2009-II-22

Ans: B
Consider $\Delta ACD$,

$\begin{array}{rcl}
\cos 60^\circ & = & \dfrac{AD}{AC} \\
\dfrac{1}{2} & = & \dfrac{AD}{AC} \\
AD : AC & = & 1 : 2
\end{array}$

Note that $\Delta ADE$ and $\Delta CAB$ are similar triangles. Hence, we have

$\begin{array}{rcl}
\dfrac{\text{area of } \Delta ADE}{\text{area of } \Delta CAB} & = & \left( \dfrac{AD}{CA} \right)^2 \\
\dfrac{1}{\text{area of }\Delta CAB} & = & \left( \dfrac{1}{2} \right)^2 \\
\text{area of }\Delta ABC & = & 4 \text{ cm}^2
\end{array}$