【文档说明】上海青浦区2022届九年级初三数学一模试卷+答案.pdf，共(11)页，934.131 KB，由baby熊上传

转载请保留链接：https://www.ichengzhen.cn/view-125368.html

以下为本文档部分文字说明：

2022年上海市青浦区中考数学一模试卷2022.1一、选择题：（本大题共6题，每小题4分，满分24分）[每题只有一个正确选项，在答题纸相应题号的选项上用2B铅笔正确填涂]1.下列图形，一定相似的是（）（A）两个直角三角形；（B）两个等腰三角形；

（C）两个等边三角形；（D）两个菱形．2.如图，已知AB//CD//EF，它们依次交直线1l、2l于点A、C、E和点B、D、F．如果AC∶CE=2∶3，BD=4，那么BF等于（）（A）6；（B）8；（C）10；（D）12．3.在Rt△ABC中，∠C=90º，那么cotA等于（）

（A）ACBC；（B）ACAB；（C）BCAC；（D）BCAB．4.如图，点D、E分别在△ABC的边AB、BC上，下列条件中一定能判定DE∥AC的是（）（A）ADBEDBCE；（B）BDBEADEC；（C）ADCEABBE；（D）BDDEBAAC

．5.如果2ab（a、b均为非零向量），那么下列结论错误．．的是（）（A）||2||ab；（B）a∥b；（C）20ab；（D）a与b方向相同．6.如图，在平行四边形ABCD中，点E在边BA的延长线上，联结EC，交边AD于点F，则下列结论一定正确的是（）

（A）EAAFABBC；（B）EAFDABAF；（C）AFEABCCD；（D）EAAFEBAD．二、填空题：（本大题共12题，每小题4分，满分48分）[请将结果直接填入答题纸的相应位置]7.已知线段b是线段a、c的比例中项，且a=1，b=3，那么c=▲．8.计算：

32(2)aab=▲．9.如果两个相似三角形的周长比为2∶3，那么它们的对应高的比为▲．10.二次函数21yxx的图像有最▲点．（填“高”或“低”）（第6题图）（第2题图）（第4题图）11.将抛物线2y

x向下平移2个单位，所得抛物线的表达式是▲．12.如果抛物线cbxaxy2（其中a、b、c是常数，且a≠0）在对称轴左侧的部分是下降的，那么a▲0．（填“＜”或“＞”）13.在△ABC中，∠C=

90º，如果tan∠A=2，AC=3，那么BC=▲．14.如图，点G为等边三角形ABC的重心，联结GA，如果AG=2，那么BC=▲．（第16题图）（第14题图）（第15题图）15.如图，如果小华沿坡度为1:3

的坡面由A到B行走了8米，那么他实际上升的高度为▲米．16.如图，在边长相同的小正方形组成的网格中，点A、B、O都在这些小正方形的顶点上，那么sin∠AOB的值为▲．17.如图，在矩形ABCD中，∠BCD的角

平分线CE与边AD交于点E，∠AEC的角平分线与边CB的延长线交于点G，与边AB交于点F，如果AB=32，AF=2BF，那么GB=▲．18.如图，一次函数(00)，yaxbab的图像与x轴，y轴分别相交于点A，点B，将

它绕点O逆时针旋转90°后，与x轴相交于点C，我们将图像过点A，B，C的二次函数叫做与这个一次函数关联的二次函数．如果一次函数(0)ykxkk的关联二次函数是22ymxmxc（0m），那么这个一次函数的解析式为▲．（第1

7题图）（第18题图）三、解答题（本大题共7题，满分78分）[请将解题过程填入答题纸的相应位置]19.（本题满分10分）计算：01sin451+2cos30tan60cot60．20.（本题满分10分,第（1）小题5分，第（2）小题5分）如图，在平行四边形A

BCD中，点E在边AD上，CE、BD相交于点F，BF=3DF．（1）求AE∶ED的值；（2）如果DCa，EAb，试用a、b表示向量CF．21.（本题满分10分,第（1）小题5分，第（2）小题5分）如图，在△ABC中，点D是BC的中点，联结A

D，AB=AD，BD=4，41tanC．（1）求AB的长；（2）求点C到直线AB的距离．（第21题图）22.（本题满分10分）如图，某校的实验楼对面是一幢教学楼，小张在实验楼的窗口C（AC∥BD）处测得教学楼顶部D的仰角为27°，教学楼底部B的俯角为13°，量得实验楼与教学楼之间的距离A

B=20米．求教学楼BD（BD⊥AB）的高度．（精确到0.1米）（参考数据：sin13°≈0.22，cos13°≈0.97，tan13°≈0.23，sin27°≈0.45，cos27°≈0.89，tan

27°≈0.51）23.（本题满分12分，第（1）小题6分，第（2）小题6分）已知：如图，在四边形ABCD中，AC、BD相交于点E，∠ABD=∠CBD，2DCDEDB．（1）求证：△AEB∽△DEC；（2）求证：BCADCEB

