$解:∵AB=2x,CD=3y$
$AC、DE的長(zhǎng)分別為AB、CD的一半$
$∴BF= AC=x,DE=\frac{3}{2}y$
$∴AD=AC+CD=x+3y$
$∴S_{涂色}$
$=S_{長(zhǎng)方形ABFC}+S_{長(zhǎng)方形CDEG}-S_{△ABF}-S_{△ADE}$
$=2x·x+3y·\frac{3}{2}y-\frac{1}{2}×2x·x-\frac{1}{2}(x+3y)·\frac{3}{2}y$
$=2x2+\frac{9}{2}y2-x2-\frac{3}{4}xy-\frac{9}{4}y2$
$=x2-\frac{3}{4}xy+\frac{9}{4}y2$
$=\frac{1}{4}(4x2-3xy+9y2)$
$=\frac{1}{4}(4x2-12xy+9y2+9xy)$
$=\frac{1}{4}(4x2-12xy+9y2)+\frac{9}{4}xy$
$=\frac{1}{4}(2x-3y)2+\frac{9}{4}xy$
$=\frac{1}{4}×4+\frac{9}{4}×4$
$=1+9$
$=10. $