四川省资阳市2025届九年级下学期中考一模数学试卷(含答案)

这是一份四川省资阳市2025届九年级下学期中考一模数学试卷(含答案)，共16页。试卷主要包含了数学家华罗庚说等内容，欢迎下载使用。
全卷分为第I卷(选择题)和第Ⅱ卷(非选择题)两部分。第I卷1至2页，第Ⅱ卷3至6页。全卷满分150分，考试时间共120分钟。
注意事项：
1. 答题前，考生务必将自己的姓名、座位号、报名号(考号)写在答题卡上，并将条形码贴在答题卡上对应的虚线框内。同时在答题卡背面第3页顶端用2B 铅笔涂好自己的座位号。
2. 第I卷每小题选出的答案不能答在试卷上，必须用2B 铅笔在答题卡上把对应题目的答案标号涂黑，如需改动，用橡皮擦擦净后，再选涂其它答案。第Ⅱ卷必须用0.5mm黑色墨水签字笔书写在答题卡上的指定位置。不在指定区域作答的将无效。
3. 考试结束，监考人员只将答题卡收回。
第Ⅰ卷(选择题 共40分)
一、选择题(本大题共10个小题，每小题4分，共40分. 在每小题给出的四个选项中，只有一个选项是符合题目的要求的.)
1. -2的绝对值是
A. 2 B. -2 C. 12 D.−12
2.下列运算正确的是
A.m2+m2=m4 B.m3⋅m2=m6 C.mn23=m3n6 D.m32=m5
3.下列立体图形中，主视图为三角形的是
4.2025年春节假期，安岳县游客接待量约913200人次，同比增长11.2%……印证着这座“中国石刻之乡”的文旅活力.请将数“913200”用科学记数法表示为
A. 0.9132×10⁵ ×105 C. 9.132×10 D. 91.32×10⁴
5. 若一多边形的内角和是900°，则它的边数为
A. 4 B. 5 C. 6 D. 7
6.猜灯谜是我国独有的富有民族风格的一种汉族民俗文娱活动形式.某校开展了猜灯谜知识竞赛活动，其中甲组学生的成绩为：82，81，83，84，81，80，则这组数据的众数和中位数分别为
A. 81, 83 B. 81, 81.5 C. 81.5, 81 D. 81, 83.5
7. 如图1, 在菱形ABCD中, AB=6, 分别以 C, D为圆心, 大于 12CD长为半径画弧，两弧交于E，F两点，作直线EF交AD于点 G，交CD于点H. 若S△DGH=6,则AG的长为
A. 1 B. 1.5 C. 2 D. 2.5
8. 如图2, 在□ABCD中, AC, BD 相交于点O, AB=4, BC=8, AC⊥CD, 以点C为圆心，CO长为半径画弧，交BC于点E，则图中阴影部分的面积为
A.43−π B.83−π C.83−2π D.163−π
9.数学家华罗庚说：“数缺形时少直观，形少数时难入微；数形结合百般好，隔离分家万事休.”由此可知方程 x2+4x−1x+3=0的实数根的个数为
A. 0个 B. 1个 C. 2个 D. 3个
10. 如图3, 正方形ABCD的边长为3, 点E, F, G分别在边 AB, BC, CD上, 且AF⊥EG. 当CF=2BF时, EF+AG的最小值为
A.210 B.310 C.25 D.35
第Ⅱ卷(非选择题 共110分)
二、填空题. (本大题6个小题，每小题4分，共24分)
11. 若式子 m+2有意义，则m的取值范围为 .
12.一袋中装有若干个球，它们除颜色外无其他差别，其中有8个白球，从盒子中任意摸出一个，摸到白球的概率是 29，则该袋中球的总个数为
13. 若a是方程. x2−2x−1=0的根，则 −2a2+4a+3的值为 .
14. 在平面直角坐标系中，将点 P(3，1)绕原点O顺时针旋转90°到点 P'处，则点P'的坐标为 .
15. 如图4-1, 点 E, F在矩形纸片 ABCD的边上, 且∠AEF=40°, 将矩形沿EF折叠成图4-2, AE与CF交于点G; 点H, K分别在线段DE, CG上, 再沿 HK折叠成图4-3, 此时 CK恰好过点 E. 若 EG=EK, 则∠EHK 的度数为 .
