# A generalized formula for finding the area of a regular polygon

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• Jan 28th 2011, 12:07 PM
evankiefl
A generalized formula for finding the area of a regular polygon
Is this possible? Assume a polygon that has all equal angles and lengths
• Jan 28th 2011, 12:20 PM
Archie Meade
Quote:

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.
• Jan 28th 2011, 01:01 PM
evankiefl
when you take TanA, is it of half of the isosceles triangle?
• Jan 28th 2011, 01:15 PM
Archie Meade
Quote:

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.

$TanA=\frac{h}{x}\Rightarrow\ h=xTanA\Rightarrow\ xh=x^2TanA$

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.