Thread: A generalized formula for finding the area of a regular polygon

Is this possible? Assume a polygon that has all equal angles and lengths

Originally Posted by evankiefl

Is this possible? Assume a polygon that has all equal angles and lengths

Sure!

A regular polygon has "n" sides.
Therefore it may be subdivided into "n" isosceles triangles, all meeting at the centre.

The angle at the central apex of each triangle is 360/n degrees.

Subtract that from 180 degrees and divide by 2 to get the other angles "A" of each isosceles triangle.
Take half the length of a side of the polygon.
Use TanA multiplied by half a side length squared to get a triangle area.

"n" of these is the polygon area.

when you take TanA, is it of half of the isosceles triangle?

Originally Posted by evankiefl

when you take TanA, is it of half of the isosceles triangle?

Here is the situation....

"x" is half the length of a side.
"h" is the perpendicular height of each isosceles triangle.



That is the area of each isosceles triangle
(twice the area of the right-angled triangle as 2 of these form the isosceles triangle).

Multiply by "n" for the polygon area.