Square Pyramid Surface Area

Unfold a square pyramid and it lies flat as a square with four triangles hinged off its edges. Each triangle has base $s$ and height equal to the slant height $l$, so the four together contribute $4 \ imes \ frac{1}{2}sl = 2sl$, and the total is $A = s^2 + 2sl$. Drop the $s^2$ term and you have the lateral area alone, which is what you want when the base sits on the ground and needs no covering.

The classic mistake is feeding in the vertical height instead of the slant height. They are different quantities, and the vertical height is always the shorter of the two. The Great Pyramid of Giza makes the gap vivid: it stood about 147 m tall but its faces slope roughly 186 m from base edge to apex, a 26 percent difference in the number you would multiply by. If you only know the vertical height, convert it first with the slant-height formula.

Going the other way, from a known total area to the base side, means solving $s^2 + 2ls - A = 0$. Only one root is positive, $s = \sqrt{l^2 + A} - l$, and that is the one a real pyramid has.