Question

Find the radius given a quadrilateral inscribed in semi-circle

attachment below:

Answer 1

diameter

Answer 2

Split up the quadrilateral into component isosceles triangles, then split each of those in half. I'll demonstrate this with the first triangle:

Each half has one leg of length \(\dfrac12\) and hypotenuse \(r\). So the central angle \(\theta_1\) subtended by the chord of length \(1\) satisfies
\[\sin\theta_1=\frac{\frac{1}{2}}{r}=\frac{1}{2r}\implies \theta_1=\arcsin\frac1{2r}\]Similarly, for the other two triangles you have, respectively, \(\theta_2=\arcsin\dfrac1r\) and \(\theta_3=\arcsin\dfrac3{2r}\).

These triangles all lie in a semicircle, so \(2\theta_1+2\theta_2+2\theta_3=\pi\). You can solve for \(r\) from here:
\[\begin{align*}
\frac\pi2&=\theta_1+\theta_2+\theta_3\\[1ex]
\sin\frac\pi2&=\sin\left(\theta_1+\theta_2+\theta_3\right)\\[1ex]
1&=\sin\theta_1\cos\theta_2\cos\theta_3+\cos\theta_1\sin\theta_2\cos\theta_3\\
&\qquad+\cos\theta_1\cos\theta_2\sin\theta_3-\sin\theta_1\sin\theta_2\sin\theta_3\\[1ex]
&=\sin\left(\arcsin\frac{1}{2r}\right)\cos\left(\arcsin\frac{1}{r}\right)\cos\left(\arcsin\frac{3}{2r}\right)\\
&\qquad+\cos\left(\arcsin\frac{1}{2r}\right)\sin\left(\arcsin\frac{1}{r}\right)\cos\left(\arcsin\frac{3}{2r}\right)\\
&\qquad+\cos\left(\arcsin\frac{1}{2r}\right)\cos\left(\arcsin\frac{1}{r}\right)\sin\left(\arcsin\frac{3}{2r}\right)\\
&\qquad-\sin\left(\arcsin\frac{1}{2r}\right)\sin\left(\arcsin\frac{1}{r}\right)\sin\left(\arcsin\frac{3}{2r}\right)\\[1ex]
&=\frac{1}{2r}\times\frac{\sqrt{r^2-1}}{r}\times\frac{\sqrt{4r^2-9}}{2r}+\frac{\sqrt{4r^2-1}}{2r}\times\frac{1}{r}\times\frac{\sqrt{4r^2-9}}{2r}\\
&\qquad+\frac{\sqrt{4r^2-1}}{2r}\times\frac{\sqrt{r^2-1}}{r}\times\frac{3}{2r}-\frac{1}{2r}\times\frac{1}{r}\times\frac{3}{2r}\\[1ex]
&=\frac{\sqrt{(r^2-1)(4r^2-9)}+\sqrt{(4r^2-1)(4r^2-9)}+3\sqrt{(4r^2-1)(r^2-1)}-3}{4r^3}
\end{align*}\]Solve this equation for \(r\) and you get your radius.