jmc

Let $F$ be the center of the square base and $M$ be the midpoint of an edge of the square, as shown. Since the pyramid is a right pyramid, triangle $PFM$ is a right triangle. We are given $PF=12$, and we have $FM=(1/2)(AB)=5$, so the Pythagorean Theorem gives us $PM=\sqrt{P{F}^2+F{M}^2}=13$. Therefore, since the four lateral faces are congruent triangles, the total surface area of the pyramid is $$[ABCD]+4[PAB]=10^2+4(13)(10)/2=\boxed{360}\text{ square centimeters}.$$