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平定县2024一2025学年九年级教学质量监测试题
数
学
注意事项：
11
1本试卷分第I卷和第Ⅱ卷两部分.全卷共8页，满分120分，考试时间120分钟.
2答题前，考生务必将自己的姓名、准考证号填写在本试卷相应的位置
3答案全部在答题卡上完成，答在本试卷上无效，
4考试结束后，将答题卡交回，
第卷选择题（共30分)
-5.
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，只有
一项符合题目要求，'请选出并在答题卡上将该项涂黑.)
山计算2行的陆果足
银好会径公头发河人：西祖《
A早2意期，臭B2时兰C二1
倍D.1
2.餐桌对于我们中国人有着非同一般的意义，它不仅体现了一个家庭的温度，还承载着家庭团
圆的欢声笑语。如左图为一张圆形木质饭桌，则其俯视图为
正面
(第2题)
3.我国是世界上最大的茶叶种植国，茶园面积最大、增速最快，同时也是世界上茶叶消费量最
大的国家.2024年我国茶叶产量为374万吨，比上年增产5.5%，强势卫冕世界第一.则数据374
万吨可用科学记数法表示为
:赋国便答·息卧明赶中界润
A.374×104吨
B.37.4x105吨C,-3.74×105吨D.374×105吨(
4.下列运算正确的是
图：洪浓全 ()
帕遥凤0数好海尚村贵色路罢00共华中合试5
A0(-6)÷(-)=2中是更食卷代是B.(2m)子÷(2m)2=m意是i55
辛途袋贺圆人西域砂计出张氏客器林西复春酸实器
C.2a3.3a2=6a6
D.-27+30=-3
数学试题第1页（共8页）
5.如图①所示，平整的地面上有一个不规则图案（图中阴影部分），为了解该图案的面积是多
少，小利采取了以下办法：用一个面积为200cm2的长方形，将不规则图案围起来，然后在适
当位置随机地朝长方形区域扔小球，并记录小球落在不规则图案上的次数（球扔在界线上或长
方形区域外不计试验结果)，他将若干次有效试验的结果绘制成了如图②所示的统计图，由此
估计不规则图案的面积大约为
A.80cm2
B.75cm2
C.70cm2
D.60cm2
0.4
小球落在不规则图案上的频率
0.35
■■
0.3…
60120180240300360420480实验次数
图0
(第5题)
图②
(第6题)
6.《千里江山图》是宋代王希孟的作品，如图，它的局部画面装裱前是一个长为2.4米，宽为
1.4米的矩形，装裱后，整幅图画宽与长的比是8：13，且四周边衬的宽度相等，则边衬的宽度
应是多少米？设边衬的宽度为x米，根据题意可列方程为
1.4-x-8
A.
B.1.4+2x8
1.4=2x8n1.4+x8
2.4-x13
2.4+2x13
C:242x13aD.
2.4+x13
7.童童的妈妈在网上销售装饰品，最近一周，童童妈妈每天销售某种装饰品的个数为：11,10，
11,13,11,13,15.关于这组数据，童童得出如下结果，其中错误的是，之
A.众数是11
B,平均数是12《
C.方差是18D.中位数是13
8.如图，将平行四边形ABCD沿对角线BD折叠，使点A落在点E处.若∠1=56°，∠2=42°
则∠A的度数为
A.108°
B.109
C.111
D.,110°
N
c2
D
00
M
B
(第8题)
(第9题y年
,翳个共（第10题）幕
9.如图，在平面直角坐标系中，
以坐标原点0(0,0)，A(0,4),B(一3,0)为项点的
Rt△AOB,其两个锐角对应的外角角平分线相交于点D,且点D恰好在反比例函数y=的图
象上，则k的值为
A.36
B.-36
C.-48
D..-64
10.如图，在Rt△ABC中，∠C=90°，BC-3,AC-6,点O,D,E是AB边上的点，以点O为
圆心，DE长为直径的半圆O与AC相切于点M,与BC相切于点N,则图中阴影部分的面积为
A,5-π
B.9-2π
C.9-π0d
D,5
数学试题第2页（共8页）平定县 2024—2025 学年九年级教学质量监测试题
数学参考答案
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 A D C B C B D D B A
20
二、填空题（本大题共 5个小题，每小题 3分，共 15 （3）2000× =800（名）·································································· 4分分） 50
11. 6（m＋1）2 12. 38 13. 2 3 答：估计此次竞赛该校获优异等级的学生人数为 800名；···························5分
（4）画树状图为：
14. 4 3 15. 9 3 5
三、解答题（本大题共 8个小题，共 75分.）
16. （1）解：原式=9＋8＋24－3.···························································4分
=41－3. ························································ 8分
=38. ············································································ 5分
(a 2)(a 2) 3(a 3) 因为共有 12种等可能的结果，······························································ 9分
（2）原式= 2 3 ················································· 8分(a 3) a 2 其中恰好抽到 A，C两人同时参赛的结果有 2种，所以恰好抽到 A，C两人同时参赛的概率
2 1
3(a 2) 3(a 3) 为 . ······················································································ 10分
= ······························································ 9分 12 6
a 3 a 3
15
= .··············································································10 19.解：（1）设商家购进 A款迷你小电扇 x台，B款迷你小电扇 y台，··········1分分
a 3
x y 100，
17. 解：（1） 根据题意，得 ························································· 3分
30x 40y 3350.
x 65，
解，得 ················································································· 4分
y 35.
