2020年上海杨浦初三数学一模试卷及答案
杨浦区2019学年度第一学期期末质量调研
初 三 数 学 试 卷 2019.12
(测试时间:100分钟,满分:150分)
考生注意:
1.本试卷含三个大题,共25题.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.
2.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) 1.把抛物线2
x y =向左平移1个单位后得到的抛物线是
A .2
1y x =
+();
B .2
1y x =
-(); C .2
1y x =+;
D .2
1y x =-.
2.在Rt △ABC 中,∠C =90°,如果AC =2,3
cos 4
A =
,那么AB 的长是 A .
52
;
B .83
;
C .
103
; D .
2
73
. 3.已知a r 、b r 和c r
都是非零向量,下列结论中不能判定//a b r r 的是
A .////a c b c r u u r r r
,
;
B .12
a c =r r
,2b c =r r ;
C .2a b =r r
;
D .a b =r r .
4.如图,在6×6的正方形网格中,联结小正方形中两个顶点A 、B ,如果线段AB 与网格线的其中两个交点为M 、N ,那么AM ∶MN ∶NB 的值是 A .3∶5∶4; B .3∶6∶5; C .1∶3∶2;
D .1∶4∶2.
5.广场上喷水池中的喷头微露水面,喷出的水线呈一条抛物线,水线上 水珠的高度y (米)关于水珠和喷头的水平距离x (米)的函数解析式是
23
6042
y x x x =-+≤≤()
,那么水珠的高度达到最大时,水珠与喷头的水平距离是 A .1米; B .2米; C .5米; D .6米.
6.如图,在正方形ABCD 中,△ABP 是等边三角形,AP 、BP 的延长线分别交边CD 于点E 、F ,联结AC 、CP ,AC 与BF 相交于点H ,下列结论中错误的是 A .AE =2DE ;
B .△CFP ∽△APH ;
C .△CFP ∽△APC ;
D .CP 2=PH ?PB .
二、填空题:(本大题共12题,每题4分,满分48分) 7.如果cot 3α=,那么锐角α= ▲ 度.
8.如果抛物线231y x x m =-+-+经过原点,那么m = ▲ . 9.二次函数2251y x x =+-的图像与y 轴的交点坐标为 ▲ .
A
D B
C
E P
F H
第6题图
第4题图
10.已知点11A x y (,)、22B x y (,)
为抛物线2
2y x =-()上的两点,如果122x x <<,那么 ▲ . (填“>”、“<”或“=”)
11.在比例尺为1:8 000 000地图上测得甲、乙两地间的图上距离为4厘米,那么甲、乙两地间的实际距
离为 ▲ 千米.
12.已知点P 是线段AB 上的一点,且2BP AP
=? 13.已知点G 是△ABC 的重心,过点G 作MN ∥BC 分别交边AB 、AC 于点M
、N ,那么
AMN
ABC
S S ??14.如图,某小区门口的栏杆从水平位置AB 绕固定点O 旋转到位置DC ,已知栏杆AB 的长为3.5米,OA 的长为3米,点C 到AB 的距离为0.3米,支柱OE 的高为0.6米,那么栏杆端点D 离地面的距离为▲ 米. 15.如图,某商店营业大厅自动扶梯AB 的坡角为31°,AB 的长为12米,那么大厅两层之间BC 的高度为 ▲ 米.(结果保留一位小数)【参考数据:sin31°=0.515,cos31°=0.867,tan31°=0.601】 16.如图,在四边形ABCD 中,∠B =∠D =90°,AB =3,BC =2,4
tan 3
A =,那么CD = ▲ .
17.定义:我们知道,四边形的一条对角线把这个四边形分成两个三角形,如果这两个三角形相似但不全
等,我们就把这条对角线叫做这个四边形的相似对角线.在四边形ABCD 中,对角线BD 是它的相似对角线,∠ABC =70°,BD 平分∠ABC ,那么∠ADC= ▲ 度.
