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2024一2025学年度第二学期期末质量监测试卷
八年级数学
本试卷共4页，23小题，满分120分。考试用时120分钟。
注意事项：1.答卷前，考生务必用黑色字迹的钢笔或签字笔将自己的学校、姓名和准考证号填写在答题卡上。
将条形码粘贴在答题卡“条形码粘贴处”
2.作答选择题时，选出每小题答案后，用2B铅笔把答题卡上对应题目选项的答案信息点涂黑；
如需改动，用橡皮擦干净后，再选涂其他答案。
3.非选择题必须用黑色字迹的钢笔或签字笔作答，答案必须写在答题卡各题目指定区域内相应位
置上：如需改动，先划掉原来的答案，然后再写上新的答案：不准使用铅笔和涂改液。不按以
上要求作答的答案无效。
4.考生必须保持答题卡的整洁。考试结束后，将试卷和答题卡一并交回。
一、选择题：本大题共10小题，每小题3分，共30分.在每小题列出的四个选项中，只有一项是
符合题目要求的，
1.下列二次根式中，是最简二次根式的是()
A.3
B.V⑧
C.阿
2.下列各组数中，能构成直角三角形三边长的是()
A.4,5,6
B.5,8,13
C.1,V5,4
D.1,1,2
3.下列计算正确的是（)
A.V2+V5=V5
B.3V2-23=1
C.,/(-2)2=-2
D.V⑧÷2=2
4.若点A(1,a)在一次函数y=2x一1图象上，则a的值是（)
A.1
B.3
C.-1
D月
5.随着互联网的发展，网络购物越来越普及。某电商平台对某款热门电子产品一周内的日销售量
(单位：台)进行了统计，数据如下：25,30,28,32,30,26,30。关于这组数据，下列说法
正确的是（)
A.中位数是28
B.众数是32
C.中位数是30
D.众数是26
6.下列命题中，是真命题的是()
A.有两边相等的平行四边形是菱形
B.有一个角是直角的四边形是矩形
C.四个角相等的菱形是正方形
D.两条对角线互相垂直的平行四边形是正方形
7.正比例函数y=a(k≠0)的函数值y随x的增大而减小，则一次函数y=:一k的图象大致是
石：女
8.如题8图，两个较大的正方形的面积分别为225和289，则字母A所代表的正方形的面积为
()
A.64
B.16
C.8
D.4
八年级数学试卷第1页（共4页）
7.如题7图所示，在图形B到图形A的变化过程中，下列描述正确的是()
A.向上平移2个单位，向左平移4个单位
B.向上平移1个单位，向左平移4个单位
C.向上平移2个单位，向左平移5个单位
D.向上平移1个单位，向左平移5个单位
B
oo
题7图
题9图
题10图
3x-1≥8
8.若关于x的不等式组
的整数解共有三个，则a的取值范围是（)
xA.-3B.2≤a<3
C.-3D.59.如题9图所示，下列推理正确的个数有()
①若∠1=∠2，则AB/1CD:②若AD/BC,则∠3+∠A=180°：
③若∠C+∠CDA=180°，则AD//BC:④若AB//CD,则∠3=∠4.
A.0个
B.1个
C.2个
D.3个
10.用大小完全相同的长方形纸片在直角坐标系中摆成如题10图所示图案，已知A(一1,5)，则
点B的坐标是()
A.（-6,4)
B.(-9,)
C.(-6,5)
D(-4,)
二、填空题：本大题共5小题，每小题3分，共15分.请将下列各题的正确答案填写在答题卡相
应的位置上
1l.若n<512.某市为了解970万市民的出行情况，科学规划轨道交通，400名调查者走入1万户家庭发放了
调查问卷，并对收回的3万份问卷进行了调查登记.该调查中的样本容量是
13.一个数的两个平方根分别是2a一1与一a+2,则这个数是
14.已知x=m是二元一次方程组x-2y=3
的解，则代数式m十6n的值为
y=n
2x+4y=5
15.如题15图，将直角三角形ABC沿BF方向平移得到直角三角形DEF,
A
已知BE=4,AG=3,AC=7,则图中阴影部分的面积为
三、解答题（一）：本大题共3小题，每小题7分，共21分，
G
16.解方程组：
3x+2y=10
B
E
4(x-1)≥3x-7
题15图
17.解不等式组：
3x<
，并求出不等式组的所有整数解的和
A
18.如题18图，在四边形ABCD中，点E是BC延长线的一点，
连接AE交CD于点F,若∠B=∠D,∠1+∠2=180°.
