Question

如圖，在平面直角坐標系中，反比例函數(shù)y=（x＞0）的圖象交矩形OABC的邊AB于點D，交邊BC于點E，且BE=2EC．若四邊形ODBE的面積為6，則k=　　　　　　．                                             

                                                                   

Answer 1

　3　．                                                                                             

                                                                                                          

【考點】反比例函數(shù)系數(shù)k的幾何意義．                                                

【專題】壓軸題．                                                                              

【分析】連接OB，由矩形的性質(zhì)和已知條件得出△OBD的面積=△OBE的面積=四邊形ODBE的面積=3，在求出△OCE的面積，即可得出k的值．                                                                     

【解答】解：連接OB，如圖所示：                                                         

∵四邊形OABC是矩形，                                                                         

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

∵D、E在反比例函數(shù)y=（x＞0）的圖象上，                                              

∴△OAD的面積=△OCE的面積，                                                           

∴△OBD的面積=△OBE的面積=四邊形ODBE的面積=3，                          

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

∴k=3；                                                                                             

故答案為：3．                                                                                   

                                                                    

【點評】本題考查了矩形的性質(zhì)、三角形面積的計算、反比例函數(shù)的圖象與解析式的求法；熟練掌握矩形的性質(zhì)和反比例函數(shù)解析式的求法是解決問題的關(guān)鍵．