△ABC中，AB=13cm，AC=15cm，高AD=12，則BC的長為（　　）             A．1...

問題詳情：

△ABC中，AB=13cm，AC=15cm，高AD=12，則BC的長為（　　）             

A．14                          B．4                            C．14或4                    D．以上都不對

【回答】

C【考點】勾股定理．                                                                        

【專題】分類討論．                                                                          

【分析】分兩種情況討論：鋭角三角形和鈍角三角形，根據勾股定理求得BD，CD，再由圖形求出BC，在鋭角三角形中，BC=BD+CD，在鈍角三角形中，BC=CD﹣BD．                                               

【解答】解：（1）如圖，鋭角△ABC中，AB=13，AC=15，BC邊上高AD=12，                 

在Rt△ABD中AB=13，AD=12，由勾股定理得                                               

BD2=AB2﹣AD2=132﹣122=25，                                                                

則BD=5，                                                                                          

在Rt△ABD中AC=15，AD=12，由勾股定理得                                               

CD2=AC2﹣AD2=152﹣122=81，                                                                

則CD=9，                                                                                         

故BC=BD+DC=9+5=14；                                                                         

（2）鈍角△ABC中，AB=13，AC=15，BC邊上高AD=12，                            

在Rt△ABD中AB=13，AD=12，由勾股定理得                                               

BD2=AB2﹣AD2=132﹣122=25，                                                                

則BD=5，                                                                                          

在Rt△ACD中AC=15，AD=12，由勾股定理得                                               

CD2=AC2﹣AD2=152﹣122=81，                                                                

則CD=9，                                                                                         

故BC的長為DC﹣BD=9﹣5=4．                                                              

故選：C．                                                                                         

【點評】本題考查了勾股定理，把三角形邊的問題轉化到直角三角形中用勾股定理解答．                 

知識點：勾股定理

題型：選擇題