Given the figure of a regular pyramid
The base of the pyramid is a hexagon with a side length = 6
The lateral area is 6 times the area of one of the side triangles
So, the side triangle has a base = 6
The height will be:
[tex]\begin{gathered} h^2=6^2+(\frac{\sqrt[]{3}}{2}\cdot6)^2=36+27=63 \\ h=\sqrt[]{63} \end{gathered}[/tex]so, the lateral area =
[tex]6\cdot\frac{1}{2}\cdot6\cdot\sqrt[]{63}=18\sqrt[]{63}[/tex]