2020年1月青浦初三数学一模试卷参考答案

这是一份2020年1月青浦初三数学一模试卷参考答案，共6页。试卷主要包含了7米．等内容，欢迎下载使用。
青浦区2019学年第一学期期终学业质量调研 九年级数学试卷      参考答案及评分说明2020.1一、选择题：1．A；      2．B；      3．C；      4．D；       5．A；        6．D．二、填空题：7．；     8．；    9．；    10．；   11．；      12．；13．；   14．；    15．；     16．；      17．；   18．．三、解答题：19．解：原式=．························································（8分）=．························································（1分）=．························································（1分）20．解：（1）∵四边形ABCD是平行四边形， ∴DC//AB，DC=AB，··········································（2分）∴．·······················································（1分）∵DE∶EC =2∶3，∴DC∶DE =5∶2，∴AB∶DE =5∶2，················（1分）∴BF∶DF=5∶2．·············································（1分）（2）∵BF∶DF=5∶2，∴．··········································（1分）∵，∴．····················································（1分） ∴．························································（1分）∵，∴．····················································（2分） 21．解：（1）∵∠ACB=90°，∴∠BCE+∠GCA=90°．            ∵CG⊥BD，∴∠CEB=90°，∴∠CBE+∠BCE=90°，∴∠CBE =∠GCA．············································（2分）又∵∠DCB=∠GAC= 90°， ∴△BCD ∽△CAG．···········································（1分）∴，·······················································（1分）∴，∴．····················································（1分）（2）∵∠GAC+∠BCA=180°，∴GA∥BC．···························（1分）∴．······················································（1分）∴．······················································（1分）∴．∴．···················································（1分）又∵，∴．·················································（1分）22．解：由题意，得∠ABD=90°，∠D=20°，∠ACB=31°，CD=13．···············（1分）在Rt△ABD中，∵，∴．···········································（3分）在Rt△ABC中，∵，∴．···········································（3分）∵CD =BD -BC，∴． ··························································（1分）解得米． ·······················································（1分）答：水城门AB的高约为11.7米．······································（1分）23．证明：（1）∵，∴．·················································（1分）又∵∠AFG=∠EFA，∴△FAG∽△FEA．··························（1分）∴∠FAG=∠E．·············································（1分）∵AE∥BC，∴∠E=∠EBC．····································（1分）∴∠EBC =∠FAG．···········································（1分）又∵∠ACD=∠BCG，∴△CAD ∽△CBG．·························（1分）（2）∵△CAD ∽△CBG，∴．·······································（1分）又∵∠DCG=∠ACB，∴△CDG ∽△CAB．·························（1分）∴．······················································（1分）∵AE∥BC，∴．·············································（1分）∴，∴，···················································（1分）∴．······················································（1分）24．解：（1）∵A的坐标为（1，0），对称轴为直线x=2，∴点B的坐标为（3，0）······（1分）将A（1，0）、B（3，0）代入，得   解得：   （2分）所以，．       当x=2时，∴顶点坐标为（2，-1）·········································（1分）．（2）过点P作PN⊥x轴，垂足为点N．过点C作CM⊥PN，交NP的延长线于点M．∵∠CON=90°，∴四边形CONM为矩形．∴∠CMN=90°，CO= MN．∵，∴点C的坐标为（0，3）·········································（1分）．∵B（3，0），∴OB=OC．∵∠COB=90°，∴∠OCB=∠BCM = 45°，············（1分）．又∵∠ACB=∠PCB，∴∠OCB-∠ACB =∠BCM -∠PCB，即∠OCA=∠PCM．·······（1分）．∴tan∠OCA= tan∠PCM．∴．设PM=a，则MC=3a，PN=3-a．∴P（3a，3-a）． ··············································（1分）将P（3a，3-a）代入，得．解得，（舍）．∴P（，）．·········································（1分）（3）设抛物线平移的距离为m．得，∴D的坐标为（2，）． （1分）过点D作直线EF∥x轴，交y轴于点E，交PQ的延长线于点F．∵∠OED=∠QFD=∠ODQ=90°，∴∠EOD+∠ODE = 90°，∠ODE+∠QDF = 90°，∴∠EOD=∠QDF， ··············································（1分）∴tan∠EOD = tan∠QDF．∴．∴．解得．所以，抛物线平移的距离为．·····································（1分）25．解：（1）∵AD//BC，∴∠EDQ=∠DBC．··································（1分） ∵，，∴．··················································（1分）∴△DEQ ∽△BCD．···········································（1分）∴∠DQE=∠BDC，∴EQ//CD．···································（1分）（2）设BP的长为x，则DQ=x，QP=2x-10．··························（1分）     ∵△DEQ ∽△BCD，∴，∴．······································（1分）（i）当EQ=EP时，∴∠EQP =∠EPQ，∵DE=DQ，∴∠EQP =∠QED，∴∠EPQ =∠QED，∴△EQP ∽△DEQ，∴，∴，解得 ，或（舍去）．···········································（2分）（ii）当QE=QP时，∴，解得 ，··················································（1分）∵，∴此种情况不存在．········································（1分）∴ （3）过点P作PH⊥EQ，交EQ的延长线于点H；过点B作BG⊥DC，垂足为点G．∵BD=BC，BG⊥DC，∴DG=2，BG，∵BP= DQ=m，∴PQ=10-2m．∵EQ∥DC∴∠PQH =∠BDG．又∵∠PHQ =∠BGD= 90°， ∴△PHQ ∽△BGD．··········································（1分）∴，∴．∴，．······················································（2分）∴，∴．·······················································（1分）