let x and y be the two polygons
x + y = 13
the formula for a diagonal in a polygon is [tex] \frac{n(n-3)}{2} [/tex]
Also since the sum of the diagonals is 25 you can write:
[tex] \frac{x(x-3)}{2} +\frac{y(y-3)}{2} =25[/tex]
Multiplying both sides by two you get,
[tex]x^2 - 3x + y^2 - 3y = 50[/tex]
Since you can transpose y to the other side you can get:
x + y = 13 can be written as: x = 13-y
Substituting,
[tex](13-y)^2 - 3(13-y) + y^2 - 3y = 50[/tex]
[tex]169 - 26y + y^2 - 39 + 3y + y^2 - 3y = 50[/tex]
[tex]2y^2 - 26y + 80 = 0[/tex]
Dividing by two you get,
[tex]y^2 - 13y + 40 = 0 [/tex]
(y-5)(y-8) = 0
Therefore the answer can be
1) x = 5, y = 8
OR
2) y=5, x=8
