help #2

Area of equilateral triangle: \(\frac{\sqrt{3}}{4}s^2\)

Area of regular hexagon: \(\frac{3\sqrt{3}}{2}s^2\)

If the perimeter of the triangle is 18, then one side is 6, therefore the area of it is,

\(\frac{\sqrt{3}}{4}\cdot36\)

\(area = 9\sqrt{3}\)

Because their areas are equal,

\(\frac{3\sqrt{3}}{2}s^2 = 9\sqrt{3}\)

Simplify,

\(3\sqrt{3}\cdot s^2 = 18\sqrt{3}\)

\(s^2 = 6\)

\(s = \sqrt{6}\)

There are 6 sides in a hexagon, so the perimeter is, \(\sqrt{6} \cdot 6 = 6\sqrt{6}\). 6 + 6 = 12, so that's the answer.

EDIT: I messed up my arithmetic