18.在Rt △ABC 中,∠A =90°,AC =4,AB =a ,将△ABC 沿着斜边BC 翻折,点A 落在点A 1处,点D 、E 分别为边AC 、BC 的中点,联结DE 并延长交A 1B 所在直线于点F ,联结A 1E ,如果△A 1EF 为直角三角形时,那么a = ▲ .
三、解答题:(本大题共7题,满分78分)
19.(本题满分10分,第(1)小题6分,第(2)小题4分)
抛物线y =ax 2+bx +c 中,函数值y 与自变量x 之间的部分对应关系如下表:
x (3)
-
2
- 1
-
1 … y
…
4-
1-
1-
4
-
…
(1)求该抛物线的表达式;
(2)如果将该抛物线平移,使它的顶点移到点M (2,4)的位置,那么其平移的方法是 ▲ . 20.(本题满分10分,第(1)小题6分,第(2)小题4分)
A
B
C
第15题图
31°
第
16题图
第14题图
如图,已知在梯形ABCD 中,AB //CD ,AB =12,CD =7,点E 在边AD 上,2
3
DE AE =,过点E 作EF //AB 交边BC 于点F .
(1)求线段EF 的长;
(2)设AB a =u u u r r ,AD b =u u u r r ,联结AF ,请用向量a r 、b r 表示向量AF u u u r
.
21. (本题满分10分,第(1)小题5分,第(2)小题5分)
如图,已知在△ABC 中,∠ACB=90o,3
sin 5
B =,延长边BA 至点D ,使AD =A
C ,联结C
D . (1)求∠D 的正切值;
(2)取边AC 的中点E ,联结BE 并延长交边CD 于点F ,求CF
FD
的值.
22.(本题满分10分)
某校九年级数学兴趣小组的学生进行社会实践活动时,想利用所学的解直角三角形的知识测量教学楼的高度,他们先在点D 处用测角仪测得楼顶M 的仰角为30?,再沿DF 方向前行40米到达点E 处,在点E 处测得楼顶M 的仰角为45?,已知测角仪的高AD 为1.5米.请根据他们的测量数据求此楼MF 的高.(结果精确到0.1m 1.414≈ 1.732≈ 2.449) 23.(本题满分12分,每小题各6分)
如图,已知在ABC △中,AD 是ABC △的中线,DAC B ∠=∠,点E 在边AD 上,CE CD =.
(1)求证:AC BD
AB AD =
; (2)求证:22AC AE AD =?.
24.(本题满分12分,每小题各4分)
第21题图
A
B
C
D
B
第20题图
第23题图
A B
C
D
E
30o 45o 第22题图
A B C D
F
E
M
已知在平面直角坐标系xOy 中,抛物线224y mx mx =-+(0)m ≠与x 轴交于点A 、B (点A 在点B 的左侧),且AB=6.
(1)求这条抛物线的对称轴及表达式;
(2)在y 轴上取点E 02(,)
,点F 为第一象限内抛物线上一点,联结BF 、EF ,如果=10OEFB S 四边形, 求点F 的坐标;
(3)在第(2)小题的条件下,点F 在抛物线对称轴右侧,点P 在x 轴上且在点B 左侧,如果直线PF 与
y 轴的夹角等于∠EBF ,求点P 的坐标.
25.(本题满分14分,第(1)小题3分,第(2)小题5分,第(3)小题6分)
已知在菱形ABCD 中,AB=4,120BAD ∠=?,点P 是直线AB 上任意一点,联结PC ,在∠PCD 内部作射线CQ 与对角线BD 交于点Q (与B 、D 不重合),且∠PCQ=30?. (1)如图,当点P 在边AB 上时,如果3BP =,求线段PC 的长;
(2)当点P 在射线BA 上时,设BP =x ,CQ =y ,求y 关于x 的函数解析式及定义域; (3)联结PQ ,直线PQ 与直线BC 交于点E ,如果△QCE 与△BCP 相似,求线段BP 的长.
