BFraction 2

Let the square have side length $s$ , and let the corner $D$ be situated at the origin of the $xy-$ plane. Congruent right triangles $ADP$ and $BCR$ each have area: $A_{1} = \frac{1}{2}(s)(s/2) = \frac{s^2}{4}.$

Let $X$ be the intersection point of segments $BR$ ( $y=2x-s$ ) and $DS$ ( $y=\frac{1}{2}x$ ) , or $x = \frac{2}{3}s, y = \frac{1}{3}s \Rightarrow X(\frac{2}{3}s, \frac{1}{3}s$ ). Triangle $DRX$ can be computed according to:

$A_{2} = \frac{1}{2}|DR||RX|\sin \angle DRX = \frac{1}{2}(s/2)\cdot \sqrt{(2/3 - 1/2)^2 + (1/3 - 0)^2}s \cdot \sin(\pi - \arcsin(2/\sqrt{5})) = \frac{1}{2} \cdot \frac{\sqrt{5}}{6} \cdot \frac{s^2}{2} \cdot \frac{2}{\sqrt{5}} = \frac{s^2}{12}.$

By exploiting symmetry, the blue area calculates to:

$A_{blue} = 2A_{1}+2A_{2} = 2(\frac{s^2}{4} + \frac{s^2}{12}) = \boxed{\frac{2}{3} s^2}.$