Let AB = 6, BC = 8, and AC = 8. What is the area of the circumcircle triangle ABC of minus the area of the incircle of triangle ABC?
+605 Drop $C$ to $AB$ at $D$. Because of congruence, $AD=BD=3$. Then $CD=\sqrt{8^2-3^2}=\sqrt{55}$. So the area of the triangle is $3\sqrt{55}$.
Then the semiperimeter is 11. So $11r=3\sqrt{55}\implies r=\frac{3\sqrt{55}}{11}$, which is the inradius. The incircle area is then $\frac{45}{11}\pi$
The area is also $\frac{abc}{4R}$. So $\frac{96}{R}=3\sqrt{55}$ so $R=\frac{32}{\sqrt{55}}$. The area of the circumcircle is then $\frac{1024}{55}\pi$.
So the answer is $\frac{1024}{55}\pi - \frac{45}{11}\pi=\boxed{\frac{799}{55}\pi}$.
