11. The area of the upper base of a frustum is $S^{\prime}$. The area of the lower base is $S$. A section parallel to the bases with an area of $Q$ divides the height of the frustum from top to bottom in the ratio $n: m$, then $\sqrt{Q}=$ $\qquad$ .
11. $\frac{n \sqrt{S}+m \sqrt{S^{r}}}{n+m}$ The top base, a section parallel to the base, and the bottom base of a frustum are similar polygons. Let their similarity ratios be $k^{\prime}: l: k$, respectively. The area ratio is the square of the similarity ratio, so $\quad \frac{k^{\prime} 2}{S^{\prime}}=\frac{l^{2}}{Q}=\frac{k^{2}}{S} \Rightarrow \frac{k^{\prime}}{\sqrt{S^{\prime}}}=\frac{l}{\sqrt{Q}}=\frac{k}{\sqrt{S}} \Rightarrow \frac{k^{\prime}}{\sqrt{S^{\prime}}}=\frac{l-k^{\prime}}{\sqrt{Q}-\sqrt{S^{\prime}}}=$ $\frac{k-k^{\prime}}{\sqrt{S}-\sqrt{S^{\prime}}}$
and $\frac{n}{n+m}=\frac{l-k^{\prime}}{k-k^{\prime}}$, so $\frac{n}{n+m}=\frac{\sqrt{Q}-\sqrt{S^{\prime}}}{\sqrt{S}-\sqrt{S^{\prime}}}$, solving for $\sqrt{Q}$ gives $\sqrt{Q}=\frac{n \sqrt{S}+m \sqrt{S^{\prime}}}{n+m}$.