遼寧省大連市金普新區(qū)2023-2024學年八年級上學期數(shù)學10月月考試卷 （含答案）

第第頁遼寧省大連市金普新區(qū)2023--2024學年八年級上學期數(shù)學10月月考試卷（含答案）2023-2024學年（上）月考試卷（十月份）

八年級數(shù)學2023.10

注意事項：

1.請在答題卡上作答，在試卷上作答無效。

2.本試卷共五大題，25小題，滿分120分。考試時間120分鐘。

一、選擇題（本題共10小題，每小題2分，共20分，在每小題給出的四個選項中，只有一個選項正確）

1.下列長度的三條線段能組成三角形的是

A.3，3，6B.3，4，7C.4，6，11D.5，6，9

2.下列圖形中具有穩(wěn)定性的是

A.三角形B.四邊形C.五邊形D.六邊形

3.△ABC中∠BAC>90°，作AB邊上的高，以下各圖作法正確的是

4.如圖，在△ABC中，點D是邊上一點，點E是邊/C上一點，且DE//BC，

∠B=40°，∠A=80°，則∠AED的度數(shù)是

A.40°B.50°C.60°D.70°

5.如圖，BC⊥AE，垂足為C，CD//AB，∠A=50°，則∠BCD的度數(shù)是

A.40°B.50°C.60°D.70°

6.用尺規(guī)作一個角等于己知角的依據(jù)是

A.SASB.SSSC.AASD.ASA

7.若一個多邊形的內(nèi)角和是它的外角和3倍，則這個多邊形是

A.六邊形B.七邊形C.八邊形D.九邊形

8.如圖，下列條件能判定△ABC≌△DEF的一組是

A.AB=DE，AC=DF，∠C=∠FB.AC=DF，BC=EF，∠A=∠D

C.∠A=∠D，∠B=∠E，∠C=∠FD.∠A=∠D，∠C=∠F，AC=DF，

9.點P在∠ABC的平分線上，點P到BC邊的距離為6，點Q是邊上的任意一點，則下列選項正確的是

A.PQ>6B.PQ≥6C.PQ<6D.PQ≤6

10.如圖，AD是△ABC的角平分線，CE⊥AD垂足為F，交AB于點E，∠CAB=30°，∠B=55°，則∠BDE的度數(shù)為

A.35°B.40°C.45°D.50°

二、填空題（本題共6小題，每小題3分，共18分）

11.如圖，△ABC≌△ADE，AD=8，AE=5，DE=6，則BE=__________.

12.等腰三角形的周長為13，其中一邊長為3，則該等腰三角形的底邊長為__________.

13.如圖，△ABC≌△ADE，∠C=80°，∠D=30°，∠BAD=40°，則∠CAD=__________.

14.如圖，∠E=∠F，CE=BF，要使△ACE≌△DBF，添加一個條件__________.

15.一個多邊形的內(nèi)角和為1260°，從這個多邊形的一個頂點出發(fā)的對角線有__________條.

16.如圖，在△ABC中，∠B=∠C，M，N，P分別是邊AB，AC，BC上的點，且BM=CP，

CN=BP，∠A=92°，則∠MPN的度數(shù)為__________°.

三、解答題（本題共4小題，其中17題6分，18、19、20題各8分，共30分）

17.尺規(guī)作圖（不寫作法.保留作圖痕跡）

己知∠AOB，

（1）作一個角等于∠AOB；

（2）作∠AOB的平分線.

18.如圖，∠B=52°，∠ACB=∠A+8°，∠ACD=60°.求證AB//CD.

19.如圖，C是AB的中點，AD=CE，∠A=∠BCE.求證CD=BE.

20.如圖，在四邊形ABCD中，∠B=∠D=90°，AE平分∠BEC，CF平分∠BCD.

求證∠BAE=∠CFD.

21.如圖，∠ACB=90°，AC=BC，AD⊥CE，BE⊥CE，垂足分別為D，E，DE=1.7cm，BE=0.8m，求AD的長.

