资源简介 建瓯市 2025—2026 学年第二学期八年级期中练习卷数学(考试时间:120 分钟 满分:150 分)一、单选题(本题共 10 小题,每小题 4 分,共 40 分)1.下列式子中,是最简二次根式的是( )A. 25 B. 17 C. 0.49 D 2.52.若 5 x在实数范围内有意义,那么 x的取值范围是( )A. x 5 B. x 5 C. x 5 D. x 53.如果下列各组数是三角形的三边长,那么能组成直角三角形的一组数是( )A.2,2,3 B.2,3,4 C.6,7,8 D. 5,12,134.下列计算,正确的是( )A. 3 5 8 B. ( 2)2 2C. 12 4 3 D. 4 9 2 35.如图,四边形 ABCD是菱形,对角线 AC,BD相交于点 O.点 E是 CD的中点,连接 OE,若OE 2,则 AB的长是( )A.2 B.3 C.4 D.5(第 5 题图) (第 7 题图)6.下列命题中正确的是( )A.对角线相等的四边形是矩形B.有一组邻边相等的四边形是菱形C.有三个角是直角的四边形是矩形D.一组对边相等,另一组对边平行的四边形是平行四边形7.如图,在正方形 ABCD的外侧,作等边三角形 ADE,则 AEB为( )A.10 B.15 C. 20 D.30 试卷第 1页,共 7页8.如图,EF过平行四边形 ABCD对角线的交点 O,交 AD于点 E,交 BC于点 F.若平行四边形 ABCD的周长为 10,OE 1.5,则四边形 EFCD的周长为( )A.6 B.7 C.8 D.99. 我国汉代数学家赵爽利用一幅“弦图”,证明了勾股定理,后人称该图为“赵爽弦图”.如图,“赵爽弦图”是用 4个全等的直角三角形与 1个小正方形镶嵌而成的正方形图案.如果该大正方形面积为 49,小正方形面积为 4,用 a,b表示直角三角形的两直角边(b>a),下列四个推断:① a2 b2 49;② 2ab 45;③b a 2.其中所有正确推断的序号是( ).A.①② B.②③ C.①③ D.①②③(第 8题图) (第 9 题图) (第 10 题图)10.如图,四边形 OABC是矩形,点 B是 y轴正半轴上一点,AC=5,A (2,1),点 C在第二象限,则点 C的坐标是( ) 9 A. 2,4 B. 2, C. 1,4 D. 3,3 2 二、填空题(本题共 6 小题,每小题 4分,共 24 分)11. 18 _______.12.如图,图形中 x的值为________.13.如图,在△ABC中, ACB 90 ,D为 AB中点,且 CA=CD,则 A的度数。为_______ .(第 12 题图) (第 13 题图)试卷第 2页,共 7页14.如图,数轴上的点 A表示的数是 -1,点 B表示的数是 1,CB AB,垂足为点 B,且 BC=2,以点 A为圆心,AC为半径画弧交数轴于点 D,则点 D表示的数为______.(第 14题图) (第 15题图) (第 16题图)15.如图,为美化环境,某小区从一块正方形空地中划出两块面积分别为 12m2和27m2的小正方形种植特色花卉,剩余部分种植草坪,则种植草坪部分的面积为______m2.16.如图,直角三角形 ABC两直角边长分别为 5和 12,以直角三角形的三边为直径作半圆,则图中两个月形图案(阴影部分)的面积之和为________.三、解答题(本题共 9小题,共 86 分.解答应写出文字说明、证明过程或演算步骤)17.(本题满分 8分)1计算: 45 5 15 20318.(本题满分 8分)如图,将平行四边形 ABCD的对角线 BD向两个方向延长,分别至点 E,F,且使 BE=DF.求证:四边形 AECF是平行四边形.试卷第 3页,共 7页19.(本题满分 8分)如图,若点 P是△ABC边 BC上的一个动点,已知 AB=6,AC=8,BC=10,求线段 AP的最小值.20.(本题满分 8分)如图,长方形纸片 ABCD中,沿 AC折叠,使点 B落在点 E处,CE交 AD于点F,AB=4,BC=6.(1)证明:AF=CF;(2)求 AF的长.21.(本题满分 8分)如图,在平行四边形 ABCD中,CE AB,垂足为点 E,CF AD,垂足为点 F,且 BE DF.(1)求证:四边形 ABCD是菱形;(2)若 BCD 110 ,求 ECF 的度数.试卷第 4页,共 7页22.(本题满分 10分)如图,点 E是矩形 ABCD的边 BC上一点,且 AE=AD.(1)尺规作图:在 BC的延长线上找到一点 F,连接 DF,使得四边形 AEFD是菱形;(保留作图痕迹,不写作法)(2)在(1)的条件下,连接 DE,若 DE=8,且四边形 AEFD的周长为 32,求 BE的长.23.(本题满分 10分)高空抛物被称为 “悬在城市上空的痛”,哪怕是微小的物品,从高空落下也可能造成严重伤害.不考虑空气阻力时,物体自由下落的时间 t(单位:秒)与h下落高度 h(单位:米)近似满足公式 t .物体下落时的冲击能量 E(单5位:焦耳)满足公式 E=10×物体质量(千克)×下落高度(米).科学常识:冲击能量达到 20焦耳时足以造成颅骨骨折,达到 65焦耳时会对无防护人体造成致命伤害.(1)一个装有水总质量为 0.2 千克的塑料水瓶从 45米高的阳台自由落下,求它落到地面的时间,并通过计算说明这个水瓶是否会对人体造成致命伤害;(2)实验表明:一个质量为 0.07千克的鸡蛋,从 11楼窗台落下产生的冲击能量足以造成颅骨骨折.求这个鸡蛋至少需要从第几层窗台落下,就会对人体造成致命伤害?(已知该小区每层楼高 3米,窗台比所在楼层地面高 1米,1楼地面高度为 0米).试卷第 5页,共 7页24.阅读材料(本题满分 12分)BC AC黄金分割点:若线段 AB上的点 C 满足 (其中 AC>BC),则称点 CAC ABAB AC 5 1为线段 的黄金分割点,比值 称为黄金比.AB 2黄金矩形:宽与长的比值等于黄金比的矩形,称为黄金矩形.(1)验证黄金分割点 (2)构造并验证黄金矩形沿用(1)中得到的线段 AB及它的已知线段 AB,按以下步骤进行尺 黄金分割点 C,继续完成以下操作规作图(对应图 1): (对应图 2):①过点 B作 DB AB①以 AB为边,在 AB的上方作正方,垂足为点1 形 ABMN;B,使得DB AB,连接 AD;2 ②连接正方形的对角线 BN;②以点 D为圆心,BD的长度为半 ③过点 C作CK AB,垂足为点 C,径画弧,交 AD于点 E; 线段 CK分别交 BN、MN于点 P、③以点 A为圆心,AE的长度为半 K;径画弧,交 AB于点 C(得到的点 ④过点 P作线段 QL//AB,分别交C 满足 AC>BC). AN、BM 于点 Q、L,得到四边形ACPQ.