2024 年秋学期初三数学期中考试参考答案与评分标准
一、选择题（本大题共 10 小题，,每小题 3 分，共 30 分．）
1．B 2．C 3．D 4．C 5．B 6．C 7．D 8．B 9．A 10．B
二、填空题（本大题共 8小题，每小题 3分，共 24分．）
1
11． 12 8． 13．1:16 14． 10
5 3
2 5 4 5 8 5
15．20 16．60° 17．4 5 18． , ,
5 5 5
三、解答题（本大题共 9 小题，共 96 分．）
19. （本题满分 16分）
5
（1） (x 2)2
25
， x 2 ········································································· 2分
4 2
1
x 91 ， x2 ······················································································· 4分2 2
（2） (m 1)2 4(m 1) 0， (m 1)(m 1 4) 0 ············································· 2分
m 1 0或m 3 0，m1 1，m2 3 ····················································4分
（3） t 2 2t 15 0 ······················································································· 1分
(t 3)(t 5) 0， t 3 0或 t 5 0 ························································ 2分
t1 3， t2 5 ·······················································································4分
（4）a＝2，b＝－4，c＝1， ( 4)2 4 2 1 8 0，······································2分
x 2 2 ······························································································ 4分
2
20. （本题满分 10分）
（1）∵四边形 ABCD是正方形，∴∠B＝∠C＝90°，∴∠AEB+∠BAE＝90°············2分
∵AE⊥EF ∴∠AEF＝90°∴∠AEB+∠CEF＝90°······································3分
∴∠BAE＝∠CEF ··················································································· 4分
∴ △ABE∽△ECF··················································································· 5分
（2）∵BE＝3，EC＝7，∴BC＝3+7＝10，···························································6分
∵四边形 ABCD是正方形，∴AB＝BC＝10····················································7分
∵△ABE∽△ECF ∴ AB BE ，10 3 ····················································· 9分
CE CF 7 CF
∴CF 21＝ ．························································································· 10分
10
21.（本题满分 10分）
（1）200，36··································································································4分
（2） ·········································································8分
1
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46
（3）2000× ＝460（名），答：估计最喜欢“E乒乓球”的人数为 460名．········· 10分
200
22. （本题满分 10分）
1 ＝（ 2k）2（ ）∵ 4 k （k 1）＝4k 2 4k 2 4k＝4k＞0 ，解得 k＞0··············3分
又 k 1 0即 k 1··············································································· 4分
∴ k 0且 1························································································· 5分
（2）∵0＜k＜5且 k 1，∴整数 k的值为 2，3，4，·············································6分
3x2 8x 4 0 2当 k＝4时，方程为 - ＝ ，解得 x1＝2，x2＝ ，·····························8分
3
当 k＝3或 2时，此时方程没有整数解．
综上所述，k的值为 4．··········································································· 10分
