I've heard that when $\pi$ was proved irrational, that squaring the circle was not proved impossible. This lead me to believe that you could construct a square with an irrational area. Is this possible?
It is possible to create a polynomial with integer coefficients that satisfies an irrational area ($x^4=2$), so that leads me to believe it is doable, although my knowledge of the precise relationship between polynomial and areas is sketchy at best.
Also, it is easy to construct a square of irrational side lengths (consider 4 1x1 squares in a grid, you can make a square side length $\sqrt{2}$ using the diagonals) but a square with an irrational area eludes me yet it seems a possible extension.
What about any shape with an irrational area, is that possible to construct? Does that shape have to be regular to be constructible?
Once you have decided on a unit length, you can construct square roots of arbitrary line segments. In particular you can create a line segment of length $\sqrt{\sqrt 2}$ and then erect its square -- which of course will have area $\sqrt 2$.
Merely showing that $\pi$ is irrational does not prove that the circle can't be squared. This was only proved as a result of $\pi$ being transcendental, since all constructible numbers are algebraic.