A triangle is inscribed in a circle. The vertices of the triangle divide the circle into three arcs of length 3,4 and 5 units, then area of the triangle is equal to

Given are AC=3 are AB =4 and arc BC= 5. Let r be the radius of the circle .
Then $\mathrm{\alpha }=\frac{3}{\mathrm{r}},\mathrm{\beta }=\frac{4}{\mathrm{r}},\mathrm{\gamma }=\frac{5}{\mathrm{r}}$ Now $3+4+5=2\mathrm{\pi r}\Rightarrow \mathrm{r}=\frac{6}{\mathrm{\pi }}\Rightarrow \frac{1}{\mathrm{r}}=\frac{\mathrm{\pi }}{6}$
$\begin{array}{rcl} & \Rightarrow & △\mathrm{ABC}=△\mathrm{OAC}+△\mathrm{OAB}+△\mathrm{OBC}\\ & =& \frac{1}{2}{r}^2(\sin \alpha +\sin \beta +\sin \gamma )\\ & =& \frac{1}{2}\frac{36}{{\mathrm{\pi }}^2}\left(\sin \frac{\mathrm{\pi }}{2}+\sin \frac{2\mathrm{\pi }}{3}+\sin \frac{5\mathrm{\pi }}{6}\right)\\ & =& \frac{36}{2{\mathrm{\pi }}^2}\left(\frac{3+\sqrt{3}}{2}\right)\\ & =& \frac{9\sqrt{3}(1+\sqrt{3})}{{\mathrm{\pi }}^2}\end{array}$