2012年上海松江区中考数学质量抽查试卷(二模)

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2012年松江区初中毕业生学业模拟考试
数学试卷
(满分150分,完卷时间100分钟)2012.4
考生注意:
1.本试卷含三个大题,共25题;
2.答题时,考生务必按答题要求在答题纸规定的位置上作答,在草稿纸、本试卷上答题一律无效;
3.除第一、二大题外,其余各题如无特别说明,都必须在答题纸的相应位置上写出证明或计算的主要步骤.
一、选择题:(本大题共6题,每题4分,满分24分)
【下列各题的四个选项中,有且只有一个选项是正确的,选择正确项的代号并填涂在答题纸的相应位置上】
1.下列二次根式中,属于最简二次根式的是
A
1
B82

2
Cxy2
Dxy
2.下列运算正确的是
235
AaaaBaa2aCaaaD(aa
2
2
3
3
2
3.在平面直角坐标系中,点A和点B关于原点对称,已知点A的坐标为(23,那么点B的坐标为
A32B23
C32D23
4.如果正五边形绕着它的中心旋转角后与它本身重合,那么角的大小可以是A36°;B45°;C72°;D90°.5.已知RtABC中,∠C=90°,那么下列各式中,正确的是AsinA
BCBCBCBC
BcosACtanADcotAABABABAB
6.下列四个命题中真命题是(A)矩形的对角线平分对角;(梯形的对角线互相垂直;
(B)菱形的对角线互相垂直平分;(平行四边形的对角线相等.


二、填空题:(本大题共12题,每题4分,满分48分)【请将结果直接填入答题纸的相应位置上】7.计算:2=___

-1-
2


8.如果关于x的一元二次方程xxm0有两个不相等的实数根,那么m的取值范围
▲.
9.方程2x13的解是____10.用换元法解方程x2x的整式方程是__.11.已知函数f(x
2
2
22
1时,如设yx2x,则将原方程化为关于y2
x2x
3
,那么f(4x1
k
k0)的图像经过点A-32,那么k=__x
12.已知反比例函数y
13.已知包裹邮资为每千克2元,每件另加手续费3元,若一件包裹重x千克,则该包裹邮y(与重量x(千克之间的函数关系式为
14.在一个不透明的口袋中,装有4个红球和6个白球,除顔色不同外其余都相同,从口袋中任意摸一个球摸到的是红球的概率为
15.已知⊙O1和⊙O2外切,O1O28,若⊙O1的半径为3,则⊙O2的半径为16.如图,在平行四边形ABCD中,对角线ACBD相交于点O,设ADaABb
DCAE那么DO
DF
BO
ACGB
(第17题图)16题图

17.如图是利用四边形的不稳定性制作的菱形凉衣架.已知其中每个菱形的边长为13cm
cosABC
5
,那么凉衣架两顶点AE之间的距离为cm13
18.将一个平面图形分成面积相等的两部分的直线叫做该平面图形的“面线”“面线”被这个平面图形截得的线段叫做该图形的“面径”,例如圆的直径就是它的“面径”.已知等边三角形的边长为2,则它的“面径”长可以是(写出2个)三、解答题:(本大题共7题,满分78分)
4a22a3a119(本题满分10分)计算:22
a3aaa1

-2-


x2xy2y20
20(本题满分10分)解方程组:
x3y2

21(本题满分10分)某公园有一圆弧形的拱桥,如图已知拱桥所在圆的半径为10米,拱桥顶D到水面AB的距离DC=4米.1)求水面宽度AB的大小;
2)当水面上升到EF时,从点E测得桥顶D的仰角,若cot3,求水面上升的高度.

-3-
D
EA
C
FB
21题图


22(本题满分10分)
随着“微博潮”的流行,初中学生也开始忙着“织围脖”,某校在上微博的280名学生中随机抽取了部分学生调查他们平常每天上微博的时间,绘制了扇形统计图和频数分布直方图,请根据图中信息,回答下列问题:
1)本次调查共抽取了名学生;将频数分布直方图补充完整;2)被调查的学生中上微博时间中位数落在这一小组内;3)样本中,平均每天上微博的时间为0.5小时这一组的频率是
4请估计该校上微博的学生中,大约有名学生平均每天上微博的时间不少于1小时.
人数


23(本题满分12分)
如图,在梯形ABCD中,ADBC,∠BCD=90°BC=DC,点E在对角线BD上,作
DA
ECF=90°,连接DF,且满足CF=EC1)求证:BDDF
2)当BCDEDB时,试判断四边形DECF的形状,并说明理由.
B
C23题图
2
1920
0.5小时
2小时
1小时
1.5小时15%
10
6
4
22题图
0.511.52时间(小时)
E
F
-4-


24(本题满分12分)
已知直线y3x3分别与x轴、y轴交于点AB抛物线yax22xc经过点AB1)求该抛物线的表达式,并写出该抛物线的对称轴和顶点坐标;2)记该抛物线的对称轴为直线l,点B关于直线l的对称点为C若点Dy轴的正半轴上,且四边形ABCD为梯形.①求点D的坐标;
②将此抛物线向右平移,平移后抛物线的顶点为P其对称轴与直线y3x3交于点E,若tanDPE求四边形BDEP的面积.

