中考精品：数学压轴题汇编(含解题过程-共67页).pdf

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冲刺中考数学压轴题汇编(含解题过程)
(2009年重庆市):如图,在平面直角坐标系xOy中,矩形OABC的边OA在y
轴的正半轴上,OC在x轴的正半轴上,OA=2,OC=∠AOC的平分线交AB
于点D,连接DC,过点D作DE⊥DC,交OA于点E.
(1)求过点E、D、C的抛物线的解析式;
(2)将∠EDC绕点D按顺时针方向旋转后,角的一边与y轴的正半轴交于点F,另一边与
6
(1)中的抛物线交于另一点M,点M的横坐标为,那么
5
EF=2GO是否成立?若成立,请给予证明;若不成立,请说明理由;
(3)对于(2)中的点G,在位于第一象限内的该抛物线上是否存在点Q,使得直线GQ与
AB的交点P与点C、G构成的△PCG是等腰三角形?若存在,请求出点Q的坐标;若不存
在,请说明理由.
y
D
AB
E
x
OC
26题图
:(1)由已知,得C(3,0),D(2,2),
ADE90°CDBBCD,
1
AEADtanADE2tanBCD21.
2
E(0,1).·····························································································(1分)
设过点E、D、C的抛物线的解析式为yax2bxc(a0).
将点E的坐标代入,得c1.
将c1和点D、C的坐标分别代入,得
4a2b12,
······················································································(2分)
9a3b10.
5
a

6
解这个方程组,得
13
b
6
513
故抛物线的解析式为yx2x1.···················································(3分)
66
文档:.
(2)EF2GO成立.············································································(4分)
6
点M在该抛物线上,且它的横坐标为,
5
12
点M的纵坐标为.···········································································(5分)
5
y
F
设DM的解析式为ykxb(k0),
1M
D
将点D、M的坐标分别代入,得AB
2kb2,1
1k,E

612解得2
kb.
515b3.
1x
OGKC
1
DM的解析式为yx3.······························································(6分)
2
F(0,3),EF2.··············································································(7分)
过点D作DK⊥OC于点K,
则DADK.
ADKFDG90°,
FDAGDK.
又FADGKD90°,
△DAF≌△DKG.
KGAF1.
GO1.····························································································(8分)
EF2GO.
(3)点P在AB上,G(1,0),C(3,0),则设P(1,2).
PG2(t1)222,PC2(3t)222,GC2.
①若PGPC,则(t1)222(3t)222,
解得t2.P(2,2),此时点Q与点P重合.
Q(2,2).····························································································(9分)
②若PGGC,则(t1)2222,
解得t1,P(1,2),此时GP⊥x轴.
GP与该抛物线在第一象限内的交点Q的横坐标为1,
7
点Q的纵坐标为.
3
文档:.
7
Q1,.························································································(10分)
3
③若PCGC,则(3t)22222,
解得t3,P(3,2),此时PCGC2,△PCG是等腰直角三角形.
过点Q作QH⊥x轴于点H,
y
则QHGH,设QHh,
Q(Q)
D(P)
AB(P)
P
Q(h1,h).
Q
E
513
(h1)2(h1)1h.
66
7x
OGHC
解得h,h2(舍去).
152
127
Q,.·····································(12分)
55
综上所述,存在三个满足条件的点Q,
7127
即Q(2,2)或Q1,或Q,.
355
(2009年重庆綦江县)26.(11分)如图,已知抛物线ya(x1)233(a0)经过点
A(2,0),抛物线的顶点为D,过O作射线OM∥
射线OM于点C,B在x轴正半轴上,连结BC.
(1)求该抛物线的解析式;
(2)若动点P从点O出发,以每秒1个长度单位的速度沿射线OM运动,设点P运动的
时间为t(s).问当t为何值时,四边形DAOP分别为平行四边形?直角梯形?等腰梯形?
(3)若OCOB,动点P和动点Q分别从点O和点B同时出发,分别以每秒1个长度单
位和2个长度单位的速度沿OC和BO运动,当其中一个点停止运动时另一个点也随之停止
yM
(s),连接PQ,当t为何值时,四边形BCPQ的面积最小?
D
C
并求出最小值及此时PQ的长.
P
A
OQBx
文档:.
*:(1)抛物线ya(x1)233(a0)经过点A(2,0),
3
09a33a··············································································1分
3
32383
二次函数的解析式为:yx2x···········································3分
333
(2)D为抛物线的顶点D(1,33)过D作DNOB于N,则DN33,
AN3,AD32(33)26DAO60°············································4分
OM∥AD
yM
D
①当ADOP时,四边形DAOP是平行四边形C
OP6t6(s)··········································5分
P
②当DPOM时,四边形DAOP是直角梯形H
A
过O作OHAD于H,AO2,则AH1OENQBx
(如果没求出DAO60°可由Rt△OHA∽Rt△DNA求AH1)
OPDH5t5(s)················································································6分
③当PDOA时,四边形DAOP是等腰梯形
OPAD2AH624t4(s)
综上所述:当t6、5、4时,对应四边形分别是平行四边形、直角梯形、等腰梯形.·7分
(3)由(2)及已知,COB60°,OCOB,△OCB是等边三角形
则OBOCAD6,OPt,BQ2t,OQ62t(0t3)
3
过P作PEOQ于E,则PEt·······························································8分
2
113
S633(62t)t
BCPQ222
33263

