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2024一2025学年度第二学期期末质量监测试卷
七年级数学
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.
1.16的平方根是(
A.4
B.+4
C.±2
D.2
2.在下列生活、生产现象中，可以用基本事实“垂线段最短”来解释的是()
B
A.平板弹墨线
B.建筑工人砌墙
C.弯河道改直
D.测量跳远
3.下列几组解中，二元一次方程x一y=1的解是()
A.
x=3
B.
x=3
x=一3
(x=2
C
D.
y=2
y=-2
y=-2
y=3
4.已知aA.a-4B.2a<2b
C.
D.-2a<-2b
5.在2025年春节联欢晚会上，16个人形机器人身着花棉袄，手持红手帕登上舞台，与舞蹈演员
默契配合，共同演绎了舞蹈《秧BOT》·小林观看节目时受到启发，将图1中机器人的手臂抽
象为图2的示意图，其中手臂AB/1CD,∠B=∠D,∠BED=104°，则∠B的度数为()
2
图1
图2
A.166°
B.128°
C.104°
D.100°
6.下列调查中，最适合采用全面调查（普查）的是()
A.对我市中学生每周课外阅读时间情况的调查
B.对我国首艘电磁弹射航空母舰福建舰各零部件质量情况的调查
C.对我市中学生观看电影《哪吒之魔童闹海》情况的调查
D.对我市新能源汽车的电池使用寿命调查
七年级数学试卷第1页（共4页)
9.如题9图，在口ABCD中，对角线AC、BD交于点O,点E是BC的中点.若OE=3cm,则CD
的长为(
)
A.3cm
B.6cm
C.9cm
D.12cm
10.已知一次函数y1=cx十b与y2=x十a的图象如题10图所示，有下列结论：①k<0;②a>0;③
关于x的方程十b=x十a的解为x=3;④当x>3时y1>2,其中正确的结论有(）个，
A.4
B.3
C.2
D.1
30o
y2=x+a
225
289
y1=kx+b
4
P
题8图
题9图
题10图
题15图
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
11.要使二次根式x十1有意义，则x应满足的条件是
12.对甲、乙两个超市在九月份每天的营业额进行调查，发现：在九月份两个超市每天营业额的平
均值相同，方差分别为S=7.5,S=2.6,则九月份每天营业额较稳定的超市是
.（填“甲”或“乙”)
13.在平面直角坐标系中，点(3，一2)到原点的距离是
14.把直线y=2x+1向上平移3个单位，平移后直线的解析式为
15.现有一张其中一个角为30°、最小边长为2的直角三角形纸片，沿如题15图所示的中位线剪
开后，将两部分拼成一个四边形，则所得四边形的周长是
三、解答题（一）：本大题共3小题，每小题7分，共21分
16.计第：a÷-×+V原
17.己知y与x一1成正比例，当x=一1时，y=4.
(1)求出y与x的函数关系式：
(2)请通过计算，判断点(3,2)是否在这个函数的图象上
18.如题18-1图所示是某校篮球架实物图，如题18-2图所示是篮球架的侧面示意图，篮板边侧
AB垂直于地面.八年级的“综合与实践”数学小组开展测量篮球架篮板AB高度的实践活动.在
不便于直接测量的情况下，小组设计了如下测量方法：如题18-3图所示，小组成员将竹竿HE
垂直固定在地面CD上，小明从竹竿上的F点处观察篮板底部B点，用测角仪测量视线FB与
竹竿HE的夹角∠FB的度数为48°，接着将观察点沿着竹竿向上移动到G点，使得从G点
观察篮板顶部A点的视线GA与竹竿HE的夹角∠HGA的度数恰好等于∠HFB的度数时，在
竹竿上标注G点的位置，测量GF的长度为1m.活动分享时，小明说：“GF的长度就是篮
板AB的高度”，你认为小明的说法是否正确，并说明理由.
八年级数学试卷第2页（共4页）2024—2025 学年度第二学期期末质量监测试卷
七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．B 2．D 3．A 4．D 5．B 6．B 7．B 8．D 9．C 10．D
二、填空题：本小题共 5小题，每小题 3分，共 15分.
