5. Let $P$ be a point inside an equilateral $\triangle A B C$, with $A P=3$, $B P=4$, $C P=5$. Then the area of quadrilateral $A B C P$ is $\qquad$
$$
\text { 5. } 6+4 \sqrt{3} \text {. }
$$

As shown in Figure 3, with point $B$ as the center of rotation, $\triangle A P B$ is rotated $60^{\circ}$ clockwise to the position of $\triangle B Q C$. Connect $P Q$.
It is easy to see that $\triangle B P Q$ is

an equilateral triangle, i.e.,
$$
\begin{array}{l}
P Q=B Q=B P=4, \angle P Q B=60^{\circ} . \\
\text { Also, } \triangle B Q C \cong \triangle A P B \Rightarrow C Q=A P=3 .
\end{array}
$$

Then $C P^{2}=5^{2}=3^{2}+4^{2}=C Q^{2}+P Q^{2}$,
$$
\angle P Q C=90^{\circ} \text {. }
$$

Therefore, $S_{\text {quadrilateral } A B C P}=S_{\text {quadrilateral } P B Q C}=S_{\triangle P B Q}+S_{\triangle P Q C}$
$$
=\frac{\sqrt{3}}{4} \times 4^{2}+\frac{1}{2} \times 4 \times 3=6+4 \sqrt{3} \text {. }
$$