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
Measurement of a Circle two legs being "c" and "r". This proposition is proved by the method of exhaustion. Proposition two states: The area of a circle is to the square on its diameter as 11 to 14. This proposition could not have been placed by Archimedes, for it relies on the outcome of the third proposition. Proposition three states: The ratio of the circumference of any circle to its diameter is greater than formula_1 but less than formula_2. This approximates what we now call the mathematical constant π. He found these bounds on the value of π by inscribing and circumscribing a circle