中考選擇填空壓軸題專項(xiàng)練習(xí)

初三中考數(shù)學(xué)壓軸題專題并會(huì)運(yùn)用多種不同的方法靈活解題.這類題目重點(diǎn)考察學(xué)生綜合分析問題、解決問題的項(xiàng)代入法和特殊值代入法等.特別注意：這些方法在通常都是要綜合靈活運(yùn)用，不能生息提示，給解題增加了一定難度，要求學(xué)生要有扎實(shí)、熟練的基礎(chǔ)知識(shí)和基本技能.還要靈活運(yùn)用多種不同的解題方法.建立模型等方法來解決問題的方法就是構(gòu)造法.在題目的相關(guān)條件或信息不夠明確具體時(shí)，則應(yīng)分情況求解，也就是分類討論法.把不易解決的問題或難點(diǎn)，通過第三個(gè)等價(jià)的量，轉(zhuǎn)化為已知的或易于解決的問題來解題的方法就是轉(zhuǎn)化法.

eq

\o\ac(△,A)

A B

l

l

A B

E

E

A B

=90

==2 EF

BEBFEF

eq

\o\ac(△,，則) BEF

A B

=

，以點(diǎn)

E．若

ED=

l

l

== eq

\o\ac(△,．)

CE

CE=

F

F

FG

GE

CEG

E，取

BE

F

DFDF=4．設(shè)

==，則

=10=60E

=BE=4eq

\o\ac(△,，將)

DE

DE

E

E

EC

C E

B

y=

(＞

A.

B.

C.

.

DE//

E

.

,

,,

.

(

P,

A.

B.

C.

//

.

,

P

.

E

P

BEED

.P

.

:

CBE

;

;

:

. .(

考點(diǎn)：坐標(biāo)與圖形變化－--旋轉(zhuǎn)．

=2=

=

=

=3

=2=2×2=4=4=4× = =4×

=

=+=4+

=

E

DE

E

B

B

E

E

=

+4

=3

=

E

B

，過

EF

，易得

eq

\o\ac(△,和)

eq

\o\ac(△,又是)

EF

EF

EF

=90==2

= = =4

eq

\o\ac(△,，)

==2

eq

\o\ac(△,S) =

=

=4

eq

\o\ac(△,S)

=2

=2

=

=

=

EF=

=2

eq

\o\ac(△,S) BEF=

EF=

=

BE

=3=5=4DE=

BE

BE=

=3=5

==3==5=4

DE=5=ED=

=

= =3= =5=

×=

×

=5

eq

\o\ac(△,CP) =

=

=

=

+=

+2

=4

l= = ==

=

=

=

=

+=

+2

=4

.

A

F

F

AB

E

AB

GE

ABC

GE

A

F

F

AB

E

AB

GE

ABC

CEG

BDE

+

DE

F

BE

+=4=16++=16BDE

BF

DFE

F

DE

=F=2=2 DFE

F

=60BE==4

eq

\o\ac(△,將)

DE

DE

DE

=F=2

=4

= = =2

=10

=106=4

=

=

=2

PE

EPC

PE

= =8

==

= =PE=

=4

=，則

=4，當(dāng)

EC

FCP=

。

EP

EPC=

=90

=1

=3

=1

=3

=1

PE=

1.答案：C；M

MGOA

Gkeq

\o\ac(△keq

\o\ac(△,S)

OMGeq

\o\ac(△,S)

OECeq

\o\ac(△,S) ODA

eq

\o\ac(△,S)

OMGS

eq

\o\ac(eq

\o\ac(△,S) OMAeq

\o\ac(△,S) AMB

keq

\o\ac(△,S) OABeq

\o\ac(△,S) OBCeq

\o\ac(△,S) OMAeq

\o\ac(△,S) AMBkkkS

OABCeq

\o\ac(△,S)

OABeq

\o\ac(△,S) OBCkAMG

kkkkS

OABCeq

\o\ac(△,S)

ODAeq

\o\ac(△,S)

OECS

ODBEk

k

B

A

AC

B

=

=3

=

=

B

B

= =

= DEeq

\o\ac(△,BC) ADEABC

=

=

=

ADE

AE

CED

CE

= = =

=

AB

AB

=5

PAB

eq

\o\ac(△,=S)

ABC

=

=

A

ABC

CBABeq

\o\ac(△,D) =

=

AB

AB

E

AB

CF

FCAABDBABA