For a rhombus, all the sides are equal and diagonals bisect each other at 90o. Hence, the triangles formed are congruent: (under RHS). Thus: \(\bar{PS}^2\) = 52 + 122 \(\bar{PS} = \sqrt{25 + 144}\) = \(\sqrt{169}\) \(\bar{PS}\) = 13cm perimeter = 4 x length of a side = 4 x 13cm = 52cm