题目

如图，△ABC中，AB=AC，BC=12cm，点D在AC上，DC=4cm．将线段DC沿着CB的方向平移7cm得到线段EF，点E，F分别落在边AB，BC上，则△EBF的周长为　 　cm．                                                                                                             答案：13【分析】直接利用平移的性质得出EF=DC=4cm，进而得出BE=EF=4cm，进而求出答案．                     【解答】解：∵将线段DC沿着CB的方向平移7cm得到线段EF，                  ∴EF=DC=4cm，FC=7cm，                                                                      ∵AB=AC，BC=12cm，                                                                             ∴∠B=∠C，BF=5cm，                                                                            ∴∠B=∠BFE，                                                                                   ∴BE=EF=4cm，                                                                                 ∴△EBF的周长为：4+4+5=13（cm）．                                                   故答案为：13．                                                                                  【点评】此题主要考查了平移的性质，根据题意得出BE的长是解题关键．                   　                                                                                                      

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