The Ratio of Squares

Join $BD$ . It will obviously pass the centre $O$ of the circle. The smaller square $EFGH$ by geometrical symmetry is positioned in such a way that $OL$ is the perpendicular bisector of side $GH$ and side $AB$ .

Let $OK = KB = 1$ unit and $HB = x$ .

So, $HK = LF = 1 - x$ , which means $FH = 2(1 - x) = KL$

Thus, $OL = OK + KL = 1 + 2(1 - x) = 3 - 2x$

But $OF^{2}$ = $OL^{2}$ + $LF^{2}$ , as joining $OF$ makes for a right angled triangle.

Now $OF = OB$ = radius of the circle. As we have assumed $OK = 1$ unit, this means $OB = \sqrt2 = OF$ .

$OF^{2}$ = $OL^{2}$ + $LF^{2}$

$(\sqrt2)^{2}$ = $(3 - 2x)^{2}$ + $(1 - x)^{2}$

Solving for $x$ , we get:

$x = 2, 0.8$

Obviously, $x = 2$ is not possible. So that leaves us with 0.8.

Now, $FH = 2(1 - x) = 2(1 - 0.8) = 0.4 = GH = EG = EF$

Also, $AB = BC = CD = DA = 2$

Thus, $\frac{Ar \square ABCD}{Ar \square EFGH}$ = $\frac{BC^{2}}{FH^{2}}$ = $\frac{2^{2}}{0.4^{2}}$ = $\frac{25}{1}$