16. 如图5, 已知△OA₁B₁, △A₁A₂B₂, △A₂A₃B₃, ……都是等边三角形, 点A₁, A₂, A₃, …都在x轴上, 点 B₁, B₂, B₃, ……都在直线 y=33x+233上. 设点 B2025的纵坐标为n, 则 n22024的值为 .
三、解答题(共8个小题，共86分. 解答应写出必要的文字说明、证明过程或演算步骤)
17.(本小题满分9分)
先化简，再求值： x+1x−1÷x2−1x2+x, 其中x=2.
18. (本小题满分10分)
2025年1月29日，《哪吒2》正式上映，该电影剧情精彩、特效震撼，还精准传递了中国传统文化，从而引起了不同年龄段观众的共鸣，特成为中国动画电影的一部杰出作品.月月为了解本校学生对该电影的关注程度，对她所在学校的学生进行了随机抽样调查，将调查结果分为：A(实时关注)、B(关注较多)、C(关注较少)、D(没有关注)四类，并将调查结果绘制成如图6所示的统计图.
请根据图中信息，解答下列问题：
(1)求本次抽样调查的学生人数，请补全条形统计图；
(2)若该校共有500名学生，请求出“B(关注较多)”的学生人数；
(3)若“A(实时关注)”中有2名男生和2名女生，现从中随机抽取2人深入了解，请用树状图或列表法求出恰好抽到1名男生和1名女生的概率.
19. (本小题满分10分)
“周礼伤心凉粉”是安岳的一大美食，它不仅口感鲜美，而且制作工艺独特，传承历史悠久，被誉为四川的传统工艺之一.现有A，B两类“周礼伤心凉粉”特受顾客喜爱.已知购买2份A类和1份B类共需38元；购买4份A类和3份B类共需86元.
(1)分别求出 A，B两类“周礼伤心凉粉”每份的价格；
(2)芮芮家为了招待远道而来的客人，准备购买A，B两类“周礼伤心凉粉”共20份，且购买的总费用不超过250元，则最多能购买A类“周礼伤心凉粉”多少份?
20. (本小题满分10分)
如图7， 已知直线y=kx+b 与双曲线 y=3x交于A，B两点， 与x轴交于D点， 且A(1, m), B(n, 1).
(1)求直线AB的解析式；
(2)连接AO 并延长交双曲线于点 C，连接BC交x轴于点E, 求△CDE的面积.
21. (本小题满分11分)
如图8, 在△ABC中, ∠ACB=90°, D, O两点分别在边AB, AC上, 过C, D两点的⊙O与AC相交于点E, 连接DE, CD, ∠ADE=∠ACD.
(1) 求证: AB 是⊙O的切线;
(2)若 BC=6,tanB=43,求⊙O的半径.
22.(本小题满分11分)
风筝起源于中国，已有2000多年的历史，它象征着希望和祝福，而放风筝则可强身健体、愉悦身心. 阳春三月，小明和好友到郊外去放风筝，由于天公作美，风筝快速飞至点 P处 (如图9).爱动脑的小明准备测量此时风筝的高度，他立即从坡底A处沿坡度i=5：12的山坡AB走了 2345m到达坡顶B处，测得P处的仰角为45°；他又沿坡面 BC 走 30m 到达坡底 C 处, 测得 P 处的仰角为60°. (点A, B, C, P在同一平面内)
(1)求坡顶B处的高度；
(2)求风筝的飞行高度(即PD的长).
23. (本小题满分12分)
【探究发现】
(1)如图10-1, 已知A, D, B, F在同一直线上, 若∠CBF=∠BDE=∠CAE,则△ABC∽△EDA, 请证明;
【灵活运用】
(2) 如图10-2, 在 Rt△ABC中, ∠ACB=90∘,ACBC=12,点D在边AB上, AE⊥CD于点E, 连接BE. 若∠BED=45°, 求 ADBD的值；
【拓展延伸】
(3) 如图 10-3, 在四边形 ABCD 中, ∠ABC=120°, ∠ADB=60°, 若 BC=2AB =221,AD=1,求CD的长.
