so hmm notice the picture below
the pyramid itself, is really, just one regular hexagon, at the bottom
and 6 triangles, stacked up at each other at the edges
now, if you just get the area of the regular hexagon, and the 6 triangles, add them up, that'd be the total surface area of the pyramid then
[tex]\bf \textit{area of a regular polygon}\\\\
A=\cfrac{1}{4}\cdot n\cdot s^2\cdot cot\left( \frac{180}{n} \right)\qquad
\begin{cases}
n=\textit{number of sides}\\
s=\textit{length of a side}\\
\frac{180}{n}=\textit{angle in degrees}\\
----------\\
n=6\\
s=4
\end{cases}\\\\\\ A=\cfrac{1}{4}\cdot 6\cdot 4^2\cdot cot\left( \frac{180}{6} \right)[/tex]
now, for the triangles, well, area of a triangle is 1/2 bh, as you'd know, and you have both
