5. In the figure below, $A B C$ is a triangle, and $D$ and $E$ are points on $A B$ and $B C$ respectively. It is given that $D E$ is parallel to $A C$, and $C E: E B=1: 3$. If the area of $\triangle A B C$ is $1440 \mathrm{~cm}^{2}$ and the area of $\triangle A D E$ is $x \mathrm{~cm}^{2}$, what is the value of $x$ ?
(A) 288
(B) 240
(C) 320
(D) 384
(E) 270
5. Answer: (E)
The area of $\triangle A B C$ is given to be $S=1440 \mathrm{~cm}^{2}$. Let $S_{1}$ and $S_{2}$ denote the areas of $\triangle A D E$ and $\triangle D B E$ respectively. Since $D E$ is parallel to $A C, \triangle D B E$ and $\triangle A B C$ are similar. Therefore
$$
\frac{S_{2}}{S}=\left(\frac{B E}{B C}\right)^{2}=\left(\frac{3}{4}\right)^{2}=\frac{9}{16} .
$$

Thus $S_{2}=\frac{9}{16} S$. As $D E$ is parallel to $A C$, we have $A D: D B=C E: E B=1: 3$. Consequently,
$$
\frac{S_{1}}{S_{2}}=\frac{1}{3} .
$$

Hence $S_{1}=\frac{1}{3} S_{2}=\frac{1}{3} \cdot \frac{9}{16} S=\frac{3}{16} \times 1440=270 \mathrm{~cm}^{2}$.