题18图
(1)证明：AB//CD.
(2)若∠E=30°，∠BAD=120°，求∠2的度数，
B
七年级数学试卷第2页（共4页)2024—2025 学年度第二学期期末质量监测试卷
八年级数学参考答案及评分标准
一、选择题：本大题共 10小题，每小题 3分，共 30分．
1．A 2．D 3．D 4．A 5．C 6．C 7．A 8．A 9．B 10．C
二、填空题：本大题共 5小题，每小题 3分，共 15分．
11．x≥－1 12．乙 13． 13 14．y＝2x＋4 15．8或 4＋2 3
三、解答题（一）：本大题共 3小题，每小题 7分，共 21分．
16．（7分）
1
解： 24÷ 3－ 2× 18＋ 32
＝ 8－ 9＋ 32························································································· 3分
＝2 2－3＋4 2··························································································6分
＝6 2－3··································································································7分
17．（7分）
（1）解：∵y与 x－1成正比例，
∴设 y＝k(x－1)(k≠0)，·······································································1分
∵当 x＝－1时，y＝4，
∴4＝k(－1－1)，················································································ 2分
解得：k＝－2，··················································································· 3分
∴y＝－2(x－1)，················································································ 4分
即 y与 x的函数关系式为：y＝－2x＋2；·················································· 5分
（2）解：将 x＝3代入 y＝－2x＋2，得 y＝－2×3＋2＝－4≠2，·························· 6分
∴点（3，2）不在这个函数得图象上．·····················································7分
18．（7分）
解：我认为小明的说法正确．理由如下：························································1分
∵HE⊥CD，AB⊥CD
∴∠HEC＝∠AKC＝90°··············································································· 2分
∴AB//GF·································································································3分
∵∠HGA＝∠HFB
∴AG//BF·································································································4分
∴四边形 AGFB是平行四边形·······································································5分
∴GF＝AB＝1m··························································································6分
∴GF的长度就是篮板 AB的高度．································································7分
八年级数学参考答案及评分标准 第 1 页 (共 5 页)
四、解答题（二）：本大题共 3小题，每小题 9分，共 27分．
19．（9分）
（1）45% 60········································································· 2分（每空 1分）
（2）解：平均睡眠时间为 8小时的人数为：60×30%＝18人；···························· 4分
补全频数分布直方图，如下图：
（3）解：根据题意得：平均睡眠时间为 7小时的人数所占的百分比最大，
∴这部分学生的平均睡眠时间的众数是 7，··············································· 5分
平均数＝6×20%＋7×45%＋8×30%＋9×5%＝7.2小时；··························· 7分
12＋27
（4）解：1200名学生中睡眠不足（少于 8小时）的学生数＝ 60 ×1200＝780人．9分
20．（9分）
（1）解：由题意可知 MN⊥AB，
在 Rt△MNB中，BN＝ BM2－MN2＝ 1502－1202＝90(m)，························2分
∴AN＝AB－BN＝250－90＝160(m)，······················································ 3分
在 Rt△AMN中，AM＝ AN2＋MN2＝ 1602＋1202＝200(m)，······················ 5分
∴供水点 M到喷泉 A需要铺设的管道长为 200m；······································6分
（2）证明：∵AB＝250m，AM＝200m，BM＝150m，
∴AB2＝BM2＋AM2，·············································································8分
∴△AMB是直角三角形，
∴∠AMB＝90°．··················································································9分
21．（9分）
（1）证明：∵四边形 ABCD是平行四边形
∴AD// BC
∴∠AFB＝∠EBF，∠FAE＝∠BEA·························································· 1分
∵O为 BF的中点
∴BO＝FO·························································································· 2分
在△AOF和△EOB中
八年级数学参考答案及评分标准 第 2 页 (共 5 页)
∠EAF＝∠AEB
∠AFB＝∠EBF
OB＝OF
∴△AOF≌△EOB（AAS）
∴BE＝FA··························································································· 3分
∴四边形 ABEF是平行四边形································································· 4分
又∵AB＝AF
∴四边形 ABEF是菱形；······································································· 5分
