What is the area of a regular hexagon with sides 1 in. long?

Answer 1

1.5a
a*a=3

Answer 2

A regular hexagon, with sides 1 inch long, can be split up into 6 equilteral triangles, with sides 1 inch. ( Draw the picture! The triangles all point in towards the center.)

There are lots of ways to figure the area of the triangles. One is to drop the height line perpendicular to the base. The height then satisfies h^2 + (1/2)^2 = 1^2 by the pythogorean theorem.

So h^2 = 3/4, so h is sqrt(3)/2, and the triangle has area 1/2 * h * b, or 1/2 * sqrt(3)/2 * 1 = sqrt(3)/4.

Remember that there were 6 of these equilateral triangles in the hexagon, so the hexagon has area 6 * sqrt(3) /4 or 3*sqrt(3)/2.

Answer 3

area = (½)(apothem)(perimeter)

Apothem = √3/2*side
Perimeter = 6*side

Area = 3√3/2 * side² = 2.5981 in²


OR


To do it with the formula:
Divide the figure into 6 equilateral triangles each with a side of 1 inch.

Area of equilateral triangle = √3/4*side²
Since there are 6 such triangles = 6*√3/4*side²
= 3√3/2 * side² = 2.5981 in²

Answer 4

The formula can be found here:
http://en.wikipedia.org/wiki/Hexagon

I would type it out, but in this simple text format, I'd lose a bit and they show it a lot better.