6. In the acute triangle $\triangle A B C$, $A D$ is the altitude, $A B+A C \cdot 4 \sqrt{\frac{2}{17}}=\sqrt{17}, 4 B C+5 A D=$ 17, then the area of $\triangle A B C$ is ().
A. I
B. $\frac{289}{250}$
C. $\frac{8}{5}$ or $\frac{10}{7}$
D. There are more than two possible values
6. B

Let $A D=h, B D=x, C D=y$, then
$$
\left\{\begin{array}{l}
\sqrt{x^{2}+h^{2}}+4 \sqrt{\frac{2}{17}} \cdot \sqrt{y^{2}+h^{2}}=\sqrt{17} \\
4(x+y)+5 h=17
\end{array}\right.
$$

It is easy to prove that $\left\{\begin{array}{l}17\left(x^{2}+h^{2}\right) \geqslant(4 x+h)^{2}(\text { when and only when } x=4 h \text { equality holds) } \\ 32\left(y^{2}+h^{2}\right) \geqslant(4 y+4 h)^{2}(\text { when and only when } y=h \text { equality holds) }\end{array}\right.$
After taking the square root of both inequalities and adding them, we get: $\sqrt{17} \cdot \sqrt{x^{2}+h^{2}}+4 \sqrt{2} \cdot \sqrt{y^{2}+h^{2}} \geq 4(x+y)+5 h$ $=17$, i.e., $\sqrt{x^{2}+h^{2}}+4 \sqrt{\frac{2}{17}} \cdot \sqrt{y^{2}+h^{2}} \geqslant \sqrt{17}$ (equality holds when and only when $x=4 h$ and $y=h$).
Therefore, $x=4 h, y=h, \sqrt{17} \cdot h+4 \sqrt{\frac{2}{17}} \cdot \sqrt{2} h=\sqrt{17}$, i.e., $h=\frac{17}{25}$.
Thus, $S_{\triangle A B C}=\frac{1}{2} \cdot A D \cdot B C=\frac{1}{2} \cdot h \cdot(x+y)=\frac{5}{2} h^{2}=\frac{289}{250}$, so the answer is B.