23. （本题满分 10分）
（1）证明：连接 OC，∵PC与半圆相切于点 C，∴∠OCD＝90°，··························1分
∴∠DCE+∠ACO＝90°，········································································· 2分
∵OA＝OC，∴∠A＝∠ACO，····································································3分
∵OD⊥AB ∴∠AOE＝90°∴∠A+∠AEO＝90°，······································· 4分
∴∠DCE＝∠DEC＝∠AEO，∴DC＝DE·······················································5分
（2）解：∵OA＝2OE，∴设 OE＝x，AO＝2x，
∴EF＝OF﹣OE＝x，∴DE＝DC＝x+3，OD＝2x+3，
∵OC2+CD2＝OD2，∴（2x）2+（x+3）2＝（2x+3）2，·····································6分
∴x＝6或 x＝0（不合题意舍去），·································································7分
∴OD＝15，OC＝OB＝12，CD＝9，
∵∠DOP＝∠OCD＝∠DOP＝90°，∴∠D+∠DOC＝∠DOC+∠COP＝90°，
∴∠D＝∠COP，∴△CDO∽△COP，···························································· 8分
∴OD CD ，∴ 15 9 ，∴OP＝20，···························································9分
OP OC OP 12
∴BP＝OP OB＝8···················································································· 10分
24. （本题满分 10分）
（1）∵2x+y＝80，∴y＝ 2x+80，
∵S＝xy，∴S＝x（ 2x+80）＝ 2x2+80x；·················································· 2分
（2）∵y≤42，∴ 2x+80≤42，∴x≥19，∴19≤x＜40，······································· 4分
当 S＝750时， 2x2+80x＝750，
x2 40x+375＝0，（x 25）（x 15）＝0，∴x＝25，········································6分
∴当 x＝25m时，矩形实验田的面积 S能达到 750m2；
（3）当 S＝850时， 2x2+80x＝850，即 x2 40x+425＝0
∵ ＝（ 40）2﹣4 1 425＝ 100 0 ∴x无解, 面积 S不能达到 850m2················ 8分
∵S＝ 2x2+80x 2 2，即 2x 80x S 0，由 ＝（ 80）﹣4 2 S 0
∴ S 800 ∴S最大值 800m2， 此时 x＝20m．···············································10分
2
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25 （本题满分 10分）
（1）①作 AC的中垂线，交射线 AQ于点 O，·······················································2分
以 O为圆心，OA为半径作⊙O交射线 AQ于点 O．·····································4分
②作∠CBQ的平分线交 CP于 M．······························································ 6分
（2） 265 ···································································································10分
2
26. （本题满分 10分）
（1）取 AC的中点 G，连接 EG
∵∠ACB＝90°，CB＝2，CA＝4∴ AB 22 42 2 5 ································ 1分
∵E是 AB的中点∴EG= 1 BC=1，EG∥AC，AE= 1 AB= 5
2 2
∴∠EGA＝∠ACB＝90°∵AE＝ 5 AD ∴AD=1，CD=3··································· 2分
∵AG=GC=2，∴DG=AG AD=1=EG
∴∠EDG＝∠DEG＝45°∴∠EDA＝135°······················································· 3分
由翻折得∠ADE＝∠FDE＝135°，AD=AF=1
∴∠CDF＝∠EDF ∠EDG＝135° 45°=90°，·············································4分
∴CF DF 2 CD2 12 32 10 ···························································· 5分
（2）过 E作 EH⊥AC于 H，设 EF与 AC相交于 M，设 AD＝x， AE 5x，
由翻折得 DF＝AD＝x，∠ADE＝∠FDE，则∠AHE＝∠ACB＝90°，
又∵∠A＝∠A，∴△AHE∽△ACB，∴ EH AH AE ，
BC AC AB
∵CB＝2，CA＝4， AB 2 5，∴ EH AH 5x ，
2 4 2 5
