6. A circle with diameter $A B$ intersects segments $A C$ and $B C$ at points $M$ and $N$, respectively, and the length of segment $M N$ is equal to the radius of the circle. Find the area of quadrilateral $A B N M$, given that $A C=12$ and $B C=8$.

![](https://cdn.mathpix.com/cropped/2024_05_06_355bebf291524cccadf1g-11.jpg?height=862&width=899&top_left_y=1057&top_left_x=470)

ANSWER: $18 \sqrt{3}$.
Solution: Arc MN is $60^{\circ}$, so angle C is $60^{\circ}$. Therefore, the area of triangle $ABC$ is $\frac{1}{2} AC \cdot BC \sin 60^{\circ} = 24 \sqrt{3}$. Triangles $ABC$ and MNC are similar with a ratio of $1 / 2$, so the area of MNC is 4 times smaller, $6 \sqrt{3}$. Therefore, the area of AMNB is $18 \sqrt{3}$.