（2）解：∵四边形 ABCD是平行四边形
∴AD＝BC
∵AF＝BE
∴DF＝CE＝2······················································································6分
∵平行四边形 ABCD的周长为 24
∴菱形 ABEF的周长为：24－4＝20
∴AB＝5····························································································· 7分
∵∠BAD＝120°
1
∴∠BAE＝ 2∠BAD＝60°······································································· 8分
又 AB＝BE
∴△ABE是等边三角形
∴AE＝AB＝5．··················································································· 9分
五、解答题（三）：本大题共 2小题，第 22题 13分，第 23题 14分，共 27分．
22．（13分）
（1）解：∵20人先下单，三种团购优惠方案的条件均不满足，
∴设这 20人中选择 A套餐的有 x人，x＜20，则选则 B套餐的有（20－x）人，20－x＜12，
∴30x＋25(20－x)＝565，······································································1分
∴x＝13，···························································································2分
∴20－x＝7························································································· 3分
答：选择 A套餐的有 13人，选择 B套餐的有 7人；
（2）解：∵两种套餐皆可的 11人中有 m人选择 A套餐，
∴当全班选择 A套餐人数不少于 20人时，即 13＋m≥20，
∴m≥7，··························································································· 4分
则选择 B套餐人数为 7＋(11－m)≤11，不满足优惠方案二的条件，
∴订餐总费用为：w＝30×0.9×(13＋m)＋25×(7＋11－m)＝2m＋801；········ 6分
（3）解：∵两种套餐皆可的 11人中有 m人选择 A套餐，
八年级数学参考答案及评分标准 第 3 页 (共 5 页)
①当 m≥7时，由（2）可知，订餐总费用为 w＝2m＋801，
∵k＝2＞0，
∴w随着 m的增大而增大，
∴当 m＝7时，订餐总费用最小为 w＝2×7＋801＝815（元）；······················8分
②当 0≤m＜7时，13＋m＜20，18－m＞11，
∴订餐总费用为 w＝30×(13＋m)＋25×0.8×(7＋11－m)＝10m＋750；
∵k＝10＞0，
∴w随着 m的增大而增大，
∴当 m＝0时，订餐总费用最小为 w＝750（元）；····································· 10分
③若选择优惠方案三，订餐总费用为 w＝30×(13＋m)＋25×(7＋11－m)＝5m＋840；
∵总费用满 850元立减 110元，
∴当 m＝2时，订餐总费用最小为 5×2＋840－110＝740（元）；··················12分
综上所述，当订购 A套餐 15 份，订购 B套餐为 16 份时，订餐总费用最低为 740 元．
·······································································································13分
23．（14分）
（1）解：∵一次函数 y＝ax＋b(a≠0)的图象经过点 A(－2，0)，点 B(2，4)
－2a＋b＝0
∴ ，················································································ 1分
2a＋b＝4
a＝1
解得 ，························································································2分
b＝2
∴一次函数的解析式为 y＝x＋2；····························································3分
（2）解：如图，作点 C关于 x轴的对称点 C′，连接 BC′交 x轴于 M，此时 MB＋MC的值最小．
········································································································ 4分
对于 y＝x＋2，
令 x＝0，则 y＝2，
∴C(0，2)
∴C′(0，－2)······················································································ 5分
设直线 BC′的解析式为 y＝kx－2(k≠0)，
则 2k－2＝4，解得：k＝3
八年级数学参考答案及评分标准 第 4 页 (共 5 页)
∴直线 BC′的解析式为 y＝3x－2，··························································· 6分
令 y＝0 2，得 x＝ 3
M 2∴ 3，0 ·························································································7分
（3）解：①当 OC为边时（如图），四边形 OCED是矩形，此时 EC⊥OC，ED⊥OA，8分
∵C(0，2)，A(－2，0)
∴OC＝OA＝2，此时 D、A重合，四边形 OCED是正方形．························ 9分
∴点 E的坐标为(－2，2)；·································································· 10分
②当 OC为对角线时（如图），四边形 OECD是矩形，则∠ODC＝90°，
由①知 OC＝OA＝2，
∴∠OCA＝∠OAC＝45°
∴∠COD＝45°＝∠OCD
∴OD＝CD························································································ 11分
∴四边形 OECD是正方形
∴OC＝DE，OC⊥DE
设 OC与 DE交于点 G
则 EG 1＝ 2DE
1
＝ 2OC＝1······································································· 12分
∴点 E的坐标为（1，1）····································································· 13分
综上所述，满足条件的点 E的坐标为（－2，2）或（1，1）······················· 14分
八年级数学参考答案及评分标准 第 5 页 (共 5 页)

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