Since, $m_1$ and $m_2$ are the roots of the equation $x^2+(\sqrt{3}+2) x+(\sqrt{3}-1)=0$
then $\quad m_1+m_2=-(\sqrt{3}+2)$,
$m_1 m_2=\sqrt{3}-1$
$\therefore m_1-m_2=\sqrt{\left(m_1+m_2\right)^2-4 m_1 m_2}$
$=\sqrt{(3+4+4 \sqrt{3}-4 \sqrt{3}+4)}$
$=\sqrt{11}$
and coordinates of the vertices of the given triangles are $(0,0),\left(c / m_1, c\right)$ and $\left(c / m_2, c\right)$. Hence, the required area of triangle
$=\frac{1}{2}\left|\begin{array}{ccc}0 & 0 & 1 \\ \frac{c}{m_1} & c & 1 \\ \frac{c}{m_2} & c & 1\end{array}\right|=\frac{1}{2} c^2\left|\left(\frac{1}{m_1}-\frac{1}{m_2}\right)\right|$
$=\frac{1}{2} c^2 \frac{\left|m_2-m_1\right|}{m_1 m_2}$
$=\frac{1}{2} c^2 \frac{\sqrt{11}}{(\sqrt{3}-1)}$
$=\frac{1}{2} c^2 \cdot \frac{\sqrt{11} \cdot(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}$
$=\left(\frac{\sqrt{33}+\sqrt{11}}{4}\right) \cdot c^2$