中考数学压轴题100题精选(1-10题)答案2013

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?33(a?0)经过点A(?2，0)，

?0?9a?33?a??3 ······································································································· 1分 3322383 ························································· 3分 x?x?333?二次函数的解析式为：y??（2）

D为抛物线的顶点?D(13，3)过D作DN?OB于N，则DN?33，

AN?3，?AD?32?(33)2?6??DAO?60° ··························································· 4分 OM∥AD

y D M C ①当AD?OP时，四边形DAOP是平行四边形 ?OP?6?t?6(s) ······················································· 5分

②当DP?OM时，四边形DAOP是直角梯形

A H P B x 过O作OH?AD于H，AO?2，则AH?1 O E N Q （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

，OC?OB，△OCB是等边三角形 （3）由（2）及已知，?COB?60°?OQ?6?2t(0?t?3) 则OB?OC?AD?6，OP?t，BQ?2t，过P作PE?OQ于E，则PE?3·················································································· 8分 t ·22?SBCPQ当t?1133?3?63??6?33??(6?2t)?t=···································· 9分 3 ·?t???2222?2?83633 ·时，SBCPQ的面积最小值为··········································································· 10分

2833?此时OQ?3，OP=，OE?24?QE?3?39?44PE?33 41

?33??9?23322 ······························································ 11分 ?PQ?PE?QE????B ?4????2???4?8【002】解：（1）1，；

5E Q B A F D C 2（2）作QF⊥AC于点F，如图3， AQ = CP= t，∴AP?3?t． 由△AQF∽△ABC，BC?5?3?4， 得

QFt414?．∴QF?t． ∴S?(3?t)?t， 45255Q D C 22E 图3

P 26即S??t2?t．

55A P （3）能．

图4 ①当DE∥QB时，如图4．

∵DE⊥PQ，∴PQ⊥QB，四边形QBED是直角梯形． 此时∠AQP=90°．

B 由△APQ ∽△ABC，得

AQAP， ?ACABA Q D P E C t3?t9即?． 解得t?． 358②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

AQAP， ?ABACQ 图5

B t3?t15即?． 解得t?． 538

G

（4）t?545或t?． 214D A P C(E) B G 【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． 34PC?t，QC2?QG2?CG2?[(5?t)]2?[4?(5?t)]2．

55534由PC2?QC2，得t2?[(5?t)]2?[4?(5?t)]2，解得t?．

255图6 Q D A P C(E)

方法二、由CQ?CP?AQ，得?QAC??QCA，进而可得

?B??BCQ，得CQ?BQ，∴AQ?BQ?图7 55．∴t?． 22②点P由A向C运动，DE经过点C，如图7．

34(6?t)2?[(5?t)]2?[4?(5?t)]2，t?45】

5514【003】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

2

8=16a+4b

得

0=64a+8b

解 得a=-

1,b=4 212

x+4x …………………3分 2PEBCPE4（2）①在Rt△APE和Rt△ABC中，tan∠PAE==,即=

APABAP811∴PE=AP=t．PB=8-t．

221∴点Ｅ的坐标为（4+t，8-t）.

21111∴点G的纵坐标为：－（4+t）2+4(4+t）=－t2+8. …………………5分

222811∴EG=－t2+8-(8-t) =－t2+t.

881∵-＜0，∴当t=4时，线段EG最长为2. …………………7分

8∴抛物线的解析式为：y=－

②共有三个时刻. …………………8分 t1=

164085， t2=，t3= ． …………………11分 3132?528?A点坐标为??4，．x??0，得x??4．0?

33【004】（1）解：由

?B点坐标为?8，．由?2x?16?0，得x?8．（2分） 0?∴AB?8???4??12．28?y?x?，?x?5，?由?解得∴C点的坐标为?5，．6?（3分） 33??y?6．??y??2x?16．11AB·yC??12?6?36．（4分） 2228?yD??8??8．（2）解：∵点D在l1上且xD?xB?8， ∴D点坐标为?8，．8?（5分）又∵点

33∴S△ABC?E在l2上且yE?yD?8，∴E点坐标为?4，．??2xE?16?8．?xE?4．8?（6分）

∴OE?8?4?4，EF?8．（7分）

①当0≤t?3时，（3）解法一：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 l2 l2 l2 l1l1l1E E E D D D C C 3 C R R R A F O G M B x A O F M G B x F A G O M B x

BGRGtRG?，，即?∴RG?2t．Rt△AFH∽Rt△AMC， BMCM36112∴S?S△ABC?S△BRG?S△AFH?36??t?2t??8?t???8?t?．

223421644．即S??t?t?（10分）

333【005】（1）如图1，过点E作EG?BC于点G．····················· 1分

A

∵E为AB的中点，

1∴BE?AB?2． E 2在Rt△EBG中，∠B?60?，∴∠BEG?30?． ··············· 2分

1B 22∴BG?BE?1，EG?2?1?3． G

2∴

即点E到BC的距离为3． ················································· 3分

D F C

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，过点P作PH?MN于H，∵MN∥AB， ∴∠NMC?∠B?60?，∠PMH?30?． N

A D ∴PH?13 PM?．22B

E

P H F C

图2

3∴MH?PMcos30??．

235则NH?MN?MH?4??．

22G M 2?5??3?22在Rt△PNH中，PN?NH?PH????? ?7．????2??2?2∴△PMN的周长=PM?PN?MN?3?7?4． ··················································· 6分 ②当点N在线段DC上运动时，△PMN的形状发生改变，但△MNC恒为等边三角形． 当PM?PN时，如图3，作PR?MN于R，则MR?NR．

3． 2∴MN?2MR?3． ·············································································································· 7分

类似①，MR?4