
$$\\\frac{sin \angle POM_1}{1}=\frac{sin45}{\sqrt5}\\\\
\frac{sin \angle POM_1}{1}=\frac{1}{\sqrt2*\sqrt5}\\\\
\frac{sin \angle POM_1}{1}=\frac{1}{\sqrt{10}}\\\\
\angle POM_1=asin\left( \frac{1}{\sqrt{10}}\right)\\\\$$
$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{1}}}{{\sqrt{{\mathtt{10}}}}}}\right)} = {\mathtt{18.434\: \!948\: \!822\: \!922^{\circ}}}$$
$$\\\angle NOM_1=30-18.434948822922=\;11.565051177078$$
$${\mathtt{30}}{\mathtt{\,-\,}}{\mathtt{18.434\: \!948\: \!822\: \!922}} = {\mathtt{11.565\: \!051\: \!177\: \!078}}$$
Now Find N
N is the intersection of
$$y=(tan30)x = \frac{x}{\sqrt3} \qquad and \qquad y=-x+2\sqrt2$$
$$\underset{\,\,\,\,{\textcolor[rgb]{0.66,0.66,0.66}{\rightarrow {\mathtt{y, x}}}}}{{solve}}{\left(\begin{array}{l}{\mathtt{y}}={\frac{{\mathtt{x}}}{{\sqrt{{\mathtt{3}}}}}}\\
{\mathtt{y}}={\mathtt{\,-\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{2}}}}\end{array}\right)} \Rightarrow \left\{ \begin{array}{l}{\mathtt{y}} = {\sqrt{{\mathtt{2}}}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{3}}}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{2}}}}\\
{\mathtt{x}} = {\mathtt{3}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{2}}}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{2}}}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{3}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{y}} = {\mathtt{1.035\: \!276\: \!180\: \!410\: \!083}}\\
{\mathtt{x}} = {\mathtt{1.793\: \!150\: \!944\: \!336\: \!107}}\\
\end{array} \right\}$$
$$N(3\sqrt2-\sqrt6,\;\sqrt6-\sqrt2)$$
Distance ON
$${\sqrt{{\left({\mathtt{3}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{2}}}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{6}}}}\right)}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\left({\sqrt{{\mathtt{6}}}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{2}}}}\right)}^{{\mathtt{2}}}}} = {\mathtt{2.070\: \!552\: \!360\: \!820\: \!166}}$$
$$Area\;\triangle\;ONM_1= 0.464\;units^2 \qquad $correct to 3 decimal places$$$
$${\mathtt{0.5}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{5}}}}{\mathtt{\,\times\,}}{\mathtt{2.070\: \!552\: \!360\: \!820\: \!166}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{11.565\: \!051\: \!177\: \!078}}^\circ\right)} = {\mathtt{0.464\: \!101\: \!615\: \!138\: \!509\: \!6}}$$
$$Area\;\triangle\;ONM_1= 0.464\;units^2 \qquad $correct to 3 decimal places$$$
NOW If I haven't made any stupid errors (which I probably have) that should be correct :)