Area of a circle as follows. Draw a perpendicular from the center to the midpoint of a side of the polygon; its length, "h", is less than the circle radius. Also, let each side of the polygon have length "s"; then the sum of the sides, "ns", is less than the circle circumference. The polygon area consists of "n" equal triangles with height "h" and base "s", thus equals ⁄"nhs". But since "h" < "r" and "ns" < "c", the polygon area must be less than the triangle area, ⁄"cr", a contradiction. Therefore, our supposition that "C" might be greater than "T" must be
Area of a circle "area" and "circumference". The analytical definitions are seen to be equivalent, if it is agreed that the circumference of the circle is measured as a rectifiable curve by means of the integral The integral appearing on the right is an abelian integral whose value is a half-period of the sine function, equal to . Thus formula_6 is seen to be true as a theorem. Several of the arguments that follow use only concepts from elementary calculus to reproduce the formula formula_1, but in many cases to regard these as actual proofs, they rely implicitly on the fact that one can