试卷第 6页,共 7页请结合上述材料,完成下列问题:(1)如图,已知线段 AB=4,点 C是线段 AB的两个黄金分割点(其中 AC>BC),点 D是线段 AB的中点,求线段 CD的长;(2)请结合上述作图步骤 1,证明图 1中的点 C是线段 AB的黄金分割点;(3)请结合上述作图步骤 2,证明图 2中的四边形 ACPQ是黄金矩形.25.(本题满分 14分)如图 1,在△ABC中,D为 BC的中点,E为 BA延长线上一点,连接 DE交AC于点 G,过点 D作 DF⊥DE,垂足为点 D,DF交 AC的延长线于点 F.延长 ED至点 H,使 DH=DE,连接 CH、FH.已知FH 2 EB2 CF 2.(1) 求证:∠BAC=90°;(2) 如图 2,当∠ACB=45°时:① 求证:AE=CF;② 若 BE=2CF,BC 8 2,求线段 AG的长.试卷第 7页,共 7页建瓯市 2025-2026 学年第二学期八年级期中练习卷数学试题参考答案及评分说明说明:(1) 解答右端所注分数为考生正确做完该步应得的累计分数,全卷满分 150分.(2) 对于解答题,评卷时要坚持每题评阅到底,勿因考生解答中出现错误而中断本题的评阅.当考生的解答在某一步出现错误时,如果后续部分的解答未改变该题的考试要求,可酌情给分,但原则上不超过后面应得分数的一半,如果有较严重的错误,就不给分.(3) 若考生的解法与本参考答案不同,可参照本参考答案的评分标准相应评分.(4) 评分只给整数分.选择题和填空题不给中间分.一、选择题(本大题共 10小题,每小题 4分,共 40分)1.B; 2.A; 3.D; 4.D; 5.C;6.C; 7.B; 8.C; 9.D; 10.A.10.解:过点 C作CE y轴于点 E,过点 A作 AF y轴于点 F,则 CEB AFO 90 ∵四边形 OABC是矩形∴OB=AC=5,BC=OA,BC∥OA∴ CBE AOF∴△BCE≌△OAF(AAS)∴CE=AF,BE=OF∵A(2,1)∴CE=AF=2,BE=OF=1∴OE=OB-BE=5-1=4∵点 C在第二象限∴点 C的坐标是(-2,4). 故选 A二、填空题(本大题共 6小题,每小题 4分,共 24分)11.3 2; 12.95; 13.60;14. 2 2 1; 15.36; 16.30.三、解答题(本大题共 9小题,共 86分)17.(本题满分 8分)45 5 1解:原式 15 4 5 ·····························3分3 9 5 2 5 ·········································· 6分 3 5 ····························································8分答案第 1页,共 7页18.(本题满分 8分)解法一:证明:连接 AC,设 AC与 BD相交于点 O∵四边形 ABCD是平行四边形∴OA=OC,OB=OD····························································4分∵BE=DF∴OB+BE=OD+DF即 OE=OF············································································6分又∵OA=OC∴ 四边形 AECF是平行四边形············································· 8分解法二:证明:∵四边形 ABCD是平行四边形∴AB∥CD,AB=CD··························································· 2分∴ ABD CDB∴180 ABD 180 CDB即 ABE CDF ································································· 3分在△ABE和△CDF中 AB CD ABE CDF BE DF∴△ABE≌△CDF(SAS)······················································ 6分∴ AEB CFD , AE CF∴AE∥CF∴四边形 AECF是平行四边形··················································8分19.(本题满分 8分)解: ∵AB=6,AC=8,BC=10∴ AB2 AC 2 62 82 100 ··························································1分BC 2 102 100 ······································································2分2 2 2∴ AB AC BC ····························································· 3分∴△ABC是直角三角形且 BAC 90 ····································· 4分答案第 2页,共 7页当 AP BC时,AP的值最小AB AC 6 8 24∴ AP BC 10 524∴线段 AP的最小值为 ························································· 8分520.(本小题满分 8分)(1)证明:∵四边形 ABCD是长方形∴AD∥BC∴∠FAC=∠BCA·································································· 1分由折叠的性质可知:△ ABC≌△ AEC∴∠BCA=∠ECA·································································· 2分∴∠FAC=∠ECA·································································· 3分∴AF=CF············································································ 4分(2)设 AF=x∵四边形 ABCD是长方形∴AD=BC=6CD=AB=4∠D=90°∴FD=AD AF=6 x································································5分由(1)知 AF=CF∴CF=x在 Rt△CDF中,根据勾股定理:∴CD2 DF 2 CF 2 ······························································ 6分42 6 x 2 x213解得 x ··········································································7分313∴AF= ···············································································8分321.