4. (10 points) As shown in the figure, $P$ and $Q$ are points on the side $AD$ and the diagonal $AC$ of square $ABCD$, respectively, and $AP: PD=1: 3$, $AQ: QC=4: 1$. If the area of square $ABCD$ is 100, then the area of triangle $PBQ$ is.
37.5

【Analysis】Connect $D Q$, and draw the height $Q E$ of $\triangle D Q P$. Calculate the areas of $\triangle A B P$, $\triangle B C Q$, $\triangle D C Q$, and $\triangle D P Q$ respectively, then subtract their sum from the area of the square.

【Solution】Solution: Connect $D Q$, and draw the height $Q E$ of $\triangle D Q P$.
The area of square $A B C D$ is 100, so its side length is 10.
Since $A P: P D=1: 3$,
we have $A P=2.5 ; D P=7.5$.
$S_{\triangle A B P}=10 \times 2.5 \div 2=12.5$.
$A Q: Q C=4: 1$,
so $S_{\triangle C Q B}=\frac{1}{5} S_{\triangle A B C}=\frac{1}{10} S_{\text {square } A B C D}=\frac{1}{10} \times 100=10$.
Similarly, $S_{\triangle D C Q}=10$.
$E Q \perp A D$, so $E Q: D C=A Q: A C=4: 5$,
$E Q=\frac{4}{5} \times 10=8, S_{\triangle O Q D}=7.5 \times 8 \div 2=30$.
$S_{\triangle P B Q}=S_{\text {square } A B C D}-S_{\triangle A B P}-S_{\triangle C Q B}-S_{\triangle D C Q}-S_{\triangle O Q D}=100-12.5-10-10-30=37.5$.