9. The vertices of a quadrilateral lie on the sides of a square with side length 1. If the sum of the squares of the four sides is $t$, then the range of values for $t$ is ( ).
(A) $[1,2]$
(B) $[2,4]$
(C) $[1,3]$
(D) $[3,6]$
9.B.

As shown in Figure 1.
$$
\begin{array}{l}
t=a^{2}+b^{2}+c^{2}+d^{2} \\
=x^{2}+(1-n)^{2}+n^{2}+(1-m)^{2}+m^{2}+ \\
\quad(1-y)^{2}+y^{2}+(1-x)^{2} .
\end{array}
$$

Since $0 \leqslant x < 1$, we have
$$
\begin{array}{l}
x^{2}+(1-x)^{2} \\
=1-2 x(1-x) \leqslant 1, \\
x^{2}+(1-x)^{2} \\
=\frac{1+(2 x-1)^{2}}{2} \geqslant \frac{1}{2},
\end{array}
$$

Thus, $\frac{1}{2} \leqslant x^{2}+(1-x)^{2} \leqslant 1$.
Similarly, $\frac{1}{2} \leqslant y^{2}+(1-y)^{2} \leqslant 1$,
$$
\begin{array}{l}
\frac{1}{2} \leqslant m^{2}+(1-m)^{2} \leqslant 1, \\
\frac{1}{2} \leqslant n^{2}+(1-n)^{2} \leqslant 1 .
\end{array}
$$

Therefore, $2 \leqslant t \leqslant 4$, which means the range of $t$ is $[2,4]$.