2. A rhombus with a side length of 5, one of its diagonals is no more than 6, and the other is no less than 6, then the maximum value of the sum of the lengths of the two diagonals of this rhombus is
(A) $10 \sqrt{2}$
(B) 14
(C) $5 \sqrt{6}$
(D) 12
(B)
2.【Analysis and Solution】Assume the diagonals of rhombus $A B C D$ are $B D \leqslant 6, A C \geqslant 6$, then
$$
\begin{array}{l}
\angle A B D=\theta, \text { so } 45^{\circ}<\arccos \frac{3}{5} \leqslant \theta<90^{\circ}, \therefore A C+B D=10(\sin \theta+\cos \theta) \\
=10 \sqrt{2} \sin \left(\theta+45^{\circ}\right) .
\end{array}
$$

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 $\theta=\arccos \frac{3}{5},(A C$ $+B D)_{\text {maximum }}=10(\sin \theta+\cos \theta)=14$.