2012-II-17

Ans: B
Join $AF$.

Since $\Delta DFG$ and $\Delta DFA$ have the same base length and the same height, then their areas are the same.

i.e. the area of $\Delta DFA=3\mbox{ cm}^2$.

Join $AC$. Since $\Delta DFA$ and $\Delta DCA$ have the same height, then we have

$\begin{array}{rcl}
\dfrac{\mbox{the area of }\Delta DFA}{\mbox{the area of }\Delta DCA} & = & \dfrac{DF}{DC} \\
\dfrac{3}{\mbox{the area of }\Delta DCA} & = & \dfrac{3}{3+4} \\
\mbox{the area of }\Delta DCA & = & 7 \mbox{ cm}^2
\end{array}$

Therefore, the area of the parallelogram $ABCD$

$\begin{array}{cl}
= & 2 \times \mbox{the area of }\Delta DCA \\
= & 2 \times 7 \\
= & 14\mbox{ cm}^2
\end{array}$