发布时间 : 星期三 文章中考数学压轴题100题精选（91-100题）答案更新完毕开始阅读27a9d365a7e9856a561252d380eb6294dd882289
∠DAP＝∠CAO， ∴△ADP∽△AOC． ∴
DPCODP432，即． ??．∴DP?DACA8551024253471． ??25410∴DE?DP2?PE2?∴Ｓ四边形ＤＥＰＦ＝34713471，即Ｓ＝． ·············································································· 14分 44（注：本卷中所有题目，若由其它方法得出正确结论，请参照标准给分．）
【094】解：（1）令二次函数y?ax?bx?c，则
2?16a?4b?c?0? ························································································································ 1分 ?a?b?c?0?c?2?1?a???2?3?·································································································································· 2分 ??b?? ·2??c?2??13?过A，B，C三点的抛物线的解析式为y??x2?x?2 ············································· 4分
220? （2）以AB为直径的圆圆心坐标为O??，?3?2??53 O?O? ··············································································································· 5分 22·························································································· 6分 QCD为圆O?切线 ?OC??CD ·
??O?CD??DCO?90°
?CO?O??O?CO?90° ??CO?O??DCO
······································································ 8分 ?△O?CO∽△CDO O?O/OC?OC/OD ·
38/2?2/OD ?OD? 23?O?C??8?····················································································································· 9分 ?D坐标为?0，? ·
?3?（3）存在 ···································································································································· 10分 抛物线对称轴为X??3 233?r，r)或F(??r，r) 22设满足条件的圆的半径为r，则E的坐标为(?而E点在抛物线y??123x?x?2上 221333?r??(??r)2?(??r)?2
2222?r1??1?2929 r2??1? 222929，1? ············ 12分 22故在以EF为直径的圆，恰好与x轴相切，该圆的半径为?1?注：解答题只要方法合理均可酌情给分
【095】（1）B（4，0），C(0，····················································································· 2分 ?2)． ·
123·················································································································· 4分 x?x?2． ·
22（2）△ABC是直角三角形． ··································································································· 5分
123证明：令y?0，则x?x?2?0．
22y??x1??1，x2?4．
································································································································ 6分 ?A(?1，0)． ·解法一：?AB?5，AC?5，BC?25． ······································································· 7分
?AC2?BC2?5?20?25?AB2．
········································································································· 8分 ?△ABC是直角三角形． ·
COAO1解法二：QAO?1，CO?2，BO?4，???
BOOC2Q?AOC??COB?90°，
·············································································································· 7分 ?△AOC∽△COB． ·
??ACO??CBO．
Q?CBO??BCO?90°，
??ACO??BCO?90°．即?ACB?90°．
········································································································· 8分 ?△ABC是直角三角形． ·
（3）能．①当矩形两个顶点在AB上时，如图1，CO交GF于H．
QGF∥AB，
?△CGF∽△CAB． GFCH??． ······························································· 9分 ABCO2解法一：设GF?x，则DE?x，CH?x，
52DG?OH?OC?CH?2?x．
52?2??S矩形DEFG?x·?2?x???x2?2x
5?5?y D A E O B x F G H C 图1
2?5?5=??x???． ················································································································· 10分
5?2?25时，S最大． 25?DE?，DG?1．
2Q△ADG∽△AOC， ADDG11??，?AD?，?OD?，OE?2． AOOC22当x?2?1??D??，0?，E(2，········································································································ 11分 0)． ·
?2?10?5x． 210?5x555················································ 10分 ?S矩形DEFG?x·??x2?5x??(x?1)2?．·
2222?当x?1时，S最大．
5?DG?1，DE?．
2Q△ADG∽△AOC， ADDG11??，?AD?，?OD?，OE?2． AOOC22解法二：设DG?x，则DE?GF??1??D??，0?，E(2，········································································································ 11分 0)． ·
?2?y ②当矩形一个顶点在AB上时，F与C重合，如图2，
QDG∥BC， ?△AGD∽△ACB． GDAG． ??BCAF解法一：设GD?x，?AC?5,BC?25，
D O A G C 图2
B G x x?GF?AC?AG?5?．
2x?1?·?5????x2?5x ?S矩形DEFG?x2?2?1x?5=?2??25?． ··············································································································· 12分 2当x?5时，S最大．
?GD?5，AG?55322，?AD?AG?GD?．?OD? 222?3??D?，0? ································································································································· 13分
2??解法二：设DE?x，QAC?5，BC?25，?GC?x，AG?5?x．?GD?25?2x．
?S矩形DEFG?x·25?2x??2x2?25x
?5?5=?2?x?·················································································································· 12分 ????2 ·2???当x?2??5时，S最大， 255322．?AD?AG?GD?．?OD?. 222?GD?5，AG??3?0? ·?D?，································································································································ 13分 ?2?0?，综上所述：当矩形两个顶点在AB上时，坐标分别为??，（2，0）；
?3?2???1?2??0? ·当矩形一个顶点在AB上时，坐标为?，········································································· 14分
【096】（1）因所求抛物线的顶点M的坐标为(2,4)，
故可设其关系式为y?a?x?2??4 ………………(1分) 又抛物线经过O(0,0)，于是得a?0?2??4?0， ………………(2分) 解得 a=-1 ………………(3分) ∴ 所求函数关系式为y???x?2??4，即y??x?4x. ……………(4分)
2222（2）① 点P不在直线ME上. ………………(5分)
根据抛物线的对称性可知E点的坐标为(4,0)， 又M的坐标为(2,4)，设直线ME的关系式为y=kx+b.
?4k?b?0?k??2于是得? ，解得?
?2k?b?4?b?8所以直线ME的关系式为y=-2x+8. ……(6分)
5555?由已知条件易得，当t?时，OA=AP?，?P??,? ……………(7分)
22?22?∵ P点的坐标不满足直线ME的关系式y=-2x+8.
5∴ 当t?时，点P不在直线ME上. ………………(8分)
2