Circles and squares and polygon 
If i replace the square with circle, i can very easily find the area of circle and get some approximation for the area of the square.
Even the square is doable, but i was thinking that can we find the area of the regular polygon which is made to sit in place of the square ?
Since, i don't have the answer to that problem, i don't know if approximating that with circle would work.
Would a regular polygon of arbitrarily large number of sides fit into that space and can we find its area (i know only upto undergrad. maths)?
I am trying to solve it by coordinate geometry but i am not getting as many equations as many variables i have.
$a^2$+b(b-7)=49=$d^2$+c(c-7)=($e^2$+$f^2$)=($g^2$+$h^2$)=(f-b)(g-c)49/(e-a)(h-d)=$(f-b)^2$+$(e-a)^2$+49-$(g-c)^2$-$(h-d)^2$=$(a-c)^2$+$(b-d)^2$+49-$(e-g)^2$-$(f-h)^2$
Thank you. 

A solution for a regular polygon with one edge a chord of the large circle and the vertices of the opposite edge on the smaller circles
Consider coordinates with $O$ as origin and the line of symmetry as the $x$-axis. Let the edge length of the regular polygon be  $2L$ and let the distance between opposite edges be $D$.

$D$ is $2L$ for a square, $2\sqrt 3L$ for a hexagon and $(2+2\sqrt 2)L$ for an octagon.
The large circle has equation $x^2+y^2=49$.
The upper small circle has equation $x^2-\frac{7}{\sqrt 2}x+y^2-\frac{7}{\sqrt 2}y=0$.

The line $y=L$ intersects the upper circle at $(x,L)$ and the large circle at $(x+D,L)$. Eliminating $x$ from the two equations for these points gives us a polynomial equation for $L$ in terms of $D$.
(1) $x^2+2Dx+D^2+L^2=49$.
(2) $x^2-\frac{7}{\sqrt 2}x+L^2-\frac{7}{\sqrt 2}L=0$.
First eliminate $x^2$.
$(7+2\sqrt 2 D)x=49\sqrt 2-\sqrt 2 D^2-7L$
Substituting into Equation 1 then gives us the required  polynomial. We shall now do this for the square.
$(7+4\sqrt 2 L)x=49\sqrt 2-7L-4\sqrt 2 L^2$
$(7+4\sqrt 2 L)(x+D)=4\sqrt 2 L^2+7L+49\sqrt 2$
$(4\sqrt 2 L^2+7L+49\sqrt 2)^2=(49-L^2)(4\sqrt 2 L+7)^2$

$64L^4+112\sqrt 2 L^3-686L^2-2058\sqrt 2 L+2401=0$

The relevant solution is $L=0.7274$ and so the length of a side is approximately $1.455$.
