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
Instant centre of rotation and P-B. The direction of speed of point A is indicated by vector V. Its instant center of rotation must be perpendicular to this vector (as V is tangentially located on the circumference of a circle). The only line that fills the requirement is a line colinear with link P-A. Somewhere on this line there is a point P, the instant center of rotation for the body BAC. What applies to point A also applies to point B, therefore this instant center of rotation P is located on a line perpendicular to vector V, a line colinear with link P-B.