Let $AB = 6$, $BC = 8$, and $AC = 10$. What is the area of the circumcircle of $/triangle ABC$ minus the area of the incircle of $/triangle ABC$?

The problem is illustrated by the figure shown below.

It is clear that ΔABC is a right triangle because

(a) it is inscribed in a circle,

(b) Its sides satisfy the Pythagorean theorem because

6² + 8² = 36 + 64 = 100, and 10² = 100.

Therefore, AC is the diameter of the circumscribing circle, and the radius is

r = 5.

The area of the circumscribing circle is

A₁ = π (5²) = 25π

The area of the triangle is

A₂ = (1/2) 8*6 = 24

The required area, of the circumcircle minus that of the triangle (shaded area) is

A₁ - A₂ = 25π - 24 = 54.54 square units.

Answer:

54.54 square units.