24. (本小题满分13分)
已知，抛物线 y=ax2+bx+3与x轴交于A(-1, 0), B(3, 0)两点, 与y轴交于C点.
(1)求抛物线的解析式；
(2)如图11, 点D 为抛物线上位于直线 BC上方的一点, DE⊥BC于点E, DF∥y轴交 BC于点 F，当△DEF的周长最大时，求点 D 的坐标；
(3)将抛物线 y=ax2+bx+3沿y轴向下平移，得到的新抛物线与y轴交于点 G，GP⊥y轴交新抛物线于点 P，射线 PO 与新抛物线的另一交点为 Q. 当PO=2OQ时，求点Q的坐标.
数学试题参考答案及评分意见
一、选择题（共 10 小题，每小题 4 分，共 40 分）
二、填空题（共 6 小题，每小题 4 分，共 24 分）
11 ．m ≥-2 12 ．36 13 ．1
14 ．( 1 ，-3) 15 ．50° 16 . ·3
三、解答题（共 86 分）
17． 解 ：原式= ···································································································· 4 分
1
= ·····································································································································7 分
x 一1
当 x =2 时，原式= 1. ······························································································································ 9 分
18．解 ：（ 1 ）此次抽样调查的人数：8÷20%=40（人），
“C（关注较少）”的人数：40-4-16-8= 12（人）.
补全条形统计图如下： ·························································································································3 分
（2 ）根据题意得：500× = 200（人），
则“B（关注较多）”的学生人数为 200 人. ······················································································ 5 分
（3 ）画图略. ········································································································································8 分
恰好抽到 1 个男生和 1 个女生的概率 . ·····································································10 分
19．解：（ 1 ）设 A 、B 两类“周礼伤心凉粉”每份的价格分别为 x 元、y 元，则
········································································································································································· 3 分
ly = 10.
解得 {[x = 14，···········································································································································5 分
答：A 、B 两类“周礼伤心凉粉”每份的价格分别为 14 元、10 元.
（2 ）设购买 A 类“周礼伤心凉粉”m 份，则
14m+10(20-m) ≤250····························································································································8 分
解之，得 m ≤ 12.5··································································································································9 分
“m 是整数，:m 的最大值为 12. 即最多能购买 A 类“周礼伤心凉粉”12 份. ····················· 10 分
20．解：（ 1 ）由题意，易知 A( 1 ，3)，B(3 ，1) · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · 1 分
:直线 AB 的解析式为：y=-x+4. 4 分
故得方程组{k++bb，，解得 { .1·, ················································································································································3 分
（2 ）由对称性得 C(-1 ，-3) ，把y =0 代入y=-x+4 得 x=4 ，:D(4 ，0)···························· 5 分
设直线 BC 的解析式为：y=k1x+b1 ，则 {l[ k一1 1 b1 , · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · 7 分
=1 = 一2
[k
lb1
:{ 1
, :直线 BC 的解析式为： y = x 一 2 ························································8 分
:E(2 ，0)···········································································································9 分
:S△CDE = 4 一 2)× 3 =3 · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · 10 分
21．（ 1 ）证明：连结 OD.
“OD=OC ， :上ACD =上ODC ，“ 上ACD =上ADE ， :上ADE =上ODC. ·································2 分
“CE 是⊙O 的直径 ， :上EDO+上ODC=90° , :上EDO+上ADE=90° . ········································4 分
:AB 是⊙O 的切线. ····························································································································5 分
（ 2 ）“BC=6 ，tanB= ，:AC=8，AB= 10. ··················································································· 6 分
易证 BC=BD=6 ，:AD=4. ··················································································································8 分
在 Rt△AOD 中，tan上AOD=tanB= ，··························································································· 10 分
: = ，:OD=3 ，即ΘO 的半径为 3. ····················································································11 分
22． 解：（ 1 ）过点 B 作 BE丄AC 于点 E.
