2010年中考数学压轴题100题精选(1-10题)答案

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2010年中考数学压轴题100题精选（1-10题）答案

【001】解：（1）

抛物线y a(x 1)2 a 0)经过点A( 2，0)，

3

0 9a a ························································································· 1分

二次函数的解析式为：y

3

x

2

3

x

3

·················································· 3分

（2）

D为抛物线的顶点 D(1过D作DN OB于N

，则DN ，

AN 3， AD

OM∥AD

·················································· 4分 6 DAO 60° ·

①当AD OP时，四边形DAOP是平行四边形

··············································· 5分 OP 6 t 6(s) ·

②当DP OM时，四边形DAOP是直角梯形

过O作OH AD于H，AO 2，则AH 1

（如果没求出 DAO 60°可由Rt△OHA∽Rt△DNA求

AH··························································································· 6分 OP DH 5t 5(s) ·

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

△OCB是等边三角形 （3）由（2）及已知， COB 60°，OC OB，

则OB OC AD 6，OP t，BQ 2t， OQ 6 2t(0 t 3)

2

过P作PE OQ于E，则PE ········································································· 8分

2

SBCPQ

32

3 6 (6 2t) t

2 2 222

11 ································ 9分

当t

时，SBCPQ32

34

···································································10分

此时OQ 3，OP=，OE QE 3

34

94

PE

4

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PQ

·····················

······11分 2【

002】解：（1）1，；

5

8

（2）作QF⊥AC于点F

，如图3， AQ = CP= t，∴AP

由△AQF∽△ABC，BC 4，

3 t

．

得

QF4

t525

．∴QF

t

2

45

t

． ∴S

12

(3 t)

45

t

图3

P

即S

65

t

．

（3）能．

①当DE∥QB时，如图4．

图4

∵DE⊥PQ，∴PQ⊥QB，四边形QBED是直角梯形． 此时∠AQP=90°． 由△APQ ∽△ABC，得即

t3 3 t5

AQAC

APAB

，

． 解得t

98

．

②如图5，当PQ∥BC时，DE⊥BC，四边形QBED是直角梯形． 此时∠APQ =90°． 由△AQP ∽△ABC，得 即

t5 3 t3

AQAB

APAC

图5

，

． 解得t或t

4514

158

．

（4）t

52

．

【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．

PC t

，QC

QC

2

QG CG [(5 t)] [4 (5 t)]

55

22

3

2

4

2

．

52

由PC2

2

，得t2

3422

[(5 t)] [4 (5 t)]55

，解得t．

方法二、由CQ

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