=t3····················································································9分
228
363
当t时,S的面积最小值为3·························································10分
2BCPQ8
文档:.
333933
此时OQ3,OP=,OEQE3PE
24444
22
33933
PQPE2QE2··············································11分

442

(2009年河北省)26.(本小题满分12分)
如图16,在Rt△ABC中,∠C=90°,AC=3,AB=
个单位长的速度向点A匀速运动,到达点A后立刻以原来的速度沿AC返回;点Q从点A
、Q的运动,DE保持垂直
平分PQ,且交PQ于点D,交折线QB-BC-、Q同时出发,当点Q到达点
B时停止运动,、Q运动的时间是t秒(t>0).
B
(1)当t=2时,AP=,点Q到AC的距离是;
(2)在点P从C向A运动的过程中,求△APQ的面积S与
t的函数关系式;(不必写出t的取值范围)
(3)在点E从B向C运动的过程中,四边形QBED能否成E
Q
为直角梯形?若能,,请说明理由;
(4)当DE经过点C时,请直接写出t的值.
..D
APC
8
:(1)1,;图16B
5
(2)作QF⊥AC于点F,如图3,AQ=CP=t,∴AP3t.
由△AQF∽△ABC,BC52324,E
Q
QFt4
得.∴QF
455AC
FP
14图3B
∴S(3t)t,
25B
26
即St2t.
55
EQ
(3)能.
E
①当DE∥QB时,
∵DE⊥PQ,∴PQ⊥QB,
DP
AC图5
此时∠AQP=90°.P
图4
AQAPB
由△APQ∽△ABC,得,
ACAB
t3t9
即.解得t.
358QG
②如图5,当PQ∥BC时,DE⊥BC,四边形QBED是直角梯形.
此时∠APQ=90°.D
C(E)
AQAPAP
由△AQP∽△ABC,得,
图6
ABACB
QG
文档
D
C(E):.
t3t15
即.解得t.
538
545
(4)t或t.
214
【注:①点P由C向A运动,DE经过点C.
方法一、连接QC,作QG⊥BC于点G,如图6.
34
PCt,QC2QG2CG2[(5t)]2[4(5t)]2.
55
345
由PC2QC2,得t2[(5t)]2[4(5t)]2,解得t.
552
方法二、由CQCPAQ,得QACQCA,进而可得
55
BBCQ,得CQBQ,∴AQBQ.∴t.
22
②点P由A向C运动,DE经过点C,如图7.
3445
(6t)2[(5t)]2[4(5t)]2,t】
5514
(2009年河南省)23.(11分)如图,在平面直角坐标系中,已知矩形ABCD的三个
顶点B(4,0)、C(8,0)、D(8,8).抛物线y=ax2+bx过A、C两点.
(1)直接写出点A的坐标,并求出抛物线的解析式;
(2),同时点Q从点C出发,沿线段CD
,⊥AB交AC于点
E
①过点E作EF⊥AD于点F,,线段EG最长?
②、Q运动的过程中,判断有几个时刻使得△CEQ是等腰三角形?
请直接写出相应的t值.
解.(1)点A的坐标为(4,8)…………………1分
将A(4,8)、C(8,0)两点坐标分别代入y=ax2+bx
8=16a+4b
得
0=64a+8b
文档:.
1
解得a=-,b=4
2
1
∴抛物线的解析式为:y=-x2+4x…………………3分
2
PEBCPE4
(2)①在Rt△APE和Rt△ABC中,tan∠PAE==,即=
APABAP8
11
∴PE=AP==8-t.
22
1
∴点E的坐标为(4+t,8-t).
2
1111
∴点G的纵坐标为:-(4+t)2+4(4+t)=-t2+8.…………………5分
2228
1
∴EG=-t2+8-(8-t)
8
1
=-t2+t.
8
1
∵-<0,∴当t=4时,线段EG最长为2.…………………7分
8
②共有三个时刻.…………………8分
164085
t=,t=,t=.…………………11分
132133
25
28
(2009年山西省)26.(本题14分)如图,已知直线l:yx与直线l:y2x16相
1332
交于点C,l、l分别交x轴于A、、E分别在直线
12
l、l上,顶点F、G都在x轴上,且点G与点B重合.
12
(1)求△ABC的面积;
(2)求矩形DEFG的边DE与EF的长;
(3)若矩形DEFG从原点出发,沿x轴的反方向以每秒1个单位长度的速度平移,设
移动时间为t(0≤t≤12)秒,矩形DEFG与△ABC重叠部分的面积为S,求S关
t的函数关系式,并写出相应的t的取值范围.
y
l
2l
1
ED
C
AOF(G)Bx
(第26题)
文档:.
28
x0,x4.A4,0.
26.(1)解:由得点坐标为
33
2x160,x8.B8,0.
由得点坐标为
∴AB8412.········································································(2分)
28
yx,x5,
C5,6.
由33解得∴点的坐标为·······························(3分)
y6.
y2x16.

