期中模拟卷02（江苏）（苏科版八上第1~3章：全等三角形、轴对称、勾股定理）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02（江苏）（苏科版八上第1~3章：全等三角形、轴对称、勾股定理）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02全解全析docx、期中模拟卷02参考答案docx、期中模拟卷02考试版测试范围第1-3章苏科版A4版docx、期中模拟卷02答题卡A4版docx等4份试卷配套教学资源，其中试卷共44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12345678BCCACBCC9．15:01            10．16cm或17cm          11．Rt△BCD　≌　Rt△CBE；HL12．36°            13．6或8           14． 515．2：3：4    16．17．（6分）解：（1）∵4（x﹣1）2＝100，∴（x﹣1）2＝25．∴x﹣1＝±5．∴x＝6或﹣4·····························································3分（2）∵（x+2）3＝﹣27，∴x+2＝﹣3．∴x＝﹣5．································································6分18．（4分）解：如图所示：（答案不唯一，每种方案1分）19．（8分）解：（1）∵DE是AB的垂直平分线，∴EA＝EB．·····································································2分∵FG是AC的垂直平分线，∴GA＝GC．∴BC＝BE+EG+CG＝AE+EG+AG＝△AEG周长＝10；·································4分（2）解：∵∠BAC＝128°，∴∠B+∠C＝180°﹣∠BAC＝52°，∵AB的垂直平分线分别交AB、BC于点D、E，∴AE＝BE，AG＝CG，∴∠BAE＝∠B，∠CAG＝∠C，····················································6分∴∠BAE+∠CAG＝∠B+∠C＝52°，∴∠EAG＝∠BAC﹣（∠BAE+∠CAG）＝76°．······································8分20．（8分）解：（1）设AB长为x米，则绳子长为（x+1）米，·····································2分AE的长度为（x﹣1）米．··························································4分故答案为：（x+1）；（x﹣1）；（2）在Rt△ACE中，AC＝x米，AE＝（x﹣1）米，CE＝8米，由勾股定理可得，（x﹣1）2+82＝（x+1）2，解得：x＝16．答：旗杆的高度为16米．··························································8分21．（8分）解：（1）∵AD＝CD，∴∠A＝∠ACD，∵CD平分∠ACB，∴∠ACB＝2∠ACD，∴∠ACB＝2∠A，∵AB＝AC，∴∠B＝∠ACB＝2∠A，∵∠A+∠B+∠ACB＝180°，∴5∠A＝180°，∴∠A＝36°，∴∠A的度数为36°；································································4分（2）△ADE，△CDB，△ADC，△DEC是等腰三角形，理由：∵DE∥BC，∴∠ADE＝∠B，∠AED＝∠ACB，∵∠B＝∠ACB，∴∠ADE＝∠AED，∴AD＝AE，∴△ADE是等腰三角形；······························································5分∵DE∥BC，∴∠EDC＝∠DCB，∵CD平分∠ACB，∴∠ACD＝∠DCB，∴∠EDC＝∠ACD，∴ED＝EC，∴△EDC是等腰三角形；······························································6分∵AD＝CD，∴△ADC是等腰三角形；······························································7分∵∠CDB＝∠A+∠ACD，∠A＝∠ACD，∴∠CDB＝2∠A，∵∠B＝2∠A，∴∠B＝∠CDB，∴CD＝CB，∴△CDB是等腰三角形，∴△ADE，△CDB，△ADC，△DEC是等腰三角形．·······································8分22．（8分）解：（1）A城受到这次台风的影响，理由：由A点向BC作垂线，垂足为M，在Rt△ABM中，∠ABM＝30°，AB＝600km，则AM＝300km，因为300＜500，所以A城要受台风影响；···············································4分 （2）设BC上点D，DA＝500千米，则还有一点G，有AG＝500千米．