Area of a regular polygon angle irregularity?

Question

(This is me trying to prove, so don't mention 1/2(apothem*perimeter))
So I tried to figure out the area of a regular polygon by splitting it up in to triangles and got {ns&sup2;*tan[90(n-2)/n)}/4 [n=#sides, s=side length, tan function in degrees] which is similar to what I found online but I found tan(pi/n) instead. 
And pi is just 180 degrees.And they aren't equal because if n=5, you get 36 and 54 degrees which are different.

Algebra proves they aren't equal:
(90n-180)/n is not equal to 180/n. Which means I probably did a error in (90n-180)/n.

I figured it out from the fact that if there are n triangles, the angles closest to the center are equally divided, and the triangles are isosceles. So one angle is 360/n, and the other two angles are (180 - 360/n)/2 each, which is (90 - 180)/n = (90n-180)/n.

Where did I go wrong?

Answer

Get some help to make a coherent presentation.

The area of a triangle can be expressed using the lengths of two sides and the sine of the included angle. Area&Delta; = &frac12; ab sin C
You may see this referred to as the SAS formula for the area of a triangle.
Source: mathbitsnotebook.com/Geometry/TrigApps/TAarea.html

In your case, a = b = radius and C = 360/n so it reduces to Area&Delta; = &frac12; r^2 sin(360/n)

Area of a regular polygon angle irregularity?

Responses (1)