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姓名　　　　　　　　　　　　　准考证号　　　　　　　　　　
5. 如图，在△ABC 中，BD 是 AC 边上的高，CE 是∠ACB 的平分线，BD，CE 交于点 F. 若
∠AEC = 80°，∠DFC = 52°，则∠ABC的度数是
2025年山西初中学业水平测试靶向联考试卷（二） A. 28° B. 38° C. 42° D. 62°
A D C
数　学 E F
F D
G
注意事项： B C A B
1. E本试卷共8页，满分120分，考试时间120分钟 . 第5题图 第7题图
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 . 6. 考古学家们破译了玛雅人的天文历，其历法非常精确 .他们计算的地球一年天数与
3. 答案全部在答题卡上完成，答在本试卷上无效 . 现代相比仅差0.000 069天 .将0.000 069用科学记数法表示为
4. - 4 - 5考试结束后，将本试卷和答题卡一并交回 . A. 0.69 × 10 B. 6.9 × 10
C. 6.9 × 10 - 4 D. 69 × 10 - 6
第Ⅰ卷 选择题（共30分） 7. 如图，在正方形ABCD中R，点E在AB边上，连接CE，过点D作DF ⊥ CE于点F，过点B
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， 作BG ⊥ CE于点G，若BG = 3，DF = 8，则FG的长为
只有一项符合题目要求，请选出并在答题卡上将该项涂黑） A. 4 B. 5 C. 7 D. 11
1. 若将下面的四个有理数表示在数轴上，则位于最左边的是 8. 某餐厅规定等位时间达到 30分钟（包括 30 分钟）可享受优惠 .现统计了某时段顾客
A. - 1 B. - 1 C. 1 D. 3 的等位时间（t 分钟），如图是根据数据绘制的统计图 .下列说法正确的是3 3 A. 此时段有1桌顾客等位时间是40分钟
2. 中国“二十四节气”已被列入联合国教科文组织人类非物质文化遗产代表作名录， B. 此时段平均等位时间不小于20分钟
下列四幅作品分别代表“立春”“立夏”“芒种”“大雪”，其中既是轴对称图形，又是中 C. 此时段等位时间的中位数可能是26
心对称图形的是 D. 此时段有5桌顾客可享受优惠
/ R频数 桌
12 12 溶解度/g
10 9 数据分成6组： a
8 10 ≤ t < 156 15 ≤ t < 20 bA B C D 6 5 20 ≤ t < 25
3. 4 2 25 ≤ t < 30下列运算正确的是 2 1 30 ≤ t < 35 c35 ≤ t < 40
A. 2 + 3 = 2 3 B. 2 3 - 3 = 3 C. 1 = 14 2 D. (-2)
2 = - 2 0 10 15 20 25 30 35 40 / O t1 t2 t3 温度/℃t 分钟
第9题图
4. 第8题图如图是由 8个相同的小立方体组成的几何体的俯视图，小正方形中的数字表示该位 9. 在化学实验中，小明研究 a，b，c三种固体物质的溶解度，如图为这三种固体物质的
置上小立方体的个数，则这个几何体的主视图是 溶解度与温度对应的图象 .下列说法正确的是
3 A. a，b，c三种物质的溶解度都随温度的增加而变大
1 2 1 B. a，b，c三种物质中，c物质的溶解度最小
1 C. 温度为 t2 ℃时，a，b，c三种物质的溶解度由大到小的顺序是a > b > c
A B C D 第4题图 D. 温度为 t1 ℃时，b，c两种物质的溶解度相等
数学试卷　第1页（共8页） 数学试卷　第2页（共8页）
10. 如图，△ABC 为等边三角形，BD 平分∠ABC，AB = 2，点 E A 三、解答题（本大题共8个小题，共75分 .解答应写出文字说明、证明过程或演算步骤）
为 BD 上一动点，连接 AE，当 AE + 12 BE 取最小值时，DE 16.（ 本题共2个小题，每小题5分，共10分）D -1
的长为 （1）计算：2sin 45° - ∣ - 18 ∣ +（ π - 4）0 + (- 13 ) ；E
B C
A. 1 B. 2 第10题图
C. 33 D. 22
第Ⅱ卷 非选择题（共90分）
二、填空题（本大题共5个小题，每小题3分，共15分）
11. x2 - 9化简：2 + 6 = 　　▲　　.x （2 ìx + 3 > 0,）解不等式组：í
12. 2( x - 1) + 3 ≥ 3
并判断 - 1， 2 这两个数是否为该不等式组
x,
“植树造林滋沃土，防风固沙护良田”.要反映一个区域近
15 的解 .
