2024—2025 学年第一学期九年级学业水平质量监测
数学试题参考答案及评分建议
一、选择题（本大题共 10个小题，每小题 3分，共 30分）
题号 1 2 3 4 5 6 7 8 9 10
答案 C B C A B D C D A B
二、填空题（本大题共 5个小题，每小题 3分，共 15分）
11 8 1．2 12． 0.9 13．55° 14． π 15．
25 2
三、解答题（本大题共 8个小题，共 75分）
16．（本题共 2个小题，每小题 5分，共 10分）
解：（1）a=1，b=1，c= -6. ··························································································（1分）
因为Δ=b2-4ac=12-4×1×（ -6）=25>0，所以方程有两个不等的实数根.································（3分）
x= b b
2 4ac = 1 25 = 1 5，即 x1=2，x2=-3．·······················································（5分）
2a 2 1 2
（2）y=-2x2-4x+1=-2（x+1）2+3.·····················································································（7分）
因为-2<0，所以当 x=-1时，该函数有最大值，最大值为 3.··················································（10分）
17. （本题 7分）
证明：因为 AB=4，BC=8 BD=2 AB 4 1 BD 2 1， ，所以 = ， = = ． ·······························（4 分）
CB 8 2 BA 4 2
AB BD
所以 ． ·······································································································（5 分）
CB BA
又∠ABD=∠CBA，所以△ABD∽△CBA． ··································································（7 分）
18.（本题 7分）
解：列表如下：
······························································································································ （4 分）
由表格知，所有可能出现的结果有 6 种，并且它们出现的可能性相等，其中抽出的两张卡片
1
均是物理变化的结果有 1 种，所以 P（抽出的两张卡片均是物理变化）= ． ···········（7 分）
6
第 1页（共 3页）
19.（本题 8分）
解：（1）将 A（ -3，0）代入 y=x+m，得 -3+m=0，解得 m=3.
所以一次函数的解析式为 y=x+3.··············································································（2 分）
将 B（n，4）代入 y=x+3，得 n+3=4，解得 n=1.
所以 B（1，4） .······································································································（4 分）
k
将 B（1，4）代入 y= ，解得 k=4.
x
4
所以反比例函数的解析式为 y= .············································································ （6 分）
x
（2）a＞1． ············································································································（8 分）
20. （本题 10 分）
解：（1）C ·······································································································（2 分）
（2）因为正三角形 ABC的边长为 2，D，E，F分别为 BC，CA，AB的中点，所以∠A=∠B=∠C=60°，
AE=AF=BD=BF=CD=CE=1.
所以阴影部分的周长为∠A，∠B，∠C C = 60π 1+ 60π 1+ 60π 1= 180π 1所对应的弧长和，即 阴 影 =π.180 180 180 180
··························································································································（6分）
（3）5π ·················································································································（10分）
21.（本题 9 分）
解：（1） y= 1000 ····································································································（2分）
x
10
a+28= 1000根据题意，得 .······················································································（7分）
10 15
解得 a=12.
所以这个空矿泉水瓶的质量为 12 g. ··········································································（9分）
22.（本题 12 分）
解：（1）因为 AD是△ABC的角平分线，所以∠BAD=∠CAD．··········································（1分）
所以∠E=∠CAD.所以 EC=AC.························································································（2分）
AB BD
所以 ＝ .···········································································································（3分）
AC CD
（2 3） ·····················································································································（5分）
2
（3）延长 BE交 AD的延长线于点 G.···············································································（6分）
因为四边形 ABCD是矩形，所以 AB∥CD．所以△GDE∽△GAB.
第 2页（共 3页）
DG = DE DG 1所以 ，即 = ，解得 DG=2．·································································· （8分）
AG AB 4 DG 3
所以 AG=AD+DG=4+2=6．····························································································（9分）
在 Rt△ABG中，由勾股定理，得 BG= AB2 AG2 = 32 62 =3 5 ．··································（10分）
AF BAG BF = AB 1 BF 1 1因为 平分∠ ，所以 = ．所以 = ．所以 BF= BG= 5 ．·····················（12分）
GF AG 2 BG 3 3
23.（本题 12 分）
解：（1）（4，2） ·····························································································（2 分）
4 ·······················································································································（3 分）
2 ·······················································································································（4 分）
（2）不能围出矩形地块 .·························································································（5 分）
当 a=6 时，一次函数的解析式为 y=-2x+6，其图象如图中 l2 所示 .
