Given that $\left|\overrightarrow{n}\right|=1,\left|\overrightarrow{m}\right|=\sqrt{3}$ And that the angle between the vectors $\left|\overrightarrow{m}\right|$ and $\left|\overrightarrow{n}\right|$ is ${30}^{\circ}$

We will define $\overrightarrow{a}\phantom{\rule{mediummathspace}{0ex}}=\overrightarrow{m}\phantom{\rule{mediummathspace}{0ex}}-\overrightarrow{n}\phantom{\rule{mediummathspace}{0ex}},\phantom{\rule{mediummathspace}{0ex}}\overrightarrow{b}=\overrightarrow{m}+\overrightarrow{n}$ calculate the area of the triangle created by vectors $\overrightarrow{a},\overrightarrow{b}$

I found out what $\left|\overrightarrow{a}\right|$ and $\left|\overrightarrow{b}\right|$ but not sure how to continue form here.

$\begin{array}{rl}{\left|\overrightarrow{a}\right|}^{2}& =(\overrightarrow{m}-\overrightarrow{n})(\overrightarrow{m}-\overrightarrow{n})\\ & ={\left|\overrightarrow{m}\right|}^{2}-2\overrightarrow{m}\cdot \overrightarrow{n}+{\left|\overrightarrow{n}\right|}^{2}\\ & =3-2\sqrt{3}\cdot 1\cdot \phantom{\rule{mediummathspace}{0ex}}\frac{\sqrt{3}}{2}+1\\ & =1\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\overline{){\displaystyle \left|\overrightarrow{a}\right|=1}}\end{array}$

$\begin{array}{rl}{\left|\overrightarrow{b}\right|}^{2}& =(\overrightarrow{m}+\overrightarrow{n})\cdot (\overrightarrow{m}+\overrightarrow{n})\\ & ={\left|\overrightarrow{m}\right|}^{2}+2\overrightarrow{m}\cdot \overrightarrow{n}+{\left|\overrightarrow{n}\right|}^{2}\\ & =3+2\sqrt{3}\cdot 1\cdot \phantom{\rule{mediummathspace}{0ex}}\frac{\sqrt{3}}{2}+1\\ & =7\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\overline{){\displaystyle \left|\overrightarrow{b}\right|=\sqrt{7}}}\end{array}$

We will define $\overrightarrow{a}\phantom{\rule{mediummathspace}{0ex}}=\overrightarrow{m}\phantom{\rule{mediummathspace}{0ex}}-\overrightarrow{n}\phantom{\rule{mediummathspace}{0ex}},\phantom{\rule{mediummathspace}{0ex}}\overrightarrow{b}=\overrightarrow{m}+\overrightarrow{n}$ calculate the area of the triangle created by vectors $\overrightarrow{a},\overrightarrow{b}$

I found out what $\left|\overrightarrow{a}\right|$ and $\left|\overrightarrow{b}\right|$ but not sure how to continue form here.

$\begin{array}{rl}{\left|\overrightarrow{a}\right|}^{2}& =(\overrightarrow{m}-\overrightarrow{n})(\overrightarrow{m}-\overrightarrow{n})\\ & ={\left|\overrightarrow{m}\right|}^{2}-2\overrightarrow{m}\cdot \overrightarrow{n}+{\left|\overrightarrow{n}\right|}^{2}\\ & =3-2\sqrt{3}\cdot 1\cdot \phantom{\rule{mediummathspace}{0ex}}\frac{\sqrt{3}}{2}+1\\ & =1\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\overline{){\displaystyle \left|\overrightarrow{a}\right|=1}}\end{array}$

$\begin{array}{rl}{\left|\overrightarrow{b}\right|}^{2}& =(\overrightarrow{m}+\overrightarrow{n})\cdot (\overrightarrow{m}+\overrightarrow{n})\\ & ={\left|\overrightarrow{m}\right|}^{2}+2\overrightarrow{m}\cdot \overrightarrow{n}+{\left|\overrightarrow{n}\right|}^{2}\\ & =3+2\sqrt{3}\cdot 1\cdot \phantom{\rule{mediummathspace}{0ex}}\frac{\sqrt{3}}{2}+1\\ & =7\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\overline{){\displaystyle \left|\overrightarrow{b}\right|=\sqrt{7}}}\end{array}$