Maximizing area of a pentagon Suppose $a,b,c,d,e$ are pairwise distinct positive integers. Consider a pentagon with sides $a,b,c,d,e$ and with angles maximizing its area (we assume that a pentagon with such sides exists). It is easy to see that its area $S$ is a positive algebraic number. 
What is the smallest possible degree of $S$? Can it be a quadratic irrational? Can it be an integer?
 A: If we allow quadratic irrational parameters, we can devise a pentagons with significantly smaller integer sides than the Robbins pentagons. The proposed Pentagon has sides $(a,b,c,d,e,)=(1,2,3,6,7)$ in rotational order.
Begin by specifying a $120°$ angle between the $a$and $b$ sides. Then the diagonal connecting their remote endpoints measures $\sqrt7$ by the Law of Cosines. Similarly, A $120°$ angle between sides $c$ and $d$ gives a diagonal measuring $3\sqrt7=\sqrt{63}$.
Now for the trick. Since the two included angles defined above each measure $120°$, the remaining arc to be bridged by side $e$ subtends an arc also measures $120°$, and so the $\sqrt7$ and $3\sqrt7$ diagonals defined previously which intercept this arc must include an angle of $60°$. The Law of Cosines then gies $e=7$ for the remaining side, as claimed and completing a pentagon with distinct integer sides.
It remains to prove the claim that the curcumradius and area are rational or quadratic-irrational (they will in fact be pure quadratic surds). For the circumradius the $e=$ side is constructed tointercept a $120°$ arc, so the circumradius us $7/\sqrt3=\sqrt{49/3}$. As for tge area, the pentagon consists of three triangles. One has sides $1,2$ with included angle $120°$ contributing an area of $\sqrt3/2$. Another has sides $3,6$ with included angle $120°$, giving an area of $9\sqrt3/2$. The remaining triangle, containing the $e$ side, has its other two sides measuring $\sqrt7$ and $3\sqrt7$, with included angle $60°$; that area is then $21\sqrt3/4$. The total area is then $41\sqrt3/4=\sqrt{5043/16}$.
