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青浦
参区
一考21．A ；2．B；3．C；4．D；5．A；6．D． 题：
答0
二、填空题： 案
1
及2 98．51；9．3e；10．a0；11．；12． 7． ；明2020.1
3 学年45
13． 第；14．229；15．2；16．
5 一学三、解答题： 期期
312
终3+8+13 研1
32 九
2 年
19．解：原式=． 级
卷
y1001x；
25
；17．1；18． 5
2
53
． 2
2
·························
···············（8分）
=32+2+31．·································································（1分）
=231．············································································（1分）
20．解：（1）∵四边形ABCD是平行四边形，
∴DC//AB，
DC=AB，·································································（2分）
∴BFAB DFDE
．········································································（1分）
∵DE∶EC=2∶3，∴DC∶DE=5∶2，∴AB∶DE=5∶2，················（1分） ∴BF∶DF=5∶2．····································································（1分） （2）∵BF∶DF=5∶2，∴5
BFBD．··············································（1分）
7
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∵BDADAB，∴BDab．
···········································（1分）
555
∴
BFBDab．
777 ∵AFABBF，∴5552
AFbabab．···················（2分）
7777
··························
····························（1分）
九年级数学第1页
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21．解：（1）∵∠ACB=90°，∴∠BCE+∠GCA=90°．
∵CG⊥BD，∴∠CEB=90°，∴∠CBE+∠BCE=90°，
∴∠CBE=∠GCA．······································
22．解：由题意，得∠分）
·····························（2分） 又∵∠DCB=∠GAC=90°，
∴△BCD∽△CAG．·································································（1分）
∴
CDBC ，
AGCA ·······································································（1分） ∴ 13 ，∴ 2 AG2
AG．
··························
3 ······························（1分）
（2）∵∠GAC+∠BCA=180°，
∴GA∥BC．······································（1分）
∴
GAAF
BCFB ．·······································································（1分）
∴ AF
2 FB 9 ．··········································································（1分） ∴ AF 2
S AB ．∴
AFC 2
．11
S 11 ABC ·················································（1分）
又∵
1
6 ················S233ABC
，∴
S． ···············（1AFC 分） 2
11 ABD=90°，∠D=20°，∠ACB=31°，CD=13．················（1
在Rt△ABD中，∵tan
D AB ，∴ ABAB
··········BD BDtan200.36 ． ·····（3分） 在Rt△ABC中，∵tan ACB AB ，∴ ABAB
·········BC BCtan310.6 ． ····（3分）
∵CD=BD-BC，
∴13
ABAB
．·································
0.360.6 ··································
···（1分）
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解得AB11.7
米．·······································································（1分）
答：AB九年级数学第2页为11.7
米．·····················································（1分）
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2
23．证明：（1）∵ AFFGFE，∴
AFFE FGAF
．··········································（1分）
又∵∠AFG=∠EFA，
∴△FAG∽△FEA．·····································
24．解：（
·（1分）
∴∠FAG=∠E．·······································································（1分）
∵AE∥BC，
∴∠E=∠EBC．······················································（1分）
∴∠EBC=∠FAG．·································································（1分）
又∵∠ACD=∠BCG，
∴△CAD∽△CBG．··································（1分）
2）∵△CAD∽△CBG，∴
CACD
．······················CBCG
······················（1分）
又∵∠DCG=∠ACB，
∴△CDG∽△CAB．··································（1分）
∴
DGCG ．
·······························
ABCB
······································（1分） ∵AE∥BC，∴
AEAG ．
·························
CBGC
····························（1分）
∴
AGGC ，∴
DGAG
，························AECB ABAE
·························（1分）
∴DGAEABAG．···························································（1分）
1）∵A的坐标为（1，0），对称轴为直线x=2，∴点B的坐标为（3，0）·（1分） 将A（1，0）、B（3，0）代入
yxbxc，得2
+ 2+ 1bc0
解得： b 4， ，
·····················
c 0.37 ·····················
93bc0.
·（2分）
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（WORD格式可编辑
243 所以，
yxx．
2 当x=2时， y 242+3=1
∴顶点坐标为（2，-1）·············································
···············（1分）．
（
∵∠CON=90°，∴四边形CONM为矩形． 2）
过∴∠CMN=90°，CO=MN． 点九P年
第3页作
PN⊥x轴，垂足为点N．过点C作CM⊥PN，交NP的延
点M．
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∵ yx
243
······································（1分）．
∵B（3，0），∴OB=OC．∵∠COB=90°，∴∠OCB=∠BCM=45°，···········（1分）． 又∵∠ACB=∠PCB，∴∠OCB-∠ACB=∠BCM-∠PCB，即∠OCA=∠PCM．（1分）．
∴tan∠OCA=tan∠PCM．∴ 设PM=a，则MC=3a，PN=3-a．
∴P（3a，3-a）．········································································（1分）
243 将P（3a，3-a）代入
yxx，得 2
3a12a33a．
11 解得
a=，a2=0（舍）．∴P（ 1
9 （3）设抛物线平移的距离为m．得
11 3 ，
yx21m，
2
16
）．····················9 ···················（1分）
1 PM
．
3 MC
∴D的坐标为（2，1m）．·······························································（1分）
过点D作直线EF∥x轴，交y轴于点E，交PQ的延长线于点F． ∵∠OED=∠QFD=∠ODQ=90°，
∴∠EOD+∠ODE=90°，∠ODE+∠QDF=90°，
∴∠EOD=∠QDF，········································································（1分）
∴tan∠EOD=tan∠QDF．∴
DEQF OEDF
29
11
m12
3
1 解得
m．所以，抛物线平移的距离为
5
1
． ·····················
·····················5
（1分）
16
m1m
．
．∴
25．解：（1）∵AD//BC，
∴∠EDQ=∠DBC．····················································（1分）
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DEBD ∵1，1 DQBC
，∴DEBD
DQBC
．········································（1分）
∴△DEQ∽△BCD．································································（1分）
∴∠DQE=∠BDC，
∴EQ//CD．···················································（1分）
九年第4页
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（B
EQQD ，∴2 P
∵△DEQ∽△BCD，∴
EQx．·················的DCCB
······（1分） 长
5 为
DQ=x，QP=2x-10．································（1分） （i）当EQ=EP时， ∴∠EQP=∠EPQ，
∵DE=DQ，∴∠EQP=∠QED，∴∠EPQ=∠QED，
∴△EQP∽△DEQ，∴
EQQP 2 ，∴ 2
DEEQ
5
x2x10x，
解得
125 x，或x0（舍
去）．···············································（2分）
23
（ii）当QE=QP时，
∴
2 25
5 x2x10，解得 x，················································（1分）
4
∵256
4 ，∴此种情况不存
在．·················································（1分）
∴ BP
125 23
（3）过点P作PH⊥EQ，交EQ的延长线于点H；过点B作BG⊥DC，垂足为点G．
∵BD=BC，BG⊥DC，∴DG=2，BG46， ∵BP=DQ=m，∴PQ=10-2m． ∵EQ∥DC∴∠PQH=∠BDG． 又∵∠PHQ=∠BGD=90°，
∴△PHQ∽△BGD．······································
··························（1分）
∴
PHPQHQ ，∴
PH102mHQ ．
BGBDGD 46102
∴102
26102m
m
PH．
·················HQ，
5
···············（25
分）
∴ 1EH202m2m 55
， 专业知识 整理分享
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∴
tanPEQ26m EH525
26102m126
PH
．
··········（1分）
九年级数学第5页
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