江苏省苏州市吴中、吴江、相城区2023年初三数学第三次模拟试卷(共3份word版pdf版含答案)

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江苏省苏州市吴中、吴江、相城区2023年初三数学第三次模拟试卷(共3份word版pdf版含答案)

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2022~2023学年初三年级模拟试卷
数学
2023.05
本卷由选择题、填空题和解答题组成,共27题,满分130分,调研时间120分钟.:
注意事项:
1答题前,考生务必将学校、班级、经名、调研号等馆息填写在答题卡相应的位置上,
2答选择题必须用2B铅笔把答题卡上对应题司的答案标号涂黑,如需改动,请用橡皮擦千净后,
再选涂其他答案;答非选择题必须用0.5毫米黑色墨水签字笔写在答题卡指定的位置上,不在
答题区域内的容案一律无效;如需作图,先用2D铅笔画出图形,再用05毫米黑色墨水签字
笔措,不得用其他笔答题,
3考生答题必须答在答题卡相应的位置上,答在试卷和草稿纸上一律无效,
一、选择题(本大题共8小题,每小题3分,共24分.在每小题所给出的四个选项中,恰有一项
是符合题目要求的,请将答案填涂在答题卡相应位置上)
1.下列四个实数中最大的是
A.-2
B.5
C.-1
2,苏州围绕打造“处处皆景、城在园中”的“公园城市”目标,扎实推进民生实事项目口袋公
园建设,2022年全年苏州各级园林绿化部门共投入资金145000000元进行新建、改建口袋公
园,为市民打造更多家门口的幸福.145000.000用科学计数法可以表示为
A.1.45×10°
B.14.5×10
C.1.45×10
D.0.145×10
3,下列运算正确的是
A.3a-a=3
B.a÷a=a.C.(a-b)2=d2-b2.D.(-ab)2=a32
4,如图,在△4BC中,AB=AC=8,BC=6,按以下步骤作图:第一步,以点A为圆心,.适当的长
为半径作弧,分别交AC,AB于红、N两点:第二步,分别以点M、N为圆心,大于1N的
长为半径作弧,两弧相交于点P;第三步,作射线AP,交BC于点E,则AE的长为
A.V55
B.8
C.√73
D.10
本班都分同学一个月的跟外阅读母折线统计图
人数人)
8
2
3
4阅读堡木)
(第4题)
第5题)
(第6题)
初三数学第1页共6
5,为激励青少年爱读书、读好书、普读书,某校积极开展全员阅读活动,小吴为了了解本班同
学一月的课外阅读量,随机选取班上部分同学进行调查,并将调查结果绘制成折线统计图(如
图)下列说法中,正确的是
A,随机选取了14名同学
B.中位数是2木
C.众数是4本
D.平均数是2.4本
6.如图,正方形ACD内接于⊙O,现有-一小球可在⊙O内自由滚动,.则小球停留在阴影部分内
(各图形的边界忽略不计)的概率是
A告
B.2W2
c.2
D.2
7,定义:两个不相交的函数图象在平行于y轴方向上的最短距离称为这两个函数的“完美距离”,
抛物线y=2x25x+3与直线y=-2x-1的“完美距离”为
A,号
B.3
c.
27
D.
2丛
8.如图I,点E为矩形ABCD中AD边的中点,点P从点A出发,.沿A→E-→B以2cs的速度
运动到点B,图2是点P运动时,△PBC的面积y(Cm)随时间t(S)变化的函数图象,则a的
值为
A.5
B.4
C.3
D.2
以cm木
12a
图1
a+5
s
图2
(第8题)
二、填空题(本大题共8小题,每小题3分,共24分,不需写出解答过程,请把答案直接填写在
答题卡相应位置上)
9.计算:十引_▲
10.若代数式-4有意义,则x的取值范为▲
2
11.若m=V2-1,则m+2m+1=▲
I2.如图,AB是⊙O的直径,弦CD交⊙O于点C,D,.连接BD
D
若∠A=34°,∠AED=87°,则∠B=▲°.
