上海奉贤区2019第一学期期末调研测试卷九年级数学含答案

奉贤区2018学年度九年级数学调研测试参考答案及评分说明（201901）
一、选择题：（本大题共6题，每题4分，满分24分）
1．C ； 2．D ； 3．A ； 4．B ； 5．A ； 6．D . 二、填空题：（本大题共12题，每题4分，满分48分）
三、解答题（本大题共7题，其中19-22题每题10分，23、24题每题12分，25题14分，满分78分） 19．解：（1）2(2)222y x x x x =-+=-+
222112(1)1x x x =-+-+=-+． ······················································· （3分）
所以抛物线的顶点坐标是(1,1) ． ························································· （2分） （2）由题意得新抛物线的顶点坐标是(1,0)， ················································ （2分）
所以新抛物线的表达式是2(1)y x =-． ························································ （3分）
20．解：（1）∵AD 是△ABC 的中线，BC b =uu u r r ，∴12
BD b =uu u r r ． ························· （1分）
∵=，∴12
AD a b =+uuu r r r
． ···································································· （1分）
∵G 是重心，∴22121()33233
AG AD a b a b ==+=+uuu r uuu r r r r r
． ···································· （2分）
∴21113333
BG AB AG a a b a b =-+=-++=-+uu u r uu u r uuu r r r r r r
． ········································ （1分）
（2）延长BG 交边AC 于点H ． ·································································· （1分） ∵∠GAC=∠GCA ，∴GA =GC ． ·································································· （1分） ∵G 是重心，AC=2，∴AH =1
12
AC =． ·
······················································ （1分） ∴BH ⊥AC ．
在Rt △ABH 中，90AHB ??，3AB =，∴BH
·
··········· （1分）
∴23BG BH ==
············································································ （1分）
7. 5a b -r r
；
8.
； 9. 1m ≠； 10. 22y x =-（等）； 11．3x =；
12.
1
2
； 13．35； 14．
1
2
； 15．6；
16．16；
17
．；
18．
25
8
．
21. 解：（1）过点A 作AH BD ^，垂足为点H ． ··········································· （1分）
在Rt △ABC 中，90BAC ??，5BC =，AC =
∴AB =． ·
···································································· （1分） ∵90BAC AHB ???，B B ??，∴△BAH ∽△BCA ．
∴
AB BH
BC AB =，即5=
1BH =． ····················································· （2分） ∴22BD BH ==． ················································································ （1分） （2）过点D 作DQ AC ^，垂足为点Q ． ∵90BAC
??，∴DQ ∥AB ．∴
DQ CD
AB CB
=
．
∵AB =3CD BC BD =-=3
5
=，即DQ =． ····················· （3分）
在Rt △ADQ 中，90AQD ??，AD AB ==
∴3
sin 5
DQ DAC AD ∠=
=． ·
··········································································· （2分） 22．解：（1）过点C 作CH AB ^，垂足为点H ， ∴过点D 作DQ AB ^，垂足为点Q ． ···················································· （1分）
在Rt △ACH 中，90AHC ??，60CAB ??，20AC =，
∴sin 20CH AC CAH =∙∠==．
·················································· （2分） ∵CH ∥DQ ，∴
CH BC
DQ BD
=
． ∵40BD =，10CD =，30BC BD CD =-=，
3040=．∴23DQ ≈（厘米）
． ··············································· （2分） 即支点D 到滑轨MN 的距离约是23厘米．
（2）在Rt △ACH 中，90AHC ??，60CAB ??，20AC =，
∴1
cos 20102
AH AC CAH =∙∠=⨯=．
在Rt △BCH 中，90BHC ??，CH =30BC =，
∴BH =
∴1034.5AB AH BH =+=+（厘米）． ··········································· （2分）
同理求得'40.6A B =（厘米）． ············································ （2分） ∴'40.634.5 6.16AA =-=≈（厘米）． ··················································· （1分） 即滑块A 向左侧移动的距离约是6厘米．
23．证明：（1）∵EF 是BD 的垂直平分线，∴EB ED =． ····························· （1分）
∵EC EA ED •=2，∴2EB EA EC =⋅， ∴
EB EC
EA EB
=
． ························· （1分） 又∵BEA CEB ∠=∠，∴△BEA ∽△CBE ． ············································ （2分）
∴EBA C ∠=∠． ············································································· （1分） （2）∵EB=ED ，∴EBD EDB ∠=∠． ························································ （1分） 即EBA ABD C DBC ∠+∠=∠+∠．∴ABD DBC ∠=∠． ······················· （1分）
∵BD CD =，∴DBC C ∠=∠． ······················································· （1分） ∴ABD C ∠=∠． ··········································································· （1分） 又BAD CAB ∠=∠，∴△ABD ∽△CAB ． ············································· （2分） ∴
AB AD
AC AB
=
．∴AC AD AB •=2． ······················································ （1分） 24．解：（1）由题意得，抛物线2
y ax bx =+经过点A (6，0)和点B (1，－5)，
代入得3660,5.a b a b ì+=ïïíï+=-ïî 解得 1,6.
