4. Given an equilateral $\triangle A B C$ with side length 1,
$$
\overrightarrow{A P}=\frac{1}{3}(\overrightarrow{A B}+\overrightarrow{A C}), \overrightarrow{A Q}=\overrightarrow{A P}+\frac{1}{2} \overrightarrow{B C} \text {. }
$$

Then the area of $\triangle A P Q$ is $\qquad$ .
4. $\frac{\sqrt{3}}{12}$.

From the problem, we know that $P$ is the centroid of $\triangle A B C$.
$$
\begin{array}{l}
\text { Given } \overrightarrow{A Q}=\overrightarrow{A P}+\frac{1}{2} \overrightarrow{B C} \\
\Rightarrow|P Q|=\frac{1}{2}|B C| . \\
\text { Then } S_{\triangle A P Q}=\frac{1}{2} P Q \cdot A P \\
=\frac{1}{2} \times \frac{1}{2} B C \times \frac{2}{3} h \\
=\frac{1}{3} S_{\triangle A B C}=\frac{\sqrt{3}}{12} .
\end{array}
$$