5.如图是一架人字梯及其侧面示意图（部分），已知AB∥CD∥EF,AC=50cm,CE=
30cm,BD=45cm,则BF的长为
2025年山西初中学业水平测试联考试卷
A.27 cm
B.50 cm
数学
C.72 cm
D.80 cm
注意事项：
1.本试卷共8页，满分120分，考试时间120分钟.
2.答卷前，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
第5题图
第6题图
3.答案全部在答题卡上完成，答在本试卷上无效
6.如图，AB为⊙0的直径，C,D为⊙0上两点，若LC=40°，则∠ABD的度数为
4.考试结束后，将本试卷和答题卡一并交回，
A.50
B.40
第I卷选择题（共30分）
C.459
D.35
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，
7.若点A(-1,y),B(2,y,),C(3,y)都在二次函数y=x2-4x-n的图象上，则y,y2y的
只有一项符合题目要求，请选出并在答题卡上将该项涂黑)
大小关系是
1.-2025的相反数是
A.y1<2B.y31
C.y3D.y2A.2025
B.-2025
C.z25
D.-2025
8.如图，将△ABC绕点C逆时针旋转得到△A'B'C.当
2.太原碑林公园作为太原的文化地标，收藏了许多珍贵碑刻，其中傅山先生的书法作品有
点B落在BA的延长线上时，恰好A'B'∥AC,若=
146幅，涵盖了篆、隶、楷、行、草五种字体，不仅具有艺术价值，还承载了丰富的历史文化
220°，则∠BCA的度数为
下面四个篆体字是轴对称图形的是
A.100
B.120°
B'
蘭麝喉粉
C.130
D.140
第8题图
9.学校劳动课上开展烘焙实践课，同学们发现烘焙某种面点时，当烘焙温度大于
150℃且小于220℃时，烘焙时间y(mim)是烘焙温度x(℃)的一次函数，部分数据如下
表所示：
3.2025年春节期间，山西文旅市场迎来“开门红”，综合通信运营商和山西省旅游大
烘焙温度（℃）
…
160170180190200…
数据联合实验室专项调研数据，2025年春节节假日期间全省共接待国内游客
烘焙时间y(min)
3027.52522.520…
2837.97万人次，同比增长22.20%；实现旅游总花费317.44亿元，同比增长30.03%.
则y与x之间的关系式为
数据2837.97万用科学记数法表示为
A.y=0.25x+30
B.y=-0.25x+40
A.2837.97×10
B.2.83797×10
C.y=-0.25x+70
D.y=-2.5x+430
C.2.83797×10
D.0.283797×108
10.在凸五边形ABCDE中，点P在BC边上，点Q在AD的延长线上，AQ与BC平行且相
4.下列运算正确的是
等，不能推出PA与CD一定平行的是
A.5a2-a2=4
B.(-a2)3=a
A.PB=OD
B.PA=CD
C.（1+x)2=1+x+x
D.m÷m3=m
C.∠BAP=∠DCO
D.∠APB=∠CDO
数学试卷第1页（共8页）
数学试卷第2页（共8页）2025年山西初中学业水平测试联考试卷
数学参考答案及评分标准
一、选择题
题号 1 2 3 4 5 6 7 8 9 10
答案 A D C D C A D B C B
二、填空题
11. (x+4y)(x-4y) 12. -4＜x≤2 13. 1 14. 5 15. 4 10
9 5
三、解答题
16. 解：（1）原式=3+2- 3 -（6+2-4 3）·············································································· 3分
=3+2- 3 -8+4 3
=3 3 -3.········································································································· 4分
（2）①一 二······················································································································ 6分
1 a
②习题 1：
a2 1 a 1
= 1 a（a-1）2 2 ·····························································································8分a 1 a 1
=1 a
2 a
2 .······································································································· 10分a 1
习题 2：方程两边同乘（x2-1），得 1+x（x-1）=x2-1.··················································7分
解，得 x=2.··························································································································8分
检验：当 x=2时，x2-1≠0.··································································································9分
∴原分式方程的解是 x=2.··································································································10分
17. k解:（1）∵反比例函数 y= （x＞0）的图象经过点 D（4，2），
x
∴2= k .
