2015-II-21

Since $AC$ and $BD$ are diameters of the circle, then the intersection point $E$ must be the centre of the circle. Therefore, $AE$, $EC$, $BE$ and $ED$ are radii of the circle and $AE=EC=BE=ED=12\text{ cm}$. Hence, the area of $\mathrm{\Delta }AEB$

$\begin{array}{cl}=& {\displaystyle \frac{1}{2}}\times AE\times BE\\ =& {\displaystyle \frac{1}{2}}\times 12\times 12\\ =& 72{\text{ cm}}^2\end{array}$