Appearances Can Be Deceiving

Relevant wiki: Regular Polygons - Problem Solving - Medium

Consider this diagram ... Let the side of the octagon be $a$

$Ar. of \triangle{ACJ} = Ar. of \triangle{BKH} =Ar. of \triangle{FMG} =Ar. of \triangle{DLE} \\ \text{Also } Ar. of (CJLD) = Ar. of (KMGH)$

$\text{Internal Angles of a Regular Octagon are equal to 135 } \\ \text{ since Angle } \angle DCJ = 90 \implies \angle ACJ = 45 \text{ and } \angle CAJ = 45 \\ \text{Which means } CJ=AJ \\ CJ^2+AJ^2=AC^2 \implies 2\cdot AJ^2=a^2 \\ AJ=\dfrac{a}{\sqrt{2}} \\ \text{Similarly } AJ=CJ=BK=KH=MG=MF=LE=DL=\dfrac{a}{\sqrt2} \\ Ar. of \triangle{ACJ} = \dfrac{1}{2} \cdot \dfrac{a}{\sqrt{2}} \cdot \dfrac{a}{\sqrt{2}} = \dfrac{a^2}{4} \\ Ar. of (CJLD) = \dfrac{a}{\sqrt{2}} \cdot a = \dfrac{a^2}{\sqrt2} \\ \text{Total Blue Area : } 4\times \dfrac{a^2}{4}+2\times \dfrac{a^2}{\sqrt2} = \boxed{a^2(1+\sqrt2)} \\ \text{Total Black Area : } AB\cdot AE \implies a\times \left( AJ+JL+LE\right) \\ = a\cdot \left( \dfrac{a}{\sqrt2}+\dfrac{a}{\sqrt2}+a\right) = \boxed{a^2\left( 1+\sqrt2 \right)} \\ \text{Area of Blue = Area of black }$