The actual problem I have is:
Calculate the area of the triangular region ABC.
I tried to relate the isosceles triangle properties with the angle-arc theorem (inscribed angle is half of the arc) but everything ends up bound to the radius. How do I solve this problem?
The distances of the circumcenter $O$ from the sides are $R-1, R-2, R-3$ and the side lengths are given by $$ a = BC = 2\sqrt{R^2-(R-2)^2} = 4\sqrt{R-1},$$ $$ b = AC = 2\sqrt{R^2-(R-3)^2} = 2\sqrt{3}\sqrt{2R-3}, $$ $$ c = AB = 2\sqrt{R^2-(R-1)^2} = 2\sqrt{2R-1} $$ and twice the area is given by $a(R-2)+b(R-3)+c(R-1)$, but also by $\frac{abc}{2R}$.
That gives a horrible equation in $R$, from which
$$R = 2\left(1+\cos\frac{\pi}{9}\right)\approx 3.87938524$$ follows.
The area is so $\Delta\approx \color{red}{17.1866}$, and none of the given options is the correct one.
An approximate construction with Geogebra also shows that the app creators are wrong:
To worsen the situation, there is the fact that $R$ is an algebraic number of degree $3$ over $\mathbb{Q}$, hence the problem cannot even be solved by straightedge and ruler only!