Use of polygons

The area of the correct polygon is half the perimeter multiplied by the apothema (height). When the number of sides increases, the polygon tends to the circle, and the apothema tends to the radius. It gives the basis to consider that the area of a circle is equal to product of half length of a circle on radius.

Archimedes’ proof

Following Archimedes, we will compare the area of the circle with the area of a rectangular triangle, the base of which is equal to the length of the circle, and the height is equal to the radius. If the area of a circle is not equal to the area of a triangle, it should be less or more. We shall exclude both variants that will leave only one possibility – the areas are equal. For proof we will use correct polygons.

No more.

Circle with inscribed square and octagon. The gap is shown.

Suppose that the area of circle C is larger than the area of triangle T = 1⁄2cr. Let’s assume that E means exceeding the area. Let’s write[en] a square in a circle so that all its four corners lie on a circle. There are four segments between the square and the circle. If their total area G4 is larger than E, we divide each arc in half, which turns the inscribed square into an octagon and forms eight segments with a smaller total gap, G8. We continue dividing until the total gap Gn becomes less than E. Now the area of the entered polygon Pn = C – Gn should be more than the triangle area.

But it leads to a contradiction. To prove it, let’s draw the height from the center of the circle to the middle of the polygon, its length h is less than the radius of the circle. Let each side of the polygon have length s, the sum of all sides will be ns, and this value is less than the length of the circle. The area of the polygon consists of n equal triangles of height h with the base s, which gives 1⁄2nhs. But h < r and ns < c, so that the area of the polygon must be smaller than the area of the triangle 1⁄2cr, got the contradiction. Read more nere https://argoprep.com/blog/area-of-a-circle/

No less.

A circle with the described square and octagon. The gap is shown.

Suppose the area of the circle is smaller than the area of the triangle. Suppose D means the difference in area. Let’s describe the square around the circle, so that the middle of the sides lie on it. If the total gap between the square and the circle G4 is larger than D, we cut the corners with tangents, turning the square into an octagon and continue such cutoffs until the gap area becomes smaller than D. The area of the polygon Pn should be smaller than T.

It also leads to a contradiction. Each perpendicular drawn from the centre of the circle to the middle of the side is a radius, i.e. it has a length of r. And since the sum of the sides is greater than the length of the circle, a polygon of n identical triangles will give an area larger than T. Again, there is a contradiction.

Thus, the area of the circle is exactly equal to the area of the triangle.

Following Sato Moshoun [4] and Leonardo da Vinci [5], we can use the inscribed correct polygons in another way. Let’s say we’ve entered a hexagon. Cut the hexagon into six triangles, making sections through the center. Two opposite triangles contain total diameters. Now let’s move the triangles so that the radial sides are adjacent. Now pair of triangles forms a parallelogram in which the sides of a hexagon form two opposite sides with length s. Two radial sides become lateral sides, and height of a parallelogram is equal h (as in Archimedes’ proof). Actually, we can collect all triangles in one big parallelogram, having in a row the received parallelograms (from two triangles). The same will be true if we increase the number of sides. For a polygon with 2n sides of a parallelogram, it will have the base ns and the height h. With growth of the number of sides the length of the base of a parallelogram increases, aspiring to half of a circle, and the height aspires to radius. In the limit of a parallelogram becomes a rectangle with width πr and height r.

Following Sato Moshoun [4] and Leonardo da Vinci [5], we can use the inscribed correct polygons in another way. Let’s say we’ve entered a hexagon. Cut the hexagon into six triangles, making sections through the center. Two opposite triangles contain total diameters. Now let’s move the triangles so that the radial sides are adjacent. Now pair of triangles forms a parallelogram in which the sides of a hexagon form two opposite sides with length s. Two radial sides become lateral sides, and height of a parallelogram is equal h (as in Archimedes’ proof). Actually, we can collect all triangles in one big parallelogram, having in a row the received parallelograms (from two triangles). The same will be true if we increase the number of sides. For a polygon with 2n sides of a parallelogram, it will have the base ns and the height h. With growth of the number of sides the length of the base of a parallelogram increases, aspiring to half of a circle, and the height aspires to radius. In the limit of a parallelogram becomes a rectangle with width πr and height r.