Triangles with side length $\frac{3\cdot (2t-1)}{(2t-1)+2t+(2t+1)},\frac{3\cdot (2t)}{(2t-1)+2t+(2t+1)},\frac{3\cdot (2t+1)}{(2t-1)+2t+(2t+1)}$ where $3t^2-3=k^2$ for some $t,k\in \mathbb{Z}^+$ satisfy the condition. Note that there exist infinitely many $t\in \mathbb{Z}^+$ by Pell's Equation $t^2-1=3\ell^2$ when $k=3\ell$.

any right angeled triangle with rational side lengths work because we can easily show that the last altitude also has a rational value and we just have to divide the all sides by a number so that the perimiter will be 3