Area of Polygons

Area of a hexagon of side length $z$ is give below where $z$ is any real number: $A=\frac{3\sqrt3}{2}\times z^2$ As for the hexagon given length is 1 therefore: $A=\frac{3\sqrt3}{2}\times 1^2=\frac{3\sqrt3}{2} \times 1=\frac{3\sqrt3}{2}=\boxed{2.598}$

A regular hexagon is composed of 6 congruent equilateral triangles and the area of an equilateral triangle is $\dfrac{\sqrt{3}}{4}x^2$ where $x$ is the side length of the triangle (and also the regular hexagon). So the area of the hexagon is $6\left(\dfrac{\sqrt{3}}{4}x^2\right)=6\left(\dfrac{\sqrt{3}}{4}\right)(1^2) \approx$ $\boxed{2.598}$