We know that a triangle say ABC is inscribed in a circle with one of its sides exactly on the diameter of the circle, so the triangle must be a right triangle. (Triangle in a semi circle is a right triangle)


Also as we are given that one of the sides of the triangle is 5 and the area of the triangle is 20 , so the 5 units side must be one of the perpendicular sides.

Note: - 5 cannot be the hypotenuse as the area of the triangle is 20 , so the product of perpendicular sides is 40 , which must give at least one of the perpendicular sides greater than 5 , which is not possible as hypotenuse is the longest side in right triangle.


Area of triangle $\(=\frac{1}{2} \times\)$ Base $\(\times\)$ Height $\(=20 \Rightarrow\)$ Base $\(\times\)$ Height $\(=40\)$, so the base $\&$ height should be 5 and 8 in any order.

Now, using Pythagoras theorem i.e. Hypotenuse $\({ }^2=\)$ Perpendicular $\(^2+\)$ Base $\(^2\)$, the hypotenuse of triangle ABC i.e. AB which is the diameter of the circle is $\(\mathrm{AB}=\sqrt{\mathrm{AC}^2+\mathrm{BC}^2}=\sqrt{5^2+8^2}=\sqrt{89}\)$, so the radius of the circle is $\\(\frac{Diameter}{2}=\frac{\sqrt{89}}{2}\)$

Finally the area of the circle is $\(\pi r^2=\frac{22}{7} \times (\frac{\sqrt{89}}{2})^2=\frac{979}{14}\)$
Hence the answer is (B).