上海市青浦区2020届九年级上学期期末(一模）质量调研数学试卷答案_0122026上海中考一模二模真题试卷_2025-2012年_2.上海中考数学一模二模（12-24）_一模_2020年上海市中考数学一模答案

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： 2 7． ； 8． 5−1； 9．−3e； 10．a0； 11．； 12．y=100(1+x)2 ； 3 4 5 2 5 5 3 13． ； 14．2 29； 15．2； 16． ； 17．1； 18． ． 5 5 2 三、解答题： 3 1 2 ( )2 19．解：原式=3 − + 8 + 1− 3 ． ······················································· （8分） 3 1 2 2 = 3−2+2+ 3−1． ······················································································ （1分） =2 3−1． ······································································································· （1分） 20．解：（1）∵四边形ABCD是平行四边形， ∴DC//AB，DC=AB， ························································································ （2分） BF AB ∴ = ． ······························································································· （1分） DF DE ∵DE∶EC =2∶3，∴DC∶DE =5∶2，∴AB∶DE =5∶2， ····························· （1分） ∴BF∶DF=5∶2． ····························································································· （1分） 5 （2）∵BF∶DF=5∶2，∴BF = BD． ·······························································（1分） 7 ∵BD= AD−AB，∴BD=a−b． ·························································· （1分） 5 5 5 ∴BF = BD = a− b． ········································································· （1分） 7 7 7 5 5 5 2 ∵AF = AB+BF ，∴AF =b+ a− b= a+ b． ························· （2分） 7 7 7 7 21．解：（1）∵∠ACB=90°，∴∠BCE+∠GCA=90°． ∵CG⊥BD，∴∠CEB=90°，∴∠CBE+∠BCE=90°， ∴∠CBE =∠GCA． ··························································································· （2分） 又∵∠DCB=∠GAC= 90°， 第 5 页∴△BCD ∽△CAG． ························································································ （1分） CD BC ∴ = ， ······························································································· （1分） AG CA 1 3 2 ∴ = ，∴AG = ． ············································································ （1分） AG 2 3 （2）∵∠GAC+∠BCA=180°，∴GA∥BC． ······················································· （1分） GA AF ∴ = ． ····························································································· （1分） BC FB AF 2 ∴ = ． ··································································································（1分） FB 9 AF 2 S 2 ∴ = ．∴ AFC = ． ··································································· （1分） AB 11 S 11 ABC 1 6 又∵S = 23=3，∴S = ． ··········································· （1分） ABC 2 AFC 11 22．解：由题意，得∠ABD=90°，∠D=20°，∠ACB=31°，CD=13． ··························· （1分） AB AB AB 在Rt△ABD中，∵tanD = ，∴BD= = ． ······················· （3分） BD tan20 0.36 AB AB AB 在Rt△ABC中，∵tanACB= ，∴BC = = ． ···················· （3分） BC tan31 0.6 ∵CD =BD -BC， AB AB ∴13= − ． ···························································································· （1分） 0.36 0.6 解得AB11.7米． ······························································································ （1分） 答：水城门AB的高约为11.7米． ········································································ （1分） AF FE 23．证明：（1）∵AF2 =FGFE，∴ = ． ························································ （1分） FG AF 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） CA CD （2）∵△CAD ∽△CBG，∴ = ． ···························································· （1分） CB CG 第 6 页又∵∠DCG=∠ACB，∴△CDG ∽△CAB． ·················································· （1分） DG CG ∴ = ． ····························································································· （1分） AB CB AE AG ∵AE∥BC，∴ = ． ········································································· （1分） CB GC AG GC DG AG ∴ = ，∴ = ， ································································· （1分） AE CB AB AE ∴DGAE = ABAG． ·············································································· （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y = x2+bx+c，得 1+b+c=0， b=−4，  解得： ························································· （2分） 9+3b+c=0. c=3. 所以，y = x2 −4x+3． 当x=2时，y =22 −42+3=−1 ∴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． ∵y = x2 −4x+3，∴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）． 1 PM ∴tan∠OCA= tan∠PCM．∴ = ． 3 MC 设PM=a，则MC=3a，PN=3-a． ∴P（3a，3-a）． ······························································································· （1分） 将P（3a，3-a）代入y = x2 −4x+3，得 (3a)2 −12a+3=3−a． 11 11 16 解得a = ，a =0（舍）．∴P（ ， ）． ···················································· （1分） 1 9 2 3 9 （3）设抛物线平移的距离为m．得y=(x−2)2 −1−m， 第 7 页∴D的坐标为（2，−1−m）． ···················································································· （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） 16 −m+1+m DE QF 2 9 ∴tan∠EOD = tan∠QDF．∴ = ．∴ = ． OE DF m+1 11 −2 3 1 1 解得m = ．所以，抛物线平移的距离为 ． ························································· （1分） 5 5 25．解：（1）∵AD//BC，∴∠EDQ=∠DBC．········································································ （1分） DE BD DE BD ∵ =1， =1，∴ = ． ······················································ （1分） DQ BC DQ BC ∴△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） EQ QD 2 ∵△DEQ ∽△BCD，∴ = ，∴EQ= x． ·································（1分） DC CB 5 （i）当EQ=EP时， ∴∠EQP =∠EPQ， ∵DE=DQ，∴∠EQP =∠QED，∴∠EPQ =∠QED， 2 EQ QP 2  ∴△EQP ∽△DEQ，∴ = ，∴  x  =(2x−10)x， DE EQ 5  125 解得 x= ，或x=0（舍去）． ······························································ （2分） 23 （ii）当QE=QP时， 2 25 ∴ x =2x−10，解得 x = ， ······························································· （1分） 5 4 25 ∵ 6，∴此种情况不存在． ·································································· （1分） 4 125 ∴BP = 23 （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=4 6， ∵BP= DQ=m，∴PQ=10-2m． ∵EQ∥DC∴∠PQH =∠BDG． 又∵∠PHQ =∠BGD= 90°， 第 8 页∴△PHQ ∽△BGD． ······················································································· （1分） PH PQ HQ PH 10−2m HQ ∴ = = ，∴ = = ． BG BD GD 4 6 10 2 10−2m 2 6(10−2m) ∴HQ = ，PH = ． ············································ （2分） 5 5 10−2m 2m ∴EH = + =2， 5 5 PH 2 6(10−2m) 1 2 6 ∴tanPEQ= =  =2 6− m． ··············· （1分） EH 5 2 5 第 9 页