2020-2021学年上宝中学初三上学期期中仿真密卷(数学学科)参考答案

2020-2021学年上宝中学初三上学期期中仿真密卷(数学学科)参考答案
2020-2021学年上宝中学初三上学期期中仿真密卷(数学学科)参考答案

第1页,共5页

2020-2021学年上宝中学初三上学期期中仿真密卷

数学学科

参考答案

一、选择题(本大题共有 6 小题,每题4分,共 24 分)

1.A 2.C 3.D 4.B 5.B 6.A

二、填空题(本大题共有 12 小题,每题4分,共 48 分)

7

.3

8.

2133b a - 9. 2

251y x x =+- 10.158 11

.2 12.34

13.16

14.24 15

1, 16.9 17

. 18

5

三、解答题(本大题共7小题,19-22题每题10分,23-24题每题12分,25题14分,共78分)

19.(1)二次函数2

y ax bx c =++的图像经过点(1,0),(0,5),(2,3)A B C -

则05423a b c c a b c ++=??

=-??++=? 165a b c =-??

=??=-?

∴二次函数的解析式为2

65y x x =-+-

(2)连接,,AC AD CD ,延长CA 交y 轴于E ,连接DE ,直线CA 的解析式为33y x =-

则(0,3)E -

则,S S 10,ACD AED AC AE AC AE ??=====

高h =DE AC ∴⊥

设DE 解析式为13y x b =-

+过点E 则为1

33

y x =-- 联立二次函数则216533x x x -+-=-

-,解得121

6,3

x x ==由题意得6x =(6,5)D ∴-

20.(1)1

(3

a b -+) (2)略

(1)由题意可得G 是BCD ?的重心,故111

()()333

CG CA AD AB a b ==-+=-+ (2)略

21.(1)过点E 作EH BC ⊥,垂足为H ,设3EH CF k ==,4BH k =,5BE k =

EF BC ∥ EFD FDC ∴∠=∠

90FDE C ∠=∠=EFD FDC ∴??∽

EF FD FD CD

=

2,FD EF CD ∴=?2

942(44)k k +=-

BE ∴=

(2)

EFD A ∠=∠

BC DE

AC DF

=

易知EHD DCF ??∽

EH DE CD DF ∴

=4

3EH CD ∴= 3423

k ∴

= 得89k =

83

CF ∴=

第2页,共5页

22.(1)延长BA 交EF 于点G ,在Rt AGE ?中,23E ∠=,

67GAE ∴∠=,又38BAC ∠=

180673875CAE ∴∠=--=

(2)过点A 作AH CD ⊥,垂足为H

在ADH ?中,60ADC ∠=,4AD =,cos DH

ADC AD

∠=

2DH ∴=,sin AH

ADC AD

∠=

,23AH ∴= 同理26AC ∴= 2

623210AB AC CD m ∴=+=≈

23.证明:(1)

BA BF BC ?=2,

BC

BA

BF BC =

, 而CBF ABC ∠=∠,

∴BCF BAC ??∽,

BC DE ∥,

∴BCF DGF ??∽,

∴DGF BAC ??∽,

BA

DG

BC DF =

, ∴DG BC AB DF ?=?;

(2)作BC AH ∥交CF 的延长线于H ,如图,

BC DE ∥,

∴DE AH ∥,

点E 为AC 的中点,

∴EG AH 2=,

DG AH ∥,

∴DGF AHF ??∽, ∴

DF AF

DG AH =

, ∴

DF

AF

DG EG =

2.

24.(1)

1tan =∠OBC ,

∴OC OB =,

()0,2x B ,

∴()0,2x C ,

将()()()0,0,0,122x C x B A 、、-代入a bx ax y 32

-+=得,

第3页,共5页

??

???-=-=-+=--222

230303x

a a bx ax a

b a , 解得:???

??=-==3212

x b a .

∴322

--=x x y .

(2)

OC OB =,CE BE =,

∴点E 在BOC ∠的平分线上,

∴直线OE 的解析式为:x y -=.

()41322

2--=--=x x x y ,

∴()4,1-D ,

()0,3B ,

∴直线BD 的解析式是:62-=x y .

∴直线BD 和直线OE 的交点为:()2,2-E .

()3,0-C ,

∴直线CE 的解析式为:312-=x y .

∴()

0,6F .

(3)当BCG ACO ∠=∠时,OCB ACM ∠=∠,

OM OC =,

∴?=∠=∠45OCB ACM ,

MCB ACM ACB ∠+∠=∠,MCB ABC AMC ∠+∠=∠,

∴AMC ACB ∠=∠,

MAC CAB ∠=∠,

∴AMC ACB ??∽, ∴

AB AC

AC AM =

. ()()3,00,1--C A 、,

∴10=AC ,

设()0,x M ,则

410

10

1=+x 解得,32=x ,

∴()0,32M .

()3,0-C ,

∴直线CM 的解析式是:32-=x y .

∴直线32-=x y 和抛物线a bx ax y 32

-+=的交点为:()()3,05,4-C G 、.

∴点()5,4G .

第4页,共5页

25.(1)

8AB =,6BC =

∴10AC ===

GD AB ⊥,90ABC ∠=? ∴90GDA ABC ∠=∠=?

DAG BAC ∠=∠ ∴ADG ABC ??∽

∴AG AD

AC AB = ∴

AG AC

AD AB =

108

x AD = ∴45AD x =

,勾股定理可得35

GD x =

GD GE = ∴35GE x =

∴85

AE x =

过点E 作EH AB ⊥,交AB 于点H .

90EHA ABC ∠=∠=?,HAE BAC ∠=∠

∴AHE ABC ??∽ ∴

AE AH

AC AB

=

AE AC

AH AB

=

∴810

58

x

AH = ∴3225AH x =

,勾股定理可得2425

EH x = ∴2424

4852522125

ADE

x x

AD EH S x ???==

= (2)

GD GE =

∴GED GDE ∠=∠

EP ED ⊥

∴90PED GDA ∠=∠=?

∴PED GED GDA GDE ∠+∠=∠+∠

AEP PED GED ∠=∠+∠,ADE GDA GDE ∠=∠+∠

∴AEP ADE ∠=∠

∴ADE AEP ??∽ ∴

AE AP

AD AE =

85AE x =

,4

5

AD x =,AP y = ∴854855x

y

x x = ∴165y x =

25(0)4

x <≤

第5页,共5页

(3)分情况讨论: ①当点F 在CB 延长线上时,

90FBP DEP ∠=∠=?,FPB DPE ∠=∠

∴DPE FPB ??∽ ∴

DE PE

FB PB =

DE FB

PE PB

=

ADE AEP ??∽ ∴

AD DE

AE PE =

AD FB

AE PB

=

∴42

58

85

x

y x =- 解得:4y =,即线段AP 的长为4. ②当点F 在线段CB 上时,

DPE FPB ??∽,ADE AEP ??∽这两对相似还是成立的. ∴

AD FB

AE PB

=

∴4258

85

x

y x =

- 解得:12y =,即线段AP 的长为12. 综上所述,线段AP 的长为4或12.

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