4
∴k=8.
8
∴反比例函数的表达式为 y= .························································································2分
x
（2）如图所示.
数学答案 第 1 页 共 6 页
········································································ 5分
第 17题答图
（3 1） ···································································································································· 7分
3
18. 解：设晋谷 21号的亩产量是 x千克.················································································ 1分
根据题意，得 30（x+1.2x）=23100.················································································ 4分
解，得 x=350.······················································································································6分
答：晋谷 21号的亩产量是 350千克.··············································································· 7分
19. 解：（1）小茗同学计算平均数的方法不恰当.·································································· 1分
因为对于三家饭店星级评价的人数不同，即权重不同，直接计算简单算术平均数会导致
结果偏差，此时应使用加权平均数，以评价人数为权重进行计算.（合理即可）····· 3分
（2）①饭店应从服务这方面提升.························································································4分
理由：三项打分中，环境和口味打分的众数都为 5分，大于服务打分的众数 4.5分，所以
为了满足大部分顾客的需求，该饭店应从服务这方面提升.（答案不唯一，合理即可）
···············································································································································6分
x 3.5 5 5 4 3 4 5 5 4.5 5② 口味 4.4（分）.·············································· 7分
10
s2 = 1 [（3.5-4.4）2+5×（5-4.4）2+2×（4-4.4）2+（3-4.4）2+（4.5-4.4）2]=0.49.口味 10
································································································································· 8分
∵0.3<0.49，
∴环境打分的分数比较稳定.
∴王老师的说法正确.········································································································· 9分
20. 解：如答图，过点 G作 GH⊥AB于点 H.········································································1分
第 20题答图
数学答案 第 2 页 共 6 页
由题意得，四边形 GFBH是矩形.
∴FB=GH，BH=FG=1.·······································································································2分
在 Rt△AGH中，∠AHG=90°，∠AGH=42°，
∴tan 42°= AH ≈0.9.
GH
AH = 9∴ .·························································································································3分
GH 10
设 AH=9x，GH=10x，
∴FB=GH=10x.
∴AB=9x+1，BC=10x+17.································································································· 4分
∵AC∥DE，
∴∠ACB=∠DEC.···············································································································5分
∵AB⊥BC，DC⊥CE，
∴∠ABC=∠DCE=90°.
∴△ABC∽△DCE.·············································································································6分
AB BC
∴ .
DC CE
9x+1 10x 17
∴ .··············································································································7分
2 3
解，得 x≈4.4.·······················································································································8分
∴AB=9x+1=9×4.4+1=40.6≈41（m）.
答：点 A到地面的距离 AB约为 41 m.··········································································· 9分
21.（1）（1）30° 3 3 ··············································································································2分
（2）证明：在△ABP中，∠1+∠ABP=180°-∠APB=180°-135°=45°.
∵∠BAC=90°，AB=AC，
∴∠ABC=∠ACB=45°.
∴∠3+∠ABP=∠ABC=45°.
∴∠3=∠1.···························································································································3分
在△CBP中，∠3+∠BCP=180°-∠BPC=180°-135°=45°.
∵∠2+∠BCP=∠ACB=45°，
∴∠3=∠2.···························································································································4分
∴∠3=∠2=∠1.
∴点 P是△ABC的布洛卡点.····························································································5分
（3）解：∵∠APB=∠BPC=135°，
∴∠APC=90°.
∵∠1=∠3,
∴△ABP∽△BCP.·············································································································· 6分
AP AB BP
∴ .
BP BC CP
∵AB=AC，∠BAC=90°，
AB 1
∴ .························································································································ 7分
BC 2
数学答案 第 3 页 共 6 页
设 AP=k，则 BP= 2 k，CP=2k.
