29. The figure below shows a triangle $\mathrm{ABC}$ where $\mathrm{AB}=\mathrm{AC} . \mathrm{D}$ and $\mathrm{E}$ are points on sides $A B$ and $A C$, respectively, such that $A B=4 D B$ and $A C=4 A E$. If the area of the quadrilateral BCED is $52 \mathrm{~cm}^{2}$ and the area of the triangle $A D E$ is $x \mathrm{~cm}^{2}$, find $x$.
29. Answer: 12

Consider $\triangle \mathrm{ADE}$ and $\triangle \mathrm{CDE}$. Since $\mathrm{CE}=3 \mathrm{AE}$, we see that area of $\triangle C D E=3 \times$ area of $\triangle \mathrm{ADE}=3 x \mathrm{~cm}^{2}$.

Similarly, we have
$$
\text { area of } \triangle B C D=\frac{1}{3} \times \text { area of } \triangle \mathrm{ACD} \text {. }
$$

Now area of $\triangle \mathrm{ACD}=$ area of $\triangle \mathrm{ADE}+$ area of $\triangle \mathrm{CDE}=4 x \mathrm{~cm}^{2}$, so area of $\triangle \mathrm{BCD}=\frac{4 x}{3} \mathrm{~cm}^{2}$.

Thus we obtain the equation $3 x+\frac{4 x}{3}=52$, which gives $x=12$.