loading
由（2）知四边形BC1DA是菱形， ∴AD?AB?2， ∴ED?AD?AE?2?23 ························································（12分） 3．（解法二）??ABC?120°，?ABE?30°，∴?EBC?90°．
在Rt△EBC中，BE?BC·tanC?2?tan30°??EA1?BA1?BE?2?2323 3． ······················································（10分） 3． ?A1C1∥AB，??A1DE??A．??A1DE??A1．∴ED?EA1?2?23·································································（12分） 3．（其它解法可参照给分）全品中 考网
26．（1）解：由
23x?83?A点坐标为??4，?0，得x??4． 0?．?B点坐标为?8，0?．由?2x?16?0，得x?8．
∴AB?8???4??12． ····················································································（2分）
28?y?x?，?x?5，?6?．由?解得∴C点的坐标为?5，·····································（3分） 33?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， 全 ·················································································（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．
yyl2 l2yE C D R l1E D C - 11 -R l1E l2 D C l1A O F M G B x （图1）
A F O G M （图2）
B x R F A G O M B x （图3）
∴
BGBM?RGCM即，t3?RG6∴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） ．- 12 -