In rhombus efgh, if he = 13 and hp=12,

what is pe?

PE=5. You need to know an important property of a rhombus: the two diagonals are perpendicular to each other. Angle EPH, same as the other four angles around the center P, is equal to 90 degrees. Since we know HE=13 and HP=12, we use Pythagorean's Theorem to solve for PE=\sqrt(13^2-12^2)=5.

Ithink it’s a or c