4. Let $a, b, c$ be the lengths of the sides of any triangle, and let $s=a+b+c, t=ab+bc+ca$. Then $(\quad)$.
(A) $t<s^{2} \leqslant 2 t$
(B) $2 t \leqslant s^{2}<3 t$
(C) $3 t<s^{2} \leqslant 4 t$
(D) $3 t \leqslant s^{2}<4 t$
4.D.
$$
\begin{array}{l}
\text { Since } s^{2}=a^{2}+b^{2}+c^{2}+2(a b+b c+c a) \\
=a^{2}+b^{2}+c^{2}+2 t,
\end{array}
$$

It is easy to know that $a^{2}+b^{2}+c^{2}-a b-b c-c a \geqslant 0$, hence $a^{2}+b^{2}+c^{2} \geqslant t$.
From $a^{2}<a b+a c, b^{2}<b c+b a, c^{2}<c a+c b$, we get $a^{2}+b^{2}+c^{2}<2(a b+b c+c a)=2 t$.
From equations (1) and (2), we have $t \leqslant a^{2}+b^{2}+c^{2}<2 t$. Therefore,
$$
3 t \leqslant s^{2}<4 t \text {. }
$$