解：连接$AC，$$BD，$过点$O$作$OH\perp CD$于点$H，$则$\angle OHC = 90^{\circ}，$$CH=\frac{1}{2}CD。$由作法可知$OA = OC = AC = OB = OD = BD，$所以$\triangle AOC$和$\triangle BOD$都是等边三角形，所以$\angle AOC=\angle BOD = 60^{\circ}，$所以$\angle COD = 180^{\circ}-\angle AOC-\angle BOD = 60^{\circ}，$所以$\triangle COD$是等边三角形，所以$CD = OC。$因为$AB = 4，$所以$CD = OC = OA=\frac{1}{2}AB = 2，$所以$CH = 1，$所以$OH=\sqrt{OC^{2}-CH^{2}}=\sqrt{3}，$所以$S_{\triangle COD}=\frac{1}{2}CD\cdot OH=\sqrt{3}。$故$\triangle COD$的面积为$\sqrt{3}。$
