2024年奉贤区初三数学一模卷参考答案和评分标准

这是一份2024年奉贤区初三数学一模卷参考答案和评分标准，共5页。试卷主要包含了选择题，解答题等内容，欢迎下载使用。
1．C ； 2．D ； 3．A ； 4．C ； 5．B ； 6．D .
填空题：（本大题共12题，每题4分，满分48分）
三、解答题（本大题共7题，其中19-22题每题10分，23、24题每题12分，25题14分，满分78分）
19．解：原式=. ···········································································（8分）
=. ·····················································································（2分）
20．（1）解：将点A（3，0），B（0，-3）代入，得：
·····································································································（1分）
解得：. ··································································································（2分）
∴抛物线表达式为. ········································································（1分）
顶点坐标为（1，-4）. ······················································································（2分）
（2）设直线AB的表达式为.
将A（3，0）代入，得：. ·····································································（1分）
∵抛物线的对称轴为直线x=1. ·························································（1分）
∴抛物线与对称轴的交点P坐标是（1，-2）. ······························································（2分）
21．（1）. ·····························································································（4分）
（2）∵点G是△ABC的重心. ∴AG=2DG. ∴.
∵AD=6. ∴AG=4. ···················································································（2分）
∵∠AGE=∠C，∠A=∠A.
∴△AEG∽△ACD. ··················································································（1分）
∴. ··························································································（1分）
∴AE=3. ··································································································（2分）
22．（1）解：由题意得：AB//A'B'，OC⊥AB，OC⊥A'B' ···············································（1分）
∴. ··················································································（1分）
∴. ···························································································（1分）
∵.
∴，厘米. ··································································（2分）
答：像A'B'的长为3.2厘米.
由题意得：四边形ACOP是矩形，
∴AP=CO=32厘米. ······················································································（1分）
∵AB//A'B'，∴.
∴. ·······························································································（2分）
∵AP//CD，∴. ······································································（1分）
∴厘米. ························································································（1分）
答：焦距OF的长为厘米.
23. （1）证明：∵AB=AC， ∴∠B=∠C. ···································································（1分）
∵∠AFD=∠B，∴∠AFD=∠C. ····································································（1分）
∵ ∠AEF=∠DEC， ∴△AEF∽△DEC. ·························································（2分）
∴. ····························································································（1分）
∴AF·CE=CD·FE. ···················································································（1分）
（2）证明：∵，∠AFD=∠B.
∴△ABC∽△AFD， ∠ADF=∠C. ··································································（2分）
∵∠CAD=∠DAE， ∠ADF=∠C，∴△ADE∽△ACD. ··········································（2分）
∴. ···························································································（1分）
∴. ·······················································································（1分）
24．（1）①解：∵，∴ y=，∴顶点A的坐标为（1，-1）. ················（2分）
∴抛物线关于y轴的镜像抛物线的顶点为（-1，-1）. ·······································（1分）
∴抛物线的表达式为. ···························································（1分）
②过点B作BM⊥y轴，垂足为点M.
∵点A和点B关于直线x=m对称. ∴点B的纵坐标等于-1. 即OM=1. ····················（1分）
∵. ∴BM=4. ································································（1分）
∴点B的坐标为（4，-1）或（-4，-1）.
∴. ························································································（2分）
（2）解：过点E作EN⊥CD，垂足为点N.
∵把点E（2，1）代入. ∴. ·························（1分）
∵抛物线，∴顶点C（-2b，c-b2）. ··········································（1分）
∵两条抛物线交于点E，△CDE是直角三角形，点C、D关于直线x=m对称，
∴△CDE是等腰直角三角形，即CE=DE. ∴CN=EN.
∵CN=2+2b，EN=1+b2+2b，∴2+2b=1+b2+2b. ···················································（1分）
∴b=1，c=-2
A
B
C
D
F
E
G
H
A(G)
B
C
D
F
E
H
∴该抛物线解析式为····································································（1分）
解：
过点D作DH⊥BC，垂足为H.
∴∠ADH=∠DHC=∠ABC=90°，∴四边形ABHD是矩形，DH=AB=4，AD=BH=6.
∵AD∥BC，∴∠ADE=∠DEC.
∵DE平分∠ADC，∴∠ADE=∠CDE.
∴∠DEC=∠CDE，∴CD=CE. ········································································（1分）
∵在Rt△DHC中，，DH=4，∴，CD=CE=5. ·············（2分）
∴BC=9，∴. ···································································（1分）
∵点G在边AD上，∠AGB=∠GBC，△CGB∽△BAG. ∴∠BGC=∠A=90°，
. ····························································································（1分）
∵在Rt△BAG中，AB=4，AG=2， ∴. ·············································（1分）
∴BC=10. ∴CH=4. ····················································································（1分）
∵DH=CH=4. ∴. ·········································································（1分）
∴. ·············································································（1分）
（3）①当点G在AD上时，
∵AG=1，∴DG=5.
∵AD∥BC，∴. ··········································································（1分）
∴DG=CE=5，∴CD=CE=5. ·········································································（1分）
②当点G在CD上时，延长DA、CG交于点M.
∴.····················································································· ·····（1分）
设AM=x，则BC=3x. ∴CH=3x-6，DM=CE=CD=x+6.
∵DH2+CH2=CD2. ∴42+(3x-6)2=(x+6)2. . ·········································（1分）
C
F
A
B
D
E
G
H
M
G
A
B
C
D
F
E
∵BC >AD. ∴. ∴. ···········································（1分）7. ；
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