(第12题)
13.已知圆锥底而损直径为18cm,母线长为15ctn,该圆锥铡面展开图扇形的圆心角度数为▲°,
14.关于x的元二次方程x2+(叶4)x+3叶3=0有一个大于-2的非正数根,那么实数a的取值范
固是▲
初三数学第2.页共6页2022~2023 学年初三年级模拟试卷
数学参考答案 2023.05
一、 选择题 (每小题 3 分,共 24 分)
题号 1 2 3 4 5 6 7 8
答案 B C D A D C A B
二、 填空题 (每小题 3 分,共 24 分)
9. 2 10. x 4 11. 2 12.53
5
13.216 14. 1 a 1 15. 6 30( , ) 16.6 2 3
13 13
三、解答题(共 11 小题,共 82 分)
17.(本题满分 5 分)
解:原式= 2 3 1 ·······················································································3 分
=1 3 . ························································································5 分
18.(本题满分 5 分)
解:方程两边同乘以 (x 1)(x 1),得: x(x 1) (x 1)(x 1) 3. ··························2 分
解方程,得: x 2. ············································································4 分
经检验, x 2是原方程的解. ······························································5 分
19.(本题满分 6 分)
解:原式= 2x2 2x 1 x2 x ············································································1 分
2
= x 2 x 1 . ···················································································3 分
2
x2 x 1 0, x2 x 1. ·······································································4 分
原式 (x2 x) 1 3 . ···············································································6 分
2 2
20.(本题满分 6 分)
解:(1) 1 ; ··································································································2 分
5
(2)树状图或表格(略); ·············································································4 分
一人选择“A:步行”,另一人选择“C:乘坐公共交通”的概率为 2 .············6 分
25
21.(本题满分 6 分)
解:(1)在△ACE 和△ BD E 中,
C D∵ AEC BED ,∴△ACE ≌△BDE .······················································2 分
AC BD
(2) △ACE≌△BDE ,
AE BE, ABC BAD , AEC 2 ABC. ···········································4 分
C 90 , AEC 90 CAE 64 . ABC 1 AEC 32 . ·····················6 分
2
九年级数学参考答案及评分标准 第 1 页(共 4 页)
22.(本题满分 8 分)
解:(1) 9 ; ·································································································2 分
(2)C 组所在扇形圆心角为: 10 360 =72 ; ·····················································4 分
50
(3) 28 650=364 (人). ················································································8 分
50
答:一周内家务劳动总时间不少于 8 小时的人数为 364 人.
23.(本题满分 8 分)
解:(1)设 4A(m,4) ,则E(m 4,).
5
把 4A(m 4,4) , kE(m 4, )代入 y ,得 4m (m 4) . ···································2 分
5 x 5
m 1. ··························································································3 分
把 A(1,4) 代入 y k ,得 k 4. ··································································4 分
x
(2)设 P(5,n),
4
PE n .·························································································5 分
5
S 1 APE AD PE 3 , 1 4 PE 2 PE 3 , 2 2 2 2
PE n 4 3 , n 31或 1 . ·····························································6 分
5 4 20 20
点P的坐标为 (5,31) 或 (5,1 ).·································································8 分
20 20
注:第二小题漏掉一解扣 2 分.
24.(本题满分 8 分)
解:(1)设每件 A 种礼盒成本价为 x 元,每件 B 种礼盒成本价为 y 元.
根据题意,得 x 2y 320 ··································································2 分

2x 3y 540
解方程组,得 x 120 ········································································4 分

y 100
答:每件 A 种礼盒成本价为 120 元,每件 B 种礼盒成本价为 100 元.
(2)设 A 种礼盒售出 m 件,B 种礼盒售出 n 件.
根据题意,得:30m 20n 1320 ,
∴ 132 2nm . ······················································5 分
3
m n 56 , 132 n 56 ,得n 36. ······················································6 分
3
m,n均为正整数, n 35时,m 62不符合题意;
3
n 34时,m 64不符合题意;n 33时,m 22符合题意. ··································7 分
3
B种礼盒最多卖出33件.··········································································8 分
25.(本题满分 10 分)
证明:(1)∵OC AO,∴ A ACO.
∵CD平分 ACE ,∴ ACO FCO ,∴ A FCO.··································2 分
∵ A E ,∴ E FCO.