-5-
(第24题图)
y
1O
1
x
3
7


25(本题满分14分)
3
,点DAB边上(点D与点AB5
1
不重合)DEBCAC边于点E,点F在线段EC上,且EFAE,以DEEF为邻
4
如图,在△ABC中,ABAC10cosB边作平行四边形DEFG,联结BG1)当EF=FC时,求△ADE的面积;
2)设AE=x,△DBG的面积为y,求yx的函数关系式,并写出x的取值范围;3)如果△DBG是以DB为腰的等腰三角形,求AD的值.
B
25题图DG
EF
C
A
-6-


参考答案及评分说明
20124
一、选择题:1D2C二、填空题:7
3B
4C
5A
6B
11
8m9x410y2y2011112-644
211
13y2x31415516ba176131823
522
(或介于23之间的任意两个实数三、解答题:19.解:原式=[
a14(a3(a1]„„„„„„„„(4分)
a(a1(a1(a1a3
(a12a3=„„„„„„„„„„„„„(4分)
a(a1(a1(a3(a1
=
1
.„„„„„„„„„„„„„„„„„„„„„„(2分)
a2a
20.解:由(1)得xy0x2y0.„„„„„„„„„„„„(2分)
原方程组可化为
xy0,x2y0,
„„„„„„„„„„„(4分)
x3y2;x3y2;
4x,1x215
解得原方程组的解为„„„„„„„„„„(4分)
y2;y2115
21.解:(1设拱桥所在圆的圆心为O,由题意可知,点ODC的延长线上,
联结OA,∵ODABACO90„„„„„„„„„„„(1分)RtACO中,OA10,OCODDC1046,AC82分)ODAB,OD是半径,AB2AC16„„„„„„„„(2分)即水面宽度AB的长为16.
2)设ODEF相交于点G,联结OEEF//AB,ODABODEF,∴EGDEGO90„„„„„„„„„(1分)RtEGD中,cot
EG
3EG3DG„„„„„(1分)DG
设水面上升的高度为x米,即CGx,则DG4xEG123x

-7-


RtEGO中,EGOGOE
123x6x102化简得x6x80
2
2
2
222
解得x14(舍去)x22„„„„„„„„„„„„„„„„(2分)答:水面上升的高度为2.„„„„„„„„„„„„„„„„„„„„(1分)22140„„„„„(2分);补全图形„„„„„„„(2分)21小时„„„„„(2分)3
19
„„„„„(2分)4147„„(2分)40
231)证明:∵BCDECF90BCEDCF„„„„(1分)
BCDC,ECCF,∴BCEDCF„„„„„„„„„„„(1分)EBCFDC„„„„„„„„„„„„„„„„„„„„„„(1分)BCDC,BCD90,∴DBCBDC45„„„„„„(1分)FDC45,∴FDB90„„„„„„„„„„„„„„„(1分)BDDF„„„„„„„„„„„„„„„„„„„„„„„„„(1分)(2四边形DECF是正方形„„„„„„„„„„„„„„„„„„„(1分)
2
BCDEDB,BCDC,∴DCDEDB
2
DCDE
„(2分)DBDC
CDEBDCCDEBDC„„„„„„„„„„„„(1分)DECDCB90„„„„„„„„„„„„„„„„„„„(1分)FDEECF90∴四边形DECF是矩形„„„„„„(1分)CECF∴四边形DECF是正方形
24.解:1)由题意得A1,0B03„„„„„„„„„„„„„„„(1分)
3∵抛物线yax2xc过点A1,0B0
2

a2c0a1
解得„„„„„„„„„„„„„„„„(1分)
c3c3
2
yx2x3„„„„„„„„„„„„„„„„„„„„„„„(1分)y(x14
∴对称轴为直线x1,顶点坐标为1,4„„„„„„„„„„„„(2分)2由题意得:AB//CD,设直线CD的解析式为y3xb„„„(1分)

-8-
2


C2,36b3b3„„„„„„„„„„(1分)∴直线CD的解析式为y3x3D0,3„„„„„„„„„„(1分)
DFPEF,PF7„„„„„„„„„„„„„„„„„(1分)RtDFP中,tanDPE
DFDF3
DF=3„„„„„(1分)PF77
x=3,y=3×3-3=6,∴点E(3,6„„„„„„„„„„„„„„(1分)
1
(BDEPDF24„„„„„„„„„„„„„(1分)2
BH3
25.1)作AHBCH,在RtAHB中,cosB
AB5
S四边形BDEP
AB10,∴BH6,∴AH8ABACBC2BH12,∴
SABC
1
12848
„„„„„„„„„(1分)2
2
SAE
DE//BC,∴ADEABC,∴ADE„„„„„„(1分)
SABCAC
1AE42AEEFFC,∴,„„„„„„„„„(1分)4AC63S464ADE,∴SADE„„„„„„„„„„„„„„„„„(1分)
4893
EF
2)设AHDEGF于点MN
AEAMDE
ACAHBC46
AEx,AMx,DEx„„„„„„„„„„„„„„„(1分)
55
11
MNAMx,∴NH8x„„„„„„„„„„„„„„(1分)
45
DE//BC,∴
SDBGS梯形DBCGS平行四边形DGFES梯形GBCFy
1616461
x128xxxx128x2525555326
xx255
y
0x8„„„„„„„„„„„„„„„(2分)
3)作FPBCPGQBCQRtFPC中,FC10PC6
53
x,cosCcosABC45
3639
xBQ12x6x6x45420
-9-



9
x„„„„„„„„„„„„„„„„„(2分)BG8x6
20
2
2
DBG中,DB10x,DG①若DBDG,10x
1
x4
1
x,解得x8„„„„„„„„„„„„„(2分)4
2
2
9
x②若DBBG,10x8x6
20
解得x10舍去x2AD8AD

560
„„„„„„„„„„„„„„„(2分)81
560
81
-10-

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