D．（第23题图）（第22题图）24.（本题满分12分，其中第（1）小题4分，第（2）小题4分，第（3）小题4分）如图，在平面直角坐标系xOy中，抛物线2yxbxc与x轴交于点A（-1，0）和点B（3，0），与y轴交于点C，顶点为点D．（1）求该抛物线的表达式及点C的坐标；（2）联结B

C、BD，求∠CBD的正切值；（3）若点P为x轴上一点，当△BDP与△ABC相似时，求点P的坐标．（第24题图）（备用图）25.（本题满分14分，其中第（1）小题4分，第（2）小题4分，第（3）小题6分）在四边形ABCD中，AD∥

BC，AB=5，AD=2，DC=25，tan∠ABC=2（如图）．点E是射线AD上一点，点F是边BC上一点，联结BE、EF，且∠BEF=∠DCB．（1）求线段BC的长；（2）当FB=FE时，求线段BF的长；（3）当点E在线段AD的延长线上时，设DE=x，BF=y，求y关于x的函数解析式，并

写出x的取值范围．（第25题图）（备用图）2022年上海市青浦区中考数学一模试卷答案一、选择题：1．C；2．C；3．A；4．B；5．D；6．D．二、填空题：7．9；8．4ab；9．2:3；10．高；11．22yx；12．；13．6；14．23；15．4

；16．1010；17．22；18．3+3yx．三、解答题：19．解：原式=102331+23223．·········································

··（4分）=21+3132．······························································（4分）=22．···············································

··································（2分）20．解：（1）∵四边形ABCD是平行四边形，∴AD//BC，AD=BC．·················································

·················（2分）∴BCBFEDDF．·········································································（1分）∵BF=3DF，∴3BFDF．∴3BCED．··············

·······························································（1分）∴3ADED．∴AE∶ED=2．·····························································

·············（1分）（2）∵AE∶ED=2∶1，∴12DEEA．∵EAb，∴12DEb．······························

·············································（1分）∵CEDEDC，∴12CEba．··········································

····························（1分）∵AD//BC，∴CFBFCEBD．························································（1分）∵BF=3DF，∴34BFBD．

∴34CFCE．∴34CFCE．········································································（1分）∴31334284

CFbaba．·················································（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）∵1tan4C，∴14AHHC，∴32AH．······································（1分）∵22ABBHAH，∴223522

2AB．··························································（2分）（2）过点C作CG⊥BA，交BA的延长线于点G．·····························

·（1分）∵sinAHCGBABBC，·····························································（2分）∴32582CG．···········································

······························（1分）∴245CG．∴点C到直线AB的距离为245．··················································（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，∵tanDH

DCHCH，∴tan272010.2DH．······（4分）在Rt△HCB中，∵tanHBHCBCH，∴tan13204.6BH．·········（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）∵2DCDEDB，∴DCDBDEDC．····················

··················································（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，∴

AEDEEBEC．·············································（1分）又∵∠AED=∠BEC，∴△AED∽△BEC．····································（1分

）∴∠ADE=∠BCE．····································································（1分）又∵∠ABD=∠DBC，∴△BDA∽△BCE．····································（1分）∴

BDDABCCE．········································································（1分）∴BCADCEBD．··

·························································（1分）24．解：（1）将A（-1，0）、B（3，0）代入2++yxbxc，得10930.，bcbc解

得：23.，bc············································（2分）所以，223yxx．······························

······························（1分）当x=0时，3y．∴点C的坐标为（0，-3）．······························（1分）（2）∵2223=14yxxx，∴点D的坐标为（1，

-4）．·············（1分）∵B（3，0）、C（0，-3）、D（1，-4），∴BC=32，DC=2，BD=25．∴222+18220BCDCDB．············································（1分

）∴∠BCD=90°．·········································································（1分）∴tan∠CBD=21332DCBC．·····································

··················（1分）（3）∵tan∠ACO=13AOOC，∴∠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ACDBCBBP时，∴102532BP．∴BP=6．∴P（-3，0）．·······························（1分）（ii）当ACBPCBDB时，∴103225BP．∴

BP=103．∴P（-13，0）．···························（1分）综上，点P的坐标为（-3，0）或（-13，0）．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=5，∴AH=2，BH=1．·····························································

··········（2分）∴DG=2．∵DC=25，∴CG=224DCDG．··········································（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=12DGGC．∵FB=FE，∴∠FEB=∠FBE．∵∠FEB=∠C，∴∠FBE=∠C．················

····································（1分）∴tan∠FBE=12．∴12EMBM，∴BM=4．·······································（1分）∵222FMEMFE，∴22242FBFB．····

··················（1分）∴BF=52．···············································································（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）∴BFBEBEBN．∴2BEBFBN

．··············································（1分）过点E作EQ⊥BC，垂足为点Q．可得EQ=2，BQ=x+3．∴22222232=613BEQEBQ

xxx．·······················（1分）∴27613yxxx．∴26137xxyx114502x．······································（2分

）