中考数学压轴题100题精选(1-10题)答案2013

∵△MNC是等边三角形，∴MC?MN?3．

此时，x?EP?GM?BC?BG?MC?6?1?3?2． ··············································· 8分

A D A A D D E B

P R

G

M

图3

C

B

G

图4

M

N F

E

P

F N C

B E F（P） N C

当MP?MN时，如图4，这时MC?MN?MP?3．

此时，x?EP?GM?6?1?3?5?3．

当NP?NM时，如图5，∠NPM?∠PMN?30?．

则∠PMN?120?，又∠MNC?60?， ∴∠PNM?∠MNC?180?．

因此点P与F重合，△PMC为直角三角形．

．∴MC?PMtan30??1

此时，x?EP?GM?6?1?1?4．

综上所述，当x?2或4或5?3时，△PMN为等腰三角形． 【006】解：（1）OC=1,所以,q=-1,又由面积知0.5OC×AB=

G

图5

M

??55,得AB=， 42 设A（a,0）,B(b,0)AB=b?a=

(a?b)2?4ab=2533，解得p=?,但p<0,所以p=?。 222 所以解析式为：y?x?（2）令y=0，解方程得x?23x?1 2311x?1?0，得x1??,x2?2,所以A(?,0),B(2,0),在直角三角形AOC222中可求得AC=5,同样可求得BC=5，显然AC2+BC2=AB2，得△ABC是直角三角形。AB为斜2555,所以??m?。 244边，所以外接圆的直径为AB=

（3）存在，AC⊥BC，①若以AC为底边，则BD//AC,易求AC的解析式为y=-2x-1,可设BD的解析式

3?25?y?x?x?1为y=-2x+b，把B(2,0)代入得BD解析式为y=-2x+4，解方程组?得D（?，9） 22??y??2x?4 ②若以BC为底边，则BC//AD,易求BC的解析式为y=0.5x-1,可设AD的解析式为y=0.5x+b，把

5

3?2153?y?x?x?1A(?，0)代入得AD解析式为y=0.5x+0.25，解方程组?得D(,) 综上，2222?y?0.5x?0.25?所以存在两点：（?

【007】

553,9）或(,)。 222

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【008】证明：（1）∵∠ABC=90°，BD⊥EC， ∴∠1与∠3互余，∠2与∠3互余，

∴∠1=∠2…………………………………………………1分 ∵∠ABC=∠DAB=90°，AB=AC

∴△BAD≌△CBE…………………………………………2分 ∴AD=BE……………………………………………………3分 （2）∵E是AB中点，

∴EB=EA由（1）AD=BE得：AE=AD……………………………5分 ∵AD∥BC∴∠7=∠ACB=45°∵∠6=45°∴∠6=∠7 由等腰三角形的性质，得：EM=MD，AM⊥DE。 即，AC是线段ED的垂直平分线。……………………7分 （3）△DBC是等腰三角（CD=BD）……………………8分 理由如下：

由（2）得：CD=CE由（1）得：CE=BD∴CD=BD ∴△DBC是等腰三角形。……………………………10分 【009】解：（1）①

AC⊥x轴，AE⊥y轴，

y N E D A B ?四边形AEOC为矩形．

BF⊥x轴，BD⊥y轴，

?四边形BDOF为矩形．

AC⊥x轴，BD⊥y轴，

············ 1分 ?四边形AEDK，DOCK，CFBK均为矩形． ·

K O C F M x 图1 OC?x1，AC?y1，x1y1?k，

?S矩形AEOC?OCAC?x1y1?k OF?x2，FB?y2，x2y2?k，

?S矩形BDOF?OFFB?x2y2?k． ?S矩形AEOC?S矩形BDOF．

S矩形AEDK?S矩形AEOC?S矩形DOCK，

S矩形CFBK?S矩形BDOF矩形?S，D

·········································································································· 2分 ?S矩形AEDK?S矩形CFBK． ·②由（1）知S矩形AEDK?S矩形CFBK．

?AKDK?BKCK．

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AKBK?． ·························································································································· 4分 CKDK?AKB??CKD?90°，

············································································································· 5分 ?△AKB∽△CKD． ·

??CDK??ABK．

··························································································································· 6分 ?AB∥CD． ·

?AC∥y轴，

?四边形ACDN是平行四边形． ··························································································································· 7分 ?AN?CD． ·

同理BM?CD．

?AN?BM． ···························································································································· 8分 （2）AN与BM仍然相等． ····································································································· 9分

S矩形AEDK?S矩形AEOC?S矩形ODKC， S矩形BKCF?S矩形BDOF?S矩形ODKC，

y 又

S矩形AEOC?S矩形BDOF?k，

E N F M O B A ···································· 10分 ?S矩形AEDK?S矩形BKCF． ·

?AKDK?BKCK．

CKDK?． ?AKBKC D K x 图2 ?K??K，

?△CDK∽△ABK． ??CDK??ABK．

························································································································· 11分 ?AB∥CD． ·

AC∥y轴，

?四边形ANDC是平行四边形．

?AN?CD．

同理BM?CD．

························································································································ 12分 ?AN?BM． ·

??3a?4a?2b?3，?【010】解：（1）根据题意，得?b ····· 2分

??1.??2ay D E ?a?1，解得??抛物线对应的函数表达式为y?x2?2x?3． 3分 ?b??2.（2）存在．

在y?x?2x?3中，令x?0，得y??3．

2N A O 1 N x F 8

C P M （第26题图）