11．2 12．3万或 30000 13．9 14．2 15．22
三、解答题（一）：本大题共 3小题，每小题 7分，共 21分．
16．（7分）
x y＋1
解： 2－ 3 ＝1，
3x＋2y＝10
3x－2y＝8①
原方程可化简为 ，···································································2分
3x＋2y＝10②
①＋②得：6x＝18，··················································································· 3分
解得：x＝3．···························································································· 4分
x 1把 ＝3代入①得：y＝ 2，··········································································· 6分
x＝3
∴方程组的解为 1．···············································································7分y＝ 2
17．（7分）
解：解不等式①：4x－4≥3x－7，解得：x≥－3， ···········································2分
解不等式②：6x＜x＋5，解得：x＜1，··························································· 4分
∴不等式组的解集为－3≤ x＜1．································································· 5分
∴整数解为－3，－2，－1，0，和为－3＋(－2)＋(－1)＋0＝－6．·····················7分
18．（7分）
（1）证明：∵∠1＝∠AFC，∠1＋∠2＝180°， ··············································· 1分
∴∠2＋∠AFC＝180°，
∴AB//CD．······················································································· 3分
（2）解：∵AB//CD，
∴∠DCE＝∠B，················································································· 4分
∵∠B＝∠D，
∴∠DCE＝∠D，·················································································5分
∴AD//BE，
∵∠E＝30°，
∴∠DAE＝∠E＝30°．·········································································· 6分
∵∠BAD＝120°，
∴∠2＝∠BAD－∠DAE＝120°－30°＝90°．·············································· 7分
七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）
解：（1）a＝0.10，c＝12，m＝35；······································································3分
（2）在图 2中，B组对应的圆心角的度数是 108度；········································· 4分
补全图 1频数分布直方图，如下：···························································6分
（3）根据题意得：200×(35%＋20%)＝110（支）·············································8分
答：估计大约有 110支队伍能进入复赛．··················································9分
20．（9分）
x＝1 x＝3
解：（1）方程 x＋2y＝5的正整数解 ， ． ··············································· 2分
y＝2 y＝1
（2）联立 x＋2y＝5与 x＋y＝0得，
x＋2y＝5
，························································································ 3分
x＋y＝0
x＝－5
解得 ，·····················································································4分
y＝5
x＝－5
把 代入方程 x－2y＋mx＋9＝0中，得－5－10－5m＋9＝0，
y＝5
m 6解得 ＝－ 5．···················································································· 5分
（3）∵x－2y＋mx＋9＝0，
∴(1＋m)x－2y＋9＝0，········································································ 6分
∵无论实数 m取何值，方程 x－2y＋mx＋9＝0总有一个公共解，
∴x＝0，···························································································· 7分
∴y＝4.5，·························································································· 8分
x＝0
即公共解为 ．············································································ 9分
y＝4.5
21．（9分）
解：问题一：1.3，474.5·····················································································4分
问题二：设建设 x个大型租赁点，则建设(8－x)个小型租赁点．
根据题意，得 8000x＋5000(8－x)≤50000，·············································· 6分
七年级数学参考答案及评分标准 第 2 页 (共 5 页)
