△ABC中，AB=13cm，AC=15cm，高AD=12，则BC的长为（　　）             A．1...

问题详情：

△ABC中，AB=13cm，AC=15cm，高AD=12，则BC的长为（　　）             

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

【回答】

C【考点】勾股定理．                                                                        

【专题】分类讨论．                                                                          

【分析】分两种情况讨论：锐角三角形和钝角三角形，根据勾股定理求得BD，CD，再由图形求出BC，在锐角三角形中，BC=BD+CD，在钝角三角形中，BC=CD﹣BD．                                               

【解答】解：（1）如图，锐角△ABC中，AB=13，AC=15，BC边上高AD=12，                 

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

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

则BD=5，                                                                                          

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

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

则CD=9，                                                                                         

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

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

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

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

则BD=5，                                                                                          

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

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

则CD=9，                                                                                         

故BC的长为DC﹣BD=9﹣5=4．                                                              

故选：C．                                                                                         

【点评】本题考查了勾股定理，把三角形边的问题转化到直角三角形中用勾股定理解答．                 

知识点：勾股定理

题型：选择题