Why is the semisimple rank of a connected reductive group equal to the rank of the commutator?

I am trying to learn the theory of linear algebraic groups over an algebraically closed field. I know that if $R(G)$ denotes it radical, then $G/R(G)$ is semisimple and is therefore equal to its own commutator. So $G/R(G)=[G/R(G),G/R(G)]=[G,G]R(G)/R(G)=[G,G]/(R(G) \cap [G,G])$. So it's a quotient of the commutator by a finite normal subgroup.

But why would the rank (dimension of a maximal torus) of $[G,G]$ be equal to the rank of the quotient of $[G,G]$ by a finite subgroup? In general, is the rank additive in short exact sequences? I know it is true for the sequence

$1 \rightarrow SL_n(k) \rightarrow GL_n(K) \rightarrow \mathbb{G}_m \rightarrow 1$.

Personally, I feel that $R(G)\cap [G, G]$ is not necessarily finite. For instance, put take $G$ to be the group of upper triangular matrices in $\operatorname{GL}_3(k)$. This $G$ is itself a connected normal solvable subgroup of $G$, and $R(G) = G$. On the other hand, $[G, G]$ is the subgroup of $G$ of the elements which have $1$ in the diagonal and $0$ in the superdiagonal.

I suppose you implicitly assume that $G$ is reductive. In this case, $R(G)$ is the maximal torus in $Z(G)$, and $Z(G)\cap [G,G]$ is finite because $G^{\mathrm{ad}} = G/Z(G)$ is semisimple and so $G^{\mathrm{ad}} = [G^{\mathrm{ad}}, G^{\mathrm{ad}}]$.

The ranks of the two groups in question are equal because the maximal torus of $[G, G]$ maps onto the one of $[G, G]$ quotiented by a finite group.