29. In the figure below, $A B C D$ is a rectangle, $E$ is the midpoint of $A D$ and $F$ is the midpoint of $C E$. If the area of triangle $B D F$ is $12 \mathrm{~cm}^{2}$, find the area of rectangle $A B C D$ in $\mathrm{cm}^{2}$.
29. Answer: 96

Let $S$ be the area of rectangle $A B C D$. Then we have area of $\triangle C D F=\frac{1}{2} \times$ area of $\triangle C D E=\frac{1}{2} \times \frac{1}{4} S=\frac{1}{8} S$.

Next we have area of $\triangle B C F=\frac{1}{2} \times$ area of $\triangle B C E=\frac{1}{2} \times \frac{1}{2} S=\frac{1}{4} S$.
Now
$$
\begin{aligned}
12 & =\text { area of } \triangle B D F=\text { area of } \triangle B C D \text { - area of } \triangle B C F \text { - area of } \triangle C D F \\
& =\frac{1}{2} S-\frac{1}{4} S-\frac{1}{8} S=\frac{1}{8} S .
\end{aligned}
$$

Hence the area of rectangle $A B C D=96 \mathrm{~cm}^{2}$.