R
年森林覆盖率的变化情况，从“扇形统计图”“条形统计
图”“折线统计图”中选择一种统计图，最适合的统计图
第12题图
是　　▲　　.
13.《 九章算术》中有这样一个题：“今有醇酒一斗，直钱五十；行酒一斗，直钱一十 .今
将钱三十，得酒二斗 .问醇、行酒各得几何？”其译文是：今有醇酒（优质酒）1斗，价
值 50钱；行酒（劣质酒）1斗，价值 10钱 .现有 30钱，买得 2斗酒 .问醇酒、行酒各能 R
买得多少？设醇酒为 x斗，行酒为 y斗，则可列二元一次方程组为　　▲　　.
14. = 4 = k 17.（ 本题6分）如图，△ABC中，AB = BC，D在BC的延长线上，连接AD，E为AD的中点 .如图，在平面直角坐标系中，点 B在函数 y x 的图象上，点 A在函数 y x 的图象 （1）尺规作图：在△ABC 内部求作一点P，使点P到点A，B，C的距离都相等（保留作
上，若OA = 2OB，∠AOB = 90°，则 k的值为　　▲　　. 图痕迹，不写作法）；
B （2）在（1）的基础上，若直线BP与线段AC交于点F，连接EF，求证：EF∥BC. y
A
E
A E
B F D
B
O C
D
x A C
第14题图 第15题图
15. 如图，在 Rt△ABC 中，∠ACB = 90°，AB = 10，BC = 8，AD 平分∠BAC，交 BC 于点 D，
点E为边AB上一点，连接CE，交AD于点F，当AE = 2BE时，DF的长为 　　▲　　.
数学试卷　第3页（共8页） 数学试卷　第4页（共8页）
18.（ 本题 9 分）为贯彻落实山西省教育厅《关于加强中小学生体质健康管理的实施意 20.（ 本题8分）阅读与思考
见》，某中学在九年级开展了“阳光体育”大课间活动，并组织全年级全体学生进行 请认真阅读材料，并完成相应任务 .
了“1分钟跳绳”体能测试 .测试成绩根据《国家学生体质健康标准》划分为四个等
级：把 1分钟跳绳完成个数用 x表示，A：x ≥ 150，B：120 ≤ x ≤ 149 C 90 ≤ ≤ 119 婆罗摩笈多是公元 7 世纪的古印度伟大数学家，曾研究过对角线，： x ， 互相垂直的圆内接四边形，这类四边形被称为“婆罗摩笈多四边形”.
D：x ≤ 89. 我们一起了解这位数学家的研究成果吧！
该校将此次“1分钟跳绳”体能测试的成绩整理成如下两幅尚不完整的统计图 . A如图 1，已知⊙O 的内接四边形 ABCD，对角
各等级人数条形统计图 线 AC ⊥ BD 于点 P. 婆罗摩笈多发现 AP 2 + BP 2 + B P D
人数 2 2
240 各等级人数扇形统计图 CP + DP 等于⊙O半径平方的4倍 . O
200 下面是他的探究思路：
160
120 B ∵AC ⊥ BD于点P，
C
80 80 60 CA ∴∠APB = 90°
图1
，∠CPD = 90°.
40 20 D 2 2 2 2
0 20% ∴AP + BP = AB ， CP + DP 2 = CD 2（. 依据1）A B C D 等级 如图2，连接CO并延长交⊙O于点Q，连接DQ， A Q
请根据上述信息，解决下列问题： 则∠CDQ = 90R°（. 依据2）
（1 P）该校九年级共有　　▲　　名学生，B等级对应的人数为　　▲　　； ∴∠1 + ∠Q = 90°. B 2 D
（2）扇形统计图中，D等级所对应扇形的圆心角度数为　　▲　　； 又∵?CD = ?CD O， 3 1
（3）已知此次测试中成绩前五名的同学（记为 A，B，C，D，E），其中A和D是女生，另 ∴∠Q = ∠2.
外 3 C人是男生 .老师从 5人中随机邀请 2人给其他同学做示范，请用列表或画树 ∵AC ⊥ BD， 图2
状图的方法，求邀请的2位同学恰好都是男生的概率 . ∴∠BPC = 90°.
