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准考证号
运城市2025~2026学年八年级第一次阶段性考试
6.2025年9月3日，我国隆重举行纪念中国人民抗日战争暨世界反法西斯战争胜利80周
第Ⅱ卷非选择题（共90分）
年阅兵式，多种自主研发新型装备首次亮相如图，以重型歼-20战斗机D,E所在的直线
二、填空题（本大题共5个小题，每小题3分，共15分）
数学
为x轴、过点A且垂直于DE的直线为y轴，建立平面直角坐标系x0y.若中型歼-35A战
11.写出一个比3大且比V15小的无理数▲
注意事项：
斗机B的坐标为(m,n),歼-16D电子战飞机C与B到x轴的距离相等
1.本试卷分为第1卷和第Ⅱ卷两部分.全卷共6页，满分120分，考试时间120分钟
12.比较大小：V写2▲之（填“>”、“<”或”=）
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，
到y轴的距离也相等，则歼-16D的坐标为
13.如图，在R1△ABC中，∠C=90°，D为边AC上一点，且满足DA=DB=8,若△DAB的面积
3.答案全那在答题卡上完成，答在本试卷上无效
A.(m,n)
为24.则AC的长为▲
4.考试结束后，将本试卷和答题卡一并交回.
B.(-m,-n)
BO
14.如图，在Rt△ABC中，∠C=90°，分别以各边为直径作半圆，图中阴影部分在数学史上被
C.(m,-n）
称为“希波克拉底月牙”当AC=8,BC=4时，阴影部分的面积为▲
第I卷选择题（共30分）
D O*
D.(-m,n)
(第6题图)
一、选择题（本大题共10个小题，每小题3分，共30分在每小题给出的四个选项中，只有
7.下列说法：①负数没有立方根：②实数和数轴上的点是一一对应的：③算术平方根等于它
(2.0)
项符合题目要求，请选出并在答题卡上将该项涂黑)】
本身的数只有0：④正数的两个平方根互为相反数其中正确的个数有
1.64的立方根是
A.1个
B.2个
C.3个
D.4个
(第13题图)
(第14题图)
(第15题图)
A.8
B.-8
C.4
D.-4
8.观察分析下列数据：0，-√2,2，-√6,2√2，-√V10,2√3，…，根据数据排列的规律
15.如图，长方形BCDE的各边分别平行于x轴或y轴，甲乙分别由A(2,0)同时出发，沿长
2.△ABC的三条边分别为a,b,c,下列条件不能判断△ABC是直角三角形的是
方形BCDE的边作环绕运动，甲按逆时针方向以1个单位/秒的速度匀速运动，乙按顺
得到的第10个数据的值是
时针方向以2个单位秒的速度匀速运动，则甲、乙运动后的第2024次相遇地点的坐标
A.=4,b=V41,c=5
B.∠B=50°，∠C=40
A.3V2
B.-3V2
C.2V5
D.-2V5
是▲
C.∠A:LB:LC=3:45
D.a:b:c=1:V2 :V3
9.如图，在平面直角坐标系中，点C(m,m)在第一象限，点B,A分别在x轴正半轴和y轴
三、解答题（本大题共8个小题，共75分解答应写出必要的文字说明、证明过程或演算步骤）】
半轴上，∠ACB=90°，则OA+0B等于
16.计算（本题共4个小题，每小题5分，共20分）》
3.已知直线MN∥x轴，M点的坐标为(3,2)，并且线段MN=4,则点N的坐标为
A.(-1,2)B.(1,2)或(7,2)C.(7,2)
D.(-1,2)或(7,2)
A.m
(1)V48-V75+-V3:
(2)V80-V45-2:
4.下列计算正确的是
B.2m
(3)(V5+3)(3-V5)+(1-V2)
(4V5-告xV5
A.V8+V2=V10
B.V(-3P=3
C.3m
17.(本题共7分)
D.4m
第9题图)
C.3V5-V5=3
D.V(-2)x(-3)=V-2xV3
如图，△ABC的顶点A(m-2,m+2)在x轴上，则点A的坐标为▲；将点A向上平移
10.如图是一个长方体仓库，在其内壁的点A(长的四等分，点)处有一只壁虎在点B(宽的三
1个单位长度，再向右平移3个单位长度得到点B,则点B的坐标为▲；BC=3,
5.在如图所示的数轴上，点B与点C关于点A对称，A,B两点对应的实数分别是√2和-1
等分点)处有一只蚊子.若该仓库长为8m(点A所在线段)，宽为6m(点B所在线段)，高
点C在x轴的上方，且BC∥y轴，则点C的坐标为▲
则点C所对应的实数是
为5m,则壁虎爬到蚊子处的最短距离应为
(1)请在图中画出△ABC
(2)将△ABC的三个顶点横坐标分别乘-1，纵坐标不变
A.1+V2
A.V85 m
依次得到点A,B1,C,请在图中画出△AB,C,并写出
B.2+V2
0V2
B.V89 m
△AB,C,与△ABC的位置关系；
C.2V2-1
(第5题图
(3)若△AB,C,内任意一点P的坐标为(m,n),那么P到
C.5V5 m
(第10题图
x轴的距离是▲
D.2V2+1
D.2+V61 m
八年级数学第1页（共6页）
八年级数学第2页（共6页）
八年级数学第3页（共6页）运城市 2025~2026 学年八年级第一次阶段性考试
数学参考答案及评分说明
一．选择题（本大题共 10 个小题，每小题 3 分，共 30 分）
题号 1 2 3 4 5 6 7 8 9 10
选项 C C D B D D B B B A
二．填空题（本大题共 5 个小题，每小题 3 分，共 15 分）
11． 11（答案不唯一） 12． < 13. 8 + 2 7 14．16 15．（-1，-1）
三．解答题（本大题共 8 个小题，共 75 分）
16．计算（每小题 5分，共 20分）
解：（1）原式= 16 × 3 25 × 3 + 3..............................................................................2分
= 16 × 3 25 × 3 + 3 .......................................................................3分
= 4 3 5 3 + 3............................................................................................4分
＝0．.................................................................................................................. 5分
2 = 4 5 3 5（ ）原式 2 ...............................................................................................3分
5
＝1﹣2·····················································································4分
＝﹣1．··················································································· 5分
（3）原式= 9 5+ (1 2 2 + 2)····························································3分
= 4 + 1 2 2 + 2 ·································································4分
= 7 2 2．··········································································5分
（4）原式= 15 × 3 5 × 3······························································ 1分
12
= 45 5 ··········································································2分
4
= 3 5 5 ·········································································· 4分
2
= 5 5 ·················································································· 5分
2
17．（本题 7分）
解：（-4，0），（-1，1），（-1，4）；·································································3分
第 1页（共 4页）
C C1
A B B1 A1
（1）如图△ABC即为所求；·········································································· 4分
