This Area of a Polygon calculator computes the area of a regular polygon with (n) equal sides of length (s). The polygon is constructed with a circle inscribed inside the polygon so that the circle touches each of the polygon's sides at exactly one point.
INSTRUCTIONS: Choose units and enter the following:
Area of Polygon (A): The area is returned in square meters. However, this can be automatically converted to other area units via the pull-down menu.
A regular n-sided polygon is a polygon with n equal length sides and is a polygon which has n equal angles at the n vertices of the polygon. The formula for the area of a polygon based on the length and number of sides is:
A = n • s² / (4 • tan(π/n))
where:
The n-sided area of a regular polygon, as can be seen in Figure 1, is comprised of n isosceles triangles. The radius of the circle is constructed to touch each side of the polygon at exactly one point, and thus the polygon's side touches the circle at a tangent point and this makes the side of the polygon perpendicular to a radius. It also can be shown that because of the symmetry of the construction, the radius to the point on the polygon splits each of n sectors of the circle into two equal sectors, and thus the triangles composing the polygon are also split into two equal right triangles.
If we imagine the sides of the polygon as the bases of these isosceles triangles, then we can see the area of each of the n triangles is given by the simple formula:
[1] `A_"(triangle)" = 1/2 * base * height = s/2 * r`
We can find the length of the polygons side s by noting first that the triangle with base s/2 and height r is a right triangle. We also note that the angle, `alpha`, is given by:
This Area of a Nonagon formula, (A =9/4•cot(π/9)•s2), computes the area of a regular nonogon, a polygon with 9 equal sides of length (s). nonogon.png
[2] `alpha = (2 * pi) /n`
We also see that line L is the hypotenuse of the right triangle and thus relates r and s/2 as:
[3] `tan(alpha/2) = (1/2s) / r`
Substituting equation [2] into equation [3] we get:
[4] `tan( ((2*pi) / n)/2 ) = s / (2*r)`
And rearranging we get the radius, r, in terms of the side length, s, and the number of sides of the regular polygon, n:
[5] `r = s / (2*tan( pi / n))`
Finally to get the area of one of the triangles, we substitute equation [5] into equation [1]:
[6] `A_"(triangle)" = s/2 * s / (2*tan( pi / n))`
And simplifying, we get the area of each of the n triangles comprising the regular polygon:
[7] `A_"(triangle)" = s^2 / (4 *tan( pi / n))`
And finally the area of the polygon is the sum of the n triangles with constant base side length s:
[8] `A_"(polygon)" = sum_0^n(s^2 / (4 *tan( pi / n))) = n * s^2 / (4 *tan( pi / n))`