Random geometry problem

A friend of mine sent me this geometry problem the other day, where the task is to find the shaded area given that l=10\text{ cm}:

sketch

As the problem is symmetrical with respect to the square diagonal, we can safely focus on just one of the two shaded regions, say the top left one, which I’m going to call S. The way I approached the solution is by noticing that the shaded region is the difference between two circular segments cut off by the same chord AB: the first segment belonging to the circle centered at O with the radius l/2 and the second belonging to the circle centered at C with the radius of l. The area A of a circular sector of radius r and central angle \theta is simply:

A = \frac{r^2}{2}(\theta - \sin\theta).

When we apply this formula to our problem, we get that

S = \frac{l^2}{8}(\alpha - \sin\alpha)-\frac{l^2}{2}(\beta-\sin\beta).

All what’s left is to find what those angles \alpha and \beta are. The central angle \theta of a chord c can be found using:

\sin\theta/2=\frac{c}{2r}.

Again, in our case we have that

\sin\alpha/2=\frac{AB}{l}, \qquad \sin\beta/2=\frac{AB}{2l}, \qquad \sin\alpha/2 = 2\sin\beta/2.

We now have a link between \alpha and \beta. Finally, we find \beta from the AOC triangle and the cosine theorem:

AO^2=AC^2+OC^2-2AC\cdot OC\cos\beta/2,

\frac{l^2}{4}=l^2+\frac{l^2}{2}-2l\frac{\sqrt2}{2}l\cos\beta/2,

\cos\beta/2=\frac{5}{4\sqrt{2}}.

The rest is trivial trigonometry:

\sin\beta/2=\frac{\sqrt{7}}{4\sqrt{2}}, \qquad \sin\beta=\frac{5\sqrt{7}}{16},

\sin\alpha/2=\frac{\sqrt{7}}{2\sqrt{2}}, \qquad \cos\alpha/2=\frac{1}{2\sqrt{2}}, \qquad \sin\alpha=\frac{\sqrt{7}}{4}.

Now we can finally obtain our answer (remember that \alpha is obtuse):

S = \frac{l^2}{8}(\pi-\arcsin\frac{\sqrt{7}}{4} - \frac{\sqrt{7}}{4})-\frac{l^2}{2}(\arcsin\frac{5\sqrt{7}}{16}-\frac{5\sqrt{7}}{16})

= \frac{l^2}{8}(\pi-\arcsin\frac{\sqrt{7}}{4}) - \frac{l^2}{2}\arcsin\frac{5\sqrt{7}}{16} + \frac{\sqrt{7}}{8}l^2.

Both shaded regions have the area of

\boxed{2S = \frac{l^2}{4}(\pi-\arcsin\frac{\sqrt{7}}{4}+\sqrt{7}) - l^2\arcsin\frac{5\sqrt{7}}{16}.}

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