在 Rt△APC中，由勾股定理，得 AC= 5 k.
∴sin ACP= AP k 5∠ = .···························································································8分
AC 5k 5
22. 解：（1）建立如答图 1所示的平面直角坐标系.
·················································································································1分
第 22题答图 1
∵窑洞顶部最高点 O离地面 3.75 m，点 A离地面 2.25 m，
∴3.75-2.25=1.5.
∴点 A，B的纵坐标为-1.5.
∵AB=4，
∴点 A的坐标为（-2，-1.5），点 B的坐标为（2，-1.5）.··········································· 2分
∵点 O为抛物线的顶点，
∴设抛物线的函数表达式为 y=ax2.···················································································3分
∵点 A（-2，-1.5）在抛物线 y=ax2上，
∴-1.5=4a.
3
解，得 a=- .
8
∴抛物线的函数表达式为 y=- 3 x2.··················································································· 4分
8
（2）∵CD离地面 3 m，
∴3.75-3=0.75.
∴点 C，D的纵坐标为-0.75.····························································································· 5分
∵点 C D 3， 在抛物线 y=- x2上，
8
∴将 y=-0.75代入 y=- 3 x2 3，得- x2=-0.75.
8 8
解，得 x1=- 2，x2= 2 .···································································································6分
数学答案 第 4 页 共 6 页
∴点 C的坐标为（- 2，-0.75），点 D的坐标为（ 2 ，-0.75）.
∴CD=2 2 .·························································································································7分
∴吊顶所需材料的面积为 8×2 2 =16 2 ≈23（m2）.
答：吊顶所需材料的面积约为 23 m2.···············································································8分
（3）如答图 2，过点 A作 AQ⊥CD，交 DC的延长线于点 Q.······································· 9分
第 22题答图 2
由题意，得 AQ=0.75，QD=2+ 2 .·················································································· 10分
∵AQ⊥CD，CM⊥CD，
∴CM∥AQ.
∴△DCM∽△DQA.
DC CM 2 2 CM
∴ ，即 .····································································································11分
DQ QA 2 2 0.75
∴CM≈0.62.··························································································································12分
∴CM+DN+CD=0.62×2+2 2 =1.24+2.828=4.068≈4.1（m）.
答：小红需要购买彩灯的总长度约为 4.1 m.···································································13分
23. 解：（1）正方形··················································································································· 1分
（2）DE+BF= 2 CD.········································································································ 2分
理由：如答图，连接 AF，过点 E作 EG⊥AD于点 G.················································· 3分
第 23题答图
∵四边形 ABCD是矩形，
数学答案 第 5 页 共 6 页
∴∠ABC=∠BAD=∠ADC=∠C=90°，AB=CD.
∵EF⊥AE，
∴∠AEF=90°.
由旋转得 AE=AB=3.
在 Rt△ABF和 Rt△AEF中，AF=AF，AB=AE，
∴Rt△ABF≌ Rt△AEF.
∴BF=EF.······························································································································4分
∵∠BAE=45°，
∴∠EAG=45°.
∴△AEG是等腰直角三角形.
∴∠AEG=45°，AG=EG= 3 2 .
2
∵AD＝3 2，
∴DG= 3 2 .
2
∴DG=EG.
∴△DEG是等腰直角三角形.
∴∠DEG=∠EDG=45°.
∴∠AED=∠AEG+∠DEG=90°.
∴∠AEF+∠AED=180°.
∴D，E，F三点在同一直线上.························································································ 6分
∴DE+BF=DE+EF=DF.······································································································· 7分
∵∠EDG =45°，
∴∠CDF=90°-∠EDG=90°-45°=45°.
在 Rt DFC CD 2△ 中，cos 45°= = ，
DF 2
∴DF= 2 CD.
∴DE+BF= 2 CD.·············································································································· 8分
（3）DM的长为 5 2或 5 2 .·············································································· 12分
【说明】以上各题的其他解法，请参照此标准评分.
数学答案 第 6 页 共 6 页
