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