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If you look at the pyramid, the pyramid is really just a hexagon with 6 triangles stacked up to each other at the edges, with the hexagon at the bottom.
now, the perpendicular distance from the center of the hexagon to a side, namely the apothem, is 6√3, and each side is 12 units long, since there are 6 of them that'd be 72 for all, namely the perimeter.
each of the triangular faces have a base of 12 and an altitude or height of 11, recall that area of a triangle is (1/2)bh, and area of a regular polygon is (1/2)(apothem)(perimeter).
so if we just get the area of the hexagon at the bottom, and the triangles, sum them up, that's the surface area of the pyramid.
![\bf \stackrel{\textit{area of the hexagon}}{\left[\cfrac{1}{2}(6\sqrt{3})(72) \right]}~~~~+~~~~\stackrel{\textit{area of the 6 triangles}}{6\left[\cfrac{1}{2}(12)(11) \right]}](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7B%5Ctextit%7Barea%20of%20the%20hexagon%7D%7D%7B%5Cleft%5B%5Ccfrac%7B1%7D%7B2%7D%286%5Csqrt%7B3%7D%29%2872%29%20%20%5Cright%5D%7D~~~~%2B~~~~%5Cstackrel%7B%5Ctextit%7Barea%20of%20the%206%20triangles%7D%7D%7B6%5Cleft%5B%5Ccfrac%7B1%7D%7B2%7D%2812%29%2811%29%20%20%5Cright%5D%7D)
now, the perpendicular distance from the center of the hexagon to a side, namely the apothem, is 6√3, and each side is 12 units long, since there are 6 of them that'd be 72 for all, namely the perimeter.
each of the triangular faces have a base of 12 and an altitude or height of 11, recall that area of a triangle is (1/2)bh, and area of a regular polygon is (1/2)(apothem)(perimeter).
so if we just get the area of the hexagon at the bottom, and the triangles, sum them up, that's the surface area of the pyramid.