··················································································（7分）
第 23题答图
8
理由：因为 l 2 与函数 y= （ x>0）的图象没有交点，所以不能围出面积为 8 m2 的矩形．
x
······························································································································ （8 分）
（3）令 -2x+a= 8 ，整理，得 2x2-ax+8=0.
x
因为一次函数与反比例函数的图象有唯一交点，所以Δ=（ -a） 2-4×2×8=0，解得 a=±8.
因为 a＞0，所以 a的值为 8． ················································································（10 分）
所以一次函数的解析式为 y= -2x+8.
y
8
， x 2，
联立 x 解得
y 2x 8，
y 4.
所以这个交点的坐标为（2，4）． ········································································· （12 分）
第 3页（共 3页）5,二次函数y=x-2x+1的图象与x轴的交点个数是
小.0个
B.I个
2024一2025学年第一学期九年级学业水平质量监测
C.2个
D.无法确定
数
学
6.如图，点A在双曲线y=上，AB1x轴于点B,若△AOB的而积为2.则k的值为
(监测内容：第二十四章24.4至第二十七章27.2.2】
A.2
B.4
C.-2
D.-4
注意事项：
1.试卷分第I卷和第Ⅱ卷两部分，全卷共8页，满分120分，考试时问120分钟
2.答案全部在答题卡上完成，答在本试卷上无效
第I卷选择题（共30分）
第6题图
第8题图
第10题图
7.象棋是起源于中国的一种两人对弈游戏，现今通行的象棋.相传为唐代牛僧需所制
一、选择题（本大题共10个小题，每小题3分，共30分.在每个小题给出的四个选项中，
刻圆木或牙、骨为棋子三十二枚，红黑各半，黑方以将统士、象、车、马、炮各二，卒五，
只有一项符合题目要求，请选出并在答题卡上将该项涂黑】
若从一套完整的象棋棋子中随机摸出一枚棋子，则该棋子为黑马的概率为
L.足球比赛常依据VAR(视频助理裁判技术)判定球是否出界，以得
到更加公正的比赛判罚.如图，把足球与场地边界线看作圆与直
B君
线，它们的位置关系是
C16
A.相切
B.相交
第1题图
8.如图，AB为⊙O的直径，C为⊙0上一点，BD平分∠ABC.若∠D=20°，则∠ABD的
C.相离
D.分离
度数为
2.我国古代数学的许多创新与发明都在世界上具有重要影响.下列图形中，风于中心
A.20
B.25
对称图形的是
C.30
D.35
9.二次函数y=a心+x和反比例函数y=。在同一平面直角坐标系中的图象可能是
猫卡尔心形绒线
赵爽弦图
刘徵创团术
中国七巧扳
A
B
C
D
3.山西，因居太行山之西而得名，简称“晋”如图，用放大镜将由“晋”
传
字设计的图标放大，则放大前后两个图形之间属于图形的
恶
A.平移
B.轴对称
10.如图，在矩形ABCD中，AB=12cm,BC=16cm,E,F分别是AB,CD上的点，且AE=
C.相似
D.旋转
第3题图
DF=8cm,两动点M,N都以2cms的速度分别从点C,F出发沿CB,FE向点B,E
4.下列诗句所描述的事件中，属于必然事件的是
运动.当矩形CFNM与矩形AEFD相似时，点M,N运动的时间为
A.黄河入海流
B.手可摘星辰
A.18或2s
B.1s或4s
C.锄禾日当午
D.大漠孤烟直
C.2s或4s
D.1s或2s或4s
九年级数学第1页（共8页）
九年级数学第2页（共8页）
