Maths Without Number.

Let the areas of the smallest square and the second smallest square be $A_1$ and $A_2$ respectively. It is noted that $A_1 = \frac {1}{2} A_2$ . Any the calculations are as follows:

$\quad A_1 = A_2 - 4 \times$ the area of corner triangle

$\quad \quad \quad = \left( \dfrac {q}{2} \right)^2 - 4 \left( \dfrac {1} {2} \right) \left( \dfrac {q}{4} \right)^2 = \left( \dfrac {1}{4} - \dfrac {1}{8} \right) q^2 = \frac {1}{8} q^2$

The radius of the circle, $r = \frac {1}{2} \left( \frac {1}{4} \right) \left( \sqrt {2} \right) q = \frac {\sqrt{2} } {8} q$

The area of the shaded region,

$A = A_1 - \pi r^2 = \frac {1}{8} q^2 - \frac {22}{7} \left( \frac {\sqrt{2} } {8} q \right)^2 = \left( \frac {1}{8} -\frac {11 } {112} \right) q^2 = \left( \frac {14 - 11 } {112} \right) q^2 = \dfrac {3q^2 } {112}$

$\Rightarrow 3L^2 + 4 = 112\quad \Rightarrow L = \sqrt {\dfrac {112-4}{3}} = \sqrt {\dfrac {108}{3}} = \sqrt {36} = \boxed {6}$

(3q^2) / ( 3L^2 + 4) = (q^2)/8 - (pi.q^2) / (32)

pi = 22/7

(3q^2) / ( 3L^2 + 4) = (3.q^2) / (8.14)

3L^2 + 4 = 14*8

3L^2 = 112 - 4

L^2 = 108/3

L = 6

i tink this problem is overrated .... it shud hav been a level 2 problem for 55 points