∴BE ∥CD.····················································································4 分
九年级数学参考答案及评分标准 第 2 页(共 4 页)
解:(2)∵BE ∥CD, B COF.
∵ B ACF ,∴ COF ACF . ······························································5 分
∵ AC AF , ACF CFO.
COF CFO , CF CO AO 2 ··························································6 分
CFO AFC, FCO A.
△CFO∽△AFC ,
CF FO
, FO 2 2FO 4 0. ········································································8 分
AF FC
∴ FO 5 1.·························································································9 分
∴BF BO OF 3 5. ·········································································10 分
方法不唯一,其他方法酌情给分.
26.(本题满分 10 分)
解:(1)把 A( 1,0) 代入 y x2 4ax 3a ,得a 1. ··············································1 分
∴抛物线的解析式为 y x2 4x 3.··························································2 分
(2)当 y 0时, x2 4x 3 0, x1 1,x 3.2 B( 3, 0).
当 x 0时, y 3, C(0,3)
y x 2 4x 3 (x 2)2 1 , D( 2, 1) . ···················································3 分
过点 D 作 DE⊥AB 于 E,连结 BD.
DE BE 1, DEB 90 , DBE 45 , BD 2.
同理可得: CBO 45 ,CB 3 2.
CBD DBE CBO 90 .
BD AO 1 , CBD COA 90 , CBD COA.
BC CO 3
ACO DCB. ·················································································5 分
∵当点 P 与点 D 重合时,符合题意.∴P( 2, 1) . ·····································6 分
延长线段 DB 至点 Q,使得 DB=QB,连结 CQ 交抛物线于点 P.
DB QB,CB BD, CQ CD .
又 CB BD, PCB DCB ACO .
B 是DQ 中点,B( 3,0),D( 2, 1), Q( 4,1) .
设 CQ 解析式为 y kx b(k 0).把C(0,3),Q( 4,1)代入 y kx b(k 0) ,得 4k b 1. b 3
解方程组,得 k 1 1
2 , y x 3. y
b 3 2
2
将 y x2 14x 3, y x 3联立,得
y x 4x 3

2 y
1 x 3 C
2
P
7 Q
解方程组,得 x x2 7 5 . 1 0 , 2 , P ( , ) Ey1 3 y 5 2 4 B A O x 2 4
D
九年级数学参考答案及评分标准 第 3 页(共 4 页)
∵ P( 2, 1)或P( 7 , 5) ·········································································8 分
2 4
方法不唯一,其他方法酌情给分.
(3)m 3 3的取值范围为: m .··························································10 分
2 2
27.(本题满分 10 分)
解:(1) ∵四边形 ABCD 是矩形,
∴AD=BC=8,AB=DC=6,∠ADC=∠C=90°. ∴∠ADG +∠FDC=90°.
∵AE⊥DF,∴∠EAD +∠ADG=90°, ∴∠FDC=∠EAD. ························1 分
又∵∠C=∠ADC,∴△DFC∽△AED.
∴ DF DC 3 .················································································2 分
AE AD 4
A
(2)过点 B,C 分别作 BG⊥AB 于 B,CG⊥AC 于 C,BG 与 CG 交于点 G,
延长 AF 交 CG 于点 H. DE
由(1)可知: AE BD, AD AB 3. 设 AB=3k,则 AC=4k.CH AC 4 B CF
点D为AC的中点, AD DC 2k.
H
在 Rt ABD 中,BD AB2 AD2 13k. G
AD 3 2k 3 8k , , CH . ·······························································4 分
CH 4 CH 4 3
在 Rt ACH 中, AH AC 2 CH 2 4 13k . ···············································5 分
3
ABG BAC ACG 90 , 四边形ABGC是矩形.
AB AF
AB CG, , 3k AF , AF 12 13 ·····························7 分k .
CH FH 8k 4 13 17
3 k AF3
AF 12
. ························································································8 分
BD 17
(3) CF 24 . ···························································································· 10 分
DE 25
九年级数学参考答案及评分标准 第 4 页(共 4 页)2022~2023学年初三年级模拟试卷
数 学 2023.05
本卷由选择题、填空题和解答题组成,共27题,满分130分,调研时间120分钟.
注意事项:
1.答题前,考生务必将学校、班级、姓名、调研号等信息填写在答题卡相应的位置上.
