13元,如果设B种饮料单价为x元/瓶,那么下面所列方程正确的是(A.2(1313xx-+=B.2(1313xx++=C.23(113xx++=D.23(113xx+-=
15.某校七年级有13名同学参加百米竞赛,预赛成绩各不相同,要取前6名参加决赛,小梅已经知道了自己的成绩,她想知道自己能否进入决赛,还需要知道这13名同学成绩的(A.中位数B.众数C.平均数D.极差
16.将宽为2cm的长方形纸条折叠成如图所示的形状,那么折痕PQ的长是(A.
B.
CD.2cm
三、解答题(每小题5分,共20分
17.在三个整式2
2
2
2,2,xxyyxyx++中,请你任意选出两个进行加(或减运算,使所得整式可以因式分解,并进行因式分解.
圆柱
正方体圆锥球60°
PQ
2cm
(第16题
18.在一个不透明的盒子里,装有三个分别写有数字6,-2,7的小球,它们的形状、大小、质地等完全相同,先从盒子里随机取出一个小球,记下数字后放回盒子,摇匀后再随机取出一个小球,记下数字.请你用画树形图或列表的方法,求下列事件的概率:
(1两次取出小球上的数字相同;(2两次取出小球上的数字之和大于10.
19.如图,,ABACADBCDADAEABDAEDEF=⊥=∠于点,,平分交于点,请你写出图中三对..全等三角形,并选取其中一对加以证明.
20.如图所示,矩形ABCD的周长为14cm,E为AB的中点,以A为圆心,AE长为半径画弧交AD于点F.以C为圆心,CB长为半径画弧交CD于点G.设ABx=cm,BCy=cm,当DFDG=时,求
xy的值.
四、解答题(每小题6分,共12分
21.下图是根据某乡2009年第一季度“家电下乡”产品的购买情况绘制成的两幅不完整的统计图,请根据统计图提供的信息解答下列问题:
G
(第20题
(第19题
B
DC
FA郜
E
电脑
电视机热水器
冰箱洗衣机冰箱
%
%35%
10%电脑
电视机
热水器
洗衣机
5%
(1第一季度购买的“家电下乡”产品的总台数为;(2把两幅统计图补充完整.
22.如图,⊙O中,弦ABCD、相交于AB的中点E,连接AD并延长至点F,使DFAD=,连接BC、BF.
(1求证:
CBEAFB△∽△;(2当58
BEFB
=时,求CBAD
的值.
五、解答题(每小题7分,共14分
23.小鹏学完解直角三角形知识后,给同桌小艳出了一道题:
“如图所示,把一张长方形卡片ABCD放在每格宽度为12mm的横格纸中,恰好四个顶点都在横格线上,已知α=36°,求长方形卡片的周长.”请你帮小艳解答这道题.(精确到1mm(参考数据:
sin36°≈0.60,cos36°≈0.80,tan36°≈0.75
24.如图,反比例函数kyx
=
的图象与直线yxm=+在第一象限交于点62P
(,,AB、为直线上的两点,点A的横坐标为2,点B的横坐标为3.DC、为反比例函数图象上的两点,且ADBC、平行于y轴.(1直接写出km,的值;(2求梯形ABCD的面积.
六、解答题(每小题8分,共16分
(第22题
F
B
(第24题
(第23题
C
25.AB、两地相距45千米,图中折线表示某骑车人离A地的距离y与时间x的函数关系.有一辆客车9点从B地出发,以45千米/时的速度匀速行驶,并往返于AB、两地之间.(乘客上、下车停留时间忽略不计
(1从折线图可以看出,骑车人一共休息次,共休息小时;(2请在图中画出9点至15点之间客车与A地距离y随时间x变化的函数图象;(3通过计算说明,何时骑车人与客车第二次相遇.
