$解：?(1)AD=\sqrt {AC^2-CD^2}=\sqrt {13^2-12^2}=5\ \mathrm {cm}?$
$?BC=\sqrt {BD^2+CD^2}=\sqrt {9^2+12^2}=15\ \mathrm {cm}?$
$?AB=AD+BD=5+9=14\ \mathrm {cm}?$
$?C_{△ABC}=AB+BC+AC=14+15+13=42\ \mathrm {cm}?$
$∵?DE//BC ?$
$∴?$△ADE∽△ABC$?$
$∴?\frac {C_{△ADE}}{C_{△ABC}}=\frac {AD}{AB}=\frac 5{14}?$
$∴?C_{△ADE}=42×\frac 5{14}=15\ \mathrm {cm}?$
$?(2)S_{△ABC}=\frac 12AB ·CD=\frac 12×14×12=84\ \mathrm {cm^2}?$
$?\frac {S_{△ADE}}{S_{△ABC}}=(\frac 5{14})^2?$
$∴?S_{△ADE}=84×(\frac 5{14})^2=\frac {75}7\ \mathrm {cm^2}?$
