10. Let $S$ be the area of a triangle inscribed in a circle of radius 1. Then the minimum value of $4 S+\frac{9}{S}$ is $\qquad$ .
$10.7 \sqrt{3}$.
Let $y=4 S+\frac{9}{S}$, it is easy to know that it is a monotonically decreasing function on $\left(0, \frac{3}{2}\right]$. Also, because among the triangles inscribed in a unit circle, the equilateral triangle has the maximum area, which is equal to $\frac{3 \sqrt{3}}{4}$, hence $0<S \leqslant \frac{3 \sqrt{3}}{4}<\frac{3}{2}$.
Therefore, $y_{\min }=4 \times \frac{3 \sqrt{3}}{4}+\frac{9}{\frac{3 \sqrt{3}}{4}}=7 \sqrt{3}$.