∴EH＝x， AH 2x，则 DH＝AH AD＝x＝EH，
∴Rt△EHD是等腰直角三角形，···································································· 6分
∴∠HDE＝∠HED＝45°，则∠ADE＝∠EDF＝135°，
∴∠FDM＝135° 45°＝90°，···································································7分
∵∠FDM＝∠EHM＝90°，∠DMF＝∠HME，DF=EH
∴△FDM≌△EHM（AAS），
1
∴DM MH x，CM AC AD DM 4
3
x，
2 2
∴ S S S 1CM EH 1CM DF 1 3 (4 x) x 2 3 (4 x) x，·········8分 CEF CME CMF 2 2 2 2 2
S 1 1 ，········································· 9分 BEC S ABC S AEC 4 2 4 x 4 2 x2 2
4 4
∴ (4 3 x)x 2(4 2x)，解得 x1 ，x2＝4（舍去），则 AD ·······················10分2 3 3
3
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27. （本题满分 10分）
（1）OM+ON＝2AP························································································· 2分
（2）①当 M在线段 AO上时，如图，延长 NM、PA交于点 G．
由（1）知 OM+ON＝2AP，设 OM＝2x，则 ON＝5x，OA＝AP＝3.5x．
∴AM＝AO OM＝1.5x，
∵∠MON＝∠MAG＝90°，∠OMN＝∠AMG，
∴△MON∽△MAG，∴OM ON ，
2x 5x
AM AG 1.5x AG
∴AG=3.75x································································································4分
∵AP∥OB，∴△ONF∽△PGF，
∴OF ON 5x 20 ，∴ OP 49 ；··············································· 6分
PF PG 3.5x 3.75x 29 OF 20
②当 M在 AO的延长线上时，如图，过 P作 PC⊥OB于 C，并延长交 MN于 G．
∵∠AOB＝90°，PA⊥OA ∴四边形 OAPC是矩形，
∵点 P在∠AOB的平分线上∴∠AOP=∠APO=45°∴PA=OA
∴四边形 OAPC是正方形
∴OA＝AP＝PC＝OC，∠APC＝90°，PC∥AO，
∵PN⊥PM，∠APM＝∠CPN＝90° ∠MPC，
又∵∠A＝∠PCN＝90°，AP＝CP，∴△APM≌△CPN，∴AM＝CN，
∴ON OM＝OC CN OM＝AO AM OM＝2AO，
∵OM＝2x，ON＝5x，∴AO＝1.5x，CN＝AM＝3.5x，
∵PC∥AO，∴△CGN∽△OMN，
∴ CG CN ，即 CG 3.5x ，∴CG＝1.4x，······················································· 8分
OM ON 2x 5x
∵PC∥AO，∴△OMF∽△PGF，
∴OF OM 2x 20 ，∴ OP 9 ； 综上， OP的值为 49 或 9 ．········· 10分
PF PG 1.5x 1.4x 29 OF 20 OF 20 20
4
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九年级数学试题
2024.11
本试卷分试题和答题卡两部分，所有答案一律写在答题卡上
考试时间为120分钟，试卷满分150分，
注意事项：
1.答卷前，考生务必用0.5毫米黑色墨水签字笔将自己的姓名、班级、考试号填写在答题卡的
相应位置上，并认真核对姓名、班级、考试号是否与本人的相符合，
2.答选择题必须用2B铅笔将答题卡上对应题目中的选项标号涂黑，如需改动，请用橡皮擦干
净后，再选涂其他答案.答非选择题必须用0.5毫米黑色墨水签字笔作答，写在答题卡上各题目指
定区域内相应的位置，在其他位置答题一律无效
3.作图必须用2B铅笔作答.
4.卷中除要求近似计算的结果取近似值外，其他均应给出精确结果
一、选择题（本大题共10小题，每小题3分，共30分，在每小题所给出的四个选项中只有一项是正
确的，请用2B铅笔把答题卡上相应的选项标号涂黑)
1.一元二次方程x2=x的解是
…（▲)
A.x1=x2=1
B.x1=1,x2=0
C.x1=x2=0
D.x1=-1,x2=0
2.已知(m-3)x2-7+2024x-2024=0是关于x的一元二次方程，则m的值为…（▲）
A.3
B.0
C.-3
D.±3
3.下列网格中各个小正方形的边长均为1，阴影部分图形分别记作甲、乙、丙、丁，其中是相似形的
为…(（▲)
甲
丙
A.甲和乙
B.乙和丁
C.甲和丙
D.甲和丁
4.已知⊙0的直径是6，点P到圆心0的距离是6，则点P与⊙0的位置关系是…（▲
A.点P在⊙0内
B.点P在⊙0上
C.点P在⊙0外
D.点P是圆心
5.两年前生产1千克甲种药品的成本为80元，随着生产技术的进步，现在生产1千克甲种药品的
成本为60元.设甲种药品成本的年平均下降率为x,下列方程正确的是…（▲)
A.80(1-x2)=60
B.80(1-x)2=60
C.80(1-x)=60
D.80(1-2x)=60
6.小亮在计算正数a的平方时，误算成a与2的积，求得的答案比正确答案小1，则a=
A.1
B.5-1
C.2+1
D.1或2+1
7.宽与长的比是5，'的矩形叫黄金矩形，黄金矩形给我们以协调的美感，世界各国许多著名建筑
2
为取得最佳的视觉效果，都采用了黄金矩形的设计.已知四边形ABCD是黄金矩形(AB
A.3
B.2
C.1
D.0
一九年级数学试卷共6页第1页一
