Finding the radius of a circle inside of a triangle What is the measure of the radius of the circle inscribed in a triangle whose sides measure $8$, $15$ and $17$ units?
I can easily  understand that it is a  right angle triangle  because of the given edges.  but I don't find any  easy formula to find the radius of the circle.

 A: Hint:
use the fact that the area $A$ (of the triangle) is given by: $A=\frac{pr}{2}$ where $p$ is the perimeter and $r$ the incircle radius. This formula can easily be proved ( divide the triangle in three triangle with a common vertex at $O$) and is valid for a convex polygon..
A: You can easily calculate the area of the triangle.
Then divide the triangle into three smaller ones: $AOB, BOC, COA$. Notice that their areas are respectively $AB\cdot r/2, BC\cdot r/2, CA\cdot r/2$. Add them up and compare to the area of $ABC$ you calculated earlier.
A: Spotting that the triangle is right-angled is a great help, but the problem can be done without.
Let the sides of the triangle be $a,b,c$ and define the semi-perimeter $s=\frac {a+b+c}2$ the inradius as $r$ to be found and the area of the triangle as $A$.
Then we have both $A=rs$ and Heron's formula for the area of a triangle given the sides $A=\sqrt {s(s-a)(s-b)(s-c)}$ from which we get $$r=\sqrt{\frac {(s-a)(s-b)(s-c)}{s}}=s\sqrt {\left(1-\frac as\right)\left(1-\frac bs\right)\left(1-\frac cs\right)}$$
A: Notice, $8^2+15^2=17^2$ hence the triangle is right angled triangle. 
area of right triangle $$\Delta=\frac{1}{2}\times 8\times 15=60$$
semi-perimeter of right triangle $$s=\frac{8+15+17}{2}=20$$
Radius of inscribed circle is given as $$r=\frac{\Delta}{s}=\frac{60}{20}=3$$ 
A: In fact, once you have spotted that it is a right triangle there is a simple formula for the diameter of the inscribed circle.
$d = a+b-c$
So your radius is:
$r = (8+15-17)/2 = 3$
It follows from two ways to compute the area of the triangle:
$A = ab/2$
$4A = 2ab = (a+b)^2 - a^2 - b^2 = (a+b)^2 - c^2 = (a+b+c)(a+b-c)$
On the other side
$2A = radius * perimeter = r (a+b+c)$
Combining the two you have
$4A = 2r(a+b+c) = (a+b+c)(a+b-c)$
$\Rightarrow 2r = (a+b-c)$
A: The nswer is $3$, because one formula for the inradius is given by:
$$\frac{\text{area of triangle}}{\text{semi-perimeter of triangle}}$$
