In the triangle ABC call the incenter point P.
The radius r of the incircle is just the altitude from P to each side of the triangle. Let a,b,c be the sides opposite A,B,C, respectively.
Draw a line from P to each vertex, A,B,C. This divides the triangle into three smaller triangles, each with height r.
So, the area of ABC is just r/2 (a+b+c)
Or, where p is the perimeter of ABC:
A = r/2 * p
r = 2A/p
This is true for all triangles, including right triangles.
The radius of the inscribed circle is equal to twice the area of the triangle divided by the perimeter of the triangle.
Prove that this relationship is true for the inscribed circle in any right triangle.
1 answer
1 answer