22.（1）如圖1，△ABC的外角∠CBD和∠BCE的平分線交于點F.用等式表示∠F與∠A的數(shù)量關(guān)系；

（2）如圖2，∠ABC的平分線和△ABC的外角∠ACG的平分線交于點H.用等式表示∠A與∠H的數(shù)量關(guān)系，并證明.

五、解答題（本題共3小題，23、24題各11分，25題12分，共34分）

23.如圖，點C在線段上，∠A=∠B=∠DCE,CE=CD.

（1）求證△ACD≌BEC；

（2）求證AD+BE=AB.

24.如圖，∠A=60°，△ABC的角平分線BD,CE交于點F.

（1）求證∠BFC=2∠DFC；

（2）求證EF=DF；

（3）用等式表示線段BE,BC,CD之間的數(shù)量關(guān)系，并證明.

25.如圖，AD⊥AB，AE⊥AC，AD=AB，AE=AC，F(xiàn)是DE的中點，連接FA并延長交BC于點G.

（1）用等式表示線段BC與的數(shù)歌關(guān)系，并證明；

（2）寫岀線段AG與BC的位置關(guān)系，并證明.月考八年級數(shù)學參考答案及評分標準2023.10

說明：試題解法不唯一，其它方法備課組統(tǒng)一意見，酌情給分。

一、選擇題（本題共10小題，每小題2分，共20分）

1．D；2．A；3．C；4．C；5．A；6．B；7．C；8．D；9．B；10．B．

二、填空題（本題共6小題，每小題3分，共18分）

11．3；12．3；13．30；14．AEDF或ACDB或ABDC或AD或AE∥DF或

ECAFBD或CE∥BF；15．6；16．44．

三、解答題（本題共4小題，其中17題6分，18、19、20題各8分，共30分）

17．（1）作一個角等于AOB；··························································································3分

（2）作AOB的平分線．·····························································································6分

18．證明：∵∠A∠B∠ACB=180°，∠B=52°，∠ACB=∠A8°，

∴∠A52°∠A8°=180°．·························································································2分

∴2∠A=120°．···········································································································3分

∴∠A=60°．··············································································································4分

∵∠ACD=60°，

∴∠A=∠ACD．··········································································································6分

∴AB∥CD．··············································································································8分

19．證明：∵C是AB的中點，

∴AC=CB．···············································································································2分

在△ACD和△CBE中

ACCB，

∠A=∠BCE，···········································································································4分

AD=CE，

∴△ACD≌△CBE．（SAS）··························································································6分

∴CD=BE．···············································································································8分

20．證明：∵在四邊形ABCD中，∠BAD∠BCD=360°∠B∠D，∠B=∠D=90°，

∴∠BAD∠BCD=180°．·····························································································2分

∵AE平分∠BAC，CF平分∠BCD，

11

∴∠BAE=∠BAC，∠FCD=∠BCD．·········································································4分

22

1111

∴∠BAE∠FCD=∠BAC∠BCD=（∠BAC∠BCD）=180°=90°．·····················6分

2222

∵在△CDF中，∠CFD∠FCD=90°，

月考八年級數(shù)學答案第1頁共4頁

{#{ABZYIUogAAQgBAAAgCEwUyCgAQkBCAAKoGhAAIoAAAwQNABAA=}#}

∴∠BAE=∠CFD．····································································································8分

四、解答題（本題共2小題，其中21題8分，22題10分，共18分）

21．解：∵AD⊥CE，BE⊥CE，垂足分別為D，E，

∴∠CEB=∠ADC=90°．·······························································································1分

∴∠ACD+∠CAD=90°．

∵∠ACB=90°，

∴∠ACD+∠BCE=90°．

∴∠CAD=∠BCE．······································································································2分

在ΔBCE和ΔCAD中，

∠CEB=∠ADC，

∠BCE=∠CAD，

BCAC，

∴ΔBCE≌ΔCAD．（AAS）····························································································4分

∴BE=CD，AD=CE．···································································································6分

∴AD=CE=CD+DE=BE+DE=0.8+1.7=2.5（cm）．······························································7分

