Finding the area of 6-sided polygon....

The area of a regular hexagon is given by $A=\dfrac{3\sqrt{3}}{2}x^2$ where $x$ is the side length of the hexagon. We have

$A=\dfrac{3\sqrt{3}}{2}(8^2) \approx \boxed{166.28}$

Each internal angles measure $120^\circ$ Inscribe a rectangle on the hexagon. Then divide the triangles into two. You now have 1 rectangle and 4 30-60-90 triangles. The side is $8$ . Using 30-60-90 triangle theorem, the measures of shorter leg and longer leg are $4$ and $4\sqrt{3}$ , respectively. Then the measures of sides of rectangle are $8\sqrt{3}$ ( $4\sqrt{3}+4\sqrt{3}$ ) and $8$ . And: $4\bigg(\frac{bh}{2}\bigg) + lw$ $= 2(4)(4\sqrt{3}) + (8\sqrt{3})(8)$ $=32\sqrt{3}+64\sqrt{3}$ $=96\sqrt{3}$ $\boxed{=166.28}$

6 x 4 Sqrt (8^2 - 4^2) = 166.27687752661222017863484878456

Note: There available with formula for area of n-polygon with a given side length.