MHB Law of Sines in Polygon Construction - Goals & Benefits

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The discussion focuses on the application of the law of sines in polygon construction to approximate the area of a circle using regular polygons. The author aims to demonstrate Archimedes' method by dividing the polygon into isosceles triangles, calculating the area based on the triangle's base and height. As the number of sides increases, the polygon more closely resembles a circle, enhancing the approximation. The law of sines is utilized to determine the length of the base of these triangles, with the vertex angle derived from dividing the full circle into n segments. This method effectively illustrates the relationship between polygonal shapes and circular area calculation.
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As the author says in the first paragraph, his goal is to demonstrate Archimedes' method of calculating the area of a circle. He does that by noting that a circle can be approximated by a regular polygon with "n" sides, all vertices on that circle- and the more sides you take, the better the approximation. He calculates the area of the polygon by drawing lines from the center of the circle to each vertex, dividing the polygon into n isosceles triangles. The area of a triangle is "1/2 base times height". In this case, the base of each triangle is the side of the polygon while height is the distance from the center of the circle to the center of the base. For n large, so that each triangle is very "skinny", that is approximately the length of the two equal sides of the triangle, the radius of the circle.

He specifically uses the sine law to calculate the length of the base. Dividing the $2\pi$ radians of a complete circle by n, the vertex angle of each triangle, at the center of the circle, is $\frac{2\pi}{n}$. Since the three angles of any triangle sum to $\pi$ radians, and the two base angles are equal, they are each $\frac{\pi- \frac{2\pi}{n}}{2}= \frac{(n-2)\pi}{2n}$.
 
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Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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