2024一2025学年度第一学期教学质量监测
九年级数学科试卷
温馨提示：请将答案写在答题卡上；考试时间为120分钟，满分120分。
一、选择题（共10题，每题3分，共30分）
1,下列关系式中，y与x是反比例关系的是()
A.y=4x
B.X=3
D.y=2x+1
y
c=是
2.方程x2一3x=2中二次项系数，一次项系数和常数项分别是（)
A.1,-3,-2
B.1,-3,2
C.1,3,-2
D.1,3,2
3.如图，该几何体的主视图是（)
A
B
!
C.
D.
4已号-号=6-0，则值为(）
A一
B.5
4
c.-
5-4
D
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5.一个不透明的袋子里装有4个白球和若干个黑球，这些球除颜色外其他都相同.从袋子中随机摸一
个球，记下颜色后放回搅匀，不断重复上面的过程，并绘制了如图所示的统计图。估计袋子里黑球
的个数为()
白球的频率
0.3
0.2
0.1
0
10002000300040005000摸球次数
A.16
B.18
C.20
D.22
6.如题6图，在菱形ABCD中，AB=6,∠A=60°，则对角线BD的长为()
A.V3
B.6
C.3V3
D.6v3
D
B
B
题6图
题7图
题8图
7.如题7图，已知D是△ABC的边AC上一点，根据下列条件，不能判定△CAB∽△CBD的是（)
A.∠A=∠CBD
B.∠CBA=∠CDB
C.AB·CD=BD·BC
D.BC2=AC·CD
8.如题8图机器狗是一种模拟真实犬只形态和部分行为的机器装置，其最快移动速度v(m/s)是载重
后总质量m(kg)的反比例函数.已知一款机器狗载重后总质量m=60kg时，它的最快移动速度v
=6m/s;当其载重后总质量m=90kg时，它的最快移动速度v是（)m/s
A.8
B.3
C.9
D.4
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9.如题9图，若函数y=:(k>0)与函数y=的图象相交于A,C两点，AB垂直x轴于B,则△MBC
的面积为()
A.1
B.2
C
D.2
B
题9图
题10图
10.如题10图，在正方形ABCD中，BD=3V2,点E为AD上一点，连接CE交BD于点F,延长CE
交BA的延长线于点G,若AG=1,则CF的长为()
12
A.2
B.
C.
7
D.
二、填空题（共5题，每题3分，共15分）
11.已知△ABC与△DEF的相似比为3：2，△ABC的周长为24，则△DEF的周长为
12.已知方程x2一2x十k=0的一个根为一2，则方程的另一个根为
13.已知点A(-6,1),B(1,y),C(3,)在反比例函数y=二(k≠0)的图象上，则y12.
(填“>”“<”或“=”)
14.如图，树AB在路灯O的照射下形成投影AC,已知树高AB=2m,树影AC=3m,树AB与路灯O
的水平距离AP=4.5m,则路灯的高度P0长是
米
Ch..
九年级数学科试卷第3页（共6页）2024—2025学年度第一学期教学质量监测
九年级数学科参考答案及评分标准
一、选择题（共 10题，每题 3分，共 30分）
1．C 2．A 3．B 4．B 5．A 6．B 7．C 8．D 9．A 10．D
二、填空题（共 5题，每题 3分，共 15分）
3 5
11．16 12．4 13．＜ 14．5 15．
2
三、解答题（一）（共 3题，每题 7分，共 21分）
16．（1）原方程移项得，x2―2x＝7，
x2―2x＋12＝7＋1，即(x―1)2＝8，······························································· 1分
x―1＝±2 2，··························································································2分
∴x1＝2 2＋1，x2＝―2 2＋1；··································································· 3分
（2）由题意可得，a＝3，b＝1，c＝―5
∴Δ＝12―4×3×(―5)＝61＞0，··································································4分
x ―1± 61∴ ＝ 2 3 ，······················································································· 6分×
x ―1＋ 61∴ 1＝ 6 ，x
―1― 61
2＝ 6 ．································································· 7分
17．解：∵DE//BC，
AD AE
∴ ＝ ，··································································································· 3分
DB EC
∵AD＝9，DB＝6，AC＝10，
9 10―EC
∴ ＝ ，································································································6分
6 EC
解得：EC＝4．······························································································· 7分
18．（1）如图，△DEF即为所求；
··································································· 5分
（2）(―2，6)··································································································7分
九年级数学科参考答案及评分标准 第 1 页 (共 4 页)
四、解答题（二）（共 3题，每题 9分，共 27分）
1
19．（1） ；········································································································· 3分
2
（2）列表如下：
黄 蓝 红 红
黄 黄黄 黄蓝 黄红 黄红
蓝 黄蓝 蓝蓝 蓝红（紫） 蓝红（紫）
红 黄红 蓝红（紫） 红红 红红
红 黄红 蓝红（紫） 红红 红红
由表可知所有可能结果有 16种，其中能配成紫色的有 4种，····························· 8分
4 1
则能配成紫色的概率为 ＝ ，····································································9分
16 4
所以小华说法错误．
20．（1）设一次函数的关系式为 y＝kx＋b， ······························································ 1分
结合表格数据图象过(45，55)，(55，45)，
45k＋b＝55
∴ ．······················································································· 2分
55k＋b＝45
k＝―1
∴ ．····························································································· 3分
b＝100
