The length of the side of the circumscribed square is $2r$, where $r$ is the radius of the circle. So, its perimeter is $8r$.

Now, I calculate the length of the inscribed square, using the Pythagoras Theorem, in a right-angled isosceles triangle, with the right angle in the centre of the circle, the two congruent sides - radii of the circle, and the a vertex of the hypotenuses - the side of the inscribed square:

$2r^2 = l^2$

$l = r\sqrt{2}$

So, the perimeter of the inscribed square is $4r\sqrt{2}$.

Knowing that the circumference of a circle is $2{\pi}r$, I can find a minimum and a maximum limit for $p$ from the condition that the circumference of the circle is lower than the perimeter of the circumscribed square and higher than the circumference of the inscribed square:

$4r\sqrt{2} < 2{\pi}r < 8r$

$4\sqrt{2} < 2\pi < 8$

$2\sqrt{2} < \pi < 4$

and approximating $\sqrt{2}$:

$2.8284 < \pi < 4$

Now, I use the same method for the hexagon. First I calculate the side of the circumscribed hexagon in a right-angled triangle with one side (the hypotenuses) the line connecting the centre of the circle with the vertex of the circumscribed hexagon, a side - half the side of the circumscribed hexagon, and the other - the radius of the circle. The side of the hexagon is perpendicular on the radius in the tangency point. The angle between the radius and the hypotenuses is $30^o$ (Here I use the Pythagoras Theorem and also the theorem that says than in a right-angled triangle with one angle of $30^o$, the side opposite to the angle of $30^o$ is 2 times smaller than the hypotenuses): $${ \left({L \over2}\right)^2 + r^2 = L^2 }$$ $${ {{L^2} \over 4} + r^2 = L^2 }$$ or $${ {3L^2 \over{4}} = r^2 }$$ so, $${ L = {2r \over{\sqrt{3}}}= {{2 \sqrt{3}r \over{3}}} }$$ The side of the inscribed hexagon is $r$. So, I found the following inequalities:

$6r < 2\pi r < 6\times (2r\sqrt{3})/3$

$6r < 2\pi r < 4\sqrt{3}r$

$6 < 2\pi < 4\sqrt{3}$

$3 < \pi < 2\sqrt{3}$

and approximating $\sqrt{3}$:

$3 < \pi < 3.4641$