2. For a rhombus with a side length of 5, one of its diagonals is no longer than 6, and the other is no shorter than 6, then the maximum value of the sum of the lengths of the two diagonals is
(A) $10 \sqrt{2}$;
(B) 14 ;
(C) $5 \sqrt{6}$;
(D) 12 .
2. (B)

Let's assume the diagonals of rhombus $A B C D$ are $B D \leqslant 6, A C \leqslant 6$, and let $\angle A B D=\theta$, then
$$
\begin{aligned}
45^{\circ} & <\arccos \frac{3}{5} \leqslant \theta<90^{\circ} \\
\therefore \quad & A C+B D \\
= & 10(\sin \theta+\cos \theta) \\
& =10 \sqrt{2}\left(\theta+45^{\circ}\right)
\end{aligned}
$$

Notice that
$$
90^{\circ}<45^{\circ}+\arccos \frac{3}{5} \leqslant \theta+45^{\circ}<135^{\circ},
$$

and $y=\sin x$ is decreasing in the second quadrant, so $\sin \left(\theta+45^{\circ}\right)$ reaches its maximum value when
$$
\theta+45^{\circ}=45^{\circ}+\arccos \frac{3}{5}
$$

At this time,
$$
\begin{array}{c}
\theta=\arccos \frac{3}{5}, \\
(A C+B D)_{\text {max }}=10(\sin \theta+\cos \theta)=14 .
\end{array}
$$