In a triangle $ABC,$ medians $AD$ and $CE$ are drawn. If $AD=5,$$DAC=\pi /8$ and $\mathrm{\angle }ACE=\pi /4$, the area of the triangle is

Let $O$ be the point of intersection of the medians of triangle $ABC$ (Fig. 12.12). Then the area of  $\mathrm{△}ABC$ is three times that of $\mathrm{△}AOC,O$ being the centroid of $\mathrm{\Delta }ABC,$ divides the median through $B$ in the ratio $2:1$

and the height of $\mathrm{\Delta }AOC$ is one- third that of $\mathrm{\Delta }ABC\text{. }$Now, in $\mathrm{\Delta }AOC,AO=(2/3)AD=10/3.$ Therefore, applying the sine rule to $\mathrm{\Delta }AOC,$ we get

$\frac{OC}{\sin (\pi /8)}=\frac{AO}{\sin (\pi /4)}\Rightarrow OC=\frac{10}{3}\cdot \frac{\sin (\pi /8)}{\sin (\pi /4)}$

Area of $\mathrm{\Delta }AOC=\frac{1}{2}\cdot AO\cdot OC\cdot \sin \mathrm{\angle }AOC$

$\begin{array}{r}=\frac{1}{2}\cdot \frac{10}{3}\cdot \frac{10}{3}\cdot \frac{\sin (\pi /8)}{\sin (\pi /4)}\cdot \sin \left(\frac{\pi }{2}+\frac{\pi }{8}\right)\\ =\frac{50}{9}\cdot \frac{\sin (\pi /8)\cos (\pi /8)}{\sin (\pi /4)}=\frac{50}{18}=\frac{25}{9}\end{array}$

Area of $\mathrm{\Delta }ABC=3\cdot \frac{25}{9}=\frac{25}{3}$