∴所求函数关系式为 y＝―x＋100．······························································ 4分
（2）由题意得 x(―x＋100)＝2600，
∴2600＝―x2＋100x．
∴x2―100x＋2600＝0． ············································································· 6分
∴Δ＝(―100)2―4×2600
＝10000―10400
＝―400＜0．····························································································8分
∴方程没有解，故该商品日销售额不能达到 2600元．······································ 9分
21．（1）证明：∵O是边 AC上的中点，
∴AO＝OC，
∵OB＝OD，
∴四边形 ABCD是平行四边形，···································································2分
∵BA＝BC，
∴四边形 ABCD是菱形；············································································ 4分
（2）∵DE⊥BC，
∴∠DEB＝90°，
∵CD＝5，DE＝4，
∴CE＝ CD2―DE2＝ 52―42＝3，·······························································5分
九年级数学科参考答案及评分标准 第 2 页 (共 4 页)
由（1）可得 BC＝CD＝5，
∴BE＝BC＋CE＝3＋5＝8，
在 Rt△DBE 中，BD＝ BE2＋DE2＝ 82＋42＝4 5，······································ 6分
S BC DE 1∵ 菱形 ABCD＝ ＝ 2AC BD，
5 1∴ ×4＝ 2×4 5×AC················································································ 8分
∴AC＝2 5．····························································································9分
五、解答题（三）（共 2题，第 22题 13分，第 23题 14分，共 27分）
22．（1）27···········································································································3分
（2）证明：∵四边形 ABCD，AEFG是正方形，
AD 2 AG 2
∴ ＝ ， ＝ ，
AC 2 AF 2
AD AG
∴ ＝ ，····························································································5分
AC AF
∵∠DAG＋∠GAC＝∠FAC＋∠GAC＝45°，
∴∠DAG＝∠CAF，···················································································7分
∴△AFC∽△AGD；···················································································8分
BF 1
（3）∵ ＝ ，
FC 2
设 BF＝k，CF＝2k，则 AB＝BC＝3k，···························································9分
∴AF＝ AB2＋BF2＝ (3k)2＋k2＝ 10k，AC＝ (3k)2＋(3k)2＝3 2k，·············· 10分
∵四边形 ABCD，AEFG是正方形，
∴∠AFH＝∠ACF，∠FAH＝∠CAF，
∴△AFH∽△ACF，················································································· 12分
AC CF
∴ ＝ ，
AF FH
FC 3 2k 3 5
∴ ＝ ＝ ···················································································· 13分
FH 10k 5
23．（1）∵点 A(1，3)，点 B(n，1)在反比例函数 y m＝ x (m≠0)上，
∴m＝1×3＝n×1，
∴m＝3，n＝3，
3
∴反比例函数为 y＝ ，点 B(3，1)，····························································· 2分
x
k＋b＝3
把 A、B的坐标代入 y＝kx＋b得 ，
3k＋b＝1
k＝―1
解得 ，
b＝4
∴一次函数为：y＝―x＋4；········································································ 4分
（2）令 x＝0，则 y＝―x＋4＝4，
∴C(0，4)，·····························································································5分
1 1
∴S△AOB＝S△BOC―S△AOC＝ ×3×4― ×1×4＝4；············································· 8分2 2
九年级数学科参考答案及评分标准 第 3 页 (共 4 页)
（3）如图 2，过 A点作 x轴的平行线 CD，作 FC⊥CD于 C，ED⊥CD于 D，
3
设 E(a， )（a＞1），················································································· 9分
a
∵A(1，3)，
3
∴AD＝a―1，DE＝3― ，·········································································10分
a
∵把线段 AE绕点 A顺时针旋转 90°得到 AF，点 E的对应点为 F，恰好也落在这个反比例函数
的图象上，
∴∠EAF＝90°，AE＝AF，
∴∠EAD＋∠CAF＝90°，
∵∠EAD＋∠AED＝90°，
∴∠CAF＝∠AED，
在△ACF和△EDA中，
∠CAF＝∠DEA
∠ACF＝∠EDA＝90°，
AF＝EA
∴△ACF≌△EDA（AAS），·······································································12分
3
∴CF＝AD＝a―1，AC＝DE＝3― ，
a
3
∴F( ―2，4―a)，·················································································· 13分
a
∵F恰好也落在这个反比例函数的图象上，
3
∴( ―2)(4―a)＝3，
a
解得 a＝6或 a＝1（舍去），
1
∴E(6， )．··························································································· 14分
2
九年级数学科参考答案及评分标准 第 4 页 (共 4 页)
