4. (3 points) As shown in the figure, $P, Q$ are points on the side $AD$ and the diagonal $AC$ of the square $ABCD$, respectively, and $PD: AP=4: 1$, $QC: AQ=2: 3$. If the area of the square $ABCD$ is 25, then the area of triangle $PBQ$ is $\qquad$ .
【Analysis】Draw $E F / / A B$, intersecting $A D$ at $F$ and $B C$ at $E$, and $Q G \perp D C$ at $G$. According to the similarity ratio, the lengths of each segment can be calculated. Then, subtract the areas of the other parts from the area of the square to get the final result.

【Solution】Solution: Connect $Q D$, draw $E F / / A B$, intersecting $A D$ at $F$ and $B C$ at $E$, and $Q G \perp D C$ at $G$. Since the area of the square $A B C D$ is 25,

so $A D=E F=5$;
Since $Q C: A Q=2: 3$, according to the symmetry of the square, $Q E=Q G=2, Q F=3$.
Since $P D: A P=4: 1$, so $A P=1, P D=4$.
$$
\begin{array}{l}
S \triangle P Q D=S_{\text {square } A B C D}-S_{\triangle C Q B}-S \triangle D Q C-S_{\triangle P Q D}-S_{\triangle P A B} \\
=25-2 \times 5 \div 2-2 \times 5 \div 2-4 \times 3 \div 2-1 \times 5 \div 2 \\
=25-5-5-6-2.5=6.5 .
\end{array}
$$

Answer: The area of triangle $P B Q$ is 6.5.
The answer is: 6.5.