11. Given that the perimeter of quadrilateral $\square A B C D$ is 28, a line through vertex $D$ is drawn perpendicular to sides $A B$ and $B C$, with the feet of the perpendiculars being $E$ and $F$ respectively. If $D E=3, D F=4$, find
(1) the lengths of sides $A B$ and $B C$;
(2) the length of $B E+B F$.
Three, 11. (1) According to the problem, $DE$ and $DF$ are the distances between the two pairs of opposite sides of $\square ABCD$.
Thus, $AB \cdot DE = BC \cdot DF$.
Let $AB = a, BC = b$.
Since $DE = 3, DF = 4$, we have $3a = 4b$.
Also, the perimeter of $\square ABCD$ is 28, so,
$a + b = \frac{1}{2} \times 28 = 14$.
From $\left\{\begin{array}{l}3a = 4b, \\ a + b = 14\end{array} \Rightarrow \left\{\begin{array}{l}a = 8, \\ b = 6\end{array}\right.\right.$
$\Rightarrow AB = 8, BC = 6$.
(2) (i) When $\angle A$ is an acute angle, as shown in Figure 5.
In the right triangle $\triangle ADE$, we have
$AD = BC = 6, DE = 3$
$\Rightarrow AE = \sqrt{AD^2 - DE^2} = 3\sqrt{3}$
$\Rightarrow BE = AB - AE = 8 - 3\sqrt{3}$.
In the right triangle $\triangle DFC$, we have
$DC = AB = 8, DF = 4$
$\Rightarrow CF = \sqrt{DC^2 - DF^2} = 4\sqrt{3}$.
Since $4\sqrt{3} > 6$, point $F$ is on the extension of $CB$.
Thus, $BF = CF - BC = 4\sqrt{3} - 6$.
Therefore, $BE + BF = (8 - 3\sqrt{3}) + (4\sqrt{3} - 6) = 2 + \sqrt{3}$.
(ii) When $\angle A$ is an obtuse angle, as shown in Figure 6.
Similarly, we get $AE = 3\sqrt{3}, CF = 4\sqrt{3}$.
Then $BE = AB + AE = 8 + 3\sqrt{3}$,
$BF = CF + BC = 4\sqrt{3} + 6$.
Therefore, $BE + BF = (8 + 3\sqrt{3}) + (4\sqrt{3} + 6) = 14 + 7\sqrt{3}$.
In summary, the length of $BE + BF$ is $2 + \sqrt{3}$ or $14 + 7\sqrt{3}$.