11. In $\triangle \mathrm{ABC}$, construct altitudes $\mathrm{AA}_{1}, \mathrm{BB}_{1}, \mathrm{CC}_{1}$, and connect the feet of the perpendiculars. If the angles of $\triangle A B C$ are known, try to calculate the ratio of the areas of $\triangle A_{1} B_{1} C_{1}$ and $\triangle A B C$.
11. First, assume that $\triangle \mathrm{ABC}$ is an acute triangle (Figure 11).

For: $S_{\triangle B C}-S_{\Delta_{1}} B_{1} c_{1}=S \boldsymbol{B}_{1} \wedge c_{1}+S_{\mathbf{c}} \boldsymbol{\Delta}{ }_{1}+S_{1} \mathrm{cs}$
Since $S, \wedge c_{1}=\frac{1}{2} A B_{1} \cdot A C_{1} \sin A=\frac{1}{2} A B \cos A$
$$
\begin{array}{l}
\text { - } \mathrm{ACcos} \mathrm{A} \sin \mathrm{A}=\frac{1}{2} \mathrm{AB} \cdot \mathrm{ACsin} \mathrm{A} \cos ^{2} \mathrm{~A} \\
=\mathrm{S}_{\mathrm{ABc}} \cos ^{2} \mathrm{~A} .
\end{array}
$$

Similarly, we have
$$
\begin{array}{l}
S_{C_{1}=1}-S_{A=E} \cos ^{2} B \text {, } \\
S_{A c, 1}=\left\{\wedge \Delta \cos ^{2}\right. \text {. } \\
\end{array}
$$

Substituting these expressions into equation (1), we get:
$$
\frac{S_{A_{1}, m_{1}, c_{1}}}{\mathrm{~S}_{\Delta B C}}=1-\cos ^{2} \mathrm{~A}-\cos ^{2} \mathrm{~B}-\cos ^{2} \mathrm{C} .
$$

If $\triangle \mathrm{ABC}$ is an obtuse triangle (Figure 12), then equation (1) becomes:
$$
\begin{array}{l}
\mathrm{S}_{\triangle B C}+\mathrm{S}_{\triangle_{1} B_{1} c_{1}}= \\
\mathrm{S}_{1} \wedge c_{i}+\mathrm{S}_{c_{1}, u_{i}}+\mathrm{S}_{\Delta_{1} c z_{1}} \\
\end{array}
$$

Accordingly, equation (2) becomes:
$$
\frac{\mathrm{S}^{\Delta}{ }_{1} 1_{1}^{c_{1}}}{\mathrm{~S}_{\mathrm{A}} \mathrm{c}}=\cos ^{2} \mathrm{~A}+\cos ^{2} \mathrm{~B}+\cos ^{2} \mathrm{C}-1 \text {. }
$$

Finally, if $\triangle A B C$ is a right triangle, then $S_{A_{1}}{ }_{1}{ }^{c}{ }_{1}=0$