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数学
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.
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，将本试卷和答题卡一并交回.
第I卷选择题（共30分）
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)
1.如果平时不注意爱护眼睛，就有可能形成近视.在验光时，验光师经常会以“***D”
的方式记录近视程度，例如，将近视50度记录为“-0.50D”,近视100度记录为
“-1.00D”,等等.通常，近视超过200度时就要持续配戴眼镜进行视力矫正，有4位
同学的验光记录如下，需要持续配戴眼镜的是
A.-1.25D
B.-0.75D
C.-1.75D
D.-2.25D
2.在中华传统春节文化中，对称、平移、旋转等几何变换常被运用于年画、窗花、10g0设
计，以体现“圆满”“和谐”“循环”等美好寓意.以下四款中央广播电视总台春节联欢
晚会主标识的图案（文字除外），最能体现平移变换的是
5为
中受广当电速它台
2○24i
22
2023r=
春节联欢晚会
33
026
春节联欢晚会：
川广只具8巴
2025
春节联欢晚会
A
B
C
D
3.五台山位于山西省忻州市，是中华十大名山之一.其由五座山峰环抱而成，因峰顶
平坦如台、无林木覆盖而得名.以下是五台山五座主峰的具体海拔数据：
东台（望海峰）
西台（挂月峰）
南台（锦绣峰）
北台（叶斗峰》
中台（翠岩峰）
2796米
2773米
2485米
3061米
2894米
根据上述五座山峰的海拔数据，下列说法错误的是
A.这五座山峰海拔的最大值与最小值相差576米
数学试卷（一）第1页（共8页）
B.这五座山峰海拔的中位数为2796米
C.这五座山峰海拔的众数为2700米
D.这五座山峰海拔的平均数为2801.8米
4.如图，在△ABC中，DE∥BC,AD=2,DB=3,DE=4,则BC的长为
A.6
B.8
D
C.10
B
D.12
第4题图
5.不等式x-3(x-2)≥4的解集在数轴上表示为
-101
2
-101
2
-101
-101
A
B
C
D
6.如图，已知BC是圆柱底面的直径，AB是圆柱的高，在圆柱的侧面上，过点A,C嵌有
一圈路径最短的金属丝.现将圆柱侧面沿AB剪开，所得的圆柱侧面展开图是
C
A
B
B
C B'B
C B'B
C B'B
C B
第6题图
A
B
D
7.为开展跨学科实践活动，某小组准备了四款功能不同的AI智能辅助软件：“智学助
手”“思维导图大师”“速记精灵”“解题通”.小聪和小明各自从这四款软件中随机选
择一款，若选择每款软件的可能性相同，则下列事件中，发生可能性最大的是
A.两人选择的软件完全相同
B.两人选择的软件完全不同
C.小明选中“思维导图大师”，小聪选中“解题通”
D.至少有一人选中“智学助手”
8.食用油的沸点远高于水的沸点(100℃).小明使用量程不超过100℃的温度计测量
某种食用油的沸点.他在锅中倒入适量该种食用油，用煤气灶均匀加热，每隔10s测
量一次锅中油温，测量数据如下表：
时间s
0
10
20
30
40
油温y/℃
10
30
50
70
90
小明发现，烧了110s时油开始沸腾.请你估计该食用油的沸点是
A.200℃
B.230℃
C.260℃
D.290℃
数学试卷（一）第2页（共8页）》数学参考答案及评分标准
一、选择题（本大题共 10 个小题，每小题 3分，共 30 分）
1.D 2.D 3.C 4.C 5.B 6.A 7.B 8.B 9.A 10.B
二、填空题（本大题共 5 个小题，每小题 3 分，共 15 分）
11. x (x 1)2
12. （-3，-2）
13. 3
14.（25 2 - 25）
15. 2 7
提示：过点 B作 BG⊥AC于点 G，过点 D分别作 DF⊥AB，DH⊥BC，垂足分别是点 F，H，
Rt BCG Rt BAG CG= 1 3 7在 △ 和 △ 中，利用勾股定理构造方程，求得 ，进而得到 BG= ，再
2 2
3 7 3
依据等面积法求得 DH= ，DC=2，DG= ，最后由△DGB∽△DCE求得 CE=2 7 .
4 2
三、解答题（本大题共 8 个题，共 75 分，解答应写出文字说明，证明过程或演算步骤）
16.解：（1）原式= 2 3-3-2 3 1-3 ························································································· 4分
=-5.····························································································································· 5分
（2）原式=a 2 a 6 a 2 a ································································································· 8分
= -６.························································································································ 10分
17.解：（1）200 ··················································································································· 2分
（2）1000×37.5%＝375（人）．························································································· 3分
答：估计每天参加体育活动时间不低于 2小时的学生有 375人．·································· 4分
（3）由调查可知，大部分同学每天参加体育活动时间低于两小时，建议：学校多提供一些
1
球场等活动场所，多提供学生活动时间．（答案不唯一，只要合理即可）·················· 7分
18.（1）证明：如图，连接 OC.······························································································· 1分
∵⊙O与 AB相切于点 C，
∴OC⊥AB.······························································································································2分
∴∠OCA=∠OCB=90°.
∵C D C E ，
∴∠AOC=∠BOC.·················································································································· 3分
在△AOC和△BOC中，
AOC BOC，