26.两个长为2cm,宽为1cm的长方形,摆放在直线l上(如图①,CE=2cm,将长方形ABCD绕着点C顺时针旋转α角,将长方形EFGH绕着点E逆时针旋转相同的角度.
(1当旋转到顶点D、H重合时,连接AG(如图②,求点D到AG的距离;(2当45α=°时(如图③,求证:
四边形MHND为正方形.
(第25题
101112131415
9x/时
图②A
DB
G
E
F
l
图①
A
D
BCH
G
EF
l
图③ADMCHGE
F
l
C
N
(第26题(H
七、解答题(每小题10分,共20分
27.某数学研究所门前有一个边长为4米的正方形花坛,花坛内部要用红、黄、紫三种颜色的花草种植成如图所示的图案,图案中AEMN=.准备在形如RtAEH△的四个全等三角形内种植红色花草,在形如RtAEH△的四个全等三角形内种植黄色花草,在正方形MNPQ内种植紫色花草,每种花草的价格如下表:
设AE的长为x米,正方形EFGH的面积为S平方米,买花草所需的费用为W元,解答下列问题:
(1S与x之间的函数关系式为S=;
(2求W与x之间的函数关系式,并求所需的最低费用是多少元;(3当买花草所需的费用最低时,求EM的长.
28.如图所示,菱形ABCD的边长为6厘米,60B∠=°.从初始时刻开始,点P、Q同时从A点出发,点P以1厘米/秒的速度沿ACB→→的方向运动,点Q以2厘米/秒的速度沿ABCD→→→的方向运动,当点Q运动到D点时,P、Q两点同时停止运动,设P、Q运动的时间为x秒时,APQ△与ABC△重叠部分....
的面积为y平方厘米(这里规定:
点和线段是面积为O的三角形,解答下列问题:
(1点P、Q从出发到相遇所用时间是秒;
(2点P、Q从开始运动到停止的过程中,当APQ△是等边三角形时
x的值是秒;(3求
y与x之间的函数关系式.
(第27题
FC
G
D
H
EQB
(第28题
吉林省2009年初中毕业生学业考试数学试卷
参考答案及评分标准
阅卷说明:
1.评卷采分最小单位为1分,每步标出的是累计分.
2.考生若用本“参考答案”以外的解(证法,可参照本“参考答案”的相应步骤给分.
一、填空题(每小题2分,共20分
1.1-
2.97a
3.1.5×105
4.x>1
5.(5,3
6.x=5
7.7-
8.120
9.710.73二、单项选择题(每小题3分,共18分
11.D12.D13.C14.A15.A16.B三、解答题(每小题5分,共20分
17.解:
222(2222(;xxyxxxyxxy++=+=+或222(2(;yxyxxy++=+
或2222(2(2((;xxyyxyxyxyxy+-+=-=+-或2222(2(2((.yxyxxyyxyxyx+-+=-=+-
说明:
选择整式正确得2分,整式加(减结果正确得1分,因式分解正确得2分,累计5分.18.解:
列表
(1P(两数相同=
13
.···················································································(3分
(2P(两数和大于10=
4
9.···········································································(5分
19.解:
(1ADBADC△≌△、ABDABE△≌△、AFDAFE△≌△、BFDBFE△≌△、
ABEACD△≌△(写出其中的三对即可.························································(3分
(2以△ADB≌ADC为例证明.
树形图
6
76-2
767
76-2-2-2
证明:
90ADBCADBADC⊥∴∠=∠=°.在RtADB△和RtADC△中,
,ABACADAD==
∴RtADB△≌RtADC△.·
·················································································(5分说明:
选任何一对全等三角形,只要证明正确均得分.
20.解:
根据题意,得
2214,2
xyxxyy+=⎧⎪
⎨-=-⎪⎩·
·····································································································(3分解得4,3.
xy=⎧⎨
=⎩··········································································································(5分
答:
x为4,y为3.说明:
不写答不扣分.
