Basically, the topic says it all. If a triangle has rational coordinates (say, in $\mathbb Q^2$), must it have rational area? I realize the side-lengths are usually irrational; that's fine.
Heron's formula seems pertinent: if the side-lengths are $a, b, c$, and you let $s=\frac{1}{2}(a+b+c)$, then the area is $\sqrt{s(s-a)(s-b)(s-c)}$. The side-lengths will always be square-roots of rational numbers, and when you simplify Heron's formula out, you get $\frac{1}{4}\sqrt{a^2b^2+a^2c^2+b^2c^2-a^4-b^4-c^4}$, and since all those inside terms have even degree, it's the square root of a rational number.
But this suggests it shouldn't be rational in general, just a square root of a rational. Is there some reason why it would be rational? Or, conversely, is there a "rational triangle" with irrational area?
Best Answer
It works. You might as well put one vertex at the origin. Then the area is half the absolute value of the cross product of the two vectors, regarded as vectors in $\mathbb R^3.$