1998 AHSME Problem 15

The hexagon can be seen as 6 smaller equilateral triangles. The ratio of the large equilateral triangle to one of these smaller triangles is $6$ . Using the fact that $($ ratio of areas $)$ = $($ ratio of sides $)^2$ , the answer is $\sqrt{6}$ .

The hexagon consists of 6 equilateral triangles of side b. If the area of one of those is B, the hexagon's area is 6 B.

The triangle consists of 4 equilateral triangles of side a/2. Call the area of each one A, so the area of the triangle is 4A. For the triangle it's ( a * a sqrt(3) / 2 ) (from Pythagorean Theorem on 30-60-90 triangle), and for the hexagon it's ( 6 * b * b sqrt(3) / 2 ) Setting those equal we have: a^2 sqrt(3) / 2 = 6 b^2 sqrt(3) / 2 a^2 = 6 b^2 a = sqrt(6) b ( take square roots ) a/b = sqrt(6).