## Elementary Geometry for College Students (6th Edition)

$162\sqrt{3}$
Half of the length of one of the sides is part of a 30-60-90 triangle with the apothem. Thus: $s/2 = \frac{9}{\sqrt3} \\ s/2 = 3\sqrt{3} \\ s = 6\sqrt3$ We now use the perimeter and the apothem to find area: $P = 6(6\sqrt{3})=36\sqrt{3}$ $A = 1/2aP = 1/2(9)(36\sqrt{3}) = 162\sqrt{3}$