jmc

If $x^2+25x+c$ is the square of a binomial, then because the coefficient of $x^2$ is $1$, the binomial must be of the form $x+a$ for some $a$.

Expanding, we have $(x+a{)}^2=x^2+2ax+a^2$. For this to be equal to $x^2+25x+c$, the coefficients of $x$ must agree, so $2a$ must be equal to $25$. This gives $a=\frac{25}{2}$, and so the constant term $a^2$ is $\boxed{\frac{625}{4}}$.