Given any rectangle $ABCD$ and we want to construct a rectangle with the same area and one side being a fixed length $l$. We do as follow: put a point $E$ on the ray $AB$ such that $AE=l$. Draw $DE$. Then draw a line through $B$ parallel to $DE$ and let it intercept the line $AD$ at $F$. Then $AE$, $AF$ would be the length of the side of the rectangle. Prove this using similar triangle.
So how to use that to do the problem? Let $a,b$ be the required length of 2 side of the triangle. First, double the square by making a copy of itself, so you get a rectangle. Now using the method above, acquire a rectangle $ABCD$ of the same area as twice the square, where $AB=a$. Let the circle centre $A$ radius $b$ intercept the line $CD$ at $E$. Then $ABE$ is the required triangle. This is because the height of the triangle with apex $E$ is the same as $AD$ and we already know the rectangle is twice the original square.
In fact, the fact that you started with a square is completely inconsequential.