That's a very good question! The area of the polygon is 14. I actually did not understand this problem at first, but then I found a way.
Solution: We are looking for positive integers (x, y) such that x^2 + y^2 = 5. We find the solutions (x,y) =
(1,2), (1, -2), (-1, 2), (-1, -2), (2, 1), (2, -1), (-2, 1), (-2, -1), which forms a convex octagon.
To calculate the area of the octagon, you can subtract the area of the right triangles from the area of the square with vertices of (+-2, +-2), where +-a represents a and -a. The area of the square is 16, while the distance between an octagon vertice and a vertice of the square is 1, so the area of each of the triangles is 1/2, so 4 of the triangles has area 2.
The answer is 16 - 2 = 14.
WARNING: Do not assume the octagon was regular! It was equiangular, with 4 sides of length sqrt2 and 4 alternating sides of length 2.
I hope you enjoy the solution! Thank you for reading it!