11
∴SAB·y12636.··················································(4分)
△ABC2C2
28
(2)解:∵点D在l上且xx8,y88.
1DBD33
∴D点坐标为8,.8··········································································(5分)
又∵点E在l上且yy8,2x168.x4.
2EDEE
∴E点坐标为4,.8··········································································(6分)
∴OE844,EF8.································································(7分)
(3)解法一:①当0≤t3时,如图1,矩形DEFG与△ABC重叠部分为五边
形CHFGR(t0时,为四边形CHFG).过C作CMAB于M,
则Rt△RGB∽Rt△CMB.
yyy
ll
22l
ll2
1El
E1EDD1
D
C
C
C
R
R
R
AOFMGBxAFOGMBxAMBx
FGO
(图1)(图2)(图3)
BGRGtRG
∴,即,∴RG2t.
BMCM36
Rt△AFH∽Rt△AMC,
112
∴SSSS36t2t8t8t.
△ABC△BRG△AFH223
41644
即St2t.··························································(10分)
333
(2009年山西省太原市)29.(本小题满分12分)
问题解决F
M
如图(1),将正方形纸片ABCD折叠,使点B落在CD边上一点EAD
E
文档
BC
N
图(1):.
CE1AM
(不与点C,D重合),时,求的值.
CD2BN
方法指导:
AM
为了求得的值,可先求BN、AM的长,不妨设:AB=2
BN
类比归纳
CE1AMCE1AM
在图(1)中,若,则的值等于;若,则的值等
CD3BNCD4BN
CE1AM
于;若(n为整数),则的值等于.(用含n的式子表示)
CDnBN
联系拓广
如图(2),将矩形纸片ABCD折叠,使点B落在CD边上一点E(不与点C,D重合),
AB1CE1AM
压平后得到折痕MN,设m1,,则的值等于.(用含
BCmCDnBN
m,n的式子表示)
F
M
AD
E

解:方法一:如图(1-1),连接BM,EM,BE.
BC
N
F
M图(2)
AD
E
BC
N
图(1-1)
由题设,得四边形ABNM和四边形FENM关于直线MN对称.
∴MN垂直平分BE.∴BMEM,BNEN.····································1分
∵四边形ABCD是正方形,∴ADC90°,ABBCCDDA2.
CE1
∵,CEDEx,则NEx,NC2x.
CD2
文档:.
在Rt△CNE中,NE2CN2CE2.
55
x22x2BN.
∴解得,即·········································3分
44
在Rt△ABM和在Rt△DEM中,
AM2AB2BM2,
DM2DE2EM2,
AM2AB2DM2DE2.·····························································5分
AMy,DM2y,y2222y212.
设则∴
11
解得y,即AM.·····································································6分
44
AM1
∴.·····················································································7分
BN5
5
方法二:同方法一,BN.································································3分
4
如图(1-2),过点N做NG∥CD,交AD于点G,连接BE.
F
MG
AD
E
B
NC
图(1-2)
∵AD∥BC,∴四边形GDCN是平行四边形.
∴NGCDBC.
5
同理,四边形ABNG也是平行四边形.∴AGBN.
4
∵MNBE,EBCBNM90°.
NGBC,MNGBNM90°,EBCMNG.
在△BCE与△NGM中
EBCMNG,

