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

- Jan 28th 2011, 12:07 PMevankieflA 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 PMArchie Meade
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 PMevankiefl
when you take TanA, is it of half of the isosceles triangle?

- Jan 28th 2011, 01:15 PMArchie Meade
Here is the situation....

"x" is half the length of a side.

"h" is the perpendicular height of each isosceles triangle.

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

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.