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!
 A: Here is a crude way. Join the vertices to the centre. Let the central angles of the little triangles be $\theta$ and let the central angles of the big triangles be $\phi$. Then $4\theta+2\phi=360^\circ$. So $\phi=180^\circ -2\theta$. 
Now we can use the Cosine Law, though I would prefer using sines. 
By the Cosine Law we have $100=2r^2-2r^2\cos\theta$ and $400=2r^2-2r^2\cos \phi$. 
It follows that 
$$4(1-\cos\theta)=1-\cos\phi.$$ 
But 
$$\cos\phi=\cos(180^\circ-2\theta)=-\cos(2\theta)=1-2\cos^2\theta.$$
  We obtain a quadratic equation for $\cos\theta$, and now only some computation is left. 
Remark: The Law of Cosines can be replaced by the observation that, for example, if the central angle is $\theta$, then the side has length $2r\sin(\theta/2)$. 
A: Hint #1:  because the hexagon is inscribed in a circle, the arrangement of the sides makes no difference:  if you draw the radii to each vertex, you obtain six isosceles triangles, two of which have sides $r$, $r$, and $20$, and four of which have sides $r$, $r$, and $10$.  Rearranging these triangles will not change $r$.
Hint #2:  If you arrange the sides in a symmetric fashion, can you then place the figure on a coordinate plane in a natural way, such that the figure is uniquely determined by the location of a single point on the circle?
A: 
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 ;).