10
解得 x≤ 3，·······················································································7分
∵x为正整数，
∴x的最大值为 3，
∴8－x＝5．························································································8分
答：至多可以建设 3个大型租赁点和 5个小型租赁点． ······························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：∵AB//CD，
∴∠FEG＝∠EFP，··············································································1分
∵EF//GH，
∴∠EFP＝∠PHG，
∴∠PHG＝∠FEG．············································································· 2分
（2）解：∵AB//CD，
∴∠EPF＝∠PEA，·············································································· 3分
∵EP平分∠AEF，
1
∴∠AEP＝ 2∠AEF，
∴∠EPF 1＝ 2∠AEF，··········································································· 4分
∵∠AEF＋∠FEG＝180°，
∴∠EPF 1＝ 2 180°－∠FEG ，
由(1)知∠PHG＝∠FEG；
EPF 1∴∠ ＝ 2 180°－∠PHG ，······························································· 5分
∵∠EPF∶∠PHG＝1∶3，
可设∠EPF＝x，则∠PHG＝3x，
x 1则 ＝ 2 180°－3x ，
解得 x＝36°，······················································································7分
∴∠PHG＝108°，
∵EF//GH，
∴∠EFD＋∠PHG＝180°，
∴∠EFD＝72°．·················································································· 8分
（3）解：∠PEM＋∠EMF＝90°；理由如下：··················································· 9分
设∠EMF＝α，∠EMG＝β，则∠HFM＝∠HMF＝α＋β，
∵EF//GH，
∴∠EFM＋∠HMF＝180°，∠FEM＝β，
∴∠EFM＝180°－ ＋β ，
∴∠EFH＝∠EFM－∠HFM＝180°－2 ＋β ，········································ 10分
七年级数学参考答案及评分标准 第 3 页 (共 5 页)
∵AB//CD，
∴∠AEF＝∠EFH＝180°－2 ＋β ，······················································11分
∵EP平分∠AEF，
PEF 1∴∠ ＝ 2∠AEF＝90°－α－β
∴∠PEM＝∠PEF＋FEM＝90°－α－β＋β＝90°－α，··································12分
∵∠EMF＝α，
∴∠PEM＝90°－∠EMF，
∴∠PEM＋∠EMF＝90°．····································································13分
23．（14分）
10
解：（1）x＝ ；······························································································ 2分
3
（2）作 AF⊥x轴于点 F，作 BE⊥x轴于点 E，(作出图形)·································· 3分
∵A －2，4 ，B 3，2 ，
∴AF＝4，OF＝2，BE＝2，OE＝3，
设 OC＝a，
1 1 1
由题意得 a＋4 ·2＋ a＋2 ·3＝ 4＋2 · 2＋3 ，························· 5分
2 2 2
5
整理得 a＝8，
2
解得 a 16＝ 5；
16
∴点 C的坐标为 0， 5 ；·····································································6分
（3）①作 AF⊥x轴于点 F，作 BE⊥x轴于点 E，设 DE＝b，(作出图形)················7分
1 1 1
由题意得 b＋2＋3 ·4－ b·2＝ 4＋2 · 2＋3 ，
2 2 2
解得 b＝5，∴OD＝5＋3＝8，
∴点 D的坐标为 8，0 ．····································································· 10分
②作 AF⊥x轴于点 F，作 PG⊥x轴于点 G，设点 P的纵坐标为yP，
当点 P在线段 AD上时，
七年级数学参考答案及评分标准 第 4 页 (共 5 页)
3
∵S△OPD＝ 5 S△OAP，
S 3 3 1
1
∴ △OPD＝ 8 S△OAD＝ 8× 2×8×4＝6，即 ×8× yP ＝6，2
y 3 PG 3∴ P ＝ 2，此时 ＝ 2，
1 3 1 3 1
设 DG＝c，由题意得 ＋4 · 2＋8－c ＋ ×c× ＝ 8＋2 ·4，
2 2 2 2 2
c 15解得 ＝ 4，
OG 15 17∴ ＝8－ 4 ＝ 4，
P 17 3∴点 的坐标为 4 ， 2 ；
当点 P在射线 AD上时，作 PH⊥x轴于点 H，作 CQ⊥PH交 PH的延长线于点 Q，
3
∵S△OPD＝ 5 S△OAP，S
1
△OAD＝ 2×8×4＝16，
∴S△OPD＝24，
1
∴ ×8 yP ＝24，2
∴yP＝－6，
设 DH＝d，
∴CQ＝OH 8 d 16＝ ＋ ，QH＝OC＝ 5，
1 16 1 1 16
由题意得 d＋8＋d · ＋ ×d×6＝ 8＋d · 6＋ ，
2 5 2 2 5
解得 d＝15，
∴OH＝8＋15＝23，
∴点 P的坐标为 23，－6 ；
17 3
综上，点 P的坐标为 4 ， 2 或 23，－6 ．············································ 14分
七年级数学参考答案及评分标准 第 5 页 (共 5 页)

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