Question

如图，在平面直角坐标系中，反比例函数y=（x＞0）的图象交矩形OABC的边AB于点D，交边BC于点E，且BE=2EC．若四边形ODBE的面积为6，则k=　　　　　　．                                             

                                                                   

Answer 1

　3　．                                                                                             

                                                                                                          

【考点】反比例函数系数k的几何意义．                                                

【专题】压轴题．                                                                              

【分析】连接OB，由矩形的性质和已知条件得出△OBD的面积=△OBE的面积=四边形ODBE的面积=3，在求出△OCE的面积，即可得出k的值．                                                                     

【解答】解：连接OB，如图所示：                                                         

∵四边形OABC是矩形，                                                                         

∴∠OAD=∠OCE=∠DBE=90°，△OAB的面积=△OBC的面积，                     

∵D、E在反比例函数y=（x＞0）的图象上，                                              

∴△OAD的面积=△OCE的面积，                                                           

∴△OBD的面积=△OBE的面积=四边形ODBE的面积=3，                          

∵BE=2EC，∴△OCE的面积=△OBE的面积=，                                        

∴k=3；                                                                                             

故答案为：3．                                                                                   

                                                                    

【点评】本题考查了矩形的性质、三角形面积的计算、反比例函数的图象与解析式的求法；熟练掌握矩形的性质和反比例函数解析式的求法是解决问题的关键．