Find the area of the red region

the area we desire is just red minus blue in the two diagrams above, so we are going to try finding both those areas. We are going to use principle of inclusion and exclusion to find these areas.

the blue area is just the sum of the areas of the two circular sections minus the area of the square. The circular sections being denoted by the green line. this means $\large [\text{blue}] =\left[\overset{\frown}{GED}\right]+\left[\overset{\frown}{HED}\right]-[\text{GDHE}]\\= \dfrac{1}{2}\times \dfrac{\pi}{2}\times 5^2+ \dfrac{1}{2}\times \dfrac{\pi}{2}\times 5^2- 5^2= 25\left(\dfrac{\pi}{2}-1\right)$

We do the same thing for the red area. by using congruent triangles we can figure out that the quadrilateral, in this case, has 2 right angles. the area, like before, is the sum of the two circular sections minus the quadrilateral. that is: $\large [\text{red}] =\left[\overset{\frown}{AFD}\right]+\left[\overset{\frown}{HFD}\right]-\left[{AFHD}\right] \\=\dfrac{1}{2}\times\theta \times 10^2+ \dfrac{1}{2}\times (\pi-\theta) \times 5^2- 5\times 10= 25\left(\dfrac{3}{2}\theta +\dfrac{\pi}{2}-2\right) \\ \to [\text{red}] -[\text{blue}]=25\left(\dfrac{3}{2}\theta -1\right)$ we can figure out using the purple right triangles that $\tan\left(\dfrac{\theta}{2}\right) = \dfrac{5}{10}=\dfrac{1}{2} \to \theta = 2 \arctan\left(\dfrac{1}{2}\right)$ . plugging this in our answer is $25( 3 \arctan\left(\dfrac{1}{2}\right)-1) \approx \boxed{ 9.77}$