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嵊州市 2025 年初中毕业生学业水平调测
数学参考答案
一、选择题（本大题共 10 小题，每小题 3 分，共 30 分．在每小题给出的四个选项中，只有
一项是符合题目要求的）
1 2 3 4 5 6 7 8 9 10
D B C B B A C C A D
二、填空题（本大题共 6 小题，每小题 3 分，共 18 分）
11． a 3x 5y 70,a 1 12 1 13 ． ．
4 5x 3y 74
14．5 15．20°或 70° 16． 6, 13
三、解答题（本题有 8 小题，第 17～21 题每题 8 分，第 22、23 题每题 10 分，第 24 题 12
分，共 72 分）
0
17 1 2025 1． π

2
2
2
1
1 1 ······························································································ 6
4
1
．···································································································· 2
4
18 x． 2 3
x 1 1 x
解：去分母得： x 2 x 1 3 ···································································· 2
去括号得： x 2x 2 3 ···································································· 2
移项得： x 5 ··································································· 2
x 5 ······································································ 1
经检验： x 5 是原方程的解．······························································ 1
19．（1） 50 25% 200 名 ·········································································· 2
答：本次调查共抽取了 200 名学生．
（2） 200 10 30 60 50 20 30 ，······················································ 3
1200 60 50 30 20 960 名 ．····················································· 3
200
答：估计能落实“中小学生每天综合体育活动时间不低于 1 小时”的学生人数为 960 人．
20．解：（1）一名工人每小时做的零件数= 56 3.5 16（个）；···························· 2
（2）设智能机器人做的零件数量 y关于工作时间 x小时的函数关系式为
1
k b 0 6 k 24
y kx b，得 1 ，解得 b 4 ，所以 y 24x 4 ·············· 2 k b 80
6
设下午工人做的零件数量 y关于工作时间 x小时的函数关系式为 y mx n，得
4.5m n 56 m 16
，解得 ，所以 y 16x 16 ．···························· 2
8m n 112 n 16
所以 24x 4 16x 16 60 ，解得 x 6．·········································· 2
21．（1）如图，过点 A作 AF∥MN，··················1
E
∵AB⊥MN，
∴AF⊥AB． CD
∵ CAB 140 ，
∴ CAF 140 90 50 ．················ 1 F G A
∵CD∥MN，
∴CD∥AF，····································· 1
∴ DCE CAF 50 ．···················· 1 M H B N
（2）作 EG⊥AF于点 F，交 MN于点 H．
∵ sin EG EAF sin 50 0.766 ，···················································· 1
AE
∴ EG AE 0.766 60 0.766 45.96 cm．············································ 1
∴ EH 45.96 60 105.96 106 cm．·················································· 2
∴点 E到 MN的距离为 106cm．
22．
（1）∵D为斜边 AC的中点，AD=CD，
A E
∴DE=BD，
∴四边形 ABCE是平行四边形，················ 1 D
∵∠ABC=90°，
∵平行四边形 ABCE是矩形，··················· 1 B C
∴AC=BE 图 1，
∴BD= 1 AC．········································ 1
2 E
（2）① 连结 BD，········································ 1 B
∵D为斜边 AC的中点，
1
∴BD= AC，
2
1 D
又∵BE= AC，
2 A C图 2
∴BD=BE，·············································1 E
∴∠EDB=∠E=18°，
∴∠A=∠ABD=36°。···································1 B
② 取 AC中点 D，连结 BD，连结 FD，······· 1
∵D，F分别为 AC BC F， 的中点，
D
∴DF∥AB，DF= 1 AB，···························· 1
2 A 图 3 C
BE= 1∵ AB，
2
∴DF∥BE，DF=BE，
∴四边形 ABCE是平行四边形．································································1
∴EF=BD= 1 AC=2．··············································································1
2
23．解：（1）当 m=1 时， y x2 2x 2 (x 1)2 1···········································1
所以顶点坐标为（1，1）．···························································· 2
（2）① y x2 2mx m2 m (x m)2 m
对称轴为直线 x m，··································································· 1
因为m 3≤x1≤m 1，且抛物线的开口向上，所以当 x m 3时， y有最大值，
即 (m 3 m)2 m 1，得m 8 ．················································· 2
②因为抛物线的开口向上，
当 y1＜y2 时， 2 3m m＞3 ，即 2 4m＞3．···································· 2
所以 2 4m＞3或 2 4m＜ 3，
1 5
解得m＜ 或m＞ ．································································· 2
4 4
24．
（1）∵∠CMF=50°， A D
∴∠CDE=25°，··································· 2
∵四边形 ABCD是正方形， G
∴ ADC 90 ，···································· 1 M
∴ ADE 90 25 65 ．······················ 1 AF N
（2）①设∠ADF=x°，··································· 1 P
则 DFM CDF 90 x ，··············· 1 B E C
∵∠DAF=50°，
R
∴ DFA DFP 180 50 x 130 x ，········································ 1
∴ MFN 130 x （90 x ） 40 ，················································· 1
②延长 FM交 DP于点 G，
∵ ADF PDF， MDF MFD，且 ADF MDF 90 ，
∴ MFD FDP 90 ，即 FG DP，
∴ GFP P 90
∵ P DAF，
∴ GFP DAF 90 ，··································································· 1
延长 AF，DC交于点 R，
∴ R DAF 90 ，
∴ MFN R，且 FMR是公共角，
∴△FMR∽△NMF．··········································································1
MF MR
∴ ，即MN·MR MF 2 4 ．
MN MF
∴当 MR的长最大时，MN的长最小，
当 AR与半圆 M相切时，MR的长最大，即 MF⊥AF，
此时△MFR∽△ADR，
FM FR MR
∴ ，
AD DR AR
FR MR 2 1
即 ，
2 MR 4 FR 4 2
2 MR 2FR，
∴ ，
4 FR 2MR
MR 10∴ ，····················································································1
3
6
∴MN的最小值为 ．········································································ 1
5嵊州市2025年初中毕业生学业水平调测
数学
考生须知：
1,全卷分试题卷和答题卷，满分120分，考试时间120分钟.
