?$解:作AH⊥BC于H,$?
?$則AH//DE,DG//BC$?
?$所以\frac {DE}{AH}=\frac {BD}{AB}=1-\frac {AD}{AB}=1-\frac {DG}{BC}$?
?$BC=\sqrt{122+162}=20\ \mathrm {cm},AH=\frac {12×16}{20}=\frac {48}{5}\ \mathrm {cm}$?
?$設DE= 3x\ \mathrm {cm},$?
?$則DG= EF = 5x\ \mathrm {cm}$?
?$所以\frac {3x}{\frac {48}{5}}=1-\frac {5x}{20}$?
?$解得x=\frac {16}{9}$?
?$所以矩形DEFG的周長= 2(DE+ DG)= 16x= \frac {256}{9}\ \mathrm {cm} .$?