试题分析:(1)由 = 及∠MON=∠ABE=90°,可得出△EOF∽△ABO. (2)证明Rt△EOF∽Rt△ABO,进而证明EF⊥OA. (3)由已知S△AEF= S四边形ABOF.得出S△FOE+S△ABE= S梯形ABOF,从而可求出t的值. 试题解析:(1)∵t=1, ∴OE=1.5厘米,OF=2厘米, ∵AB=3厘米,OB=4厘米, ∴ ,![](http://img.shitiku.com.cn/uploads/allimg/20191027/20191027193356-65648.png) ∵∠MON=∠ABE=90°, ∴△EOF∽△ABO. (2)在运动过程中,OE=1.5t,OF=2t. ∵AB=3,OB=4. ∴ . 又∵∠EOF=∠ABO=90°, ∴Rt△EOF∽Rt△ABO. ∴∠AOB=∠EOF. ∵∠AOB+∠FOC=90°, ∴∠EOF+∠FOC=90°, ∴EF⊥OA. (3)如图,连接AF,
![](http://img.shitiku.com.cn/uploads/allimg/20191027/20191027193356-31428.png) ∵OE=1.5t,OF=2t, ∴BE=4﹣1.5t ∴S△FOE= OE•OF= ×1.5t×2t= t2,S△ABE= ×(4﹣1.5t)×3=6﹣ t, S梯形ABOF= (2t+3)×4=4t+6 ∵S△AEF= S四边形ABOF ∴S△FOE+S△ABE= S梯形ABOF, ∴ t2+6﹣ t= (4t+6),即6t2﹣17t+12=0, 解得t= 或t= . ∴当t= 或t= 时,S△AEF= S四边形ABOF. |