湖南省岳阳市城区2022-2023学年八年级下学期期末考试数学试题及参考答案

这是一份湖南省岳阳市城区2022-2023学年八年级下学期期末考试数学试题及参考答案，文件包含八年级数学参考答案docx、八年级数学印刷版docx、八年级数学参考答案pdf、八年级数学印刷版pdf等4份试卷配套教学资源，其中试卷共26页， 欢迎下载使用。
一、选择题
二、填空题
11. x≥1 12. （3，0） 13. y=3x+1
14. 6 15. 8 16. （-21011,−21012）
三、解答题
17. 解：（1）如图略·····································································2分
点A1的坐标为 （3，0）；······················································3分
（2）如图略······································································5分
点C2的坐标为（-1，-1）．······················································6分
解：（1）△AOB是直角三角形；理由如下：
∵32+22＝（）2，
∴AO2+BO2＝AB2，···························································2分
∴∠AOB＝90°，
∴AC⊥BD，·································································3分
∵四边形ABCD是平行四边形，
∴▱ABCD是菱形．···························································4分
（2）∵0A=3，OB=2
∴.········································6 分
解：（1）观察函数图象可知：不等式2x﹣4＞kx+5的解集为x＞3.·························2分
（2）把C（3，a）代入y=2x﹣4中，得：
a=3×2-4=2
∴点C的坐标为（3，2）．······················································3分
把C（3，2）代入y=kx+5中，得
3k+5=2 ∴k=-1
∴y=-x+5·······································································4分
（3）当y＝2x﹣4＝0时，x＝2，
∴点D的坐标为（2，0），
∴S△ACD＝（xA﹣xD）•yC＝×（5﹣2）×2＝3．································6分
解：（1）∵在Rt△ABC中，∠ACB＝90°，∠B=30°，AC＝2 ，
∴AB=2AC=2×2=4 ，···························································2分
∵CD是斜边上的中线，
∴CD＝AB＝2；······························································4分
（2）证明：∵CD是斜边上的中线，
∴CD＝DB＝AD＝AB，
∵∠B＝30°，
∴∠CAB＝60°，
∴△ACD是等边三角形，
∴∠CDA=60°···································································5分
∵CE⊥AD，
∴AE＝DE，∠ECD＝30°，
∴DE＝DC，·····································································6分
∵F是CD的中点，
∴EF＝CD，
∴EF＝DF，·······································································7分
∴△EDF为等边三角形．···························································8分
解：（1）a＝60，b＝0.15；·································································4分
（2）补图略，·········································································6分
（3）成绩在80分以上（包括80分）的频率为0.3+0.4＝0.7，
∴参加这次比赛的3000名学生中成绩“优”等约有3000×0.7＝2100（人）．···············8分
解：（1）证明：∵AD∥BC，AB∥CD，
∴四边形ABCD是平行四边形，·····················································1分
∴BD＝2OB，······································································2分
∵AC＝2OB，
∴AC＝BD，·······································································3分
∴平行四边形ABCD是矩形；························································4分
（2）解：∵四边形ABCD是矩形，
∴∠DAB＝∠ADC＝90°，OA＝OD，BD＝2OD，
∵AE平分∠BAD，
∴∠DAE＝45°，
∴△ADE是等腰直角三角形，
∴AD＝DE，·······································································5分
∵BD＝2AD，
∴OA＝OD＝AD，
∴△AOD是等边三角形，···························································6分
∴∠ADO＝60°，
∴∠ODE＝∠ADC﹣∠ADO＝90°﹣60°＝30°，····································7分
∵AD＝DE，AD＝OD，
∴DE＝OD，·······································································8分
∴∠DOE＝∠DEO＝（180°﹣∠ODE）＝×（180°﹣30°）＝75°．··············9分
解：（1）乙；甲，16；·····································································3分
（2）设线段AB、DE的解析式分别为：y1＝kx+b，y2＝mx+n，
∵AB经过点（0，14）和（7，0），DE经过（0，4）和（4，16）
∴AB解析式为y＝-2x+14，DE解析式为y＝3x+4，·····································7分
令 -2x+14＝3x+4，解得x＝2，
∴注水2分钟时，甲、乙两个水槽中水的深度相同；·································9分
解：（1） ③④ ·····································································2分
（2）①连接EG
∵四边形ABCD是矩形，
∴∠B＝∠C＝90°．
∵E是BC的中点，EB＝EC，
∵将△ABE沿AE折叠后得到△AFE，
∴∠AFE＝∠B＝90°，EF＝EB，
∴∠EFG＝180°﹣∠AFE＝90°＝∠C············································3分
在Rt△EFG和Rt△ECG中
∴Rt△EFG≌Rt△ECG（HL），···················································4分
∴四边形FECG沿EG折叠完全重合
∴四边形FECG是“忧乐四边形”················································5分
②：①中结论仍然成立，理由如下:
证明：连接EG、FC
∵E是BC的中点
∴BE＝CE
∵将△ABE沿AE折叠后得到△AFE
∴BE＝EF，∠B＝∠AFE
∴EF＝EC
∴∠EFC＝∠ECF
∵四边形ABCD是平行四边形
∴AB∥CD
∴∠ECD＝180°﹣∠B，且 ∠EFG＝180°﹣∠AFE＝180°﹣∠B
∴∠ECD＝∠EFG
∴∠GFC＝∠GFE﹣∠EFC＝∠ECG﹣∠ECF＝∠GCF
∴∠GFC＝∠GCF
∴FG＝CG···································································6分
在△EFG和△ECG中
∴△EFG≌△ECG（SAS）······················································7分
∴四边形FECG沿EG折叠完全重合
∴四边形FECG是“忧乐四边形”···············································8分
（3）····················································10分
解：（1）∵k＝﹣1，
∴y＝﹣x+4，
∴B（0，4），
∴OB＝4，
∵BE＝3，
∴OE＝，··································································1分
由题意可知：△BEO≌△AOD（K型全等），
∴AD＝OE=；······························································2分
（2）k＝﹣时，y＝﹣x+4，
∴A（3，0），
∵△ABM是以AB为斜边的等腰直角三角形
∴过点M作MH⊥x轴，MG⊥y轴，
∴△BMG≌△AHM（AAS），·················································3分
∴BG＝AH，GM＝MH，
∴OB-OG=OH-OA
即 4﹣MH＝MH﹣3，
∴MH＝，·······························································4分
∴M（，）；···························································5分
①当k=1时，y=x+4
令 y=0，x+4=0 解得 x=-4
∴A（-4，0）
由旋转可得A’（4，0）
∴OB=OA’=4
∴△OBA’为等腰直角三角形
∴A’（4，0），k=1···························································6分
②当k≠1时
过A’点作A’H⊥OB交于点H，
且AB⊥A’B ，OA⊥OB
∴△AOB≌△BHA’（AAS）
∴OB=A’H=4
∵△OBA’为等腰三角形
∴OA’=BA’
∵A’H⊥OB
∴OH=BH= 12OB=2
∴A’（4，2）··································································7分
∵△AOB≌△BHA’
∴OA=BH=2
∴A（-2，0）
将A（-2，0）代入y=kx+4中，得：-2k+4=0 解得 k=2··························8分
∴A’（4，2）， k=2
②由题意可得：0＜k＜1························································10分
题号
1
2
3
4
5
6
7
8
9
10
答案
A
B
D
A
C
C
D
D
B
B