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2024~2025学年度第一学期期中检测
九年级数学试题
(全卷共4页，满共140分，考试时间90分钟：答案全部涂、写在答题卡上)
一、选择题（本题共8题，每小题3分，共24分）
1.如图，点A、B、C在⊙O上，∠AOB=72°，则∠ACB=
A.28
B.54
C.18
D.36
2.下列图形中，是中心对称图形的是
(第1题)
B.
3.一元二次方程4x2-4x+1=0的根的情况是
A.有两个不相等的实数根
B.有两个相等的实数根
C.没有实数根
D.无法判断
4.用配方法解方程x2一6x一6=0时，配方后得到的方程是
A.(+3)2=15
B.（x+3)2=3
C.(x-3)2=15
D.（x-3)2=3
5.若x,是一元二次方程x2-2x-3=0的两个根，则x2的值是
A.-3
B.-2
C.3
D.2
6.将抛物线y=2x2向右平移3个单位，再向下平移5个单位，得到的抛物线的表达式为
A.y=2(x-3)2+5B.y=2x+3)2+5C.y=2x-3)2-5D.y=2x+3)2-5
7.下列命题正确的是
A.三点确定一个圆
B.三角形的内心是三边垂直平分线的交点
C.长度相等的弧是等孤
D.等弧所对的圆心角相等
8.已知二次函数y=ax2+bx+C的自变量x与函数y的部分对应值列表如下：
-2
0
1
2
一3
一3
0
5
九年级数学试题第1页（共4页）
下列说法：①>0：②函数图像的顶点坐标是(-1，-4)；③函数图像与x轴的交点坐标
是(1,0)、（-3,0)；④若（-5，y1),(2.5,2)是函数图像上两点，则y1>y2,
其中说法正确的有()个
A.1
B.2
C.3
D.4
二、填空题（本题共10题，每小题3分，共30分）
9.⊙0的半径为5，OP=4,则点P在⊙0
▲
(填“内”、“外”、“上”).
10.方程x2一2x=0的解是▲
11.已知圆锥底面圆半径是8，圆锥的母线长为6，则这个圆锥的侧面积是▲，
12.抛物线y=x2-2x-3的顶点坐标是▲
13、如图，P是⊙0外的一点，PA、PB分别与⊙O相切于点A、B,C是劣弧AB上的任
意一点，过点C的切线分别交PA、PB于点D、E,若PA=4,则△PED的周长为▲
14.己知m是一元二次方程x2-x-2=0的一个根，则20-m2+m的值为▲
15.如图，AB是⊙O的直径，C,D是⊙O上的两点，若∠ABD=54°，则∠BCD的度
数是·▲一
16.某商店今年1月份的销售额是2万元，3月份的销售额是4.5万元，从1月份到3月
份，该店的销售额平均每月的增长率是▲一、
17.直角三角形两直角边长分别为6和8，那么这个三角形内切圆半径等于▲
18.如图，在四边形ABCD中，∠ABC=60°，∠DCB=30°，AD=2,BC=4,E
为AD的中点，连接BE,CE,则△BEC面积的最小值为
D
0
(第18题)
(第13题)
(第15题)
三、解答题（本大题有7小题，共86分）
19.(本题12分)解方程：
(1)22-3x+1=0:
(2)(x-1)2=3x-3.
九年级数学试题第2页（共4页）2024~2025学年度第一学期期中检测
九年级数学参考答案
题号 1 2 3 4 5 6 7 8
选项 D B B C A C D D
9． 内 10． x1 0 x2 2 11．48 π 12．（1，-4） 13．8
14．18 15．36 16．50% 17．2 18． 2 3 2
19．（1） a 2，b 3，c 1．······································································ 1分
b2 4ac 9 8 1．················································································2分
b b2 4ac 3 1．····································································· 4分x
2a 4
x 1 1 ．················································································· 6分1 x2 2
（2） (x﹣1)(x﹣4)＝0（10分）x1＝1，．x2＝4．··············································12分
20．（1）3．·······························································································3分
（2） AB是 O的直径， AB CD， CE DE 8 ································ 6分
在 Rt△ODE中，∠OED=90 ，
OD2 OE 2 DE 2 ，OD2 OD 4 2 82 ,··········································· 10分
OD 10 ··························································································· 12分
故 O的半径为10
21．（1）∵二次函数 y＝ax2+2x+c图象经过点 A（1，4）和点 C（0，3），
∴ a 2 c 4．························································································1分

