# Radius of a circle intercepting 3 corners of a triangle

So I have a Triangle, ABC, and I've found BC from $BC^2 = AB^2+AC^2$ , and I've found AD (height) with Heron's Formula.

Now, what if I draw a circle which touches all three corners of the triangle like so:

Is there any way I can find the radius (OB)? Would $R = (a*b*c/4) = (BC*AC*AB)/4 = (10*8*4)/4 = 80$ work? Is $80$ the answer?

• it is $$OB=\frac{a}{2}=5$$ Mar 5 '17 at 17:04
• Which c are you referring to? Mar 5 '17 at 17:06
• oh sorry it must be $$a$$ Mar 5 '17 at 17:06
• The figure is misleading (probably on purpose). As the triangle is a right one, the hypothenuse is a diameter of the circle.
– user65203
Mar 5 '17 at 17:17
• You have misquoted the formula for the circumradius.
– robjohn
Mar 5 '17 at 18:27

Since $BC^2=AB^2+AC^2$, we have $\angle BAC=90^{\rm o}$,

Also, the angle subtended by a diameter is always $\,90^{\rm o}\,$, so $$BC\ \,\text{is the diameter of the circle}$$

Hnece the radius$\,OB=BC\,/\,2=5$

Obviously radius of circumscribed circle can't be greater than the length of the longest side, so you result is wrong.

Actually you forgot to divide by area (and what for did you find it?), and substituted wrong value of $AB$.

But the most useful note is that the center of right-angled triangle lies in the middle of hypotenuse.

Hint: The radius of the circumcircle of a triangle, the circumradius, is $$\frac{abc}{4A}$$ where $a,b,c$ are the lengths of the sides of the triangle and $A$ is the area of the triangle.