2.答选择题必须用2B铅笔把答题卡上对应题目的答案标号涂黑,如需改动,请用橡皮擦干净后,再选涂其他答案;答非选择题必须用0.5毫米黑色墨水签字笔写在答题卡指定的位置上,不在答题区域内的答案一律无效;如需作图,先用2B铅笔画出图形,再用0.5毫米黑色墨水签字笔描黑,不得用其他笔答题.
3.考生答题必须答在答题卡相应的位置上,答在试卷和草稿纸上一律无效.
一、选择题(本大题共8小题,每小题3分,共24分.在每小题所给出的四个选项中,恰有一项是符合题目要求的,请将答案填涂在答题卡相应位置上)
1.下列四个实数中最大的是
A.-2 B. C.-1 D.
2.苏州围绕打造“处处皆景、城在园中”的“公园城市”目标,扎实推进民生实事项目口袋公园建设.2022年全年苏州各级园林绿化部门共投入资金145000000元进行新建、改建口袋公园,为市民打造更多家门口的幸福.145000000用科学计数法可以表示为
A.1.45×109 B.14.5×107 C.1.45×108 D.0.145×109
3.下列运算正确的是
A.3a-a=3 B.a6÷a2=a3 C.(a-b)2=a2-b2 D.(-a3b)2=a6b2
4.如图,在△ABC中,AB=AC=8,BC=6,按以下步骤作图:第一步,以点A为圆心,适当的长为半径作弧,分别交AC,AB于M、N两点:第二步,分别以点M、N为圆心,大于MN的长为半径作弧,两弧相交于点P;第三步,作射线AP,交BC于点E.则AE的长为
A. B.8 C. D.10
5.为激励青少年爱读书、读好书、善读书,某校积极开展全员阅读活动.小吴为了了解本班同学一月的课外阅读量,随机选取班上部分同学进行调查,并将调查结果绘制成折线统计图(如图)下列说法中,正确的是
A.随机选取了14名同学 B.中位数是2本
C.众数是.4本 D.平均数是2.4本
6.如图,正方形ABCD内接于⊙O,现有一小球可在⊙O内自由滚动,则小球停留在阴影部分内(各图形的边界忽略不计)的概率是
A. B. C. D.
7.定义:两个不相交的函数图象在平行于y轴方向上的最短距离称为这两个函数的“完美距离”.
抛物线y=2x2-5x+3与直线y=-2x-1的“完美距离”为
A. B.3 C. D.
8.如图1,点E为矩形ABCD中AD边的中点,点P从点A出发,沿A→E→B以2cm/s的速度运动到点B,图2是点P运动时,△PBC的面积y(cm2)随时间t(s)变化的函数图象,则a的值为
A.5 B.4 C.3 D.2
二、填空题(本大题共8小题,每小题3分,共24分.不需写出解答过程,请把答案直接填写在答题卡相应位置上)
9.计算: = _____.
10.若代数式有意义,则x的取值范围为_____.
11.若m=-1,则m2+2m+1=_____.
12.如图,AB是⊙O的直径,弦CD交⊙O于点C,D,连接BD.
若∠A=34°,∠AED=87°,则∠B=_____°.
13.已知圆锥底面圆直径为18cm,母线长为15cm,该圆锥侧面展开图扇形的圆心角度数为_____°.
14.关于x的一元二次方程x2+(a+4)x+3a+3=0有一个大于-2的非正数根,那么实数a的取值范围是_____.
15.如图,直线AB与x轴,y轴分别交于点A(-2,0),B(0,3),已知点C坐标为(3,0),点P是线段AB(不与点A,B重合)上一点;连结线段PC,PO.若∠CPO=45°,则点P坐标为_____.
16.如图,在△ABC中,∠ACB=90°,AC=BC,以AC为边在△ABC下方作△ADC,连接BD,已知AD=3,DC=6,则BD的最大值为_____.
三、解答题(本大题共11小题,共82分.请在答题卡指定区域内作答,解答时应写出文字说明、证明过程或演算步骤)
17.(本题满分5分)
计算: °-(π-1)0.
