2021年上海市普陀区九年级数学一模试卷

22普陀区学年第一学期初质量调研数学试卷考注：1．试共25题．2．卷分150分．试间100分钟3．题，生必答要求答纸定位上答在稿、试上题一无．4．第、大外其各题无别明都须答纸相位上出明或算主步．一选题本题6题，题4分满）【列题四选中有只一选是确，择确项代并涂答题的应置】1．列函数中，关的次函数是（）y2；（B）y

x

；（）y(；（D）

yx2．2．果点(3,m)在x轴上，那么点B(mm在的象限是（）第一象限；

（）第二象限；C）第三象；（）四象限．3．知在eq\o\ac(△,Rt)eq\o\ac(△,)ABC中，AB，，么tan的值等于（）

；

5（）；（C）；D）

25

．4．下列对抛物线y2

的描述中，正确的是（）开口向上；（）点在轴上；（）对称轴是直线（）与

轴的交点是，5．是零向量，，下列说法中错误的是（）b与a平；（B）与a互相反向量；（）a；（）1

1ab．2

6．1，四边形的角线与BD相于，论是

OD，由此推得的正确结（）

AD；（）；ODCDOCBC

A

D（）

ABAD；（）．ODCDCDBC

OB

C二填题本题12题，题分满分）

图1x7．知，那么y2y

▲

．8．果正比例函数ykx的图像经过第一、三象限，么y的随着的增大而

．（填“增大”或“减小9．着正方向看，如果抛物线左侧的部分是下降的，那么的值范围是．▲10．次函数yx

图的顶点坐标为

▲

．

y11．图2，已知二次函数y

bx的图像经过点那么f▲0“或“．12．eq\o\ac(△,在)eq\o\ac(△,)中，▲

O

A

1

图2

x13．图，DE别是△的AB、的点且AED如果AB，，EC，么的长等于

▲

．14．图4，在eq\o\ac(△,Rt)eq\o\ac(△,)ABC中、E别在边BC、上CD，，AD与交点，如果AB，么CF的等

▲

．15．图5，小明在教学楼的楼顶测得：对面实验大D的顶端C的角

，底部D的角为

果学楼的高度为米么两栋教学楼高度CH为

▲

米．A

CAD

A

HEE

FB

图3

C

C

D图4

B

B

图5

D2

16．图6△为等边三角形，点D、E分在边BC、，果DC2

，

，那么DE的等于

▲

．17．股定理是世界文明宝库中的一颗璀璨明珠，我国汉代数学家赵爽将四个全等的直角三角形拼成了一个大正方形ABCD同留下一个小正方形的（如图面证明了勾股定理果正方形EFGH的积是4，GBC

那大正方形ABCD的面积等于

▲

．18．图，eq\o\ac(□,)ABCD中，点E在边BC上，eq\o\ac(△,将)eq\o\ac(△,)ABE沿着直线AE翻折得eq\o\ac(△,到)eq\o\ac(△,)，点B的应点F恰好落在线段上AF的延长线交边CD于G果BE:，那么AF:FG值等于

▲

．A

A

DE

H

ADFGE

F

GB

D

图6

C

B

图7

C

BE图8

C三解题本题7题，分78分19本满分10分计算：

cos302

2sin60

．20.（本题满分分，(小满分6分第2)题满分4分如图9，已知点、E、C在一条直线上，AGDG1交于点G，，．GE（）BF的长；

DE，AC

DF与DE相（设

b

那么

BF

▲

，

DF

▲

（用向量、b表ADGBE

图9

CF3

21.（本题满分分，第(小满分4，(2)题满分6分在平面直角坐标系xOy中反比例函数y于横坐标为3的A．（）这个一次函数的解析式；

x

的图像与一次函数ykx的图相交（）图10，已知点B在这个一次函数图像上，点在比例函数y

x

的图像上，直线//x轴，且在点A上，并与轴交于点D．果恰是的中点，求点的坐标．

yDC

BO22本满分10分第1)小题满分5分第小题满分5分）

A3图10

x如图，eq\o\ac(△,在)eq\o\ac(△,)ABC中，BC上一点D边的垂直平分线上，AB．（）证；（）果AB10

，，的值．ABD图11

C4

23本满分12分第1)小题满分6分第小题满分6分）已知：如图12，AD∥，ABD，AEBD，BC，、F分为垂足．（）证：

AEBD；（）结EF，ADBBDF，证：DFEF．A

DEBF

C24本满分12分，每小题满分各4分）

图12在平面直角坐标系

xOy

中（如图13知抛物线y

与交于点A顶点B的标（）接写出点A的标，并求抛物线的表达式；（）点C在轴，且CAB90，直线AC与物线的另一个交点为点D．①求点C、D坐标；②将抛物线bx

沿着射线BD的向平移：平移后的抛物线顶点仍在线段BD上的应点为点P线AB与x轴交点为Qeq\o\ac(△,果)eq\o\ac(△,)与CBQ相，求点P的标．

y1-

O

1

x图135

25本满分14分，第(小满分5分第小满分6分第小题满分3分如图14，矩形ABCD中，，是边上一个动点（不与点、重合AE的线AF交CD的长线于点．G在段上满足FG:．设．（）证：