如图，BE即为所求；··········································································· 3分
答：商家购进 A款迷你小电扇 65台，B款迷你小电扇 35台；····················· 5分
（2）①CD，②90，③两直线平行，内错角相等，④两角分别相等的两个三角形相似． （2）设该商家这次购进 A款迷你小电扇 a台，则购进 B款迷你小电扇(150－a)台，
······································································································· 7分 ······································································································· 6分
18. 解：（1）50，144；······································································ 2分
根据题意，得 30a＋40（150－a）≤5200，··············································· 8分
（2）补全条形统计图如下：································································· 3分
数学试题答案第 1页 （共 6页） 数学试题答案第 2页 （共 6页）
解，得 a≥80.······················································································ 9分 ∴ac2=d. ∴y2=d－m.……………………………………………………………… 4分
答：该商家这次至少购进 A款迷你小电扇 80台．····································· 10分 ∴点 Q（c，d－m）在 y2=ax2－m上.
20.解：如图，延长 OB交 AC于点 D，····················································1分 ∴猜想 2正确；.…………………………………………………………………… 5分
3
任务三：AB的长为 .………………………………………………………………7分
4
22.（1）y=15－0.5（x－10）=－0.5x+20；（不化简不扣分）…………………… 3分
（2）解：w=xy+9（20－x）…………………………………………………………6分
（第 20题） =x（20－0.5x）+180－9x
由题意可知 BD⊥CA. =－0.5x2+11x+180.……………………………………………………………………7分
设 BC=x，则 BO=OA－BC=75－x.···························································2分
（写成一般式或顶点式都可以）
BD
在 Rt△CBD中，∵sin∠ACB= ，
BC （3）方法一：∵－0.5<0，
∴BD=BC·sin∠ACB=x·sin37°≈0.6x.························································· 3分 11∴当 x 11时，w有最大值.…………………………………………8分
2 0.5
∴DO=OB＋BD≈75－x＋0.6x=75－0.4x.····················································4分
当 x=11时，20－x=9．………………………………………………………………9分
DO
在 Rt△AOD中，∵cos∠AOD= ，
OA ∴每日加工 A款 11只，B款 9只，可使日销售利润最大．…………………… 10分
∴DO=OA·cos∠AOD=75·cos37°≈75 0.8=60.·············································5分 方法二：w=－0.5x2+11x+180=－0.5（x－11）2+240.5．
∴75－0.4x=60.……………………………………………………………………… 6分 ∵－0.5<0,
解，得 x=37.5.···················································································· 7分 ∴当 x=11时，w有最大值．……………………………………………………… 8分
∴BD≈0.6x=0.6 37.5=22.5（cm）.
当 x=11时，20－x=9．……………………………………………………………… 9分
答：点 B到 AC的距离约为 22.5 cm.······················································· 8分
∴每日加工 A款 11只，B款 9只，可使日销售利润最大．…………………… 10分
21.解：任务一：方程思想；…………………………………………………………1分 23．解：（1）AC=BC.········································································· 1分
任务二：证明：设点 P（c，d）为二次函数 y1=ax2（a≠0）图象上的任意一点， 理由如下：
由旋转得点 C′在 AB上，∠BAC=∠BAB′.·················································· 2分
则该抛物线沿 y轴向下平移 m个单位后的对应点为 Q（c，d－m），……………2分 ∵B′D⊥AC，
将 x=c代入 y ∴∠CAB′=90°.2=ax2－m（a≠0，m>0），得 y2=ac2－m，………………………… 3分
∴∠BAC=∠BAB′=45°.·········································································· 3分
∵点 P（c，d）为二次函数 y1=ax2图象上的点， ∵∠ACB=90°，
数学试题答案第 3页 （共 6页） 数学试题答案第 4页 （共 6页）
∴∠ABC=45°. 又∵∠BB′E=∠AB′B，
∴AC=BC.···························································································4分 ∴△BB′E∽△AB′B.………………………………………………………………………10分
（2）AD=BC′.····················································································· 5分 BB B E
∴ .
理由如下： AB B B
由旋转得点 C′在 AB上，AB′=AB. BB AB BB
∴ .
∴∠ABB′=∠AB′B. AB B B
∵∠B′AD=∠ABB′， 5 1
∴AB= BB′.···············································································11分
∴∠B′AD=∠AB′B. 2
∴AD∥BB′.·························································································6分 ∵∠DAB′=∠C′BB′，∠ADB′=∠BC′B′，
∴∠BAC=∠ABB′. ∴△ADB′∽△BC′B′.············································································· 12分
∵∠ABB′=∠DAB′， AB AD
∴ .
由旋转，得∠BAC=∠BAB′， BB BC
1 又 AB′= AB.
∴∠DAB′=∠BAC=∠BAB′= ×180°=60°.
3 5 1 5 1
∴AD= ·BC′= x .································································· 13分
∴△ABB′是等边三角形. 2 2
∴AB′=BB′.··························································································7分
在△ADB′和△BC′B′中，
D BC B 90 ,

DAB C BB ,
注：如果考生的解法与本解答不同，可根据试题的主要考查内容参照评分标准制定相应的
AB BB , 评分细则后评阅.
∴△ADB′≌△BC′B′（AAS).
∴AD=BC′.··························································································8分
（3）如图所示，
由旋转，得点 C′在 AB上，∠BAC=∠BAB′.
∴∠ABB′=∠DAB′=∠BAC+∠BAB′=2∠BAB′.
由旋转，得 AB′=AB，
∴∠ABB′=∠AB′B=2∠BAB′.
∵∠BAB′+∠ABB′+∠AB′B=180°，
∴∠BAB′+2∠BAB′+2∠BAB′=180°.
∴∠BAB′=36°.················································ 9分
作 BE平分∠ABB′，交 AB′于点 E，
∴∠BAB′=∠EBB′=36°.
∴AE=BE=BB′.
∴B′E=AB′－AE=AB′－BE=AB′－BB′.
数学试题答案第 5页 （共 6页） 数学试题答案第 6页 （共 6页）

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