8. AB = 2√13, R= 6,5. S_{ABCD}=?

The problem provides the length of side AB as $$2\sqrt{13}$$ and a radius R as 6.5. It asks for the area of ABCD. Without knowing the type of quadrilateral ABCD, or what R represents (e.g., radius of a circumscribed circle, altitude, etc.), it is impossible to solve. If R is the radius of a circumscribed circle and ABCD is a cyclic quadrilateral, then the diagonal would be $$2R = 13$$. If ABCD is a rectangle, then $$AB = CD = 2\sqrt{13}$$ and $$BC = AD$$. The diagonal $$d = \sqrt{AB^2 + BC^2} = 13$$. So, $$(2\sqrt{13})^2 + BC^2 = 13^2$$, which means $$4 \times 13 + BC^2 = 169$$, so $$52 + BC^2 = 169$$, and $$BC^2 = 117$$. $$BC = \sqrt{117} = 3\sqrt{13}$$. The area would be $$AB \times BC = 2\sqrt{13} \times 3\sqrt{13} = 6 \times 13 = 78$$. However, this assumes ABCD is a rectangle and R is the circumradius.