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2022 年上海市青浦区中考数学一模试卷
官方标答
一、选择题：
1．C； 2．C； 3．A； 4．B； 5．D； 6．D．
二、填空题：
7． ； 8． ； 9． ； 10．高； 11． ； 12． ；
13． ； 14． ； 15． ； 16． ； 17． ； 18．
．
三、解答题：
19．解：原式= ．···········································（4分）
= ．······························································（4分）
= ．··················································································（2分）
20．解：（1）∵四边形ABCD是平行四边形，
∴AD//BC，AD=BC．···································································（2分）
∴ ．··········································································（1分）
∵BF=3DF，∴ ．
∴ ．··············································································（1分）
∴ ．
∴AE∶ED=2．··············································································（1分）
（2）∵AE∶ED=2∶1，∴ ．
∵ ，
∴ ．············································································（1分）∵ ，
∴ ．·······································································（1分）
∵AD//BC，∴ ． ··························································（1分）
∵BF=3DF，∴ ．∴ ．
∴ ．·········································································（1分）
∴ ．··················································（1分）
21．解：（1）∵过点A作AH⊥BD，垂足为点H．
∵AB=AD， ∴BH=HD．·······························································（1分）
∵点D是BC的中点， ∴BD=CD．
∵BD=4，∴CD=4．
∴HC=6．··················································································（1分）
∵ ，∴ ，∴ ．········································（1分）
∵ ，
∴ ．···························································（2分）
（2）过点C作CG⊥BA，交BA的延长线于点G． ·······························（1分）
∵ ，······························································（2分）
∴ ． ··········································································（1分）
∴ ．
∴点C到直线AB的距离为 ．····················································（1分）
22．解： 过点C作CH⊥BD，垂足为点H．······················································（1分）由题意，得∠DCH=27°，∠HCB=13°，AB=CH=20（米）．
在Rt△DHC中，∵ ，∴ ．·········（4分）
在Rt△HCB中，∵ ，∴ ．·············（4分）
∴BD =HD+HB 10.2 +4.6=14.8（米）．················································（1分）
答：教学楼BD的高度约为14.8米．
23．证明：（1）∵ ，
∴ ．·······································································（1分）
又∵∠CDE=∠BDC，∴△DCE∽△DBC．············································（1分）
∴∠DCE=∠DBC．······································································（1分）
∵∠ABD=∠DBC，
∴∠DCE=∠ABD．······································································（1分）
又∵∠AEB=∠DEC，∴△AEB ∽△DEC．···········································（2分）
（2）∵△AEB ∽△DEC，∴ ．··················································（1分）
又∵∠AED=∠BEC，∴△AED ∽△BEC．···········································（1分）
∴∠ADE=∠BCE．······································································（1分）
又∵∠ABD=∠DBC，∴△BDA ∽△BCE．···········································（1分）
∴ ．·········································································（1分）
∴ ． ····························································（1分）
24．解：（1）将A（-1，0）、B（3，0）代入 ，得
解得： ············································（2分）
所以， ． ····························································（1分）
当x=0时， ．∴点C的坐标为（0，-3）．······························（1分）
（2）∵ ，∴点D的坐标为（1，-4）．············（1分）
∵B（3，0）、C（0，-3）、D（1，-4），∴BC= ，DC= ，BD=．
∴ ．·············································（1分）
∴∠BCD=90°． ·········································································（1分）
∴tan∠CBD= ．·························································（1分）
（3）∵tan∠ACO= ，∴∠ACO=∠CBD．········································（1分）
∵OC =OB，∴∠OCB=∠OBC=45°．∴∠ACO+∠OCB =∠CBD+∠OBC．
即：∠ACB =∠DBO．································································（1分）
∴当△BDP与△ABC相似时，点P在点B左侧．
（i）当 时，
∴ ．∴BP=6．∴P（-3，0）．································（1分）
（ii）当 时，
∴ ．∴BP= ．∴P（- ，0）．····························（1分）
综上，点P的坐标为（-3，0）或（- ，0）．
25．解：（1）过点A、D分别作AH⊥BC、DG⊥BC，垂足分别为点H、点G．
可得：AD=HG=2，AH=DG．∵tan∠ABC=2，AB= ，
∴AH=2，BH=1．········································································（2分）
∴DG=2．
∵DC= ，∴CG= ．············································（1分）
∴BC=BH+HG+GC=1+2+4=7．·······················································（1分）
（2）过点E作EM⊥BC，垂足为点M．可得EM=2．
由（1）得，tan∠C= ．
∵FB=FE，∴∠FEB=∠FBE．
∵∠FEB=∠C，∴∠FBE=∠C．·························································（1分）∴tan∠FBE= ．∴ ，∴BM=4．··········································（1分）
∵ ，∴ ．························（1分）
∴BF= ．················································································（1分）
（3）过点E作EN//DC，交BC的延长线于点N．
∵DE//CN，∴四边形DCNE是平行四边形．
∴DE=CN，∠DCB=∠ENB．
∵∠FEB=∠DCB，∴∠FEB=∠ENB．·················································（1分）
又∵∠EBF=∠NBE，
∴△BEF ∽△BNE．·······································································（1分）
∴ ．∴ ．················································（1分）
过点E作EQ⊥BC，垂足为点Q．可得EQ=2，BQ=x+3．
∴ ．························（1分）
∴ ．
∴ ．·······································（2分）
