30. As shown in Figure 7, extend the three sides of $\triangle ABC$ such that $BD=\frac{1}{2} AB$, $CE=\frac{1}{2} BC$, and $AF=\frac{1}{2} CA$. Then the ratio of the area of $\triangle DEF$ to the area of $\triangle ABC$ is ( ).
(A) $3: 1$
(B) $13: 4$
(C) $7:2$
(D) $14: 15$
(E) $10: 3$
30.B.

As shown in Figure 8, draw the altitude from point $C$ to side $AB$ of $\triangle ABC$, with the foot of the altitude being point $X$. Draw the altitude from point $F$ to side $AD$ of $\triangle ADF$, with the foot of the altitude being point $Y$. Then,
$$
\begin{array}{l}
\triangle A X C \backsim \triangle A Y F . \\
\text { Therefore, } \frac{F Y}{C X}=\frac{A Y}{A C}=\frac{1}{2} . \\
S_{\triangle A D F}=\frac{1}{2} \times A D \times F Y=\frac{1}{2} \times \frac{3}{2} \times A B \times \frac{1}{2} C X \\
=\frac{3}{4}\left(\frac{1}{2} \times A B \times C X\right)=\frac{3}{4} S_{\triangle A B C} .
\end{array}
$$

Similarly, $S_{\triangle B H D}=S_{\triangle C F E}=\frac{3}{4} S_{\triangle A B C}$.
Thus, $S_{\triangle D E F}=S_{\triangle A B C}+S_{\triangle B E D}+S_{\triangle C F E}+S_{\triangle A D F}$ $=\left(1+\frac{3}{4}+\frac{3}{4}+\frac{3}{4}\right) S_{\triangle A B C}=\frac{13}{4} S_{\triangle A B C}$.
Therefore, the required ratio is $\frac{13}{4}$.