How do I find the radius of the circle which touches three sides of a right angled triangle?

Question

I think it may have something to do with the radius-tangent theorem although I am not sure what steps I would follow to figure it out.

Answer 1

Hint. Note that the segments from the centre of the circle to the vertices of the given triangle, divide it into three smaller triangles. Then the area of the triangle, which is equal to $4\cdot 3/2=6$, can be also found as the sum of the areas of these three smaller triangles (what are their heights?). Can you find length of the radius now?

Answer 2

The center of the incircle of a triangle is the intersection of the angular bisectors.To compute its radius $r$, note that the center partitions the triangle into three subtriangles each of height $r$, but with bases $a,b,c$. We conclude that the total area of the triangle (which happens to be $\frac12ab$ for a right triangle) is also expressable as $\frac 12ar+\frac12br+\frac12cr$.

Answer 3

use that $$A=\frac{ab}{2}=\left(\frac{a+b+c}{2}\right)r$$

Answer 4

There is a neat formula: $$r = \dfrac{a+b-c}{2},$$ where $c$ is the hypothenuse - $c = \sqrt{a^2+b^2}.$