$解:(2)∵在Rt△ABC中,∠ABC=90°,AB=AC=3,$
$∴S_{△ABC}=\frac{1}{2}×AB×BC$
$=\frac{1}{2}×3×3=\frac{9}{2}$
$∵在Rt△ACD中,∠ACD=90°,AC²=18,CD²=8,$
$∴AC=3\sqrt{2},CD=2\sqrt{2}$
$∴S_{△ACD}=\frac{1}{2}×AC×CD$
$=\frac{1}{2}×3\sqrt{2}×2 \sqrt{2}=6. $
$∴S_{四边形ABCD}=S_{△ABC}+S_{△ACD}$
$=\frac{9}{2}+6$
$=\frac{21}{2} $
$∴四边形ABCD的面积为\frac{21}{2}$
