# Finding the radius of excircles from a right angled triangle

Right angled triangles have 3 excircles, I'm struggling to find a formula which gives me the radius of all three excircles, I've been struggling with this for a while. I've done some googling and I think I have parts of the correct formula,

$$s =\frac{(a+b+c)}{2}$$

$$A = \sqrt(s*(s-a)*(s-b)*(s-c))$$

$$r_1 =\frac{A}{(s-a)}$$

$$r_2 =\frac{A}{(s-b)}$$

$$r_3 =\frac{A}{(s-c)}$$

$A$ is the area of the right angled triangle. $a, b$ and $c$ are sides of the right angled triangle. $s$ is the semi-perimeter of the right angled triangle. $r_1, r_2$ and $r_3$ are the radius of the excircles.

Can anyone find the formula?

Preferably I would like a formula without using any angles

• You can express the ex-radii in terms of inradius but that's the simplest you can get, for e.g., $r_1=r\frac{s}{s-a}$ Edit: oh it's a right angle triangle didn't notice. – Mann May 11 '15 at 9:56
• Would $r_1=\frac{ab}{2(s-a)}$ , $r_2=\frac{ab}{2(s-b)}$ work? Where i assumed c is hypotenuse. – Mann May 11 '15 at 9:59
We know that $\Delta = \frac{abc}{4R}$
Now I will assume that $c$ is the hypotenuse. Draw a circum-circle around your triangle you can easily observe by Thales theorem that $c$ is the diameter of the circle . Hence $c=2R$, Therefore $\Delta = \frac{ab}{2}$ or Area is triangle = $\frac{1}{2}$ * base * height.