This is a classical Sangaku problem, also known as old Japanese geometry problems, that I found out just recently. The figure shows a semicircle with a smaller circle and an equilateral triangle inscribed inside it. Note that the semicircle can be any general semicircle, but in this case it has a radius of $1$ unit. Also note that one of the vertices of the equilateral triangle lies on the center of the semicircle. I have solved this problem and I'll post my approach as an answer, in order to not clutter up this space. Please share your own approaches as well!
Best Answer
A slightly quicker step (2) skips the quadratic formula: Observe that the hypotenuse shared by your two 30-60-90 triangles has length $\frac 2{\sqrt 3} r$ (since the side lengths are proportional to $1:\sqrt 3:2$), hence $$r + \frac 2{\sqrt 3}r = 1$$ which yields $r=2\sqrt 3-3$.