(本小题满分 8分)解: (1) 证明:∵四边形 ABCD是平行四边形,∴∠B=∠D·············································································1分∵CE⊥AB,CF⊥AD∴∠CEB=∠CFD=90°答案第 3页,共 7页在△CEB和△CFD中: B D CEB CFD BE DF∴△ CEB≌△ CFD ···································································· 2分∴BC=CD···············································································3分又∵四边形 ABCD是平行四边形∴平行四边形 ABCD是菱形·······················································4分(2)∵四边形 ABCD是平行四边形∴∠A=∠BCD=110° ································································5分∵CE⊥AB,CF⊥AD∴∠AEC=∠AFC=90°在四边形 AECF中,内角和为 360° ············································6分∠ECF=360° ∠A ∠AEC ∠AFC···············································7分=360° 110° 90° 90°=70° ···········································································8分22.(本小题满分 10分)解:(1) 如图所示,菱形 AEFD即为所求···············································5分(2)∵四边形 AEFD是菱形,周长为 32∴AE=AD=EF=DF=8································································· 6分∵DE=8∴DE=DF=EF=8∴△DEF是等边三角形····························································· 7分∵四边形 ABCD是矩形∴∠BCD=90°,BC=AD=8···························································8分即 DC⊥EF,由三线合一得BC 1∴ EF 4 ······································································ 9分2∴BE=BC-EC=4······································································ 10分答案第 4页,共 7页23.(1) h 45mh 45 t 9 3s...............................................................................................2分5 5 E 10 0.2 45 9(0 焦耳) 90 65...........................................................................................................................3分 它落到地面的时间为3s,这个水瓶会对人体造成致命伤害........................................4分(2)设这个鸡蛋至少需要从第x层窗台落下则h 3(x 1) 1 3x 2....................................................................................5分 E 10 0.07 (3x 2) 2.1x 1.4......................................................................................................6分当E 65时即2.1x 1.4 65....................................................................................................7分解得x 3113 .................................................................................................8分21 x为正整数 x 32....................................................................................................................9分 这个鸡蛋至少需要从第32层窗台落下,就会对人体造成致命伤害................10分24(. 