安徽省合肥市行知中学2024—2025学年下学期八年级期末数学试卷(扫描版,含答案)

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名称 安徽省合肥市行知中学2024—2025学年下学期八年级期末数学试卷(扫描版,含答案)
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更新时间 2025-07-28 17:27:29

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八年级数学期末反馈问卷
参考答案
一、选择题(本大题共10小题,每小题4分,满分40分)
题号 1 2 3 4 5 6 7 8 9 10
答案 C B B A D B A C A C
二、填空题(本大题共4小题,每小题5分,共20分)
11. x. 12. 3  .13. 180  米.
14.(1) 8 ; (2) 2
解:(1)∵四边形ABCD是矩形,
∴AB=8=CD,∠D=∠DAB=90°,
∵∠DCA=60°,
∴∠DAC=30°
∴AC=2CD=16,∠BAC=60°,
∵AE平分∠BAC,
∴∠BAE=∠CAE=30°,
又∵CE⊥AE,
∴CEAC=8,
故答案为:8;
(2)如图2,延长AB,CE交于点H,
∵AB=8,AC=12,
∴BC4,
在△CAE和△HAE中,

∴△CAE≌△HAE(ASA),
∴CE=EH,AC=AH=12,
∴BH=AH﹣AB=4,
∴CH4,
∴CE,
故答案为:2.
三、(本大题共2小题,每小题8分,满分16分)
15.解:原式 ··················(5分)
. ·····················(8分)
16.解:∵x2+2x+1=4,
∴(x+1)2=4, ·····················(2分)
则x+1=2或x+1=﹣2, ·····················(6分)
解得:x=1或x=﹣3. ·····················(8分)
四、(本大题共2小题,每小题8分,满分16分)
17.解:(1)把x=1代入方程可得1﹣(m+1)+2m﹣2=0,
解得m=2,
当m=2时,原方程为x2﹣3x+2=0,
∴(x﹣1)(x﹣2)=0,
解得x1=1,x2=2,
即方程的另一根为2; ·····················(4分)
(2)∵a=1,b=﹣(m+1),c=2m﹣2,
∴Δ=[﹣(m+1)]2﹣4×1×(2m﹣2)
=m2﹣6m+9
=(m﹣3)2≥0,
∴不论m为何值时,方程总有两个实数根.·····················(8分)
18.解:(1)由题意得,∠CDB=90°,
∴CD2=BC2﹣BD2=252﹣152=400,
∴CD=20(负值舍去),
∴CE=CD+DE=20+1.6=21.6(米),
答:风筝的高度CE为21.6米;····································(4分)
(2)∵CM=12,
∴DM=20﹣12=8,
∴(米),
∴BC﹣BM=25﹣17=8(米),
∴他应该往回收线8米.·································(8分)
五、(本大题共2小题,每小题10分,满分20分)
19.解:(1):;···············(3分)
(2)第n个等式为,·······(6分)
证明:∵n为正整数,
∴左边右边,
∴等式成立.·····················(10分)
20.(1)证明:∵四边形ABCD是平行四边形,
∴AD∥BC,
∴∠ADB=∠CBD,
∵BD平分∠ABC,
∴∠ABD=∠CBD,
∴∠ADB=∠ABD,
∴AB=AD,
∴四边形ABCD是菱形.·····················(4分)
(2)解:∵AD∥BC,点E在BC的延长线上,
∴AD∥CE,
∵DE∥AC,
∴四边形ACED是平行四边形,
∴CE=AD=BC,
∵四边形ABCD是菱形,
∴AC⊥BD,
∴∠BDE=∠BFC=90°,
∵AC=6,CD=3,
∴DE=AC=6,CD=BC=CEBE=3,
∴BE=2CD=6,
∴BD12,
∴S△BDEBD DE12×6=36,
∴△BDE的面积为36.·····················( 10 分)
六、(本题满分12分)
21.
解:(1)由题意得:a=6÷12%=50,
∵b%100%=20%,
∴b=20,
中位数c78;
故答案为:a= 50  ,b= 20  ,c= 78  ;·················(3分)
(2)C组的人数为11人,D组的人数为50×30%=15(人),
补全频数分布图如下:
·····················(7分)
(3)1500×(30%+16%)=690(名),
答:估计该校八年级学生中暑假安全教育相关知识掌握合格的人数为690名.·····(12分)
七、(本题满分12分)
22.解:(1)由题意可得AB=CD=x,AC=BD,
∴BF=x﹣8,EC=AC﹣4,
由于篱笆长为28m,
∴x﹣8+x+BD+BD﹣4=28,
∴BD=20﹣x;
故答案为:(20﹣x);··············································(2分)
(2)由题意得:x(20﹣x)=75,
即(x﹣15)(x﹣5)=0,
解得x1=15,x2=5,
∵AF=8,
∴x>8,
∴x=15.·····················( 6 分)
(3)由题意可得BF=x﹣8,EC=AC﹣4
由于篱笆长为28m,
∴x﹣8+x﹣2+BD+BD﹣4=21﹣x,
∴BD=21﹣x
∴x(21﹣x)=110
解得x1=10,x2=11.
当CD=10或11时,生态园的面积能达到110m2.·····················(12分)
八、(本题满分14分)
解:(1),45. ························································· (2分)
(2)①过点E作EM⊥BC于M,EN⊥CD于点N,如图1所示:
则四边形EMCN为矩形,
∵∠ACB=45°,
∴△EMC为等腰直角三角形,
∴EM=CM,
∴矩形EMCN为正方形,
∴EM=EN,∠EMF=∠END=∠MEN=90°,
∴∠MEF+∠FEN=90°,
∵四边形DEFG为矩形,
∴∠DEF=90°,
∴∠NED+∠FEN=90°,
∴∠MEF=∠NED,
在△MEF和△NED中,

∴△MEF≌△NED(ASA),
∴EF=ED,
∴矩形DEFG是正方形; ·······················································(6分)
②连接EG,如图2所示:
∵四边形ABCD和四边形DEFG都是正方形,
∴AD=CD,DE=DG,∠ADC=∠EDG=90°,∠DAE=∠ACD=45°,
∴∠ADE+∠EDC=∠EDC+∠CDG=90°,
∴∠ADE=∠CDG,
在△ADE和△CDG中,

∴△ADE≌△CDG(SAS),
∴AE=CG,∠DAE=∠DCG=45°,
∴∠ECG=∠ACD+∠DCG=90°,
∵AC,
∴EC=AC﹣AE,
在Rt△ECG中,由勾股定理得:EG,
在Rt△DEG中,由勾股定理得:DE2+DE2=EG2,
∴,
∴DE; ············································································(10分)
(3). ···········································································(14分)
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