Fenced Semicircle 2

If the angles subtended by the radii to the two slant sides of the triangle are $x$ and $y$ as marked, then the triangle has base length $\sec x + \sec y$ . A little coordinate geometry (solving the equations of the two slant lines simultaneously) gives us that the triangle has height $\frac{\cos x + \cos y}{\sin(x+y)}$ .

Thus the triangle has area $\begin{aligned} A & = \; \frac{(\sec x + \sec y)(\cos x + \cos y)}{2\sin(x+y)} \; = \; \frac{(\cos x + \cos y)^2}{2\cos x \cos y \sin(x+y)} \\ & = \; \frac{1}{2\sin(x+y)}\left(\sqrt{\frac{\cos x}{\cos y}} + \sqrt{\frac{\cos y}{\cos x}}\right)^2 \; \ge \; \frac{2}{\sin(x+y)} \; \ge \; 2 \end{aligned}$ with the minimum value of $\boxed{2}$ achieved when $x=y=45^\circ$ .