2020年1月青浦初三数学一模试卷参考答案

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）