Question

Divide the polygon into triangles and using trig, solve for the area.

1. A regular heptagon with a side of 7 inches. (The answer is 178.06 sq in, I just need to know how to solve this using trig.)

2. A square with a side of 2 feet  (answer is 4 sq. ft)

Answer 1

If you take the center of a regular n-polygon and you connect it with all the vertices, you will get n isosceles triangles with the angle between two equal sides equal to $\frac{360^\circ}{n}$ and the opposite side equal to the side of a polygon. Thus, you can solve this triangle to get that the equal sides of this triangle are $\frac{a}{\sqrt{2-2\cos(\frac{360^\circ}{n})}}$, where a is the polygon side (we appply simply the law of cosines to the side a). Thus the area of one triangle is $\frac{1}{2} \frac{a^2}{2-2\cos(\frac{360^\circ}{n})} \cdot \sin(\frac{360^\circ}{n})$, and so the whole area is $S=n\cdot \frac{1}{2} \frac{a^2}{2-2\cos(\frac{360^\circ}{n})} \cdot \sin(\frac{360^\circ}{n})=\frac{n}{4}\cdot\frac{a^2\sin\frac{360^\circ}{7}}{1-\cos\frac{360^\circ}{7}}$. Applying this formula we get :

1. $S = \frac{7}{4} \cdot \frac{7^2}{1-\cos(\frac{360^\circ}{7})} \cdot \sin(\frac{360^\circ}{7}) = 178.06 \text{ inch}^2$
2. $S=\frac{4}{4}\cdot \frac{2^2 \cdot 1}{1-0}=4 \text{ ft}^2$