因为DA＝AG，所以△ADG是等腰三角形，因为AM⊥BC，所以AM是DG的垂直平分线，MD＝GM，在Rt△ADM中，DA＝500千米，AM＝300千米，由勾股定理得，MD400（千米），则DG＝2DM＝800千米，遭受台风影响的时间是：t＝800÷200＝4（小时），答：A城遭受这次台风影响时间为4小时．··············································8分23．（8分）（1）解：如图1中，设∠C＝x．∵∠ABC＝2∠C，∴∠ABC＝2x，∵BD平分∠ABC，∴∠ABD＝∠CBD＝x，∵AB＝BD，∴∠A＝∠ADB＝∠DBC+∠C＝2x，∵∠A+∠ABC+∠C＝180°，∴2x+2x+x＝180°，∴x＝36°，∴∠A＝2x＝72°，故答案为：72．······································································2分 （2）证明：如图1中，∵∠ABD＝∠DBC＝∠C，∴BD＝CD，在△ABD和△ECD中，，∴△ABD≌△ECD（AAS），∴AB＝EC．··········································································4分 （3）证明：如图2中，延长BD到T，使得CD＝CT．∵CD＝CT，∴∠T＝∠CDT＝∠ADB，∵BD＝CD，∴BD＝CT，在△ABD和△ECT中，，∴△ABD≌△ECT（AAS），∴AB＝EC．··········································································8分24．（8分）（1）解：∵AO平分∠BAD，∴∠DAO＝∠OAB，∵BO平分∠EOA，∴∠EBO＝∠OBA，∵∠ACB＝50°，∴∠CBA+∠CAB＝130°，∴∠EBA+∠BAD＝360°﹣130°＝230°，∴∠OBA+∠OAB＝115°，∴∠AOB＝360°﹣50°﹣115°﹣130°＝65°；··········································2分（2）解：过O点作OM⊥AD于M，ON⊥BE于N，OP⊥AB于P，∵AO，BO分别平分∠DAB，∠EBA，∴OM＝OP，OP＝ON，∴OM＝ON，∴CO平分∠ACB，∵∠ACB＝50°，∴∠BCK＝∠ACK＝25°；·····························································4分（3）证明：∵∠BAC＝105°，∠ACB＝50°，∴∠ABC＝25°，∵∠KCB＝25°，∴∠KBC＝∠KCE，∴KB＝KC，过A点作AH∥BC交CO于H，∴∠AHK＝∠KCB，∠HAK＝∠KBC，∴∠AHK＝∠HAK，∴KA＝KH，∴AB＝CH，∵∠AHK＝∠ACH，∴AH＝AC，∵AF⊥CO，∴HF＝CF，∴CH＝2CF，∴AB＝CH＝2CF．·································································8分25．（10分）解：（1）DE2＝BD2+EC2；···························································2分 （2）关系式DE2＝BD2+EC2仍然成立．证明：将△ADB沿直线AD对折，得△AFD，连FE∴△AFD≌△ABD，·································································3分∴AF＝AB，FD＝DB，∠FAD＝∠BAD，∠AFD＝∠ABD，又∵AB＝AC，∴AF＝AC，∵∠FAE＝∠FAD+∠DAE＝∠FAD+45°，∠EAC＝∠BAC﹣∠BAE＝90°﹣（∠DAE﹣∠DAB）＝45°+∠DAB，∴∠FAE＝∠EAC，··································································5分又∵AE＝AE，∴△AFE≌△ACE，∴FE＝EC，∠AFE＝∠ACE＝45°，∠AFD＝∠ABD＝180°﹣∠ABC＝135°∴∠DFE＝∠AFD﹣∠AFE＝135°﹣45°＝90°，∴在Rt△DFE中，DF2+FE2＝DE2，即DE2＝BD2+EC2；··································································7分解法二：将△EAC绕点A顺时针旋转90°得到△TAB．连接DT．∴∠ABT＝∠C＝45°，AT＝AE，∠TAE＝90°，∵∠ABC＝45°，∴∠TBC＝∠TBD＝90°，∵∠DAE＝45°，∴∠DAT＝∠DAE，∵AD＝AD，∴△DAT≌△DAE（SAS），∴DT＝DE，∵DT2＝DB2+EC2，∴DE2＝BD2+EC2； （3）当AD＝BE时，线段DE、AD、EB能构成一个等腰三角形．如图，与（2）类似，以CE为一边，作∠ECF＝∠ECB，在CF上截取CF＝CB，可得△CFE≌△CBE，△DCF≌△DCA．∴AD＝DF，EF＝BE．∴∠DFE＝∠1+∠2＝∠A+∠B＝120°．若使△DFE为等腰三角形，只需DF＝EF，即AD＝BE，∴当AD＝BE时，线段DE、AD、EB能构成一个等腰三角形，且顶角∠DFE为120°．········10分