How to solve this problem using either Ptolemy's Theorem or Law of Cosines? A hexagon is inscribed in a circle of radius r. Suppose that four of the edges of the hexagon are 10 feet long and two of the edges are 20 feet long, but the exact arrangement of the edges is unknown. What is the value of r to three decimal places?
At first I split up the hexagon into 6 triangles and found the vertex angles of these triangles to be either 45 or 90 but apparently this is incorrect and I have to use one of the above methods to try to solve this. Any help would be appreciated!


Consider the following diagram where I have rearranged the sides such that $EF = 20, FC = ED = 10, DC = 2r$ and $A$ is the center. 
Let $H$ be the altitude to $DC$ from $F$. Note that $\angle DFC = \angle AGF - \angle DFG + \angle HFC$. However, since $CF' = DE = 10, \angle DFG = \angle HFC$ so $\angle DFC = 90^\circ$. (This relationship always holds in a cyclic isosceles trapeziod.)
Then $DF = \sqrt{(DC)^2-(FC)^2} = \sqrt{4r^2-100}$. From Ptolemy's theorem of $DEFC$,
$$ED \cdot FC+EF \cdot DC = DF \cdot EC$$ or equivalently,
$$10 \cdot 10 + 20 \cdot 2r = \left( \sqrt{4r^2-100} \right)^2.$$
Then the task is easy ;).
