4. The vertices of an inscribed triangle in a circle divide the circumference into three arcs of lengths $3, 4, 5$. Then the area of this triangle is $\qquad$ .
$$
\text { 4. } S=\frac{9}{\pi^{2}}(\sqrt{3}+3) \text {. }
$$

Divide into three arcs of $90^{\circ}, 120^{\circ}$, and $150^{\circ}$.
$$
\begin{array}{l}
S_{\triangle A B C}=\frac{1}{2} r^{2}\left(\sin 90^{\circ}+\sin 120^{\circ}+\sin 150^{\circ}\right) \\
=\frac{1}{4} r^{2}(\sqrt{3}+3) \text {, } \\
2 \pi r=3+4+5=12, r=\frac{6}{\pi} \text {. } \\
S_{\triangle A B C}=\frac{1}{4} \times \frac{36}{\pi^{2}}(\sqrt{3}+3) \\
=\frac{9}{\pi^{2}}(\sqrt{3}+3) \text {. } \\
\end{array}
$$