∴∠3 + ∠2 = 90°.
∴∠1 = ∠3.
…… R
任务：
19.（ 本题 10分）某电器商场从厂家购进了A，B两种型号的电烤箱，已知一台A型电烤 （1）填空：材料中的依据1是指：　　▲　　，依据2是指：　　▲　　；
箱的进价比一台B型电烤箱的进价多400元，用7600元购进A型电烤箱和用6000元 （2）请完成材料中的剩余证明；
购进B型电烤箱的台数相同 . （3）如图 3，⊙M 的半径为 5，四边形 EFGH 内接于⊙M，且 EG ⊥ FH 于点 P，EF = 4，
（1）求一台A型电烤箱和一台B型电烤箱的进价各为多少元？
2 则HG的长为　　▲　　.（）在销售过程中，A型电烤箱因为造型精致，噪音小而更受消费者的欢迎 .该商场 E
决定停止购进B型电烤箱，并对库存货品进行降价销售，力求尽快清空库存货 H
品 .经市场调查，当B型电烤箱的售价为 2400元时，每天可售出 4台，在此基础 M P F
上，售价每降低50元，每天将多售出1台，如果每天该商场销售B型电烤箱的利
润为5600元，请问该商场应将B型电烤箱的售价定为多少元？ G
图3
数学试卷　第5页（共8页） 数学试卷　第6页（共8页）
21.（ 本题 8分）随着手机普及率不断提高，其对社会和个人产生多维度影响 .在经济领 问题解决：
域，不仅推动消费升级，还重塑传统商业模式，“手机直播”也随之成为新兴潮流 . （3）在（2）的情况下，计划调整出水口 A的高度，使落水点恰好在石栏底部，已知出
常见的悬臂式手机直播支架便是这一潮流下的产物，如图 1 所示，它外观简洁实 水口高度调整后，喷头喷出的水流形状和对称轴不变，则OA的高度如何变化 .
用 .从图 2 和图 3 的几何示意图可见，支架底座 AB平稳置于水平地面，立杆CD ⊥ y
AB，高度为 100 cm；支杆DE可绕点D旋转，长度为 30 cm，与可调节的悬臂EF相互 A
配合，为手机直播提供稳定支撑 . A
（1）如图 2，C，D，E三点共线，先将支杆 DE绕点 D 逆时针旋转 60°，再将悬臂 EF绕 O O x
E 点旋转，同时调节悬臂 EF 的长度（如图 3），此时∠DEF = 75°，求点 E 到地面 图1 图2
的距离；
（2）在（1）的条件下，点F到地面的距离为150 cm，求调节后悬臂EF的长 .
（结果精确到1 cm，参考数据： 2 ≈ 1.41， 3 ≈ 1.73）
F 23.（ 本题13分）综合与探究
E
F E 问题情境：学习完轴对称的性质后，同学们开展了折叠矩形纸片的实践探究活动 .
D D 如图 1，点E，F分别是R矩形ABCD的边BC，AD上的点，将△ABE，△CDF分别沿AE，
CF折叠，使点B，点D的对应点分别落在矩形对角线AC上的点B'，D'处 .
A C B A
猜想证明：
C B
1 2 3 （1）判断四边形AECF的形状，并说明理由；图 图 图
深入探究：
（2）如图 2，在矩形ABCD中，E为线段BC上一动点，设AE = mAB，将△ABE沿AE折
叠，得到△AB'E，延长AB'交CD于点F，若AF = mAE，猜想线段BE与CE的数量
关系，并说明理由;
拓展应用： R
（3）如图 3，一张直角三角形纸片记作 Rt△ABC，∠B = 90°，AB = 4 cm，BC = 6 cm，
点E为BC的中点，沿AE折叠该纸片，点B的对应点为点B'，B'E与AC相交于点G，
22. 11 则△AB′E与原三角形纸片重叠部分的面积为　　▲　　cm2.（本题 分）综合与实践 F A
问题情境：某购物广场计划在门前的广场中央新建造一个圆形喷水池，并在水池中 A D A B' DD' B'
央垂直于地面处安装一根石柱，在石柱顶端A处安装一个喷头向外喷水，水流在各 F G
个方向上沿形状相同的抛物线路径落下，且在过OA的任一平面上抛物线路径如图 B'
1 B C B C B C所示，已知石柱在水面以上部分OA的高度为 4米，为使水流形状较为漂亮，要求 E E E
图1 图2 图3
设计水流在距离石柱2米处达到最高，此时离水面6米 .
