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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!

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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 $\dfrac{17}2$.

enter image description here

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  • $\begingroup$ [+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. $\endgroup$ – Jean Marie Jan 4 '18 at 23:52
  • $\begingroup$ 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. $\endgroup$ – Gustavo Mezzovilla Jan 5 '18 at 19:05
  • $\begingroup$ 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 !)... $\endgroup$ – Jean Marie Jan 5 '18 at 20:43
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8.5, the sides make up a right angled triangle by Pythagoras and the radius is half the longest side.

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What you asked is the circumradius

$$R= \frac{abc}{4Δ}=\frac{8. 15 . 17}{4.Δ}$$ $$=\frac{8. 15 . 17}{4. 4. 15}= \frac{17}{2}$$

This is the required radius.

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I think that your problem is about to see that the hypotenuse of the triangle are the diameter of the circunference. The intersection of the perpendicular bisectors of the sides are the center of circunference.

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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. :)

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