2022-2023學(xué)年度第一學(xué)期八年級(jí)數(shù)學(xué)學(xué)科期中抽測(cè) 評(píng)分標(biāo)準(zhǔn)一、選擇題(本大題有8小題,每小題3分,共24分)題號(hào)12345678選項(xiàng)BABCCDAD二、填空題(每題共8小題,每題4分,共32說(shuō)明:第16題只看對(duì)1個(gè)2分,對(duì)2個(gè)3分,對(duì)3個(gè)4 9   30        10     100         11   10        12B=C(答案不唯一)          13  15        14     20        15   90      16  70°或40°或55°    三、解答題(本大題9小題,共84分)
17每空1分,共8分)已知,ABO=CDO,ABO=∠CDO,AOB=∠COD,對(duì)頂角相等AAS,全等三角形的對(duì)應(yīng)邊相等?????????????  18AD平分BAC,·····························1DEABDFAC,垂足為EF·····························2DE=DF,······················3BED=∠CFD=90°·············4RtBEDRtCFD中,BED=∠CFD=90°,·····························6RtBEDRtCFD············7BE=CF,··················8191)如圖所示··············4(每個(gè)點(diǎn)1    2)如圖所示················7316···················10          20 1證明:AECFAEF=∠CFE,················1AEB=∠CFD················2DE=BF,DF=BE······················4在△ABE和△CFD中,·····························7ABE≌△CFD,···············82AB=CDABCD·············1021證明:在等邊ABC中,AB=BCABC=∠ACB=60°, 2BD是邊AC上的高,∴∠ABD=∠CBD=30°,···············4CE=CD,∴∠CDE=∠CED,··················6∵∠ACBCDE的外角,∴∠CED=30°,··················8∴∠DBC=∠CED,··················9BDDE 10221)在ABC中,,······························································3,································································4ABC是直角三角形,且∠ACB=90°,····································5點(diǎn)DAB的中點(diǎn),CD=BD=,···························································7B=BCD=50°,···················································8DCA=ACB-∠BCD=90°50°=40°·································92······························································12
231)如圖·································62)解:                                ······································8           ·····································10             ······································12   24在△ABC中,∵AB=AC,∠BAC=90°,∴∠B=C=45°,  1又∵點(diǎn)DBC的中點(diǎn),AD=BD=,且ADBC,∠BAD=CAD= 2∴∠ADN+BDN=90°,又∵△DEF是直角三角尺,∴∠EDF =90°,即∠ADN+ADM=90°,∴∠BDN=ADM  3BDNADM∴△BDN≌△ADM, ···············5DN=DM;··························································6
2BDN≌△ADMBN= AM,BND=AMD,DN=DM········································7BNF=AME,且由于△DEF是含45°直角三角尺,DF=DE,DFDN=DEDMFN=EM····························································8BNFAME∴△BNF≌△AMEAE=BF;···························································9
3)作圖正確(如圖所示)·············································10猜想:AEBF,理由如下:··············································11BNF≌△AME,∴∠BFN=AEM, ∵∠FDE=90°,∴∠AEM+APD=90°又∵∠APD=FPQ,···················································12∴∠FPQ+BFN=90°,∴∠FQP =90°,······················································13AEBF··························································14 

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