杨浦区2019学年度第一学期初三数学期末质量调研试卷答案
第24题图 A B
C D
P
Q
第25题图
备用图
A B
C
D
2019.12
一、选择题:(本大题共6题,每题4分,满分24分)
1.A ; 2.B ; 3.D ; 4.C ; 5.B ; 6.C 二、填空题:(本大题共12题,每题4分,满分48分)
7.
8.1; 9.0(,-1)
;
10.320; 12
13 14.2.4; 15.6.2; 16.145; 18.、4
(本大题共7题,满分78分) 19.解:(1)∵二次函数2y ax bx c =++图像过点10(-,)、 (01)-,和(14)-,
, ∴01 4.a b c c a b c -+=??
=-??++=-?
,
, ··········································································· (3分) ∴121.a b c =-??
=-??=-?
,,∴二次函数解析式为221y x x =---. ·································· (3分) (2)平移的方法是先向右平移3个单位再向上平移4个单位
或先向上平移4个单位再向右平移3个单位. ······················· (4分)
20.解:(1)过D 作DH //BC 交AB 于H ,交EF 于G .
∵DH //BC ,AB //DC ,∴四边形DHBC 是平行四边形. ································· (1分) ∴BH =CD ,∵CD=7,∴BH =7.······························································ (1分) 同理GF =7. ······················································································· (1分) 又AB=12,∴AH =5. ············································································ (1分)
∵EF //AB , ∴
EG DE
AH DA
=
. ···································································· (1分) ∵23DE AE =,∴25DE DA =. ∴255EG =,2EG =,∴9EF =. ·
························································· (1分) (2)3345a b →→
+ ··················································································· (4分)
21. 解:(1)过C 作CH ⊥AB 于H . 在Rt △ABC 中,∵3sin =5B ,∴3
=5
AC AB . ·
········································· (1分) ∴设AC =3k ,AB =5k ,则BC =4k . ∵1122ABC S AC BC AB CH ?=
?=?,∴12
5
AC BC CH k AB ?==. ·
·············· (1分) ∴9
=5
AH k . ················································································ (1分)
∵AD=AC ,∴DH =924
355
k k k +=. ·
················································ (1分)
在Rt △CDH 中,1215tan =2425
k
CH CDH DH k ∠==. ·
·································· (1分) (2)过点A 作AH//CD 交BE 于点H.
∵AH//CD ,∴AH AE
CF EC =
. ···································································· (1分) ∵点E 为边AC 的中点,∴AE CE =.∴AH CF =. ···································· (1分) ∵AH//CD ,∴
AH AB
DF BD
=
. ···································································· (1分) ∵AB =5k ,BD =3k ,∴
5
8
AB BD =.∴
58AH DF =. ·············································· (1分) ∴
5
8
CF DF =. ·
······················································································ (1分) 22.解:由题意可知∠MCA =90°,∠MAC =30°,∠MBC =45°,AB =40,CF =1.5.
