资源简介 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.-2B.5C.-12,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.14.5×10C.1.45×10D.0.145×103,下列运算正确的是A.3a-a=3B.a÷a=a.C.(a-b)2=d2-b2.D.(-ab)2=a324,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.V55B.8C.√73D.10本班都分同学一个月的跟外阅读母折线统计图人数人)8234阅读堡木)(第4题)第5题)(第6题)初三数学第1页共65,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.2W2c.2D.27,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.3c.27D.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.5B.4C.3D.2以cm木12a图1a+5s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▲211.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BDD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 12.53513.216 14. 1 a 1 15. 6 30( , ) 16.6 2 313 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2 220.(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2521.(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 5m 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;3n 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 22将 y x2 14x 3, y x 3联立,得 y x 4x 3,2 y 1 x 3 C 2P 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 4D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2 227.(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 4A(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 13 173 k AF3AF 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.0.145×1093.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=a6b24.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.105.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 5B 6≤t <7 7C 7≤t<8 10D 8≤t<9 19E t≥9 a1 展开更多...... 收起↑ 资源列表 初三数学参考答案.pdf 苏州市吴中、吴江、相城区2023年初三数学第三次模拟试卷.doc 苏州市吴中、吴江、相城区2023年初三数学第三次模拟试卷.pdf