11.2. On the sides $A B$ and $A C$ of an equilateral triangle $A B C$ with side length 10, points $P$ and $Q$ are taken such that the segment $P Q$ is tangent to the inscribed circle of the triangle and its length is 4. Find the area of triangle $A P Q$.
Answer: $\frac{5}{\sqrt{3}}$.

Solution: Let the lengths of segments $A P$ and $A Q$ be $x$ and $y$, respectively, and the point of tangency of segment $P Q$ with the inscribed circle of the triangle be $S$. The points of tangency of $A B$ and $A C$ with the inscribed circle are denoted as $R$ and $T$. By the cosine rule for triangle $A P Q$, we have: $x^{2}+y^{2}-x y=16$. On the other hand, due to the equality of the segments of tangents $P R=P S, Q T=Q S$, $x+y=A P+A Q=A R-P R+A T-Q T=A R+A T-(P S+Q S)=10-4=6$. Squaring the equality $x+y=6$, we get $x^{2}+y^{2}+2 x y=36$, subtracting $x^{2}+y^{2}-x y=16$ from it, we have $x y=\frac{20}{3}$. Then $S_{A P Q}=\frac{1}{2} x y \sin 60=\frac{1}{2} \frac{20}{3} \frac{\sqrt{3}}{2}=\frac{5}{\sqrt{3}}$.

Criterion: Cosine rule for triangle $A P Q: 1$ point.

Sum of $A P$ and $A Q$ equals 6: 3 points. Area through $x y=\frac{20}{3}-3$ points.