6. The ratio of the sides of a triangle, which is inscribed in a circle of radius $2 \sqrt{3}$, is $3: 5: 7$. Find the area of the triangle.
(1 mark)
6. A triangle is inscribed in a circle with a radius of $2 \sqrt{3}$. The ratio of its side lengths is $3: 5: 7$. Find the area of the triangle.
6. $\frac{135}{49} \sqrt{3}$
6. Let $A B C$ denote the triangle, and the lengths of the sides opposite $A, B, C$ be $3 k, 5 k, 7 k$ respectively. By cosine law, we have
$$
\cos C=\frac{(3 k)^{2}+(5 k)^{2}-(7 k)^{2}}{2(3 k)(5 k)}=-\frac{1}{2}
$$

Hence $C=120^{\circ}$. By sine law,
$$
\frac{7 k}{\sin 120^{\circ}}=2(2 \sqrt{3})
$$
from which we get $k=\frac{6}{7}$. It follows that the area of $\triangle A B C$ is equal to
$$
\frac{1}{2}(3 k)(5 k) \sin 120^{\circ}=\frac{15}{2}\left(\frac{6}{7}\right)^{2}\left(\frac{\sqrt{3}}{2}\right)=\frac{135}{49} \sqrt{3} .
$$