模型建构：
（1）以点O为原点建立如图 2所示的平面直角坐标系，求水流抛物线在第一象限内
对应的函数表达式（不要求写自变量的取值范围）；
模型分析：
（2）为了防止小朋友误入喷水池范围，计划在距喷头出水口的水平距离 5米处修建
高度为1米的一圈石栏，请通过计算判断此时喷出的水流会不会落到石栏外；
数学试卷　第7页（共8页） 数学试卷　第8页（共8页）2025年山西初中学业水平测试靶向联考试卷（二）
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 B D C A C B B B D C
二、填空题
11. x-3
x y=2，
12. 13. 14. -8 15. 9 5折线统计图
2 50x 10y=30 19
三、解答题
16. 2解：（1）原式=2× -3 2 +1+（-3）·············································································· 4分
2
= 2 -3 2 +1-3
=-2 2 -2.········································································································5分
x 3＞0，①
（2）
（2 x-1）+3≥3x.②
解不等式①，得 x＞﹣3.······································································································· 6分
解不等式②，得 x≤1.··········································································································· 7分
∴原不等式组的解集为﹣3＜x≤1.······················································································ 8分
∴﹣1是该不等式组的解，·································································································· 9分
2 不是该不等式组的解．··································································································· 10分
17. （1）解：如答图所示，点 P即为所求作.
······························································································3分
第 17题答图
（2）证明：由（1）中所作的图形可知：PA=PC，
数学答案 第 1 页 共 6 页
∵AB＝BC，
∴BP为线段 AC的垂直平分线.··························································································· 4分
∴AF＝CF.
∵E为 AD的中点，
∴AE＝DE.
∴EF是△ADC的中位线.····································································································· 5分
∴EF∥CD，即 EF∥BC．···································································································· 6分
18. 解：（1）400 240············································································································· 2分
（2）18°··································································································································4分
（3）列表如下：
A B C D E
A (A，B) (A，C) (A，D) (A，E)
B (B，A) (B，C) (B，D) (B，E)
C (C，A) (C，B) (C，D) (C，E)
D (D，A) (D，B) (D，C) (D，E)
E (E，A) (E，B) (E，C) (E，D)
·················································································································································· 6分
由列表可知，总共有 20种结果，每种结果出现的可能性都相同.其中邀请的 2 位同学恰好
都是男生的结果有 6种.········································································································· 8分
6 3
所以 P（邀请的 2位同学恰好都是男生）= = .·························································· 9分
20 10
19. 解：（1）设一台 B型电烤箱的进价为 x元，则一台 A型电烤箱的进价为（x+400）元.
················································································································································· 1分
6000 = 7600根据题意，得 .·······························································································2分
x x 400
解，得 x＝1500. ··················································································································· 3分
经检验，x＝1500是原方程的解.························································································· 4分
∴x+400＝1900．
答：一台 A型电烤箱的进价为 1900元，一台 B型电烤箱的进价为 1500元．············5分
（2）设该商场应将 B型电烤箱在 2400元的基础上降价 m元.
m
根据题意，得（2400-m-1500）（4+ ）＝5600. ·····························································7分
50
数学答案 第 2 页 共 6 页
解，得 m1＝200，m2＝500．································································································8分
因为力求尽快清空库存，所以应降价 500元.···································································· 9分
2400-500=1900（元）.
答：该商场应将 B型电烤箱的售价定为 1900元．···························································10分
20. 解：（1）勾股定理（或直角三角形两直角边的平方和等于斜边的平方）··················· 1分
直径所对的圆周角是直角······························································································· 2分
（2）如答图，连接 OA，OB，OD，··················································································3分
第 20题答图
∴∠DOQ=∠AOB.
∴DQ=AB.······························································································································· 4分
在 Rt△CDQ中，DQ2+CD2=CQ2，
∴AB2+CD2=CQ2.
∴AP2+BP2+CP2+DP2=CQ2.···································································································· 5分
设⊙O的半径为 r，则 CQ=2r.
∴AP2+BP2+CP2+DP2=(2r)2=4r2.·····························································································6分
（3）2 21 ······························································································································ 8分
21. 解：（1）如答图 1，过点 E作 EG⊥AB于点 G，过点 D作 DH⊥EG于点 H.············1分
第 21题答图 1
数学答案 第 3 页 共 6 页
∵CD⊥AB，
∴四边形 HGCD为矩形.
