$
\eqalign{
  & \Delta ABS:O = \frac{1}
{2} \cdot z \cdot \frac{2}
{3}z = \frac{1}
{3}z^2   \cr
  & \Delta BPS:O = \frac{1}
{2} \cdot \frac{1}
{2}z \cdot \frac{1}
{3}z = \frac{1}
{{12}}z^2   \cr
  & ABPS = \frac{{\frac{1}
{3}z^2  + \frac{1}
{{12}}z^2 }}
{{z^2 }} = \frac{5}
{{12}} \cr}
$

$\eqalign{\frac{5}{12}}$-deel van het vierkant is gekleurd.