答：AD的長為2.5cm．·································································································8分

1

22．（1）∠F=90°∠A．·····························································································2分

2

（2）∠A=2∠H．·······································································································4分

證明：∵△ABC的外角∠ACG的平分線和∠ABC的平分線交于點H，

∴∠ACG=2∠HCG，∠ABC=2∠HBC．··································································6分

∵∠ACG=∠ABC+∠A，∠HCG=∠HBC+∠H，

∴∠ABC+∠A=2（∠HBC+∠H）=2∠HBC+2∠H．····················································8分

∴2∠HBC+∠A=2∠HBC+2∠H．···········································································9分

∴∠A=2∠H．··································································································10分

五、解答題（本題共3小題，23、24題各11分，25題12分，共34分）

23．（1）證明：∵∠A+∠D=∠DCB=∠DCE+∠BCE，∠A=∠DCE，

∴∠D=∠BCE．···································································································2分

在△ACD和△BEC中，

∠A=∠B，

∠D=∠BCE，·····································································································4分

CDCE，

∴△ACD≌△BEC（AAS）．····················································································6分

（2）證明：∵△ACD≌△BEC，

∴AD=CB，BE=AC．·····························································································8分

∴AD+BE=CB+AC=AB．····················································································11分

月考八年級數(shù)學答案第2頁共4頁

{#{ABZYIUogAAQgBAAAgCEwUyCgAQkBCAAKoGhAAIoAAAwQNABAA=}#}

24．（1）證明：∵△ABC的角平分線BD，CE交于點F，

11

∴∠ABD=∠CBD=∠ABC，∠ACE=∠BCE=∠ACB．

22

∴∠BFC=180°∠CBD∠BCE=180°

1

∠ABC

1

∠ACB

22

=180°11（∠ABC∠ACB）=180°（180°∠A）

22

=180°1（180°160°）=180°120°=180°60°=120°．········································1分

22

∴∠DFC=180°∠BFC=180°120°=60°．·································································2分

∴∠BFC=2∠DFC．······························································································3分

（2）證明：在BC上截取BG=BE，連接FG．

在△BFE和△BFG中，

BEBG，

∠ABD=∠CBD，

BF=BF，

∴△BFE≌△BFG（SAS）．··············································4分

∴∠BFG=∠BFE=∠DFC=60°，EF=GF．

∴∠CFG=∠BFC∠BFG=12060°=60°=∠DFC．······················································5分

在△CFG和△CFD中，

∠CFG=∠DFC，

CFCF，

∠BCE=∠ACE，

∴△CFG≌△CFD（ASA）．····················································································6分

∴GF=DF．·········································································································7分

∴EF=DF．·········································································································8分

（3）證明：∵△BFE≌△BFG，△CFG≌△CFD，

∴BE=BG，CD=CG．····························································································9分

∴BE+CD=BG+CG=BC．····················································································11分

25．（1）BC=2AF．···········································································································1分

證明：延長AF至H，使FH=AF，連接EH．

∵F是DE的中點，

∴DF=EF．···································································2分

在△AFD和△HFE中，

DFEF，

∠AFD=∠HFE，

AF=HF，

∴△AFD≌△HFE（SAS）．·············································3分（第25題）

∴∠ADF=∠HEF，HE=AD．

月考八年級數(shù)學答案第3頁共4頁

{#{ABZYIUogAAQgBAAAgCEwUyCgAQkBCAAKoGhAAIoAAAwQNABAA=}#}

∴EF∥AD．

∴∠HEA+∠DAE=180°．

∵AD⊥AB，AE⊥AC，

∴∠DAB=90°，∠EAC=90°．

∵∠DAE+∠BAC+∠EAC+∠DAE=360°，∠DAB=90°，∠EAC=90°，

∴∠BAC+∠DAE=180°．

∴∠BAC=∠HEA．································································································4分

∵HE=AD，AD=AB，

∴HE=AB．········································································