2024-2025学年鞍山市九年上数学期末试卷（pdf版，含答案）

2024—2025 学年度第一学期期末质量检测
九年级数学参考答案及评分标准
一、 选择题：（每题 3分，共 30 分）
题号 1 2 3 4 5 6 7 8 9 10
答案 D C A C A B C D C B
二、填空题：（每题 3分，共 15分）
2
11. y=(x－4)(x－5)或 y=x －9x＋20 12. 13. 8 14. 115° 15. ，
三、解答题：（本题共 8小题，共 75分）
16. （每小题 5 分，共 10 分）
解：（1） ·································································· (2 分)
， ······························································· （5 分）
（2）解：(y－1)(y－2)=0 ··································································· (2 分)
， ································································· （5 分）
17. （本题 8 分）
解：（1）把点 A（3，1）代入 ·························································· （1 分）
解得 k=3 ················································································· （2 分）
∴所求的函数解析式为 ··························································· （3 分）
（2）由对称可得，点 C（－3，1）（或 AO=CO） ······································ （4 分）
把 x=3 代入
∴点 B 坐标（3， ） ································································· （5 分）
∴ ······················································· （7 分）
解得 m=－5 ·············································································· （8 分）
18. （本题 8 分） y M
D' B
C
解：如图所示
D
O A x
（1）所图，△OCD′即为所求························································· （1 分）
点 D′坐标为（－1，6） ······················································· （2 分）
作图··················································································· （4 分）
（2）如图，△OBM 即为所求 ··························································· （5 分）
△OM D′为等腰直角三角形 ························································ （6 分）
作图··················································································· （8 分）
19.（本题 8 分）
A E D
解：（1）∵四边形 ABCD 为矩形
F
∴AD∥BC，∠A=90°
在 Rt△ABC 中， O
BD= B H G C
又∵AD 是⊙O 切线，OE 是半径
∴OE⊥AD，
∴OE∥AB ·················································································· (2 分)
∴ ，即 ····························································· （3 分）
解得 r= ，即⊙O 的半径是 ···················································· （4 分）
（2）延长 EO 交 BC 边于点 H 点 ·················································· （5 分）
由 AD∥BC，∴EH⊥BC
∴EH=AB，∠EHG=∠A=90°
∵四边形 ABGE 是⊙O 内接四边形
∴∠AFE=∠BGE，
∴△ABC∽△ABC ··································································· （6 分）
∴ ················································································· （7 分）
由（1） ，即
∴AE=3， ································································ （8 分）
20.（本题 8 分）
解：（1）由表格可知；当 50≤x≤85 时
y= ········································· (1 分)
整理，得 2y=－4x +540x－12600 ···················································· (3 分)
当 30≤x＜50 时
y=220(x－30)=220x－6600 ······························································ (4 分)
∴所求函数关系式 ················· (5 分)
（2）当 2y=5000 时，由－4x +540x－12600=5000
解得 x1=80，x2=55 ····································································· (6 分)
由 220x－6600=5000
解得 >50， 不合题意 ························································ (7 分)
所以，若每天销售利润不低于 5000 元，其价格的合适范围是 55≤x≤80 ···· (8 分)
21.（本题 8 分）
解：如下图，用描点法画函数图象 ·························································· （2 分）
（1） 取全体实数） ························································· （4 分）
（2）过点 P 作 PB⊥y 轴于点 B，连接 PA ·················································· (5 分)
∵l1 是 AM 的垂直平分线
∴PA=PM ························································································· (6 分)
∴PB=x，PA=y，AB=
在 Rt△ABP 中，
∴ ··········································································· （7 分）
即 ··············································································· （8 分）
y
8 y l1
7
6 B P
54 A
3
2 l2
1 O x
–6–5–4–3–2–1 1 2 3 4 5 6
O M x
–1
–2
–3
22.（本题 12 分）
解：（1）由旋转可得 CD=CB，∠DCB=90°
B
∴∠DCE+∠BCA=90°
D
∵∠A=90°
∴∠CBA+∠BCA=90°
∴∠DCE=∠CBA E C A
又∵DE⊥CA
∴∠DEC=∠A=90°
∴△ABC≌△ECD ···································································· （3 分）
（2）延长 FD，过点 C 作 CG⊥FD 交 FD 延长线于 G
∴∠G=90° F
M H
∵CH⊥BM，DF⊥BM
D B
∴∠DFH=∠CHB=90°
∴四边形 FGCH 是矩形
G
∴∠GCH=90°
E AC
∴∠DCG=∠HCB
又∵CD=CB
∴△HBC≌△GDC ········································································ (5 分)
∴CG=CH
∴矩形 FGCH 是正方形
∴四边形 CBFD 的面积等于正方形 FGCH 面积
∴四边形 CBFD 的面积 2S=6 =36 ··················································· (7 分)
（3）过点 E 作 EN⊥CD 交 CD 于点 N
由（1）△ABC≌△ECD
M P
且 AB=AC，AE= D B
∴CE=ED=
N
在 Rt△ECD 中
A
E C
∴CD= （ ） （ ）
∴DN=CN= EN=3 ······································································· （9 分）
设 MD=x，则 MN=3+x
在 Rt△ECD 中
∴ME=
∵3MD=CP，2ME=MP
∴CP=3x，MP=2 ，MC=x+6
在 Rt△MCB 中
∵ 2 2 2 MC +CP =MP
∴
解得 x= (负数不合题意)
∴MD= ········································································ （12 分）
23.（本题 13 分）
解：（1）当 a>0 时，k<0；当 a<0 时，k>0 ················································· （2 分）
（2）根据题意， 2y=a(x－h) +2h ······························································ （3 分）
把点 A（0，－2）代入解析式
－ 22=ah +2h，解得
∵a>0，∴－2－2h>0
∴h 的取值范围是 h< －1 ································································ (6 分)
2
（3）设抛物线解析式为 y=a(x－h) +2h+1
把点 A（0，－2）代入解析式
－ 2
2=ah +2h+1，解得
∴抛物线解析式为 ····································· (7 分)
由题意
解得
∴点 B 的横坐标是 ···································································· (8 分)
根据抛物线对称性，将点 B 向左平移 6 个单位为 x=h
∴ －6=h
解得 h=－3
∴抛物线解析式为 ················································ (10 分)
根据题意 ······················································ （11 分）
解得 ， ， 两交点成中心对称，设第一象限交点为 M，过点
M 作 MN⊥x 轴与点 N，由勾股定理得 OM=
∴抛物线与直线 l 两交点之间的距离为 ····································· （13 分）