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5.据文旅部2026年4月7日公布的数据，2026年清明节假期3天，全国国内出游1.35
亿人次，国内出游总花费613.67亿元，创历史新高.数据“613.67亿元”用科学记数法
表示为
A.6.1367×101元
B.6.1367×1010元
数学
C.6.1367×102元
D.6.1367×109元
6.为规范小区车辆通行、提升出入口管理效率，某居民小区大
A
门安装了车牌识别智能升降挡车杆，车辆驶入时设备自动识
别车牌，控制挡车杆绕固定支点旋转升降，实现快速通行.如
注意事项：
图是其工作示意图，智能箱与挡车杆AB交于点O,挡车杆AB
1.本试卷共8页，满分120分，考试时间120分钟.
为4m,0B为0.5m,要使挡车杆右端从水平位置下降的垂直
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
距离B'D为0.4m,则栏杆左端垂直上升的距离A'C的长为
3.答案全部在答题卡上完成，答在本试卷上无效
A.2.8m
B.3.2m
第6题图
4.考试结束后，将本试卷和答题卡一并交回.
C.3.5m
D.4m
7.一元一次不等式组+5>3，的解集为
3x-4≤8
第I卷选择题（共30分）
A.x>-2
B.x≤4
C.-2D.x<-2
、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，
8.2026年多项重大体育赛事陆续
举办.现有四张外观、大小、质地
只有一项符合题目要求，请选出并在答题卡上将该项涂黑)
完全相同的赛事会徽卡片，背面
1.小满为了记录自己对零花钱的积攒和消费情况，建立了专属小账本，如果积攒100
朝上洗匀，正面分别印有四项赛
1连云港2026
元记作+100元，那么“-20元”表示小满
事：甲：2026年第六届亚洲沙滩
活力山日·长动袋出
A.积攒80元
B.积攒20元C.消费20元
D.消费80元
运动会（中国·三亚）；乙：2026
甲
丙
入
2.随着航天、人工智能、精密制造等现代科技飞速发展，对称美学已经成为设计的核
年山西省第十七届运动会（中国·长治）：丙：2026年江苏省第二十一届运动会（中国·
心语言.现有四幅源自前沿科技领域的设计图标，均体现了科技产品的视觉美感，
连云港)；丁：2026年米兰冬季奥林匹克运动会（意大利·米兰）.将卡片背面朝上洗
匀后，从中任意抽取两张（不放回），则抽到的两张卡片对应的赛事举办地都在中国
请根据对称性判断下列图标是轴对称图形但不是中心对称图形的是
的概率为
A行
B
D.i
9.如图，已知菱形ABCD的面积为14，对角线AC与BD相交于点O,对角线BD的长为
7,AE⊥BC于点E,连接OE,则OE的长为
A.1
B.4
C.2
D.5
3.下列运算正确的是
A.a2+b2 =2a2b2
B.a2·a3=a6
C.(2ab2)3=6a3b5
D.a3b5÷ab=a2b4
4.如图是一个正五棱柱实物图，它的俯视图是
第9题图
第10题图
10.如图，BC为⊙O的直径，点A在圆上，口ABCD中边AD与⊙O交于点E,已知BC=
√3，点E恰好为AD的中点，则阴影部分的面积为
正面
第4题图
A
B
c号
D.T数学参考答案与评分标准
一、选择题
1 2 3 4 5 6 7 8 9 10
C D D B B A C A C B
二、填空题
x 1，
11. 3 14（或 2） 12. 9 13. 14. 5 15.
y 3 5
三、解答题
16.解：（1）原式 1 4 3 3 ·······················································································4分
3. ·································································································· 5分
3x 2y 8，①
（2）
x 2y 4.②
①+②，得 4x=12.············································································································· 6分
解得 x=3.··························································································································· 7分
将 x=3代入②，得 3-2y=4.
解得 y= 1 .·······················································································································9分
2
x 3，

所以，原方程组的解为 1 ···············································································10分
y . 2
17.解：（1）如图 1所示，点 D即为所求作.·······························································3分
图 1
（2）由（1）可知∠DAC=∠ACB，
∴AD∥BC.
∵AD=BC，
∴四边形 ABCD是平行四边形.················································································· 5分
1
如图 2，过点 A作 AE⊥BC于点 E.
E
图 2
∴∠AEB=∠AEC=90°.
在 Rt△ABE中，AB=4，∠B=60°，
∴AE=AB·sin B=4× 3 =2 3，BE=AB·cos B=4× 1 =2.
2 2
在 Rt△ACE中，CE= AC 2 AE 2 =6.
∴BC=CE+BE=8.·········································································································· 6分
∴S□ABCD=BC·AE=8×2 3 =16 3 .···············································································7分
18.解：（1）80 80.5 80 ······················································································· 3分
4
（2） 200 300 1 30% 10% 40% 100（架）.······································· 4分20
答：甲、乙两个品牌一等品无人机共有 100架.·······················································5分
（3）甲品牌无人机的除冰效率更高.·············································································6分
在甲、乙两个品牌无人机每日除冰距离的平均数相同的情况下，甲品牌无人机
的方差为 25.2小于乙品牌无人机的方差 34.09，从方差角度来看，甲品牌无人
机在进行除冰工作时更加稳定；················································································8分
甲品牌无人机每日除冰距离的众数为 80米，高于乙品牌无人机的众数 74米，
从众数的角度来看，甲品牌无人机除冰效率更高.···················································9分
19.解：设一个工人平均每天可采摘黄花菜 x千克，则一台智能采摘机器人平均
每天可采摘黄花菜 5x千克.························································································1分
600 600
根据题意，得 1. ······················································································· 3分
4x 5x
解得 x=30.····················································································································· 4分
经检验， x=30是原分式方程的解.············································································5分
5x=5 30 150（千克）. ···························································································6分
答：一台智能采摘机器人平均每天可采摘黄花菜 150 千克.···································7分
2
20.解：如图，延长 BD交 EP于点 M，则 BM⊥EF.················································ 1分
∴∠FMB=90°.