c 3
解得： a 1．························································································· 3分

c 3
（2）略．··································································································6分
（3）0＜y≤4 x≤0或 x≥2．······························································· 12分
第 1页（共 3页）
{#{QQABLYAEgggoABJAAAhCAQ3iCgKQkgAAAYgGxEAAsAABSANABAA=}#}
22．（1）解：设每件衬衫降价 x元，
根据题意得(40-x)(20+2x)=1200，·································································· 3分
解之得 x1 10, x2 20，············································································· 5分
根据题意要尽快减少库存，所以应降价 20元···················································· 6分
答：每件衬衫应降价 20元．········································································· 7分
（2）解：设每天利润为w元，每件衬衫降价 x元，
w 40 x 20 2x 2(x 15)2 1250，················································ 9分
所以当 x=15元时，w有最大值，此时w =1250················································ 11分
答：每件衬衫降价 15元时，商场平均每天盈利最多共 1250元····························12分
23．（1）证明：连接 OC，
∵CE⊥AB，∴∠ACE+∠CAE=90 ．························································· 1分
∵OA=OC，∴∠OCA=∠CAE·································································· 2分
∵∠ACD=∠ACE，∴∠OCA+∠ACD=90 ．················································4分
OC⊥CD，∴DC是⊙O的切线．······························································· 6分
（2）解：在 Rt△OCD中，点 A为 OD的中点，∴CA=OA=OC．··················· 7分
∴△AOC为等边三角形，∴∠DOC=60 ．··················································8分
∵OA=2，∴OD=4，在 Rt△OCD中，CD OD2 OC 2 2 3．················· 10分
．················································12分
24.（1)解：（1）B（4，0），C（0，4）···························································4分
4k b 0
(2)设直线 BC的解析式为 y＝kx+b，将 B（4，0）和 C（0，4）代入得
b 4
k 1
解得： ···················································································· 5分
b 4
∴直线 BC的解析式为 y＝﹣x+4，
设 P（t，﹣t2+3t+4），G（t，﹣t+4），
∴PG＝﹣t2+3t+4﹣（﹣t+4）＝﹣t2+4t，
∵S△PCB＝6，
1
∴ PG OB＝6 1，即 （﹣t2+4t）×4＝6．··················································· 6分
2 2
解得：t1＝1，t2＝3
∴点 P的坐标为（1，6）或（3，4）．··························································8分
第 2页（共 3页）
{#{QQABLYAEgggoABJAAAhCAQ3iCgKQkgAAAYgGxEAAsAABSANABAA=}#}
（3）∵OB＝OC＝4，∴△OBC是等腰直角三角形，∴∠BCO＝45°，
∵PE⊥x轴，即 PE∥y轴，∴∠PGF＝∠BCO＝45°，
∵PF⊥BC，∴∠PFG＝90°，
2
∴△PFG是等腰直角三角形，∴PF＝FG PG，
2
∴△PFG的周长＝PF+FG+PG＝（ 2 1）PG．············································· 9分
当 PG最大时，△PFG的周长最大.
PG＝﹣t2+2t= （t 2）2 4．···································································10分
∴当 t 2时，△PFG的周长最大，此时点 P的坐标为（2，6）．····················· 12分
25．（1）
······················································································ 4分
（2）①
·······················································································8分
② ≤m≤4 ．············································································ 11分
(3) 3≤m 3 3 2∴ < ···············································································14分
2
第 3页（共 3页）
{#{QQABLYAEgggoABJAAAhCAQ3iCgKQkgAAAYgGxEAAsAABSANABAA=}#}

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