（2）如图△ 1 1 1即为所求；······································································ 5分
△ABC与△ 1 1 1关于 y轴对称····························································· 6分
（3）n．····································································································7分
18．（本题 8分）
解：（1）将 h 50 50＝ 代入得： 1 = = 10（秒）············································ 1分5
h 100 = 100将 ＝ 代入得： 2 = 20 = 2 5（秒）······································ 2分5
答：50m所需时间 1是 10秒，100m所需时间 2是 2 5秒．··························· 3分
（2）∵2 5 ÷ 10 = 2·········································································· 4分
答： 2是 1的 2倍·················································································· 5分
（3）将 t＝1.5代入得：1.5= ······························································· 6分
5
解得：h＝11.25····················································································· 7分
答：物体下落的高度是 11.25m．·······························································8分
19．（本题 7分）
解：（1）在 Rt△ABD中，∠A=90°，由勾股定理得：
BD= 2 + 2 = 42 + 32 = 5······························································ 1分
答：B、D之间的距离为 5m······································································2分
（2）∵52 + 122 = 132，
∴ 2 + 2 = 2················································································ 3分
∴△BCD是直角三角形，∠CBD=90°························································4分
第 2页（共 4页）
1 1 1四边形 = + = × 4 × 3 +
1 × 12 × 5 = 36.····················· 6分
2 2 2 2
答：四边形 ABCD的面积是 36 m2·················································································7分
20．（本题 9分）
解：（1）A（10，0），B（10，8），C（0，8）······················································· 3分
（2）∵△BCE沿 CE折叠，
∴△BCE≌△FCE
∴BC=CF=10，BE=EF·············································································4分
在 Rt△COF中，∠O=90°， = 2 2 = 102 82 = 6.
AF=OA-OF=10-6=4················································································· 5分
设 EF的长为 x，则 BE=x，AE=8-x，·························································· 6分
在 Rt△AEF中， 2 = 8 2 + 42···························································· 7分
解得 = 5，·························································································· 8分
答：线段 EF的长度为 5米．···································································· 9分
21．（本题 6分）
解：
每种情况正确画图并标注得 2分，任意画出三种不重复的情况得 6分
22．（本题 8分）
解：（1）三角形的任意两边之和大于第三边；······················································· 1分
（2）D······································································································2分
（3） 17- 2 < 13·········································································· 3分
.........................................................................................4分
∵三角形任意两边之差小于第三边，
∴AB-AC第 3页（共 4页）
∵ = 12 + 42 = 17， = 12 + 12 = 2， = 22 + 32 = 13
∴ 17- 2 < 13·························································································6分
（4）A点即为所求表示 14的点········································································ 8分
（做法不唯一）
23．（本题 10分）
解：（1）在 Rt△ABC中，∠C=90°
由勾股定理得: = 2 2 = 52 32 = 4.···········································1分
答：BC的长为 4cm.··················································································· 2分
（2）∵△ACP沿 AP折叠，∴△ CP≌△ ' ∴ = '=3， = '
BC’=AB-AC’=5-3=2···················································································· 3分
设 BP的长为 x，则 CP=4-x·········································································· 4分
在 Rt△BC’P中， 2 = 4 2 + 22······························································ 5分
5 5 5
解得 = ， ÷ 1 = 秒··············································································· 6分
2 2 2
C AB 5答： 恰好落在 边上 C’处时 t的值为 秒.
2
25
（3） .........................................................................................................................................7分
4
25
（4） 或 5 或 8....................................................................................................................... 10分
8
第 4页（共 4页）

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