A cyclic quadrilateral has side lengths $$3, 5, 5, 8$$ in this order. If $$R$$ is its circumradius, find $$3R^2$$.

For a cyclic quadrilateral with sides $$a, b, c, d$$ and area $$K$$, the circumradius satisfies $$R = \frac{\sqrt{(ab+cd)(ac+bd)(ad+bc)}}{4K}$$. Here $$s = \frac{21}{2}$$ and $$K = \sqrt{(s-a)(s-b)(s-c)(s-d)} = \sqrt{\frac{9075}{16}}$$, while $$(ab+cd)(ac+bd)(ad+bc) = 55 \times 55 \times 49 = 148225$$. Hence $$R^2 = \frac{148225}{9075} = \frac{49}{3}$$ and $$3R^2 = 49$$.