18.(本题满分5分)
解方程: ·
19.(本题满分6分)
已知x2+x-1=0,求(2x+1)2-x(x+1)的值.
20.(本题满分6分)
为缓减校园周边道路的交通压力,及时调整学生上学时间,某校需要了解本校学生的上学方式,学生可以从“A:步行,B:骑自行车,C:乘坐公共交通,D:家用汽车接送,E:其他方式”五个选项中进行选择.
(1)学生甲随机选择“C:乘坐公共交通”方式的概率为_____.
(2)若两名学生分别从A,B,C,D,E五种上学方式中随机选择一种,求两名学生一人选择“A:
步行”,另一人选择“C:乘坐公共交通”的概率(请用画树状图或列表等方法说明理由).
21.(本题满分6分)
如图,AD,BC交于点E,AC=BD,∠C=∠D=90°.
(1)求证:△ACE≌△BDE;
(2)若∠CAE=26°,求∠ABC的度数.
22.(本题满分8分)
适当的劳动对青少年的成长和发展具有十分重要的意义.为了解八年级学生每周家务劳动的总时长,某校数学社团成员采用随机抽样的方法,抽取了八年级部分学生,对他们一周内家务劳动总时间t(单位:小时)进行了调查,并将数据整理后得到下列不完整的统计图表:
请根据图表信息回答下列问题:
(1)频数分布表中,a=_____;
(2)扇形统计图中,C组所在扇形的圆心角的度数是_____°;
(3)请估计该校650名八年级学生中一周内家务劳动总时间不少于8小时的人数.
23.(本题满分8分)
如图,反比例函数(k≠0,x>0)的图像经过边长为4的正方形ABCD的顶点A,与正方形的边CD交于点E,且EC=.
(1)求k的值;
(2)若点P是正方形CD边上不与点E重合的点,连接AE,
AP,当△APE的面积为时,求点P的坐标.
24.(本题满分8分)
为迎接五一假期的到来,某景区一商户准备了两种当地特产礼盒,按成本价1件A种礼盒和2件B种礼盒共需320元,2件A种礼盒和3件B种礼盒共需540元.
(1)求A、B两种礼盒每件的成本价分别是多少元?
(2)若A种礼盒的售价为每件150元,B种礼盒的售价为每件120元.商户原计划在五一当天将现有的A、B两种礼盒共56件按售价全部售出,但在实际销售过程中56件商品没有全部售完,两种礼盒的实际销售利润总和为1320元.五一当天商户最多卖出B种礼盒多少件?
25.(本题满分10分)
如图,已知AB,CD是⊙O的两条直径,直径CD平分∠ACE,∠ACE的一边CE与⊙O和直径AB分别交于点E,F,连接BE,且AC=AF.
(1)证明:BE//CD;
(2)若CF=2,求BF的长.
26.(本题满分10分)
如图,抛物线y=x2+4ax+3a(a是常数且a≠0)与x轴交于点A,B两点(点A位于点B右侧),与y轴交于点C,点D为抛物线的顶点,且点A的坐标为(-1,0),连结AC,BC,CD.
(1)求该抛物线的表达式:
(2)若点P为抛物线上的点,连结CP,当∠ACO=∠PCB时,求点P的坐标;
(3)若在x轴上总存在一点Q,且点Q的横坐标为m(m>-3),
当∠DCB<∠QCB<∠CAO时,直接写出m的取值范围.
27.(本题满分10分)
[问题探究]
课外兴趣小组活动时,同学们正在解决如下问题:
如图1,在矩形ABCD中,点E,F分别是边DC,BC上的点,连结AE,DF,且AEDF于点G,若AB=6,BC=8,求的值.
(1)请你帮助同学们解决上述问题,并说明理由.
[初步运用]
(2)如图2,在△ABC中,∠BAC=90°,,点D为AC的中点,连结BD,过点A作AEBD于点E,交BC于点F,求的值.
[灵活运用]
(3)如图3,在四边形ABCD中,∠BAD=90°,,AB=BC,AD=CD,点E,F分别在边AB,AD上,且DECF,垂足为G,则 .
组别 家务劳动总时间分组 频数
A t<6 5
B 6≤t <7 7
C 7≤t<8 10
D 8≤t<9 19
E t≥9 a
1

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