ADDF；AB（）点G在△ADF的部时，用x的数式表ADG的余切；（）当=AFE，求线段长．FGADADBE

C

B

C图14

备用图6

普区2020学年度一期三量研学卷参答及分明一选题本大题共6题，每题分满分24分1．

．；

．

4．；

5．；

6．．二、填空题大题共题每题，满分48分）7．；10．．416．；

；

8．增大；．；14．；17．；

9．a＞；12．；15．m．18．

；三解题（大题共7题其中第19---22题题10分第23、题每题12分，第25题14分满分分19解：原式

+2

+1

····················································4）

3·················································································（）23+1

32

················································································（分）

332

．·····················································································（）20．：∵AB//DE，∵

．··························································（分）∵

，BE，∵EC．························································2）2同理CF．···················································································1）∵BF．······················································································（）7

3（2BF，．······················································（分2分221．）∵比函数

x

的图像经过点，∵把x代入y

x

，可得2．···················································（）∵点A的标为

·····································································（分）由一次函数图像经过点

k，得k．····1）所以这个一次函数的解析式是．···············································（分）（）题意可设点的标为

m，点的标为

,

．··（分∵在比例函数y

的图像上，∵把，ym代xx

．得

12m

．···················································································（分整理，得m20．解得

（舍

．····························································（分）∵的标为·······································································（分）22．）∵BD，∵

ABBCBDAB

．···········································（分）∵为共角，∵∵∵∵．···················································（分∵C．··············································································（分）∵D在AB的垂直平分线上∵ADBD．∵．·············································································1）∵．·················································································（）（）点作AH∵BC，足为点H.··············································（）由

BD，AB10

，BC，得．∵AD．···········································································（分）∵，∵AC．又∵∵，∵

12

BC．··················································（）∵．··················································································（）8

在Rt∵中cos

DH1．············································1）AD423．明∵AD∵，∵DBC············································（）∵，∵∵ADB∵∵DBC.·····················································（）∵AEBD，∵是∵的高，同理

DF是∵DBC的高，································································（分）∵

AEBDDFBC

．···················································································（）（）∵BC∵DFC90∵∵，∵ADF∵ADB∵

．·····················································（）∵AEBD，∵AE．··································································（）∵

AEBDDEBD，∵．DFBCDF∵

DEDFBD

．···················································································（分）又∵EDF，∵∵DEF∵∵BDC．··········································（分）∵

DFDC

．·················································································（分∵·········································································（分）24．）A，．···········································································（分）由抛物线yax

的顶点B的标为可设此抛物线的表达式为···········································1）由抛物线经过点(0，，得1a

解得a

12

·····································································1）∵抛线的表达式是y

12

················································（分即：y

12

2．（）由B

，(0，，可得BAO45由，得CAO．∵点C的坐标为····································································（分）9

1由直线与物的另一个交点为点D，可设点的坐标为,2

．过点D作，H为足．可得AH

12

m2m，DH．∵CAO∵AHDH．∵

12

m

m······················································（分∵m，得m．·······································································（分）∵点的坐标为·····································································（分）过点作BGCO，为足．1在Rt∵中tan，在Rt∵ADB中tan3∵ADB．

AB1．AD3由平移可得

，∵．∵PAD．··········································································（分）在∵CBQ中BQC135因为平移后顶点仍在线段BD上∵最等于∵ADP与∵CBQ相，只能APD135··（分由

12，得APAD5

5．·························································（分）∵QCBPAH∵ABCPAH过点P作PMAH，M为足．在Rt∵ABC中tanABC

11，∵在∵APM中PAM．2212可得点的坐标为(,)5525．解）∵形ABCD，

．·······························································（）∵CDA90ADCD．···········（分由BAD90得EAD90由AE，得DAFEAD

．∵BAEDAF．············································································（分由得10

∵．················································································（1）∵∵ABE∵∵．···········································································（）∵

ADDFAB

．···················································································（分）（）

ADDF