设 BE=5x ，则 AE= 12x ···························································································································2 分
由勾股定理得 AB= 13x= . ···············································································································4 分
:x= ，:BE= 18. 即坡顶 B 处的高度为 18m. ·············································································5 分
（ 2 ）过点 B 作 BF丄PD 于点 F.
在 RtΔBCE 中 ，CE= ·BC2 - BE2 = ·302 - 182 =24. ······································································6 分
设 CD=a ，则 DE=BF=24+a. ···············································································································7 分
在 RtΔPBF 中 ， 由上PBF=45° , 得 PF=BF=24+a. ···········································································8 分
在 RtΔPCD 中 ，tan60°= ，: = ·、i3 ，:a= 21v3 + 21 . · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · 10 分
:PD= ·i3a = 63 + 21i3 . 即风筝的飞行高度为 (63 + 21i3 )m. · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · 11 分
23．（ 1 ）证明：“ 上CBF=上BDE ， :上ABC=上ADE. ········································································1 分
“ 上BDE=上EAD+上E ， 上CAE=上EAD + 上BAC ， 上BDE=上CAE ， :上BAC=上E. ················2 分
:△ABC ∞△EDA. ······························································································································ 3 分
（ 2 ）解 ：作 BF⊥CD 交 CD 的延长线于点 F.
由（ 1 ）同理可证△ACE ∞△CBF ， : = = = . ·························································· 4 分
设 CE=x ，则 BF=2x ，“ 上BEF=45° , :EF=BF=2x ， :CF=3x ，“CF=2AE ， :AE= x .
·································································································································································6 分
3
x
易证 AE ‖BF ，: = ，: = = . ·············································································7 分
（ 3 ）在线段 BD 上取点 E ，使上CED= 120° , 过点 C 作 CF⊥DB 交 DB 的延长线于点 F.
由（ 1 ）同理可证△ABD ∞△BCE，: = = = . “AD= 1，:BE=2，设 BF=x，则 EF=2+x.
········································································································································································· 8 分
在 RtΔCEF 中 ，易得上CEF=60° , :CF= (x + 2) ，CE= 2 (x + 2) . ··········································· 9 分
在 RtΔBCF 中，由勾股定理得：BF2+CF2=BC2 ，: x2 + ·i3 (x + 2)2 = (2s21)2 ，解得 x=3 或-6（舍
去） . ····························································································································································10 分
:EF=5 ，CE= 10 ，“ = ，:BD=5. · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · 11 分
在 RtΔCDF 中 ， 由勾股定理得： CD=
=
DF2 + CF2
82 +
5 3
( )
= ·139 . ······························12 分
24． 解：（ 1 ）由题意 ，得 3，······························································································1 分
.1，·············································································································································· 2 分
∴此抛物线的解析式为： y = -x2 + 2x + 3 . ······················································································· 3 分
（ 2 ）易求 C(0 ，3) ，又∵B(3 ，0) ， ∴yBC= -x + 3 ， ∠OCB=45° . ················································ 4 分
∵DF∥y 轴 ， ∴ ∠DFE=∠OCB=45° , ∴△DEF 为等腰直角三角形， ∴当边 DF 最大时，△DEF 的
周长最大. ······················································································································································5 分
设 D（ m ， -m2 + 2m + 3 ），则 F（ m ， -m + 3 ）.
∴DF= -m2 + 3m = - . ····································································································· 6 分
∴当 m= 时，DF 最大 ，此时 D（ ， ）. ················································································· 7 分
（ 3 ）设新抛物线的解析式为： y = -x2 + 2x + 3 - n ，则 G（ 0 ，3-n），P（ 2 ，3-n）. ······· 8 分
过点 Q 作 QD⊥y 轴于点 D.
易证△POG ∽△QOD ， ∴ = 2 ， ∴ OD = ，QD= 1. ······························ 10 分
①当点 G 在y 轴正半轴上时 = -1- 2 + 3 - n ， ∴n=1 ，Q
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②当点 G 在y 轴负半轴上时 = -1+ 2 + 3 - n ， ∴n=5 ，Q
综上所述 ，点 Q 的坐标为（-1 ，-1 ）或（ 1 ，-1 ）. ····································································13 分题号
1
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5
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答案
A
C
D
B
D
B
A
B
D
C