##\text{Area} \left(E\right) = \text{Area} \left( \text{Square}\right) - \text{Area} \left( \text{semi-circle}\right) - \text{Area} \left(\text{quarter-circle} \right) + \text{Area} \left( B \right) + \text{Area} \left( C \right)##
Recall the area of a segment of a circle is ## \frac{1}{2} r^2 \left( \theta - \sin \theta \right)## so the above becomes
##\text{Area} \left(E\right) = 100 - \frac{25 \pi}{2} - 25 \pi + 50 \left(\theta_1 - \sin \theta_1 \right) + \frac{25}{2} \left( \theta_2 - \sin \theta_2 \right) ##
So we have to solve for ##\theta_1## and ##\theta_2## we have the following system of linear equations (which doesn't have a singular solution which is why I'll use the law of Sines and half angle identities later). First equation is the sum of the angles of a quadrilateral. The next two equations are the sums of the angles of triangles. The fourth equation is a statement that ##\theta_3## and ##\theta_4## are compliments. The 5th and last equation can be derived from the other 4.
##\theta_1 + \theta_2 + 2 \theta_3 + 2 \theta_4 = 2 \pi##
##\theta_1 + 2 \theta_3 = \pi##
##\theta_2 + 2 \theta_4 = \pi##
##\theta_3 + \theta_4 = \frac{\pi}{2}##
##\theta_1 + \theta_2 = \pi ##
Use Law of Sines
##\frac{ \sin \theta_1}{S} = \frac{\sin \theta_3}{10} \Rightarrow S = \frac{10 \sin \theta_1}{\sin \theta_3}##
##\frac{ \sin \theta_2}{S} = \frac{\sin \theta_4}{5} \Rightarrow S = \frac{5 \sin \theta_2}{\sin \theta_4}##
Setting ##S = S##
## \frac{10 \sin \theta_1}{\sin \theta_3} = \frac{5 \sin \theta_2}{\sin \theta_4}##
## \frac{2 \sin \theta_1}{\sin \theta_3} = \frac{\sin \theta_2}{\sin\theta_4}##
Recall from the linear systems of equations that ##\theta_3 = \frac{\pi - \theta_1}{2}## and ##\theta_4 = \frac{\pi - \theta_2}{2}## so....
## \frac{2 \sin \theta_1}{\sin \left(\frac{\pi - \theta_1}{2} \right) } = \frac{\sin \theta_2}{\sin \left( \frac{\pi - \theta_2}{2}\right)}##
Now use half angle identity for sine in both denominators
## \frac{2 \sin \theta_1}{\sqrt{\frac{1 - \cos \left(\pi - \theta_1 \right)}{2}}} = \frac{\sin \theta_2}{\sqrt{\frac{1 - \cos \left(\pi - \theta_2 \right)}{2}}}##
Of course ##-\cos \left(\pi - \theta_1 \right) = \cos \left(\theta_1 \right)## and ##-\cos \left(\pi - \theta_2 \right) = \cos \left(\theta_2 \right)##
##\frac{2 \sin \theta_1}{\sqrt{\frac{1 + \cos \theta_1}{2}}} = \frac{ \sin \theta_2}{\sqrt{\frac{1 + \cos \theta_2}{2}}}##
Recall ## \theta_2 = \pi - \theta_1##
##\frac{2 \sin \theta_1}{\sqrt{\frac{1 + \cos \theta_1}{2}}} = \frac{ \sin \left(\pi - \theta_1 \right)}{\sqrt{\frac{1 + \cos \left(\pi - \theta_1 \right)}{2}}}##
Of course ##\cos \left( \pi - \theta_1 \right) = - \cos \theta_1##
##\frac{2 \sin \theta_1}{\sqrt{\frac{1 + \cos \theta_1}{2}}} = \frac{ \sin \theta_1}{\sqrt{\frac{1 - \cos \theta_1}{2}}}##
Cancelling out like factors
##\frac{2}{\sqrt{1 + \cos \theta_1}} = \frac{1}{\sqrt{1- \cos \theta_1}}##
Squaring both sides and moving the cosines to one side and constants to the other we get ##\cos \theta_1 = \frac{3}{5} \Rightarrow \theta_1 = \arccos \left(\frac{3}{5} \right) \Rightarrow \theta_1 = \arcsin \left(\frac{4}{5} \right)##
##\theta_1 = \arcsin \left(\frac{4}{5} \right)##
##\theta_2 = \pi - \arcsin \left(\frac{4}{5} \right)##
Plugging the two angles above into the original expression for ##\text{Area} \left( E \right)## we get
##\text{Area} \left(E \right) = 100 - \frac{25 \pi}{2} - 25 \pi + 50 \left(\theta_1 - \sin \theta_1 \right) + \frac{25}{2} \left( \theta_2 - \sin \theta_2 \right)##
##\text{Area} \left(E \right) = 100 - \frac{25 \pi}{2} - 25 \pi + 50 \left(\arcsin \left( \frac{4}{5} \right) - \sin \arcsin \left( \frac{4}{5} \right) \right) + \frac{25}{2} \left( \pi - \arcsin \left( \frac{4}{5} \right) - \sin \left(\pi -\arcsin \left( \frac{4}{5} \right) \right) \right)##
Simplifying all of this
##\text{Area} \left(E\right) = 50 - 25 \pi + \frac{75}{2} \arcsin \left( \frac{4}{5}\right)##
Multiplying this by ##4## we get the answer
@docnet got in #2