10. A. Given positive real numbers $a$, $b$, $c$ satisfy $9a + 4b = abc$. Then the minimum value of $a + b + c$ is $\qquad$.
10. A. 10 .

From the given, we know that $c=\frac{4}{a}+\frac{9}{b}$.
Therefore, $a+b+c=\left(a+\frac{4}{a}\right)+\left(b+\frac{9}{b}\right)$ $=\left(\sqrt{a}-\frac{2}{\sqrt{a}}\right)^{2}+\left(\sqrt{b}-\frac{3}{\sqrt{b}}\right)^{2}+10 \geqslant 10$.
When $a=2, b=3, c=5$, the equality holds.
Thus, the minimum value of $a+b+c$ is 10.