中考数学压轴题评分标准
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(2)如图2,将△ABE绕点A顺时针旋转90°得到△AB′C,连结B′B,则BE=B′C
当∠ABC变化时,点B′的轨迹是以点B为圆心,BB′为半径的圆
∴当点B′在CB的延长线上时(如图3),B′C取得最大值,即BE取得最大值
此时BE=B′C=BB′+BC=2+4=6,即BE的最大值为6
当点A在BC上时,△ABC不存在
此时BE=B′C=(4-2)2(2)2=25-22
故BE的取值范围是Fra Baidu bibliotek25-22<BE≤6···················································8分
EE
DD
E
D
B′A
A
AB′
BC
图1
BC
图2
B′
BC
图3
302.解:(1)①方法一:
∵四边形ABCD是矩形,AB∥x轴,B(-3,3)
22<
2
故有△=(-15)-4(36+a)>0,解得:a
81
4
又a>0,∴0<a<
9
2
故当0<a<
9
2
时,在线段BC上存在不同的两点P1、P2满足条件
304.解:(1)在正方形ABCD中,AB=AD········································1分
∵DF=BE,∠1=∠2····················································3分
mnmn
k
m
-
k
n
∴直线AC的解析式为y=-
k
mn
(mn)k
x+
mn
∴E(m+n,0),∴EM=m-(m+n)=-n
∵CN=-n,∴EM=CN
∵EM∥BA∥CN,∴∠AEM=∠FCN
又∵∠AME=∠FNC=90°,∴△AEM≌△FCN
∴AE=CF
303.解:(1)过D作DF⊥BC于F
当CP=3时,由已知可得四边形ADFB是矩形,则CF=3
k
3
+3,AD=
k
3
+3,∴AB=AD,∴四边形ABCD是正方形
∴∠AEO=∠ACD=45°,∴OE=OF=b
∴E(-b,0),∴-ab+b=0
∵b≠0,∴a=1
②∵S=S△ABC-S△OAC=S△ACD-S△OAC=S△AOM+S△CON+S
矩形ONDM
=
1
2
×
k
3
×3+
1
2
×3×
k
3
+
k
3
×
k
3
kk
∴
AE
AC
=
AM
AD
=
m
k
m
k
n
=
m
k
m
k
n
=
n
n
m
∵FN∥AD,∴△CFN∽△CAD
∴
CF
AC
=
CN
CD
=
n
m
=
nn
n
m
∴
AE
AC
=
CF
AC
,∴AE=CF
方法三:设A(m,
k
m
),C(n,
k
n
),则M(m,0)、N(0,
k
n
)
ma
k
+b=
m
从而∴(m-n)a=
k
na+b=
n
k(mn)k
∴a=-,∴b=
=
1
9
k2+k=
2+k=
1
9
(k+
9
2
2
)-
9
4
∴当k>-
9
2
时,S随着k的增大而增大
又∵k>0,k没有最小值,∴S没有最小值
(2)答:AE=CF,理由如下:
方法一:如图,连接MN,设AB交y轴于点P,BC交x轴于点Q
∵S矩形APOM=S
矩形CQON=
k
3
×3=k,∴DN·AD=DM·CD
∴
DN
CD
kk
∴A(,3),C(-3,-)
33
y
∵y=ax+b经过A、C两点
k
a+b=3
3
∴
-3a+b=-
k
3
∴(
k
3
k
3
∵k>0,∴
+)a=3
+3≠0,∴a=1
k
3
+3
P
B
F
Q
E
OM
D
C
N
x
A
方法二:∵四边形ABCD是矩形,AB∥x轴,B(-3,3)
∴A(
k
3
,3),C(-3,-
k
3
),D(
k
3
,-
k
3
)
∴AB=
AD
∴△ADF≌△ABE···························································4分
3
1
4F
(2)由(1)有△ADF≌△ABE
∴AF=AE,∠3=∠4·····················································5分
所以,y关于x的函数关系式为:
1
2
(x-15x+36)(0<x≤3)
a
y=
1
2
(x-15x+36)(3<x≤12)
-
a
1
a
2
(x-15x+36)
(3)当点E与A重合时,y=EB=a,此时点P在线段BF上
由a=-
1
a
22=0①
2
(x-15x+36)得:x-15x+36+a
