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# help

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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

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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

Aug 7, 2018