OC OC，

OCA OCB，
∴△AOC≌△BOC（ASA） .
∴OA=OB.······························································································································· 4分
1
（2）解：由（1）可得 AC=BC= AB= 2 3 .
2
∵OA=OB，OA=4，
∴OB=4.
∴在 Rt△BOC中，OC= OB 2 BC 2 =2，sin∠BOC= BC = 3 .
OB 2
∴∠BOC=60°.·························································································································5分
∵S 1△BOC = BC OC =
1
2 3 2=2 3 ，·············································································6分
2 2
S = 60 2
2
= 2 ,········································································································7分
扇形EOC 360 3
2
∴S阴影=S△BOC S =2 3 .······················································································· 8分扇形EOC 3
19.解：设一个杯垫的进价是 x元，则一个帆布包的进价是（x+5）元.······························1分
2
200 150
根据题意，得 . ·····································································································4分
x 5 x
解，得 x=15．·························································································································5分
经检验，x=15是原方程的根．·····························································································6分
当 x=15时，x+5=20．
答：一个杯垫的进价为 15元，一个帆布包进价为 20元.················································· 7分
20.解：(1)示意图如图所示.······································································································· 1分
根据题意可知，当α=75°时，安全攀爬的高度最大．························································ 2分
如图，在 Rt△ABC中，∠C=90°，AB=3，∠B=α=75°，
AC
∴ sin B .
AB
∴AC=AB·sin75°≈3×0.97≈2.9(米)．···············································································3分
答：使用这架梯子，最高可以安全攀上 2.9米高的墙面．··············································· 4分
（2）如图，根据题意，得在 Rt△ABC中，∠C=90°，AB=3，BC=1.5，
cosB BC 1.5 1∴ ，
AB 3 2
∴∠B=60°．··························································································································· 5分
在 Rt△ABC中， sin 60 AC ，
AB
∴AC=AB·sin60 3°=3× ≈2.595（米）．······································································6分
2
∵1.5+2.595=4.095＞3，········································································································ 7分
∴小东能安全使用这架梯子将春联贴在 3米高的大门上方．·········································· 8分
21.解：（1）角平分线上的点到角两边的距离相等······························································· 2分
（2）如图 1，延长 CB到 E，使 BE=BA，连接 EA，则∠E=15°.···································3分
在 Rt△ABC中，AB=2，∠ABC=30°，∠C=90°，
∴AC= 1 AB=1，BC=AB·cos30°=2 3× = 3．···························································4分
2 2
3
∵BE=BA=2，
∴EC=2+ 3．························································································································ 5分
Rt AEC tanE AC 1 1 (2 3)∴在 △ 中， 2 3．
EC 2 3 (2 3)(2 3)
∴ tan15 2 - 3 1 （结果写成 tan15 也是正确的）．·········································· 6分
2 3
图 1
（3）如图 2，∠DBC=67.5°即为所求．
图 2
（作图正确，留有痕迹，有结论）······················································································ 8分
tan 67.5 2 1. ····················································································································10分
22.解：（1）由题意，得抛物线顶点 A的坐标为（20，10）．·········································· 1分
设抛物线的函数表达式为 y a(x 20)2 10 .······································································ 2分
由题意，得点 B的坐标为（40，0）.
将点 B的坐标为（40，0）代入 y a(x 20)2 10，得
0 a(40 20)2 10 .·················································································································3分
解，得 a 1 ．··················································································································· 4分
40
1
∴沙丘轮廓线的抛物线的函数表达式为 y (x 20)2 10．····································· 5分
40
（2）①沙障高度为 6 米，令 y=6，
4
6 - 1则 （x 20)2 10 .········································································································6分
40
解，得 x1 20 4 10, x2 20 4 10 ．················································································ 7分
∴（20 4 10）（- 20 - 4 10） 8 10 25（米）.
∴草方格沙障的长度约为 25米．························································································ 8分
3
②设左侧立柱底端到原点 O的水平距离为 x米，根据施工要求，左侧立柱顶端的高度为 x
4
米，
3
∴左侧立柱顶端的坐标为（x， x）．·············································································· 9分
4
3 1
将（x， x）代入 y (x 20)2 3 1 10，得 x (x 20)2 10 .·····························10分
4 40 4 40
解，得 x1= 0（不合题意，舍去），x2=10．·······································································11分
∴左侧立柱底端的坐标为（10，0）.
∵两侧立柱关于对称轴直线 x=20对称，
∴右侧立柱底端的坐标为（30，0）.
∴两根立柱之间的水平距离为 30 - 10=20（米）．··························································· 12分
23.解：（1） FG 2EF .········································································································· 2分
证明：∵四边形 ABCD是矩形，
∴∠B=90°．
∵AB=BC，E，F分别是 AB，BC的中点，
∴BE=BF=AE=FC．··············································································································· 3分
在 Rt△BEF中，由勾股定理，得 EF= 2 BF．··································································4分
CG AB
∵ ，AB=BC，
AE BC
∴CG=AE=FC．
∵点 G在 BC边的延长线上，
∴FG=2BF．····························································································································5分
FG 2BF
∴ 2，即 FG 2EF ．··············································································6分
EF 2BF
（2）①∵四边形 ABCD是矩形，
∴∠B=90°．
∵AB=6，BC=8，E，F分别是 AB，BC的中点，
5
∴AE=BE=3，BF=FC=4．·····································································································7分
在 Rt△BEF中，由勾股定理，得 EF=5．··········································································· 8分
CG AB
∵ ，
AE BC
CG 6
∴ .
3 8
∴CG= 9 ．······························································································································ 9分
4
∵FC=4,点 G在 BC边的延长线上，
∴FG=FC+CG=4 9 25 .
4 4
25
FG
∴ 4 5 ．···················································································································10分
EF 5 4
②△DEF 9 73 9平移的距离是 ， 或 37．····································································· 13分
4 16 4
提示：如图 1，若 E′G=E′D′ 9，此时 DD′= ；
4
图 1
如图 2，若 GE′=GD′ 73，此时 DD′= ；
16
图 2
3 D′E′=D′G DD′= 9如图 ，若 ，此时 37．
4
图 3
【说明】以上各题的其他解法，请参照此标准评分.
6

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