假设在线段BC上存在不同的两点P1、P2满足条件,即方程①有两个不相等的正根
2010年全国各地中考数学压轴题
参考答案及评分标准(三)
301.解:(1)如图1,将△ABE绕点A顺时针旋转90°得到△AB′C,连结B′B
则△ABE≌△AB′C,BE=B′C,AB=AB′,∠BAB′=90°
∴∠ABB′=45°,∴∠B′BC=90°
∴BB′=2AB=2
∴BE=B′C=BB2+BC2=22+42=25············································3分
∴点P与点F重合
AD
又∵BF⊥FD,∴此时点E与点B重合
(2)当点P在BF上,即3<x≤12时
E
∵∠EPB+∠DPF=90°,∠EPB+∠PEB=90°
∴∠DPF=∠PEB,∴Rt△PEB∽Rt△DPFBC
PF
∴
BE
BP
=
PF
DF
,即
y
12
x
=
x3
a
∴y=-
1
a
2
(x-15x+36)
当点P在CF上,即0<x≤3时,同理可求得y=
=
DM
AD
,又∵∠D=∠D,∴△DNM∽△DCA
∴∠DNM=∠DCA,∴MN∥AF
又∵AM∥FN,∴四边形AFNM是平行四边形,∴AF=MN
同理CE=MN,∴AF=CE
∴AE=CF
方法二:设A(m,
k
m
),C(n,
k
n
)
则AM=
k
m
,AD=
k
m
-
k
n
,CN=-n,CD=m-n
∵EM∥CD,∴△AEM∽△ACD
EO
2
在正方形ABCD中,∠BAD=90°
BC∴∠BAF+∠3=90°
图①
∴∠BAF+∠4=90°
∴∠EAF=90°···································································································6分
∴△EAF是等腰直角三角形
∴EF
2=AE2+AF2····························································································7分
∴EF
2=2AE2
∴EF=2AE···································································································8分
当∠ABC变化时,点B′的轨迹是以点B为圆心,BB′为半径的圆
∴当点B′在CB的延长线上时(如图3),B′C取得最大值,即BE取得最大值
此时BE=B′C=BB′+BC=2+4=6,即BE的最大值为6
当点A在BC上时,△ABC不存在
此时BE=B′C=(4-2)2(2)2=25-22
故BE的取值范围是Fra Baidu bibliotek25-22<BE≤6···················································8分
EE
DD
E
D
B′A
A
AB′
BC
图1
BC
图2
B′
BC
图3
302.解:(1)①方法一:
∵四边形ABCD是矩形,AB∥x轴,B(-3,3)
22<
2
故有△=(-15)-4(36+a)>0,解得:a
81
4
又a>0,∴0<a<
9
2
故当0<a<
9
2
时,在线段BC上存在不同的两点P1、P2满足条件
304.解:(1)在正方形ABCD中,AB=AD········································1分
∵DF=BE,∠1=∠2····················································3分
mnmn
k
m
-
k
n
∴直线AC的解析式为y=-
k
mn
(mn)k
x+
mn
∴E(m+n,0),∴EM=m-(m+n)=-n
∵CN=-n,∴EM=CN
∵EM∥BA∥CN,∴∠AEM=∠FCN
又∵∠AME=∠FNC=90°,∴△AEM≌△FCN
∴AE=CF
303.解:(1)过D作DF⊥BC于F
当CP=3时,由已知可得四边形ADFB是矩形,则CF=3
k
3
+3,AD=
k
3
+3,∴AB=AD,∴四边形ABCD是正方形
∴∠AEO=∠ACD=45°,∴OE=OF=b
∴E(-b,0),∴-ab+b=0
∵b≠0,∴a=1
②∵S=S△ABC-S△OAC=S△ACD-S△OAC=S△AOM+S△CON+S
矩形ONDM
=
1
2
×
k
3
×3+
1
2
×3×
k
3
+
k
3
×
k
3
kk
∴
AE
AC
