The diagonals of a rhombus measure 18 feet and 12 feet. What is the perimeter of the rhombus? Express your answer in simplest radical form.
The diagonals of a rhombus intersect at a 90-degree angle, partitioning the rhombus into four congruent right triangles. The legs of  one of the triangles are 6 feet and 9 feet, so the hypotenuse of the triangle - which is also the side of the rhombus - is $\sqrt{(6^2 + 9^2)} = \sqrt{(36 + 81)} = \sqrt{117}$ feet. Since $117 = 9 \times 13$, we can simplify this as follows: $\sqrt{117} = \sqrt{(9 \times 13)} = \sqrt{9} \times \sqrt{13} = 3\sqrt{13}$ feet. The perimeter of the rhombus is four times this amount or $4 \times 3\sqrt{13} = \boxed{12\sqrt{13}\text{ feet}}$.