∴HG=CD=100 ,∠CDH=90°.·································································································2分
由旋转得∠CDE=180°-60°=120°，
∴∠EDH=∠CDE -∠CDH=30°.
∵在 Rt△EDH中，∠EHD=90°，DE=30，∠EDH=30°，
∴sin∠EDH= EH .
DE
1
∴EH=DE sin 30°=30× =15.·································································································3分
2
∴EG=EH+HG=115（cm）.
答：点 E到地面的距离为 115 cm.························································································4分
（2）如答图 2，过点 F作 FM⊥EG，交 GE的延长线于点M.········································5分
第 21题答图 2
∴∠M=90°.
∵∠DHE=90°，∠EDH=30°，
∴∠DEH=60°.
∵∠DEF=75°，
∴∠MEF=180° -∠DEH -∠DEF=45°.·················································································6分
∵GM=150，
∴EM=GM - EG=35.··············································································································· 7分
∵在 Rt△EMF中，∠M=90°，∠MEF=45°，
cos MEF= EM∴ ∠ .
EF
∴EF= EM 35 =35 2 ≈49(cm).
cos MEF cos45
答：调节后悬臂 EF的长约为 49 cm.··················································································· 8分
22. 解：（1）由题意可知抛物线的顶点坐标为（2，6），····················································· 1分
∴设水流抛物线在第一象限内的函数表达式为 y＝a（x﹣2）2+6.·································· 2分
∵OA=4，
∴点 A的坐标为（0，4）.····································································································3分
∵点 A（0，4）在抛物线 y＝a（x﹣2）2+6上，
数学答案 第 4 页 共 6 页
∴4＝4a+6．
1
解，得 a=- ．
2
1
∴水流抛物线在第一象限内的函数表达式为 y＝- （x﹣2）2+6．································4分
2
1
（2）由（1）可知：y＝- （x﹣2）2+6．
2
当 y＝1 1时，- （x﹣2）2+6=1.
2
解，得 x1=2+ 10，x2=2- 10（舍去）.············································································· 6分
∵2+ 10＞5，
∴在离喷头出水口的水平距离 5米处，水流高度高于石栏高度，水流会落到石栏外.
················································································································································· 7分
1
（3）设出水口高度调整后，在第一象限内水流所在抛物线的函数表达式为 y=- （x﹣2）2
2
+h．··········································································································································8分
∵抛物线经过点（5，0），
1
∴- ×9+h=0.
2
9
解，得 h= .··························································································································· 9分
2
1 9
∴y=- （x﹣2）2+ .
2 2
当 x=0 5时，y= .
2
4- 5 = 3 （米）.······················································································································· 10分
2 2
3
∴OA要降低 米.······················································································································ 11分
2
23. 解：（1）四边形 AECF是平行四边形.··············································································1分
理由如下：
∵四边形 ABCD是矩形，
∴AD∥BC，AB∥DC.
∴∠BAC=∠DCA.·················································································································· 2分
由折叠可知∠BAE=∠CAE= 1 ∠BAC 1，∠DCF=∠ACF= ∠DCA，·····························3分
2 2
∴∠CAE=∠ACF.
∴AE∥FC.
∴四边形 AECF是平行四边形.·····························································································4分
（2）BE=CE.···························································································································5分
理由如下：
数学答案 第 5 页 共 6 页
如答图，连接 EF，
第 23题答图
∵四边形 ABCD是矩形，
∴∠B=∠C=90°.······················································································································6分
由折叠可知∠BAE=∠B′AE，∠AEB=∠AEB′，∠B=∠AB′E=90°,BE=B′E.···················7分
∴∠EB′F=90°.
∵AE=mAB，AF=mAE，
AE AF
∴ =m.
AB AE
∴△AEF∽△ABE.
∴∠AEF=∠B=90°.·················································································································8分
∴∠AEB′+∠FEB′=90°，∠AEB+∠FEC=90°.
∴∠FEB′=∠FEC.
∵EF=EF，∠EB′F=∠C=90°,
∴△B′EF≌△CEF.··················································································································10分
∴B′E=CE.
∴BE=CE.·································································································································11分
（3 150） ·····································································································································13分
43
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数学答案 第 6 页 共 6 页

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