$解:∵ 四边形ABCD是菱形$
$∴ AB=AD=CD=BC,AC⊥BD,$
$OA=OC,OB=OD$
$∵ 菱形ABCD的周长为20\ \mathrm {cm}$
$∴ AD=\frac 1 4×20=5\ \mathrm {cm}$
$∵ AC=8\ \mathrm {cm}$
$∴ OA=\frac 1 2AC=4\ \mathrm {cm}$
$在Rt△OAD中,∵ AD=5\ \mathrm {cm},OA=4\ \mathrm {cm}$
$∴ OD={\sqrt {{AD}^2-{OA}^2}}=3\ \mathrm {cm}$
$∴ BD=2OD=6\ \mathrm {cm},{S}_{菱形ABCD}={\frac {1} {2}}AC×BD={\frac {1} {2}×6×8=24{cm}^2}$