0?．?B点坐标为?8，由?2x?16?0，得x?8．
∴AB?8???4??12． ·························································································· （2分）
28?x?，?x?5，?y?6?．由?∴C点的坐标为?5， ······································· （3分） 33解得?y?6．??y??2x?16．?∴S△ABC?12AB·yC?12?12?6?36． ······························································· （4分）
23?8?83?8．
?yD? （2）解：∵点D在l1上且xD?xB?8，8?． ∴D点坐标为?8， ····························································································· （5分）
??2xE?16?8．?xE?4．又∵点E在l2上且yE?yD?8，
8?．∴E点坐标为?4， ····························································································· （6分）
∴OE?8?4?4，EF?8． ················································································· （7分）
（3）解法一：①当0≤t?3时，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．
yl2 yE C D R l1l2 yE D C R l1E l2 D C R l1A O F M G B x （图1）
∴
BGBM?RGCMA F O G M （图2）
，即
t3?RG6B x F A G O M B x （图3）
，∴RG?2t．
?Rt△AFH∽Rt△AMC，
∴S?S△ABC?S△BRG?S△AFH?36?即S??43t?212?t?2t?12 ?8?t???8?t?．32163t?443． ········································································· （10分）
7、（2009年山西省太原市）29．（本小题满分12分） 问题解决
如图（1），将正方形纸片ABCD折叠，使点B落在CD边上一点
E（不与点C，D重合），压平后得到折痕MN．当
CECD?12A M F D
时，
E
B
N 图（1）
C
求
AMBN的值．
方法指导：
AM 为了求得的值，可先求BN、AM的长，不妨设：AB=2
BN 类比归纳
在图（1）中，若于 ；若
CECD?CECD1n?13，则
AMBN的值等于 ；若
AMBNCECD?14，则
AMBN的值等
（n为整数），则的值等于 ．（用含n的式子表
示）
联系拓广 如图（2），将矩形纸片ABCD折叠，使点B落在CD边上一点E（不与点C，D重合），压平后得到折痕MN，设
m，n的式子表示）
ABBC?1m?mCE1AM?1?，?，则的值等于 ．（用含
CDnBN
29．问题解决
解：方法一：如图（1-1），连接BM，EM，BE．
B
N 图（1-1）
C E
A M F D F A
M D E
B
N
图（2）
C
由题设，得四边形ABNM和四边形FENM关于直线MN对称．
∴MN垂直平分BE．∴BM?EM，BN?EN． ·············································· 1分 ∵四边形ABCD是正方形，∴?A??D??C?90°,AB?BC?CD?DA?2．
∵
CECD?12，?CE?DE?1．设BN?x，则NE?x，NC?2?x．
222 在Rt△CNE中，NE?CN?CE． ∴x??2?x??1．解得x?22254，即BN?54． ···················································· 3分
在Rt△ABM和在Rt△DEM中，
AMDM22?AB?BM?DE222222，
2?EM，
?DE． ············································································· 5分
2222?AM?AB?DM142 设AM?y，则DM?2?y，∴y?2??2?y??1． 解得y? ∴
AMBN?，即AM?14． ······················································································· 6分
15． ··········································································································· 7分
54． ················································································ 3分
方法二：同方法一，BN? 如图（1－2），过点N做NG∥CD，交AD于点G，连接BE．
F G M A D
E
B C N
图（1-2）
∵AD∥BC，∴四边形GDCN是平行四边形． ∴NG?CD?BC． 同理，四边形ABNG也是平行四边形．∴AG?BN?54．
??EBC??BNM?90°． ∵MN?BE，
??MNG??BNM?90°，??EBC??MNG． ?NG?BC，
在△BCE与△NGM中
??EBC??MNG，? ?BC?NG，∴△BCE≌△NGM，EC?MG． ································５分
??C??NGM?90°．?∵AM?AG?MG，AM=∴
类比归纳
2554?1?14．··································································· 6分
AMBN?15． ········································································································· 7分
（或
410）；
917；
?n?1?n?122 ················································································ 10分
联系拓广
nm?2n?1nm?12222 ············································································································· 12分
评分说明：1．如你的正确解法与上述提供的参考答案不同时，可参照评分说明进行估分．
2．如解答题由多个问题组成，前一问题解答有误或未答，对后面问题的解答没
有影响，可依据参考答案及评分说明进行估分．
8、（2009年安徽省）23．已知某种水果的批发单价与批发量的函数关系如图（1）所示． （1）请说明图中①、②两段函数图象的实际意义．
金额w（元） 【解】
批发单价（元）
① 5 300 ② 4
200
100
60 O 20 批发量（kg） O 第23题图（1）
（2）写出批发该种水果的资金金额w（元）与批发量m（kg）之间的
函数关系式；在下图的坐标系中画出该函数图象；指出金额在什 么范围内，以同样的资金可以批发到较多数量的该种水果．
【解】
（3）经调查，某经销商销售该种水果的日最高销量与零售价之间的函
20 40 60 批发量m（kg）
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