如图所示，直线l过正方形ABCD的顶点B，点A、C到直线l的距离分别为1和3，则正方形ABCD的边长是（　　）...

问题详情：

如图所示，直线l过正方形ABCD的顶点B，点A、C到直线l的距离分别为1和3，则正方形ABCD的边长是（　　）                                                                        

A．2                       B．3                            C．                       D．4

【回答】

C【分析】设BE=x，BF=y，先*Rt△BEA∽Rt△CFB，由相似的*质得xy=3…①，再由勾股定理得1+x2=32+y2…②，联立①②解方程组即可．                                                                       

【解答】解：设BE=x，BF=y，                                                               

∵易*Rt△BEA∽Rt△CFB，                                                                   

∴，                                                                                      

∴xy=3…①                                                                                        

∵正方形ABCD中：AB=BC                                                                     

∴1+x2=32+y2…②                                                                               

由①可知x=，将其代入化简得：y4+8y2﹣9=0                                              

解之、检验符合题意的：y=1，                                                                

∴x=3，y=1                                                                                        

AC2=1+x2=10，                                                                                  

∴AC=                                                                                         

即：正方形的边长为：                                                                     

故：选C                                                                                            

【点评】本题考查了正方形的*质与相似三角形的判定与*质，解题的关键是分析图形中存在的等量关系及有数形结合的思想意识．                                                                                        

知识点：相似三角形

题型：选择题