Find the area of the blue zone

$\text{area of yellow zone}=4\left(\dfrac{135}{360}\right)(\pi)(3^2)=13.5\pi$

By cosine law, we have

$x^2=2x^2-2x^2\cos 45$

$36=x^2\left[2-2\left(\dfrac{\sqrt{2}}{2}\right)\right]$

$x^2=\dfrac{36}{2-\sqrt{2}}$

$\text{area of octagon=area of 8 isosceles triangles}=8\left(\dfrac{1}{2}\right)(x^2)(\sin 45)=4\left(\dfrac{36}{2-\sqrt{2}}\right)\left(\dfrac{\sqrt{2}}{2}\right)=\dfrac{72\sqrt{2}}{2-\sqrt{2}}$

$\text{area of blue zone inside=area of octagon - area of yellow zone}=\dfrac{72\sqrt{2}}{2-\sqrt{2}}-13.5\pi$

$\text{area of blue zone outside}=4\left(\dfrac{225}{360}\right)(\pi)(3^2)=22.5\pi$

$\text{area of blue zone=area of blue zone inside + area of blue zone outside}=\dfrac{72\sqrt{2}}{2-2\sqrt{2}}+9\pi$

Formulas used:

$A=\dfrac{\theta}{360}\pi r^2$ $\implies$ $\text{area of a circular sector}$

$A=\dfrac{1}{2}ab \sin C$ $\implies$ $\text{area of a triangle}$

The blue area = the octagon area – 4(3π/8)3^2 + 4(5π/8)3^2 . The area of octagon = 8(r^2)sin(45). The radius r can be found from sin rule: r/sin(135/2) = 6/sin 45, r^2 = 36(1 + 1/2√2). The blue area = the octagon area – 4(3π/8)32 + 4(5π/8)32 = (72√2 + 72) + 9π