# How to find the area of ​​a polygon?

How to find the area of ​​a polygon?

1. There is no universal formula, because an arbitrary polygon
not a "rigid" figure, its shape and area can be changed by changing
angles. Only by triangulation.
2. In general, at work, I was confused with my own deduction of the square ... and wondered how much for more number of sides and the same perimeter - the area increases.

I decided to derive the formula of the polygon S = Pr ^ 2 / (4N * tg (180 / n)), where Pr is the perimeter, and N is the number of sides. So I found out that N * tg (180 / N) -gt; PI, ie, tends to the number of PI, for infinitely high N, thereby increasing the area to the area of ​​the ideal circle.
Transforming S = Pr ^ 2 / 4PI.

3. divide it into triangles and add up the sum of the areas of all triangles
4. In the first case, it will be a triangle, and you can use one of the formulas: S = 1 / 2 * a * n, where a is the side, n is the height to it; S = 1 / 2 * a * b * sin (A), where a, to the sides of the triangle, And the angle between the known sides; S = (p * (p - a) * (p - c) * (p - c)), where c is the side of the triangle, to the already marked two, p is a semi-perimeter, that is, the sum of all three sides, divided into two.
5. Only along the lengths of the sides it is impossible. The coordinates of the vertices can be:

where

6. Any or any regular polygon? ?
The area of ​​any regular polygon can be calculated from formula
S = n * a ^ 2 / (4tg 180 / n) where a is the length of the polygon side, n is the number of sides
7. Can you draw a triangle or quadrilateral in which ALL corners are stupid?
8. no such. For polygons with number of sides there are formulas, but they include angles or diagonals. for example, the brahmagupta formula for the 4-square or the formulas 5 and 6-gons