6. An isosceles triangle $P Q T$ with base $P Q$ is inscribed in a circle $\Omega$. Chords $A B$ and $C D$, parallel to the line $P Q$, intersect the side $Q T$ at points $L$ and $M$ respectively, and $Q L=L M=M T$. Find the radius of the circle $\Omega$ and the area of triangle $P Q T$, if $A B=2 \sqrt{14}, C D=2 \sqrt{11}$, and the center $O$ of the circle $\Omega$ is located between the lines $A B$ and $C D$.
Answer: $R=\frac{15}{4}, S=18$.

Solution. The line $T O$ is perpendicular to the chords $C D, A B, P Q$ and bisects each of them. Let points $N$, $H$, and $E$ be the midpoints of $C D, A B$, and $P Q$. Denote the radius of the circle $\Omega$ as $R ; T N=N H=H E=x$.

Then $O H=T H-O T=2 x-R, O N=O T-T N=R-x$ and by the Pythagorean theorem for triangles $O A H$ and $O C N$ we get $R^{2}=14+(2 x-R)^{2}, R^{2}=11+(R-x)^{2}$, from which after expanding the brackets and combining like terms, we have

$$
\left\{\begin{array}{l}
2 x^{2}-2 R x+7=0 \\
x^{2}-2 R x+11=0
\end{array}\right.
$$

Subtracting the second equation from the first, we find that $x^{2}-4=0, x=2$. Then $R=\frac{2 x^{2}+7}{2 x}=\frac{15}{4}$; $O E=T E-T O=3 x-R=\frac{9}{4}, Q E^{2}=O Q^{2}-O E^{2}=\left(\frac{15}{4}\right)^{2}-\left(\frac{9}{4}\right)^{2}=9$. Therefore, $S_{P Q T}=T E \cdot Q E=6 \cdot 3=18$.