江苏省徐州市2022-2023学年八年级上学期期中检测数学试卷（含答案）

这是一份江苏省徐州市2022-2023学年八年级上学期期中检测数学试卷（含答案），共7页。试卷主要包含了选择题，解答题等内容，欢迎下载使用。
 
 
 
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 2022-2023学年度第一学期八年级数学学科期中抽测 评分标准一、选择题（本大题有 8 小题，每小题 3 分，共 24 分）题号选项1B23B4C5C678ADAD二、填空题（每题共 8 小题，每题 4 分，共 32 分．说明：第 16 题只看对 1 个 2 分，对 2个 3 分，对 3 个 4 分）9．3010．14．1002011．15．109012．∠B=∠C（答案不唯一）16． 70°或 40°或 55°13． 15三、解答题（本大题有 9 小题，共 84 分）17．（每空 1 分，共 8 分）已知，∠ABO=∠CDO，∠ABO=∠CDO，∠AOB=∠COD，对顶角相等，AAS，全等三角形的对应边相等18．∵AD 平分∠BAC，··································································1 分DE⊥AB，DF⊥AC，垂足为 E，F··································································2 分∴DE=DF，················································3 分∠BED=∠CFD=90°，·····························4 分在 Rt△BED 和 Rt△CFD 中，∠BED=∠CFD=90°，DE = DF·············································6 分BD = CD∴Rt△BED≌Rt△CFD，···························7 分∴BE=CF，·········································8 分19．（1）如图所示···································4 分（每个点 1 分）（2）如图所示·································7 分（3）16 ············································10 分lACA'C'D'DBF'P20． （1）证明：∵AE∥CF，∴∠AEF=∠CFE， ·································1 分∴∠AEB=∠CFD，·································2 分∵DE=BF，∵DF=BE， ················································4 分
 在△ABE 和△CFD 中，BE = DFAEB = CFD ··································7 分AE = CF∴△ABE≌△CFD，··································8 分（2）AB=CD，AB∥CD·····························10 分21．证明：在等边△ABC 中，AB=BC，∠ABC=∠ACB=60°， 2 分∵BD 是边 AC 上的高，∴∠ABD=∠CBD=30°, ·······4 分∵CE=CD，∴∠CDE=∠CED，····································6 分∵∠ACB 是△CDE 的外角，∴∠CED=30°，······································8 分∴∠DBC=∠CED， ···································9 分∴BD＝DE． 10 分22．（1）在△ABC 中，∵ AC∴ AC2++BCBC2==62+82=100 ，AB2=102=100 ·····································································3，分分22AB ··················································································································42，故△ABC 是直角三角形，且∠ACB=90°，·················································································5 分∵点 D 是 AB 的中点，1∴CD=BD= AB , ····························································································································7 分2∴∠B=∠BCD=50°，····················································································································8 分∴∠DCA=∠ACB－∠BCD=90°－50°=40°．···········································································9 分（2） 4.8 ········································································································································12 分23．（1）如图 ·············································································6 分1（2）解：∵ S大正方形 =4 ab + (b − a)22= 2ab + b2− 2ab + a2= a2+ b ······················································82分又∵ S大正方形 =c2·····················································10 分·····················································12分∴ a2+b2=c224．在△ABC 中，∵AB=AC，∠BAC=90°，∴∠B=∠C=45°， 1 分FQ又∵点 D 是 BC 的中点，BC1EA∴AD=BD=，且 AD⊥BC，∠BAD=∠CAD= BAC = 45 2 分P22NMDCB
 ∴∠ADN+∠BDN=90°，又∵△DEF 是直角三角尺，∴∠EDF =90°，即∠ADN+∠ADM=90°,∴∠BDN=∠ADM 3 分在△BDN 和△ADM 中 = =BDAM 45BD = ADBDN = ADM∴△BDN≌△ADM··5··，···分········∴DN=DM；·6·分（2）∵△BDN≌△ADM∴BN= AM，∠BND=∠AMD，DN=DM·7·分∴∠BNF=∠AME，且由于△DEF 是含 45°直角三角尺， ∴DF=DE，∴DF－DN=DE－DM 抖音：即 FN=EM8·分在△BNF 和△AME 中BN = AMBNF = AMEFN = EM∴△BNF≌△AME，∴AE=BF； ······································································································································9 分（3）作图正确（如图所示）·······································································································10 分猜想：AE⊥BF，理由如下： ········································································································11 分∵△BNF≌△AME，∴∠BFN=∠AEM，∵∠FDE=90°，∴∠AEM+∠APD=90°又∵∠APD=∠FPQ， ·····················································································································12 分∴∠FPQ+∠BFN=90°，∴∠FQP =90°，···························································································································13 分∴AE⊥BF． ····································································································································14 分