A geometry problem by Aziz Alasha

We note that:

$\begin{aligned} \frac x2 & = \text{Area of sector }AOC - \text{Area of sector }ADE - \text{Area of }\triangle ODE \\ & = \frac 18 \pi (8^2) - \frac 14 \pi (4^2) - \frac 12 (4^2) \\ & = 4\pi - 8 \end{aligned}$

$\begin{aligned} \frac y2 & = \text{Area of sector }ODE - \text{Area of }\triangle ODE \\ & = \frac 14 \pi (4^2) - - \frac 12 (4^2) \\ & = 4\pi - 8 \end{aligned}$

$x+y = 4\left(4\pi - 8\right) = 16\pi - 32 = 16 \times \frac {22}7 - 32 \approx \boxed{18.29}$

Since the area of both halves of two-sided arc $FC$ is the area of the combination of two inner arcs $AF$ and $FB$ , the sum of the shaded region is equivalent to determining the area of the outer sliced arc in the quarter circle with radius $8$ .

Thus, the answer is $\begin{array}{rl} \text{Area}_{\text{Shaded}} &= \text{Area}_{\text{Quarter circ }ABC} - \text{Area}_{\Delta ABC}\\ &= \dfrac{\pi}{4}\cdot 8^2 - \dfrac{1}{2} \cdot 8^2\\ &= 16\pi - 32 \approx \boxed{18.29} \end{array}$