BCNG,∴△BCE≌△NGM,ECMG.·························5分

CNGM90°.
文档:.
51
∵AMAGMG,AM=1.·····················································6分
44
AM1
∴.···················································································7分
BN5
类比归纳
n12
249
(或);;·································································10分
51017n21
联系拓广
n2m22n1
······················································································12分
n2m21
评分说明:,可参照评分说明进行估分.
,前一问题解答有误或未答,对后面问题的解答没
有影响,可依据参考答案及评分说明进行估分.
(2009年安徽省)(1)所示.
(1)请说明图中①、②两段函数图象的实际意义.
金额w(元)
【解】
批发单价(元)
①
5
②300
4
200
100
O2060
批发量(kg)
O204060
批发量m(kg)
第23题图(1)
(2)写出批发该种水果的资金金额w(元)与批发量m(kg)之间的
函数关系式;在下图的坐标系中画出该函数图象;指出金额在什
么范围内,以同样的资金可以批发到较多数量的该种水果.
【解】
(3)经调查,某经销商销售该种水果的日最高销量与零售价之间的函
数关系如图(2)所示,该经销商拟每日售出60kg以上该种水果,
且当日零售价不变,请你帮助该经销商设计进货和销售的方案,
使得当日获得的利润最大.
【解】
文档:.
日
(6,80)
80
40(7,40)
最高销量(kg)
O2468
零售价(元)
第23题图(2)
23.(1)解:图①表示批发量不少于20kg且不多于60kg的该种水果,金额w(元)
可按5元/kg批发;……3分
图②表示批发量高于60kg的该种水果,可按4元/kg批发.
………………………………………………………………3分
300
5m(20≤m≤60)
(2)解:由题意得:w,
4m(m>60)200
………………………………………………………………7分100
由图可知资金金额满足240<w≤300时,以同样的资金可
批发到较多数量的该种水果.……………………………8分O204060批发量m(k
(3)解法一:
设当日零售价为x元,由图可得日最高销量w32040m
当m>60时,x<
由题意,销售利润为
y(x4)(32040m)40[(x6)24]………………………………12分
当x=6时,y160,此时m=80
最大值
即经销商应批发80kg该种水果,日零售价定为6元/kg,
当日可获得最大利润160元.……………………………………………14分
解法二:
设日最高销售量为xkg(x>60)
320x
则由图②日零售价p满足:x32040p,于是p
40
320x1
销售利润yx(4)(x80)2160………………………12分
4040
当x=80时,y160,此时p=6
最大值
即经销商应批发80kg该种水果,日零售价定为6元/kg,
当日可获得最大利润160元.……………………………………………14分
文档:.
(2009年江西省),在等腰梯形ABCD中,AD∥BC,E是AB的中点,过
点E作EF∥4,BC6,∠B60.
(1)求点E到BC的距离;
(2)点P为线段EF上的一个动点,过P作PMEF交BC于点M,过M作MN∥AB
交折线ADC于点N,连结PN,设EPx.
①当点N在线段AD上时(如图2),△PMN的形状是否发生改变?若不变,求出△PMN
的周长;若改变,请说明理由;
②当点N在线段DC上时(如图3),是否存在点P,使△PMN为等腰三角形?若存在,
请求出所有满足要求的x的值;若不存在,请说明理由.
N
ADADAD
N
PP
EFEFEF
BCBCBC
MM
图1图2图3
AD(第25题)AD
EFEF
BCBC
图4(备用)图5(备用)
25.(1)如图1,过点E作EGBC于点G.····················1分
∵E为AB的中点,
AD
1
∴BEAB2.
2
EF
在Rt△EBG中,∠B60,∴∠BEG30.············2分
1
∴BGBE1,EG22123.
2BC
G
即点E到BC的距离为3.·····································3分图1
(2)①当点N在线段AD上运动时,△PMN的形状不发生改变.
∵PMEF,EGEF,∴PM∥EG.
∵EF∥BC,∴EPGM,PMEG3.
同理MNAB4.··················································································4分
如图2,