Smallest area that can be formed by 3 non intersecting diagonal of a hexagon We have a regular hexagon: a polygon with six sides, in which all internal angles have 120 degrees and length is the 5 cm of each side. We are going to draw three non-intersecting diagonals in some way. These will divide the hexagon into four triangles. We will then compute their areas. Find the smallest of those areas.
Let $ABCDEF$ be our hexagon and $O$ be a circumcircle of the hexagon.

Thus, the radius of the circle is equal to $5$ and $\Delta AFE\cong\Delta AOE,$ which gives $$S_{\Delta AFE}=\frac{1}{2}S_{\Delta AED}$$ and since $\Delta AFE\cong \Delta ABC$ and $\Delta AED\cong ACD,$ we see that $S_{\Delta AFE}$ is a need area.

Id est, $$S_{\Delta AFE}=\frac{5\cdot5\cdot\sin120^{\circ}}{2}=\frac{25\sqrt3}{4}$$ gives the answer.
Let $ABCDEF$ be our hexagon and $O$ be a circumcircle of the hexagon.

Thus, the radius of the circle is equal to $5$ and $\Delta AFE\cong\Delta AOE,$ which gives $$S_{\Delta AFE}=\frac{1}{2}S_{\Delta AED}$$ and since $\Delta AFE\cong \Delta ABC$ and $\Delta AED\cong ACD,$ we see that $S_{\Delta AFE}$ is a need area.

Id est, $$S_{\Delta AFE}=\frac{5\cdot5\cdot\sin120^{\circ}}{2}=\frac{25\sqrt3}{4}$$ gives the answer.