2,答题时所有试题卷的答案请填在答题卷相应的位置上，做在试题卷上无效.
3,本次考试不使用计年器，
一、选择题（本大题共10小题，每小题3分，共30分.在每小题给出的四个选项中，只有
一项是符合题目要求的】
1.下列四个数，最大的数是
A.-2
B.0
C.2
D.-2
2.下列用七巧板拼成的图案轮廓中，是轴对称图形的是
3.据某平台统计，2025年“五一”期间，嵊州市网红街“东前街”共接待游客约175000人
次.数字175000用科学记数法表示为
A.1.75×104
B.17.5×10
C.1.75×103
D.0.175×106
4.为纪念“五·四”运动106周年，某校举办歌咏比赛，某班演唱后五位评委给出的分数为：
9.5,9.2,9.6,9.4,9.5,则这组数据的中位数是
A.9.6
B.9.5
C.9.4
D.9.2
5.估计3V7-√28的值应在
A.3和4之间
B.2和3之间
C.1和2之间
D.0和1之间
6.己知菱形的两条对角线长分别是6cm和8cm,则菱形的边长为
A.5cm
B.6cm
C.8cm
D.10cm
7.已知m是一元二次方程x2-x-2=0的一个根，则代数式2m2-4m+2025的值为
A.2027
B.2028
C.2029
D.2030
8.如图，在平面直角坐标系中，四个点分别表示甲，乙，丙，丁四
件商品的数量y与单价x的情况，且乙，丁两件商品所表示的
甲
点在同一反比例函数图象上，则四件商品中，总价（总价=单价
、丙
×数量)最多的是
0
A.甲
B.乙
C.丙
D.丁
第8题田
数学试题卷第1页（共6页）
9.如图，平面直角坐标系中有四个点E(4,-4),F(-3,0),M
(-2,-4),0(0,0),二次函数y=ar2+bx+c（a,b,c为常数，
且a≠0)的图象经过这四个点中的其中三个点，若要使a取得最
小值，则抛物线y=ax2+bx+c经过的三个点是
A.E,F,M
B.E,F,O
M
C.E,M,O
D.F,M,O
第9题图
10.如图，在△ABC中，∠ACB=90°，分别以△ABC的三边向
外作正方形ACFG,正方形BDEC,正方形ANB.连结
DN,若DN=x,AC=y,BC=a(a为常数)，则下列
C
D
各式为定值的是
A.x+y
B.x2+y2
C.￥
D.x2-y2
二、填空题（本大题共6小题，每小题3分，共18分）
M
11.分解因式：a2+a=▲
第10题图
12.一个不透明的袋子里装有1个红球和3个白球，它们除颜色外均相同，从袋中任意摸
出一个球是红球的概率为▲·
13.《算法统宗》中有这样一个问题：今有上禾三束，下禾五束，共价七十钱；上禾五束，下
禾三束，共价七十四钱.问上、下禾每束价各几何？小明设上禾每束x钱，下禾每束y钱，
则符合题意的二元一次方程组是▲·
14.如图，AB是⊙O的直径，弦CD⊥AB于点E,若AE=CD=8,连结OC,则OC的长为
▲，
15.如图，在△MBC中，AB=AC,分别以点B,C为圆心，大于BC长为半径作弧，两
弧交于E,F两点；再以点A为圆心，AB长为半径作弧，交直线EF于点P,连结BP,
则∠BPA的度数是▲·
y
0
E
B
C
第14题田
第15题田
第16题田
数学试题卷第2页（共6页）

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