+239 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 $\triangle ABC$?
+984 Looking at the numbers, we can tell that this triangle is right, as \(6^2+8^2=10^2\).
The area must be: \(\frac12\cdot6\cdot8=24\)
The semiperimeter is: \((6+8+10)/2=12\)
The formula to find inradius, \(r\), is \([ABC]=s\cdot{r} \)
Where \(s\) is the semiperimeter and \([ABC]\) is the area.
Filling in the numbers, we get: \(24=12\cdot{r}\Rightarrow r=2\)
The area of the incircle is \(2^2\pi=4\pi\)
The circumcircle's diameter is the hypotenuse of the triangle.
The radius is \(10\div2\), so the area is: \(5^2\pi=25\pi\)
The difference is \(25\pi-4\pi=\boxed{21\pi}\)
I hope this helped,
Gavin
