静安区2020学年第一学期期末教学质量调研九年级数学试卷

静安区2020学第一学期末教质量调研九年级学试

2021.1（完成时间分钟

满分：150分）考生注意：1．本试卷含三个大题共25题．题时，考生务必答题要求在答题纸规定的位置上答，在草稿纸、试卷上答题一律无效；2．第一、二大题外其余各题如无特别说明，都必须答题纸的相应位置上写出证明或计算的主要步；3.答时需使用函数型计算器．一、选择题本大题共6题每题4分，满分分）1．如果下列计算正确的是（A）()

=0；（B；（）

0

；（）-

=．2．A

2

111；B2x+；Cx；D2．24443．将抛物线2(

平后抛线y

重合，那么平移方法可以是（A）向右移1个单，再向上平移3个单；（B）向右平移个单位，再向下平移单位；（）向左平移个单位，再向上平3个位；（D向左平移1单位，再向下平移3个单．4．在△ABC中点DE分在边BACA的长线上，下列比例式中能判DEBC的为（A）

ABACAC；（）=；（）=；（D）=．DEABCE5．如果锐角的正切为，那么下列结论中正的是（A）

；（）

；（C）

；（）

．6．在eq\o\ac(△,Rt)，C，是高，如果=m∠A=

那CD的长为（A）

；（B

；（）m

；D）

．第页共页

二、填空题本大题共12，每题，满分48分）37．的反数是2

▲

．8．函数f)

的定义域为

▲

．9．方程x根为

▲

．10．二次函数x图的开口方向是

▲

．．抛物线x的点坐为

▲

．12．如果一次函数ym的像经过第一、二、四象限，么常数m的取范围为

▲

．13．在二次函数y

2

x图的上升部分所对应的自量x的值围

▲

．14．如图，在ABC中点分别在边、上AED=∠，

如果AD=2AE，=1，那么为

▲

.

D15．在，点重心，°，BC=8，那么AG的长为▲.如，在ABC中点D、分别边ABAC上//，

C（第题图）如果，BC，，四边形BCED的长为，那么

D

的长为

▲

．

（第题图）．17．如图，在梯形，AD//，BD与相于点，OB，设，么▲

O

D（用向量a、b的式子示）在eq\o\ac(△,Rt)中，∠C，AB，tan

23

（如图

（第题图）

△ABC绕C旋后，点落在斜边AB上点A，点落点，’与边BC相于点D那么

CDA'D

的值为

▲

．

（第题图）

三、解答题本大题共7题满分19本题分）

分）第页共页

，，计算：

cot30sin6045

．20本题分）已知线段、满

2xxx的值21本题分分,其每小题分）如图，点A、B第一象限的反比例函数图像上的长与y轴于点C已知点、B的坐标分别为6、，AB2（1）求∠ACO的弦值；（2）求这个反比例数的解析式．

22本题分）

O

（第21题图）

如图，一处地铁入口的无障碍通道是转折的斜，沿着坡度相同的斜CD共走可到出入口，出入口点距地面的高DA为8米求无障通道斜坡的坡度与坡角(角度精确到1'，他近似数取四个有效数).D

D

（第题图）23本题分，其中每小题分）已知：如图，在中点D分别在边、上DE∥BC，AD2．求证）BCD∽CDE

第页共页

D（第图）

22CDAD（2）．BC24本题分，其中每小题分）1如图，在平面直坐标系xOy中直线轴轴分交于点2、．物

2

bx

（）经过点，与y轴相交于，=．（1）求直线AB的表达式；（）如果点在段的长线上，且

AD=AC经过点的抛物线表达式；

yax

的（3果抛物

y

2

的对称轴与线段ABAC分别相交于点，=1，求抛物线的顶点坐．

O

（第24题图）25本题分，其中第（）小题5分，）小题5分第3小题4分）已知∠是角，点B在AM上点D在边上∠∠MAN，且CE//BD∠MAN=

，，=9（1）如图1，当CE与相于点F时求：DF··；（2）当点E在上，求AD的；（）当点E在MAN外部，设ADeq\o\ac(△,，)的积为y，求y与之的函数析式，并写出定域．

D

MA（图1第页共页

（备用图）

M（第题图）

静区学第学期末学量调九级学试参考答案评分说明2021.1一、选择题：．；2A．A；．；5．C；6．．二、填空题：

．；．下；．；．6；

8；110,1441217．a+b；3

．x12．；15．8；13．．三、解答题：19．解：原式＝

············································································（4分＝

(232)(32)(3

································································（3分＝+223

．····························································（3）20．解：xyx2xy,···········································································（2分y2,···········································································（）∵y，∴

x2····························································2分）y2∴

313，·············································································（2分2∵负值不符合题，∴

313．··················································（）221．解）分别点、作AD⊥y轴BE⊥x轴垂足分别为D、E，ADBE相交于点H．·············································································（）∵轴，∴∠ACO=∠∠ADC=90°∵点的坐标分别为62∴．····································（）在eq\o\ac(△,Rt)中，∵=

2

AH

2

(2

2

2

．·················（）第页共页

ADADcosACO=cosABH

BH．·····································（2分55（）设反比例函数的解析式为

(k0)

，···········································（1分设点（，

B，)，·······················································（）62∴，···············································································（分6∴k．······················································································（1分）∴反比例函数解式为

．··························································（分22．解）延长DC、相于点E．·························································（）∵斜坡BC、CD坡度相同，∴∠=∠，=CB．···················（）∴＋CB7米···································································（）在中AE=AE

0.8

6.9541（米················（）∴iAD：AE：≈1：8.693．·················································（）∵tan∠AD：AE4，∴∠AED'．·································（）答：无障碍通道坡度约为∶，坡角为6°34'．·······························（1）23．证明）∵AD

2

，

．··············································（1分又∵∠=∠A，∴△ADE∽△．············································∴∠=ACD，··································································∵DEBC，∴∠EDC∠，∠，·····························∴∠=∠ACD······································································∴△BCD．·······························································

（1分（1分（1分（1分（1分CDDE（）BCD∽△CDE∴．·········································（2分）∴

CD

22

CDCDDE=BCBCCD

．·············································（）∵DE∥，∴

DEAD

．························································（）第页共页

22CDAD∴．·········································································（）BC124．解）∵直与x轴y轴别交于点m，0)、B(0，m．···（1分）2∴OAm=m．∵=∠OAB，∴∠OCA=tan∠OAB=

OAOB1=．··················1分OCOA∵

yax

2

（a过点(0，4)∴OA，OB．··········································································（1分1∴直线的达式为．···················································（）2（2）过点D作⊥轴，足为，∵∠DGA∠︒，∠DAG=∠ACO，，∴△DGA△．·····································································（1分）∴DG=AO=2，=，OG=2∴点D（，··························（1）∵抛物线

2

经点A、D∴

=aba

··········································································（1分a,3∴∴物线的达式为2．····························（）14b.（3）设抛物线的对轴FE与OA交点H，∵//，

AHEF14，AH，OH=．························（）AOBC3∴

4ab423

··········································································（）∴∴抛物线的表达式为yx

2

．··································（1分第页共页

.x.x当

4时，y，物线的顶点坐标，)．·························（1分325．解）∵//，∴CEB∠DBE，=∠BCE························（）∠DBE，∴∠A∠BEC．∴ABD．·························（）∴∵∴

EBABECABBCEBECBC

．···············································································（1），···············································································1分），∴DF．····························