普陀区2019学年度第一学期初三质量调研数学答案练习题

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这是一份普陀区2019学年度第一学期初三质量调研数学答案练习题，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。

普陀区2019学年度第一学期初三质量调研数学试卷参考答案及评分说明一、选择题：（本大题共6题，每题4分，满分24分）1．(B)； 2．(C)； 3．(A)； 4．(D)； 5．(B)； 6．(C)．二、填空题：（本大题共12题，每题4分，满分48分）7．； 8．；9． 7； 10．； 11．1； 12．6； 13．； 14．（等）；15．2 ；16．； 17．； 18．．三、解答题（本大题共7题，其中第19---22题每题10分，第23、24题每题12分，第25题14分，满分78分）19．解：原式·························································（4分） ································································（3分）．······························································（3分） 20．解：（1）∵//，//，∴．·····························································（1分）∵，∴．··························································（1分）∵//，∴．·····························································（1分）∵，∴．··························································（1分）解得，··························································（1分）．································································（1分）（2），．····························································（2分+2分） 21．解：（1）∵，，∴．·····························································（1分）在Rt△与Rt△中，，∴Rt△∽Rt△．·····················································（1分）∴．·····························································（1分）∵，，得．····························································（2分）（2）∵Rt△∽Rt△．∴．∴．·····························································（1分）∵，，∴．························································（1分）∵，，∴△∽△．························································（1分）∴．·····························································（1分）得．····························································（1分） 22．解：由点在函数的图像上，可得．···························································（1分）整理，得．·······················································（1分）解得，．························································（2分）∵正比例函数的值随的值增大而减小，∴．·····························································（2分）得，点．·························································（2分）由点在函数的图像上，可得．···························································（1分）∴．·····························································（1分）两个函数的解析式分别为，． 23．证明：（1）过点作⊥，垂足为点. ··············································（1分）∵S△AOD=, S△AOB=,∴.·····························································（2分）同理，．·························································（1分）∵，∴．···························································（1分）（2）∵，，∴△∽△．·······················································（1分）∴．···························································（1分）．····························································（1分）∵△的面积为，∴.················································（1分）又∵，∴．······················································（1分）同理，. ·······················································（1分）∴ ．···················································（1分） 24．解：（1）由抛物线经过点和点，得解得·························································（2分） ∴抛物线的表达式是．···············································（1分）点的坐标是．·····················································（1分）（2）联结交于点，过点作，为垂足．∵，，∴．由对称性可得．···················································（1分）∵，∴．在Rt△中，．······················································（1分）在Rt△中，，∵，∴．∴．··························································（1分）∵，∴．∴点的坐标是．·····················································（1分）（3）∵△是以为直角边的直角三角形， ∴或．设点点的坐标为．①当时，点只能在的下方．过点作，为垂足．∴，． ∵，，∴．∴．∴．∴．························································（1分）解得，．∵不合题意舍去，∴．∴点的坐标是．···················································（1分）②当时．同理可得点的坐标是．·············································（2分） 25．解：（1）过点作，为垂足．∵，∴． ∵，∴．∴//．∵//，∴． ························································（1分）同理可得．·······················································（1分）在Rt△中，，，∴．∴． ····························································（1分）∵，∴．·························································（1分）（2）延长、交于点．···············································（1分）∵//，∴．由，，，得，解得．·······················································（1分）∴．····························································（1分）∵//，∴．在Rt△中，，，，可得．···························································（1分）由，化简，得（）．···················································（2分）（3）①当点在线段上时，．··········································（2分）②当点在线段的延长线上时，点在线段的延长线上，．······················（2分）