Question

如圖，在矩形ABCD中，AD=4cm，AB=10cm，在邊AB上有一點P以2cm/s的速度由A點向B點運(yùn)動，設(shè)P點運(yùn)動了t秒．                                                                       

（1）用含t的代數(shù)式表示BP的值；                                                        

（2）當(dāng)t為何值時，△APD與△BPC相似．                                           

                                                    

Answer 1

【考點】矩形的性質(zhì)；相似三角形的性質(zhì)．                                             

【專題】動點型；分類討論．                                                                  

【分析】（1）設(shè)P點運(yùn)動了t秒．則AP=2t；BP=10﹣2t．                             

（2）畫出圖形可知，要分三種情況討論．                                              

【解答】解：（1）BP=10﹣2t；                                                              

                                                                                                          

（2）①②當(dāng)位于P和P₂時，△DAP∽△PBC時， =，                          

∴=，解得t=1秒或4秒；                                                       

③當(dāng)位于P₁位置時，AP₁=BP₁，2t=10﹣2t，解得t=2.5．                                 

∴t=1或t=4或t=2.5時兩個三角形相似．                                                

                                                    

【點評】此題是一道動點問題，需要通過數(shù)形結(jié)合來進(jìn)行計算．要注意，位于P₁位置時兩三角形全等，是相似的特殊情況．