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 $\triangle ABC$?

 Aug 7, 2018

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,



 Aug 7, 2018

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