四、解答题(每小题6分,共12分21.解:
(1500.································································································(1分(2
······························································································································(6分说明:
第(2问中每图补对一项得1分,条形图中不标台数不扣分.22.(1证明:
,AEEBADDF==
ED∴是ABF△的中位线,
ED∴,BF∥··································································································(1分
CEBABF∴∠=∠·
························································································(2分又,CA∠=∠···································································································(3分,CBEAFB∴△∽△·······················································································(4分(2解:
由(1知,CBEAFB△∽△,
冰箱
%
%35%
10%电脑
电视机
热水器
洗衣机
注意:
将答案写在横线上
50
电视机
热水器
冰箱20305%
5.8
CBBEAF
FB
∴
=
=
·
·································································································(5分又2,AFAD=
54
CBAD
∴=.·········································································································(6分
五、解答题(每小题7分,共14分
23.解:
作BEl⊥于点E,DFl⊥于点F.·····················································(1分
18018090909036.
DAFBADADFDAFADFαα+∠=-∠=-=∠+∠=︒∴∠==︒°°°°,,
根据题意,得BE=24mm,DF=48mm.··································································(2分在RtABE△中,sinBEAB
α=
·············································································(3分
2440sin360.60
BEAB∴=
=
=°
mm············································································(4分
在RtADF△中,cosDFADFAD
∠=
····································································(5分
4860cos360.80
DFAD∴=
=
=°
mm.········································································(6分
∴矩形ABCD的周长=2(40+60=200mm.·························································(7分24.解:
(1k=12,m=4-.···············································································(2分
(2把x=2代入y=
12x
得y=6.D∴(2,6.
把x=2代入4yx=-,得2.y=-A∴(2,2-.
6(28.DA∴=--=·
····························································································(4分
C
l
把x=3代入4yx=-,得y=1-,B∴(3,1-.
BC∴=4-(-1=5.························································································(6分(581
13.2
2
ABCDS+⨯∴=
=梯形·
···············································································(7分六、解答题(每小题8分,共16分25.解:
(1两.两.·························································································(2分(2
······························································································································(4分(3设直线EF所表示的函数解析式为.ykxb=+把(10,0,(11,45EF分别代入ykxb=+,得
100
1145kbkb+=⎧⎨
+=⎩
·······································································································(5分解得45450.
kb=⎧⎨
=-⎩,
∴直线EF所表示的函数解析式为45450.yx=-······················································(6分把30y=代入45450,yx=-得
4545030.x-=····································································································(7分210
3
x∴=.
答:
10点40分骑车人与客车第二次相遇.·······························································(8分说明:
第(3问时间表达方式可以不同,只要表达正确即可得分,不写答不扣分.26.解:
(12CDCEDE===cm,
CDE∴△是等边三角形.60CDE∴∠=°.·································································································(1分36029060120ADG∴∠=-⨯-=°°°°.
1011121314159x/时
(H
A图②ADB
CG
E
F
l
K
又AD=DG=1cm,∴∠DAG=∠DGA=30°·························································································(2分).如图②作DK⊥AG于点K.11∴DK=DG=cm.221∴点D到AG的距离为cm.··························································································(4分2
(2)Qα=45°∴∠NCE=∠NEC=45°,∴∠CNE=90°.···············································································································(5分∴∠DNH=90°Q∠D=∠H=90°,∴四边形MHND是矩形.····························································································(6分)又CN=NE,∴DN=NH,··············································································································(7分)∴矩形MHND是正方形.····························································································(8分)七、解答题(每小题10分,共20分)27.解:
(1)x2+(4−x2或2x2−8x+16.·································································(2分)
(2)W=60×4S△AEB+80(S正方形EFGN-S正方形MNPQ+120S正方形MNPQ=60×4×21x(4−x+80[x2+(4−x2−x2]+120x2.···················································(4分)2=80x−160x+1280.·····································································································(5分)配方,得W=80(x−12+1200.····················································
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