4. As shown in the figure, $P, Q$ are points on the side $AD$ and the diagonal $AC$ of the square $ABCD$, respectively, and $PD: AP=4: 1, QC$: $\mathrm{AQ}=2: 3$. If the area of the square $\mathrm{ABCD}$ is 25, then the area of triangle $\mathrm{PBQ}$ is $\qquad$.
Parse: Connect $\mathrm{QD}$, draw $\mathrm{QE} \perp \mathrm{BC}$ at $\mathrm{E}$, $\mathrm{QF} \perp \mathrm{AD}$ at $\mathrm{F}$, $\mathrm{QG} \perp \mathrm{CD}$ at $\mathrm{G}$, the area of square $A B C D$ is 25, so $A D=E F=5, Q C: A Q=2: 3$, according to the symmetry of the square, so $Q E=Q G=2, Q F=3, P D: A P=4: 1, A P=1, P D=4$.
$$
\begin{aligned}
\mathrm{S} \triangle \mathrm{PQB} & =\mathrm{S}_{\text {square }}-\mathrm{S} \triangle \mathrm{CQB}-\mathrm{S} \triangle \mathrm{DQC}-\mathrm{S} \triangle \mathrm{PQD}-\mathrm{S} \triangle \mathrm{PAB} \\
& =25-2 \times 5 \div 2 \times 2-4 \times 3 \div 2-1 \times 5 \div 2 \\
& =25-10-6-2.5 \\
& =6.5
\end{aligned}
$$