If we require the pentagon to have integer side length, then the answer is no. The area of a regular pentagon with side length $a$ is $$5\cdot\left(\frac12a h\right)=\frac54a^2\tan(54^\circ)$$ But $\tan(54^\circ)$ is irrational, because the only rational values of $\tan k\pi/n$ are $0$ and $\pm1$.