期中模拟卷（福建）2023-2024学年八年级数学上学期期中模拟考试试题及答案

这是一份期中模拟卷（福建）2023-2024学年八年级数学上学期期中模拟考试试题及答案，文件包含期中模拟卷全解全析docx、期中模拟卷参考答案docx、期中模拟卷考试版测试范围人教版第11-13章A4版docx、期中模拟卷考试版测试范围人教版第11-13章A3版docx、期中模拟卷答题卡A3版docx等5份试卷配套教学资源，其中试卷共38页， 欢迎下载使用。
2023-2024学年上学期期中模拟考试八年级数学一、选择题：（本题共10小题，每小题4分，共40分）。12345678910BBBDCCABCD二、填空题：（本题共6小题，每小题4分，共24分）。11．                12．0            13．1 < AD < 514．45                 15．           16．①②三、解答题：（本题共9小题，共86分。其中：17-21每题8分，22-23每题10分，24题12分,25题14分）。17．【解析】设这是边形，则，，．····················4分.所以这个多边形的边数是12，它的对角线的条数是54．····················8分18.【解析】∵△ABC是等边三角形，∴∠B=∠BAC=∠C=60°，又∵AD =AE ，∠DAE =100°，∴∠ADE=∠E =40°.···················································4分∵DE⊥AC，∴ ∠DAC =∠EAC =50°，∴ ∠BAD=60°-50°=10°，又∵∠ADC=∠B +∠BAD =70°， ∴∠EDC =∠ADC -∠ADE =30°.·······································8分19．【解析】，，在和中，，，．·······································8分20．【解析】（1）∵，，∴，∵平分，∴，∴，∴的度数为.········································4分（2）结论：．证明：∵平分，∴，又∵，∴，即．········································8分21．【解析】（1）如图，即为所求.········································3分（2）如图，点P即为所求.········································5分（3）如图，即为所求，共有3个．········································8分22．【解析】（1）是等边三角形.········································1分理由：是等边三角形，，，，，，是等边三角形.········································4分（2）连接，是等边三角形，，，是线段的垂直平分线，平分，，，，，，，．········································10分23．【解析】（1）如图所示，BM即为所求.········································4分（2）如图所示，延长BM交AC于N，∵AD⊥BN，∴∠AMB=∠AMN，∵AD平分∠BAD，∴∠BAM=∠NAM，又∵AM=AM，∴△ABM≌△ANM（ASA），∴∠ABN=∠ANB，NM=BM=4，AN=AB=10，∴BN=8，NC=AC-AN=8，∴BN=CN=8，∴∠C=∠NBC，∴∠ABN=∠ANB=∠C+∠NBC=2∠C，∴∠ABC=∠ABN+∠NBC=3∠C．········································10分24．【解析】（1）∵△ABC中，∠ACB=90°，AB=5，BC=3，由勾股定理得AC=，连接BP，如图所示，当PA=PB时，PA=PB=t，PC=4t，在Rt△PCB中，PC2+CB2=PB2，即（4t）2+32=t2，解得t=，∴当t=时，PA=PB.········································3分（2）如图1，过P作PE⊥AB，又∵点P恰好在∠BAC的角平分线上，且∠C=90°，AB=5，BC=3，∴CP=EP，在Rt△ACP和Rt△AEP中，，∴Rt△ACP≌Rt△AEP（HL），∴AC=AE=4，∴BE=1，设CP=EP=x，则BP=3-x，在Rt△BEP中，BE2+PE2=BP2，即12+x2=（3-x）2，解得x=，∴CP=，∴CA+CP=4+=，∴t=；当点P沿折线A-C-B-A运动到点A时，点P也在∠BAC的角平分线上，此时，t=5+4+3=12.综上，若点P恰好在∠BAC的角平分线上，t的值为或12.··································8分（3）①如图2，点P在CA上，当CP=CB=3时，△BCP为等腰三角形，则t=4-3=1.········································9分②如图3，当BP=BC=3时，△BCP为等腰三角形，∴AC+CB+BP=4+3+3=10，∴t=10.········································10分③如图4，若点P在AB上，当CP=CB=3时，△BCP为等腰三角形；作CD⊥AB于D，则根据面积法求得：CD=，在Rt△BCD中，由勾股定理得，BD=，∴PB=2BD=3.6，∴CA+CB+BP=4+3+3.6=10.6，此时t=10.6.········································11分④如图5，当PC=PB时，△BCP为等腰三角形，作PD⊥BC于D，则D为BC的中点，∴PD为△ABC的中位线，∴AP=BP=AB=2.5，∴AC+CB+BP=4+3+2.5=9.5，∴t=9.5.综上所述，t为1或10或10.6或9.5时，△BCP为等腰三角形．·································12分25．【解析】（1）补全图形如下图，①∵C、D两点关于y轴的对称的两点，∴横坐标互为相反数，纵坐标不变，∵，∴.故答案为.········································2分②∵C、D两点关于y轴的对称，，∴，∴，∴，∵是等边三角形，∴，∴，∴，∴．故答案为60．········································5分（2）如图，延长交于点G，∵，∴，∴，∴，∴，∴，∴，∴，∴，∵，∴垂直平分．········································9分（3），证明如下：如图：作，连接，∵C、D两点关于y轴的对称，∴，∴，∵，∴，∴，∴，∴，∴，∵，∴，∵，，∴是等边三角形，∴，∴，∵，∴，在和中，，∴，∴，∵，∴，∴．········································14分