a b ì=ïïí
ï=-ïî ∴抛物线的表达式是26y x x =-. （4分） 由题意得，设直线AB 的表达式为y kx b =+，它经过点A (6，0)和点B (1，－5)， 代入得60,5.k b k b ì+=ïïíï+=-ïî 解得 1,6.k b ì=ïïíï=-ïî
∴直线AB 的表达式是6y x =-. ··· （2分） （2）过点O 作OH AB ^，垂足为点H ．
设直线AB 与y 轴交点为点D ，则点D 坐标为()0,6-.
∴45ODA
OAD ???，cos45DH OH OD ==∙︒=
∵BD =
BH =.
在Rt △OBH 中，90OHB ??，3
tan 2
OH OBH
BH ?=． ·
························· （2分） ∵∠BOC 的正切值是
3
2
，∴BOC CBO ??. ·········································· （1分） ①当点C 在点B 上方时，BOC CBO ??.∴CO CB =．
设点C (,6)x x -，
解得 174x =，1776644
x -=-=-. --------------------------------------------------------------------（2
分）
所以点D
坐标为17
7,44⎛⎫- ⎪⎝⎭.
②当点C 在点B 下方，BOC CBO ??时，OC //AB . 点C 不在直线AB 上. ·
······ （1分） 综上所述，如果∠BOC 的正切值是32
，点C 的坐标是177,44⎛⎫- ⎪⎝⎭．
25．解：（1）∵CD ∥EF ，DF ∥CE ，
∴四边形DFEC 是平行四边形． ·································································· （1分） ∴EF=DC ． ···························································································· （1分） ∵26AB CD ==，∴3CD EF ==． ∵AB ∥CD ，∴AB ∥EF ． ∵点G 与点C 重合，∴
1
2
EF CE AB BC ==．∴:1CE BE =． ·
································ （2分） （2）过点C 作CQ ∥AG ，交AB 于点Q ，交EF 于点P ． 过点C 作CM ⊥AB ，交AB 于点M ，交EF 于点N ． 在Rt △BCM 中， 90CMB
??，4CM AD ==，3BM AB CD =-=，∴5BC =．
∵AB ∥EF ∥CD ，∴GC=PF =AQ ． ∴EP CE
BQ BC
=
． 又3EF =，∴
365GC m
CG -=-．
∴1565m
GC m
-=-． ··················································································· （2分）
∴35m
DG DC GC m
=-=-． ····································································· （1分）
∵NE ∥MB ，∴
CN CE
CM BC
=
． 又4CM AD ==，∴45CN m =，45
m
CN =
． ················································· （1分） ∴2
113462254255DFG
m m m S DG CN m m
∆=∙∙=∙∙=
--． ·········································· （1分） （3）当AFD ∆∽ADG ∆时，
∵∠DAB =90°，∴ADG ∆是直角三角形，∴AFD ∆也是直角三角形． ∵90DAF 泄?，90FDA 泄?，∴90DFA
??． ·
····································· （1分） ∵90FAD
ADF ???，90FDC ADF
???，∴FAD FDC ??．
∵AB ∥EF ，∴B
CEF ??．
∵四边形DFEC 是平行四边形，∴FDC CEF ??．
∴B
FDC FAD ???． ·
······································································ （1分） 在Rt △BCM 中， 90CMB ??，3BM AB CD =-=，5BC =，
∴3
cos 5
BM B BC =
=． ·
··············································································· （2分） ∴3
cos 5
DAG ∠=．··················································································· （1分）
即∠DAG 的余弦值是3
5
．。