Solution :
In triangles AOB and COD,
\(\angle\) AOB = \(\angle\) COD (vertically opposite angles)
\(\angle\) OAB = \(\angle\) OCD (corresponding angles)
\(\therefore\) By AA similarity,
\(\triangle\) AOB ~ \(\triangle\) COD
\(\implies\) \(area(\triangle AOB)\over area(\triangle COD)\) = \({AB}^2\over {DC}^2\)
\(\implies\) \(area(\triangle AOB)\over area(\triangle COD)\) = = \({2DC}^2\over {DC}^2\) = \(4\over 1\)
Hence, Area of triangle AOB : Area of triangle COD = 4 : 1