Figure : The Elemental Triangle We know that the area of this triangle is: We will now manipulate this relationship so that it can be expressed entirely in terms of r. To do this, we will use some trigonometry. As reasoned earlier, an polygon is made up of n identical, elemental triangles. The angle subtended at the triangle’s apex is there are n since triangles making up the polygon, the value of must be 360/n. For our  purposes, we are interested in the angle between r and h; since this is half of we are dealing with an isosceles triangle), its value must be remember, 360/2n, or 180/n once reduced. We can derive the following expressions from the characteristics of the triangle: ; and ; Recalling that the area of the triangle is , we can then make the substitutions for s and h and get: There is a trigonometric identity that can be used to further reduce this expression, as follows: For an angle , Here, if we set , then Applying this identity to our expression for, we get:
Figure : The Elemental Triangle We know that the area of this triangle is: We will now manipulate this relationship so that it can be expressed entirely in terms of r. To do this, we will use some trigonometry. As reasoned earlier, an polygon is made up of n identical, elemental triangles. The angle subtended at the triangle’s apex is since there are n triangles making up the polygon, the value of  must be 360/n. For our purposes, we are interested in the angle between r and h; since this is half of remember, we are dealing with an isosceles triangle), its value must be 360/2n, or 180/n once reduced. We can derive the following expressions from the characteristics of the triangle: ; and ; Recalling that the area of the triangle is , we can then make the substitutions for s and h and get: There is a trigonometric identity that can be used to further reduce this expression, as follows: For an angle , Here, if we set , then Applying this identity to our expression for, we get:
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