jmc

Let $K$ be the area of the triangle, and let $p$ be the semi-perimeter. Then by Heron's formula, $$K^2=p(p-r)(p-s)(p-t).$$By Vieta's formulas, $r+s+t=4,$ so $p=2.$ Also, since $r,$ $s,$ $t$ are the roots of $x^3-4x^2+5x-\frac{19}{10},$ $$x^3-4x^2+5x-\frac{19}{10}=(x-r)(x-s)(x-t).$$Setting $x=2,$ we get $$(2-r)(2-s)(2-t)=\frac{1}{10}.$$Then $$K^2=2(2-r)(2-s)(2-t)=\frac{1}{5},$$so $K=\sqrt{\frac{1}{5}}=\boxed{\frac{\sqrt{5}}{5}}.$