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`P = 2* "n" * "r" * sin(pi/ "n" )`

Enter a value for all fields

The **Perimeter of a Polygon (Outer Radius)** calculator computes the length of the perimeter of a regular polygon of (**n**) sides that is circumscribed by an outer circle of radius (**r**).

**INSTRUCTIONS:** Choose units and enter the following:

- (
**n**) This is the number of sides in the regular polygon. - (
**r**) This is the radius of the circle containing the polygon where the corners of the polygon are all on the circle.

**Polygon Perimeter (P):** The calculator return to total perimeter of the regular polygon in meters. However this can be automatically converted to compatible units via the pull-down menu.

- Area of a Polygon based on the length and number of sides.
- Area of a Polygon based on the number of sides and the outer radius.
- Area of a Polygon based on the number of sides and the inner radius.
- Perimeter of a Polygon based on the number and length of sides.
- Perimeter of a Polygon based on the number of sides and the outer radius.
- Perimeter of a Polygon based on the number of sides and the inner radius.
- Length of a Polygon Side based on Circumscribed Circle Radius
- Length of a Polygon Side based on Inscribed Circle Radius

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. Because of the symmetry of this construction, all the vertices of the regular polygon lie on the circle and the sides of the regular polygons form **n** chords of the circle. The formula for the perimeter of a polygon based on the number of sides and the outer radius is:

P = 2⋅n⋅r⋅sin(π/n)

where:

- P is the perimeter of the polygon
- n is the number of sides
- r is the outer radius of the polygon

The n-sided area of a regular polygon, as can be seen in Figure 1, is comprised of **n** isosceles triangles. The regular polygon is constructed to have all its vertices on the circle, and thus a radii of the circle intersects a vertex of the polygon and bisects the angle of the polygon. It also can be shown that the radius to the polygon's vertex forms the side of **n** triangles whose third side is the polygon's side. Since these triangles with base **s** are isosceles triangles, the radii intersecting the polygon's side at s/2 splits these triangles into two 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] `"Perimeter"_"(Polygon)" = n * s`

We can find the length of the polygons side **s** by noting first that the triangle with base **s/2** and height **L **is a right triangle. We also note that the angle, `alpha`, is given by:

[2] `alpha = (2 * pi) /n`, because all the **n** equal angles `alpha` must sum to `2pi` radians

We also see that the circle's radius, **r**, is the hypotenuse of a right triangle and thus relates** r **and **s/2** as:

[3] `s/2 = r * sin (alpha/2)`

Substituting equation [2] into equation [3] we get:

[4] `s/2 = r *sin( ((2*pi) / n)/2 )`

And rearranging we get the polygon side, **s**, in terms of the circle's radius, **r**, and the number of sides of the regular polygon, **n**:

[5] `s = 2 * r * sin( pi / n)`

We finally compute the total** **length of the polygon's perimeter by substituting equation [5] into equation [1]:

[6] `"Perimeter"_"(Polygon)" = n * 2 * r * sin( pi / n) `