由题易得四边形 ABDC和四边形 ABMF是矩形.
∴MF=AB=1.6 m，BD=35 m，∠EBM=90°-55°=35°，∠EDM=90°-45°=45°.····· 2分
设 EM=x m，则 EF=(x+1.6) m.
在 Rt△EDM中，∠EDM=45°，tan∠EDM= EM ，
DM
∴DM=EM=x.················································································································3分
∴BM=35+DM=x+35.
在 Rt△EBM EM中，∠EBM=35°，tan∠EBM= ，
BM
∴EM=BM·tan∠EBM=(x+35)·tan 35°≈0.7(x+35).······················································ 4分
∴x=0.7(x +35).··············································································································5分
解得 x≈81.7.·················································································································· 6分
∴EF=x+1.6=81.7+1.6≈83(m).······················································································ 7分
答：文峰塔的高度 EF约为 83 m.···············································································8分
21.解：（1）1·················································································································· 2分
（2）如图，连接格点 A，Q和 B，Q.·········································································· 3分
由图可知 DC∥QB.
∴∠DPA=∠QBA.········································································································· 4分
由勾股定理得 AQ=2 2 ，BQ= 2，AB= 10 .
3
∵（2 2）2+（ 2）2=（ 10）2，
∴AQ2+BQ2=AB2.
∴△QAB是等腰直角三角形，∠AQB=90°.·····························································5分
在 Rt△QAB中，cos BQ∠QBA= = 5 .
AB 5
∴cos∠DPA= 5 .·········································································································7分
5
13
（3） ····················································································································· 9分
13
22.解：（1）一次函数···································································································· 1分
y=-0.8x+180 ················································································································ 3分
（2）设月利润为W元，则
W=(x-50)(-0.8x+180)-1000
=-0.8x2+220x-10 000································································································· 5分
=-0.8（x-137.5）2+5125.
∵-0.8＜0，
∴当 x=137.5时，W最大，为 5125元.···································································· 6分
答：当销售单价为 137.5元时，该文创企业销售这款刻花瓷的月利润最大，最
大利润是 5125元.·········································································································7分
（3）①令W=5000，得-0.8x2+220x-10000=5000.···················································8分
解得 x1=150，x2=125.·································································································· 9分
由（2）知W=-0.8（x-137.5）2+5125，
∵-0.8＜0，∴抛物线开口向下.
∴当W≥5000时，125≤x≤150.···············································································10分
②∵尽可能提高销量，
∴单价应最小，即 x=125元.······················································································ 11分
此时销量 y=-0.8×125+180=-100+180=80（件）. ··················································· 12分
23.解：（1）BD=CF.······································································································· 1分
理由：如图 1，连接 CE.····························································································· 2分
图 1
4
∵AB=AC，
∴∠ABC=∠ACB=α.
∴∠BAC=180°-2α.
由旋转可知∠DAE=180°-2α，AD=AE，
∴∠BAC=∠DAE.
∴∠BAC-∠DAC=∠DAE-∠DAC，即∠BAD=∠CAE.
∴△ABD≌△ACE.······································································································ 3分
∴BD=CE，∠ABD=∠ACE.
∴∠ACB=∠ACE.
∵EF // AC，
∴∠ACB=∠F，∠ACE=∠CEF.
∴∠CEF=∠F.
∴CE=CF.
∴BD=CF.······················································································································ 4分
（2）DF=2BC.··················································································································5分
理由:如图 2，在 CD上取点 G，使 CG=CB，连接 AG，BE.································6分
图 2
∵∠ACB=90°，CG=CB,
∴AC垂直平分 BG.
∴AG=AB.
∴∠AGB=∠ABC=α.
∴∠BAG=180°-2α.······································································································ 7分
由旋转可知 DA=EA，∠DAE=180°-2α.
∴∠BAG=∠DAE.
5
∴∠DAG=∠EAB.
∴△DAG≌△EAB(SAS).····························································································8分
∴DG=EB，∠AGD=∠ABE=180°-α.
∵EF∥AB，
∴∠BFE=∠ABF=α，∠BEF=180°-∠ABE=180°-(180°-α)=α.
∴∠BFE=∠BEF.
∴BF=BE=DG.·············································································································· 10分
∴DF=DG+GF=BF+GF=GB=2BC.···············································································11分
3 7 3（ ） 或 .·····················································································································13分
2 2
提示：分两种情况讨论:
①当点 D在线段 BC 的延长线上时，如图 3所示，四边形 ABEF是平行四边
形.在 CD上取点 P，使 CP=CB，连接 AP.
由(2)可知 BE=DP，DF =2BC=2.
设 CF=x，则 CD =CF +DF =x+2，∴DP=CD-CP=x+2-1=x+1.
∵四边形 ABEF是平行四边形，∴AF=BE=DP=x+1.
在△ACF中，由勾股定理，得 AF2=CF2+AC2，即(x+1)2=x2+22，解得 x= 3 .
2
∴CD=x 7+2= .
2
②当点 D在线段 CB的延长线上时，如图 4所示，四边形 ABFE 是平行四边
形.
设 CD=x，则 CF=x+2，BF=x+2-1=x+1.
∵四边形 ABFE是平行四边形，∴BF=AE=AD=x+1.
在△ACD 3中，由勾股定理，得 AD2=CD2+AC2，即(x+1)2=x2+22，解得 x= .
2
∴CD= 3 .
2
7 3
综上所述，线段 CD的长为 或 .
2 2
6
图 3 图 4
7

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