Can area be irrational?

If $ABC$ is an equilateral triangle and $A=(0,0),B=(x,y)$, then $C$ lies in: $$ C=\left(\frac{x\mp \sqrt{3}y}{2},\frac{y\pm \sqrt{3}x}{2}\right),$$ so assuming that $B$ is a point with rational coordinates, $C$ is not.

Moreover, the area of an equilateral triangle is just $\frac{\sqrt{3}}{4}l^2$ where $l$ is the length of a side.

In the same way, if the side length is a rational number, the area is not.

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With an alternative approach, if all the three vertices of an equilateral triangle have rational coordinates then the area is a rational number by the shoelace formula. But in such a case the squared length of the side is also a rational number, hence the area is at the same time a rational number and $\sqrt{3}$ times a rational number, contradiction.

Your counterexample doesn't work, because $x+y=2$ doesn't imply that $x$ and $y$ are rational. In fact, we have also that $x^2+y^2=8$. By substituting $y=2-x$ into the latter equation, you will get a quadratic equation with two irrational roots.

You should re-read the book's proof...

Briefly,

1. Assume that the third point has rational co-ordinates
2. Then by the formula you cite, the area is rational
3. But the area is an irrational number times the length of a side squared
4. But by the original assumptions, two vertices have rational coordinates, so the side length squared is also rational
5. A rational times an irrational is irrational; therefore in this case, the area is irrational
6. Thus the assumption (Point 1) leads to a contradiction (Point 2 vs Point 5)

It's Points 4 and 5 that the book fails to state explicitly...