Question 1369
 Sol: Let the radius of the circumscribed circle be r,
      By area of the circle = pi r^2 = 196 pi, so r^2 = 196, r = 14
      Since the bisector of any angle of a regular polygon must pass through 
      the center of the circumscribed circle. Hence,the bisectors of angles 
     A and D intersect at P which is just the center of the circle. 
     So,we have AP = r = 14...Answer