let the heagon be ABCDEF and O be the centre of te circle.
see. in a regular hexagon all the sides are equal. so the angle formed at centre by joining the vertexs of the hexagon with the centre is equal to
theta = 360/6 = 60o.
thus. if we consider any 1 triangle. let it be
ABO. then OA = OB =r.
also, angle, AOB = 60o
thus, by angle sum property we get
angle OAB = OBA = AOB = 60.
this implies that AOB is an equilateral triangle. so,
AB = r cm.
thus, side of hexagon = r
perimeter = 6r.