What is the radius of a circle that has a 8 15 17 inscribed within?

A triangle has sides 8, 15, 17. It's inscribed in a circle. What's the radius of the circle?

The triangle is inscribed in the circle, not the circle is inscribed in the triangle. I understand how the triangle is a right triangle, but not anything else. Thanks!

If you see that $17^2=8^2+15^2$, we can conclude that it's a right triangle. BUT, if you did not saw this, you can apply the law of cosines at the biggest side and you get $\cos(\theta)=0$, meaning this triangle has a $90^\circ$ angle.

So, the hypotenuse is the diameter of circumscribed circle. thus the radius is $\dfrac{17}2$.

• [+1] Good observation of a "Pythagorean triple". I have taken the liberty to correct numerous little vocabulary and grammatical errors. Be more careful next time. – Jean Marie Jan 4 '18 at 23:52
• I'm sorry, English is not my official language and I'm a bit rusty. I realized that you adjusted the cosine theorem to the law of cosines. Where I study, this approach is not frequent. Thanks for correcting me. – Gustavo Mezzovilla Jan 5 '18 at 19:05
• English is not more than you my native language... I make remarks because I have always appreciated to be corrected by others (and I am happy that you receive it in the same way), improving step by step a written proficiency that I hadn't at the beginning. The only advice I can do is to check your texts before pressing the "send" button (this advice is valid for e-mails !)... – Jean Marie Jan 5 '18 at 20:43

8.5, the sides make up a right angled triangle by Pythagoras and the radius is half the longest side.

$$R= \frac{abc}{4Δ}=\frac{8. 15 . 17}{4.Δ}$$ $$=\frac{8. 15 . 17}{4. 4. 15}= \frac{17}{2}$$
Consider a right triangle $ABC$ and $M$ the intersection of the perpendicular ($DM$ and $EM$) bisectors of the sides of $ABC$. Since $DM//BC$ and $EM//AB$ (because $ABC$ are right), we have that $ADM\equiv ABC$. $$\therefore \dfrac{AM}{AC}=\underbrace{\dfrac{AD}{AB}}_{1/2} \Longrightarrow AC=2\cdot AM$$ But the circunference of center $M$ has $AM$ as the radius. Therefore $AC$ are the hypotenuse. :)