=
AM
AD
=
m
k
m
k
n
=
m
k
m
k
n
=
n
n
m
∵FN∥AD,∴△CFN∽△CAD
∴
CF
AC
=
CN
CD
=
n
m
=
nn
n
m
∴
AE
AC
=
CF
AC
,∴AE=CF
方法三:设A(m,
k
m
),C(n,
k
n
),则M(m,0)、N(0,
k
n
)
ma
k
+b=
m
从而∴(m-n)a=
k
na+b=
n
k(mn)k
∴a=-,∴b=
=
1
9
k2+k=
2+k=
1
9
(k+
9
2
2
)-
9
4
∴当k>-
9
2
时,S随着k的增大而增大
又∵k>0,k没有最小值,∴S没有最小值
(2)答:AE=CF,理由如下:
方法一:如图,连接MN,设AB交y轴于点P,BC交x轴于点Q
∵S矩形APOM=S
矩形CQON=
k
3
×3=k,∴DN·AD=DM·CD
∴
DN
CD
kk
∴A(,3),C(-3,-)
33
y
∵y=ax+b经过A、C两点
k
a+b=3
3
∴
-3a+b=-
k
3
∴(
k
3
k
3
∵k>0,∴
+)a=3
+3≠0,∴a=1
k
3
+3
P
B
F
Q
E
OM
D
C
N
x
A
方法二:∵四边形ABCD是矩形,AB∥x轴,B(-3,3)
∴A(
k
3
,3),C(-3,-
k
3
),D(
k
3
,-
k
3
)
∴AB=
AD
∴△ADF≌△ABE···························································4分
3
1
4F
(2)由(1)有△ADF≌△ABE
∴AF=AE,∠3=∠4·····················································5分
所以,y关于x的函数关系式为:
1
2
(x-15x+36)(0<x≤3)
a
y=
1
2
(x-15x+36)(3<x≤12)
-
a
1
a
2
(x-15x+36)
(3)当点E与A重合时,y=EB=a,此时点P在线段BF上
由a=-
1
a
22=0①
2
(x-15x+36)得:x-15x+36+a
假设在线段BC上存在不同的两点P1、P2满足条件,即方程①有两个不相等的正根
2010年全国各地中考数学压轴题
参考答案及评分标准(三)
301.解:(1)如图1,将△ABE绕点A顺时针旋转90°得到△AB′C,连结B′B
则△ABE≌△AB′C,BE=B′C,AB=AB′,∠BAB′=90°
∴∠ABB′=45°,∴∠B′BC=90°
∴BB′=2AB=2
∴BE=B′C=BB2+BC2=22+42=25············································3分
∴点P与点F重合
AD
又∵BF⊥FD,∴此时点E与点B重合
(2)当点P在BF上,即3<x≤12时
E
∵∠EPB+∠DPF=90°,∠EPB+∠PEB=90°
∴∠DPF=∠PEB,∴Rt△PEB∽Rt△DPFBC
PF
∴
BE
BP
=
PF
DF
,即
y
12
x
=
x3
a
∴y=-
1
a
2
(x-15x+36)
当点P在CF上,即0<x≤3时,同理可求得y=
=
DM
AD
,又∵∠D=∠D,∴△DNM∽△DCA
∴∠DNM=∠DCA,∴MN∥AF
又∵AM∥FN,∴四边形AFNM是平行四边形,∴AF=MN
同理CE=MN,∴AF=CE
∴AE=CF
方法二:设A(m,
k
m
),C(n,
k
n
)
则AM=
k
m
,AD=
k
m
-
k
n
,CN=-n,CD=m-n
∵EM∥CD,∴△AEM∽△ACD
EO
2
在正方形ABCD中,∠BAD=90°
BC∴∠BAF+∠3=90°
图①
∴∠BAF+∠4=90°
∴∠EAF=90°···································································································6分
∴△EAF是等腰直角三角形
∴EF
2=AE2+AF2····························································································7分
∴EF
2=2AE2
∴EF=2AE···································································································8分