area of shaded region

Consider diagram $A$ above. First we get the area of the red region plus the area of the green region. Then we subtract the area of the red region. The combined area is

$A_{red~+~green}=\dfrac{10(20)-2(3.14) (5^2)}{2}=\dfrac{43}{2}=21.5$

Consider diagram $B$ above. The area of the red region is equal to the area of the big right triangle minus the area of the small isosceles triangle and minus the area of the sector.

$\tan~\theta=\dfrac{5}{10}$ $\implies$ $\theta \approx~26.57$

It follows that $2\theta \approx~53.14$ and $\beta \approx~126.86$ . So

$A_{red}=\dfrac{1}{2}(5)(10)-\dfrac{1}{2}(5^2)(\sin~126.76)-\dfrac{53.14}{360}(3.14)(5^2) \approx~3.41$

Finally.

$A_{green}=21.5-3.41 \approx~$ $\boxed{18}$