$ 解：由平移的性質(zhì)可得：BC=EF=5\ \mathrm {cm}，AD=BE=2\ \mathrm {cm}$
$∠E=∠ABC=90°，S_{△ABC}=S_{△DEF}$
$∴BH=BC-CH=5-2=3\ \mathrm {cm}$
$∵S_{△ABC}=S_{△陰影}+S_{△DBH}，S_{△DEF}=S_{梯形BEFH}+S_{△DBH}$
$S_{△DEF}=S_{梯形BEFH}+S_{△DBH}$
$∴S_{陰影}+S_{△DBH}=S_{梯形BEFH}+S_{△DBH}$
$∴S_{陰影}=S_{梯形BEFH}$
$=\frac {1}{2}(BH+EF)·BE=\frac {1}{2}×(3+5)×2=8\ \mathrm {cm}2$
