Suppose we use hexagons rather than squares:
Let the radius of the circle be 1, so that its area is $\displaystyle \pi$ units squared. The area $\displaystyle A_S$ of the smaller hexagon is:
$\displaystyle A_S=6\left(\frac{1}{2} \sin\left(\frac{2\pi}{6}\right)\right)= \frac{3\sqrt{3}}{2}\approx2.6$
And the area $\displaystyle A_L$ of the larger hexagon is:
$\displaystyle A_L=6\left(\frac{1}{2} \left(\frac{2}{\sqrt{3}}\right)^2\sin\left(\frac{2 \pi}{6}\right)\right)= 2\sqrt{3}\approx3.5$
What do you think would happen if we used $\displaystyle n$-gons having more and more sides? Let's let $\displaystyle A_n$ be the area of an $\displaystyle n$-gon circumscribed by the circle...we have:
$\displaystyle A_n=\frac{n}{2}\sin\left(\frac{2\pi}{n}\right)$
Using a computer, we find:
$\displaystyle n$ | $\displaystyle A_n$ |
10 | 2.938926261462366 |
100 | 3.1395259764656687 |
1000 | 3.1415719827794755 |
10000 | 3.141592446881286 |
1000000 | 3.141592653569122 |
It appears that as $\displaystyle n\to\infty$ we have $\displaystyle A_n$ approaching some fixed finite value.