设MC =x 米,则在Rt △MBC 中,由 tan MC
MBC BC
∠=得BC =x . ················· (2分)
又Rt △ACM 中,由cot AC
MAC MC ∠=得AC
=. ···································· (2分)
∴40x -=. ············································································· (2分) ∴x
=20. ··············································································· (1分) ∴MF =MC+CF
=56.1≈米. ····················································· (2分) 答:此楼MF 的高度是56.1米. ······························································ (1分)
23.证明:(1)∵CD =CE ,∴∠CED =∠CDA . ········································ (1分) ∴∠AEC =∠BDA . ······························································· (1分) 又∵∠DAC =∠B ,∴△ACE ∽△BAD. ········································ (1分)
∴
AC CE
AB AD
=
. ····································································· (1分) ∵AD 是ABC △的中线,∴BD CD =. ········································ (1分)
∵CD =CE ,∴BD CE =.∴AC BD
AB AD
=
. ······································· (1分) (2)∵∠DAC =∠B ,又∠ACD =∠BCA ,∴△ACD ∽△BCA. ······················· (1分)
∴AC CD BC AC
=,∴2AC CD CB =?. ················································· (1分) ∵AD 是ABC △的中线,∴2BC CD =,∴222AC CD =. ·················· (1分)
∵△ACE ∽△BAD ,∴CE AE
AD BD
=
. ················································ (1分) 又∵CD =CE=BD ,∴2CD AD AE =?. ············································ (1分)
∴22AC AD AE =?. ································································ (1分)
24.解:(1)抛物线对称轴212m
x m
-=-=... ................................................................. (1分)
∵AB =6,∴抛物线与x 轴的交点A 为(20),
-,B (40),.................................................. (1分) ∴4440m m ++=(或16840m m -+=).. ................................................................ (1分)
∴1
2
m =-.∴抛物线的表达式为2142y x x =-++. ..................................................... (1分)
(2)设点F 21
(4)2
x x x ,-++. ...................................................................................... (1分) ∵点E 02-(,)
,点B 4(,0),∴OE = 2,OB = 4. ∵=+10OEF OBF OEFB S S S ??=四边形, ∴2111
24(4)10222x x x ??+??-++=.. .................... (1分)
∴12x =或,∴点F 9
12
(,)、24(,)
.. ............................................................................... (2分) (3)∵=+10OBE BEF OEFB S S S ??=四边形,又11
42422
OBE S OB OE ?=?=??=,∴6BEF S ?=.
过F 作FH BE ⊥,垂足为点H .
∵162BEF S BE FH ?=?=
,又BE =
FH =............................... (1分)
又BF ==
BH ∴在Rt BFH ?中,tan ∠EBF=3
584
FH BH ==.................................................................. (1分)
设直线PF 与y 轴的交点为M ,则∠PMO=∠EBF ,过F 作FG x ⊥轴,垂足为点G.
∵FG//y 轴,∴∠PMO=∠PFG . ∴tan ∠PFG=tan ∠EBF ................................................ (1分)
∴tan ∠PFG=3
4PG FG =.
又FG =4,∴PG =3.
∴点P 的坐标10(-,). .......................................................................................................... (1分)
25.解:(1)过P 作PH BC ⊥,垂足为点H.
在Rt BPH ?中,∵BP =3,∠ABC =60°
,∴32BH PH =,................................. (2分)
在Rt PCH ?
中,35422CH PC =-
==,................................... (1分) (2)过P 作PH BC ⊥,垂足为点H. 在Rt BPH ?
中,12BH x PH =
,. ∴在Rt PCH ?
中,142CH x PC =-
==,........... (1分)
设PC 与对角线BD 交于点G .
∵AB//CD ,∴4
BP PG BG x
CD GC GD ===.
∴BG CG =··················································· (1分) ∵∠ABD =∠PCQ ,又∠PGC =∠QGC ,∴△PBG ∽△QCG .
∴PB BG CQ CG =
,∴x y ··················································· (1分)
∴y =08x ≤<). ······················································ (2分)
(3)i )当点P 在射线BA 上,点E 在边BC 的延长线时.
∵BD 是菱形ABCD 的对角线,∴∠PBQ =∠QBC=1
302
ABC ∠=?.
∵△PBG ∽△QCG ,∴
PG BG
QG CG
=
,又∠PGQ =∠BGC ,∴△PGQ ∽△BGC . ∴∠QPG =∠QBC 30=?, 又∠PBQ =∠PCQ 30=?,∴60CQE QPC QCP ∠=∠+∠=?. ∴ 60CQE PBC ∠=∠=?. ···································································· (1分) ∵PCB E ∠>∠,∴ PCB QCE ∠=∠.
又180PCB QCE PCQ ∠+∠+∠=?,∠PCQ 30=?,∴ 75PCB QCE ∠=∠=?. 过C 作CN BP ⊥,垂足为点N ,∴在Rt CBN ?
中,2BN CN ==,∴在Rt PCN ?
中,PN CN ==
∴2BP = . ................................................................................................................. (2分) ii )当点P 在边AB 的延长线上,点E 在边BC
上时,同理可得2BP = . ...... (3分)