1) AB 4,点C是线段AB的黄金分割点.AC 5 1 AB 2 AC 2 5 2......................................................1分 点D是线段AB的中点 AD 1 AB 2.....................................................2分2 CD AC AD 2 5 4....................................3分(2)解:设DB x DB 1 AB2 AB 2DB 2x.....................................................4分 DB AB AD DB2 AB2 5x......................................5分 DE DB x AC AE AD DE ( 5 1)x...........................6分AC ( 5 1)x 5 1 AB 2x 2 点C是线段AB的黄金分割点...................................7分答案第 5页,共 7页(3) 四边形ABMN是正方形 NAB MBA 90 , MBN ABN 45 CK AB ACP BCP 90 QL // AB QPC CPL PLB 90 四边形AQPC和四边形PCBL是矩形...............................9分 PB PB PCB PLB(AAS ) CB LB 四边形PCBL是正方形......................................................11分 AQ PC BC (2) BC AC 5 1由 知 AC AB 2AQ 5 1 AC 2 四边形ACPQ是黄金矩形.....................................................12分25.(1) D是BC的中点 CD BD CDH BDE,DH DE CDH BDE(SAS )...................................................................1分 CH EB, HCD EBD CH // EB.........................................................................................2分 FH 2 EB2 CF 2 FH 2 CH 2 CF 2 FCH是直角三角形,即 FCH 90 ............................................3分 BAC FCH 90 .........................................................................4分(2) 连接AD BAC 90 ,D是BC的中点 AD 1 BC CD BD......................................................................5分2 ACB 45 FCD 180 ACB 135 , B 90 ACB 45 DAB B 45 EAD 180 DAB 135 , ADB 180 DAB B 90 .......6分答案第 6页,共 7页 FCD EAD, ADC 180 ADB 90 DF DE GDF 90 FDC GDF GDC, EDA ADC GDC FDC EDA.......................................................................................8分 FCD EAD(ASA) AE CF ...................................................................................................9分 BC 8 2 由勾股定理得AB AC 8 由 知AE CF ,BE 2CF AE AB AC CF 8...........................................................................10分 DA是 CAB的角平分线 点D到直角边AB, AC的距离均为4S 1 ADC AC 4 16,S1 DAE AE 4 16................................................11分2 2 S CGD S ADC S ADG ,S EAG S DAE S ADG S EAG S CGD ..............................................................................................12分设AG y,则CG 8 y1 1 AE AG CG 42 2y 8解得 38 线段AG的长为 ...........................................................................................14分3答案第 7页,共 7页 建瓯市 2025-2026 学年第二学期八年级期中练习卷 18.(8分) 20.(8分)数学 答题卡 解:(1) 缺考标志。考生严禁填涂,由监考教师填涂。姓名:________________班级:________________座位号:______________注 意 事 项1.答题前,考生先将自己的姓名、班级、座位号填写清楚。2.考生作答时,按照题号顺序在答题卡各题目的答题区域内作答,超区域书写的 (2)答案无效;在草稿纸、试卷上答题无效。3.选择题部分使用 2B 铅笔填涂;非选择题部分用 0.5 毫米黑色签字笔书写,字体工整、笔迹清楚。选择题修改时用橡皮擦干净后,再选涂其他答案标号;非选择题答题区域修改禁用涂改液和不干胶条。4.保持卡面清洁,不折叠、不破损。考试结束后,将答题卡交回。5.正确的填涂示例:正确▄一、选择题(本大题共 10小题,每小题 4分,共 40分)1 6 2 7 3 8 4 9 5 10 二、填空题(本大题共 6小题,每小题 4分,共 24分)21.(8分)11 12 19.(8分). . . .解:(1)13. . 14. .15. . 16. .三、解答题(本大题共 7小题,共 86分)注意:作图或画辅助线可先用铅笔画,确定后再用 0.5 毫米的黑色签字笔画好17.(8分)1 (2)计算 : 45 5 15 203数学 答题卡 第 1 页 共 1 页 22. (10分) 24. (12分) 25. (14分)解:(1) 解:(1) 解:(1)(2) (2)(2)123. (10分)解:(1)(3)②(2)数学 答题卡 第 2 页 共 2 页 展开更多...... 收起↑ 资源列表 2025-2026八年级第二学期期中数学答题卡.pdf 建瓯市2025-2026学年第二学期八年级期中练习卷数学试题参考答案(1).pdf 建瓯市2025—2026学年第二学期八年级期中练习卷(数学学科).pdf