# How to find the inradius of a triangle with given side lengths?

I need to find the inradius of a triangle with side lengths of $20$, $26$, and $24$.

I know the semiperimeter is $35$, but how do I find the area without knowing the height? Thank you.

By Heron's Formula the area of a triangle with sidelengths $a,b,c$ is $K = \sqrt{s(s-a)(s-b)(s-c)}$, where $s = \frac{1}{2}(a+b+c)$ is the semi-perimeter. You can then use the formula $K = rs$ to find the inradius $r$ of the triangle.

Solution:

Semiperimeter is given

$$K = RS$$ so

$$\text{K} = \sqrt{35(15)(9)(11)} = \sqrt{51975}$$

concluding that

$$R = \frac{\sqrt{51975}}{35} = \frac {3 \sqrt {231}} 7 .$$

Please correct me if I got something wrong.

Adding this as an addendum: since a triangle is uniquely determined (up to a direct or indirect congruence) by its side lengths, you can, in principle, express the inradius (and, indeed, any triangle quantity) in terms of these quantities. Using the $$p,q$$ method, there is a systematic method to do this. We write $$A_1$$ and so on for the vertices, $$e_{12}$$ and so on for the side lengths. We choose coordinates so that we have the vertices at $$(0,0)$$, $$(e_{12},0)$$ and $$(p,q)$$. Then we have the two equations $$(p-e_{12})^2+q^2=e_{23}^2,\,p^2+q^2=e_{31}^2$$ which can easily be solved for $$p$$ and $$q$$ in terms of the side lengths. Once you know that, it remains only to compute any triangle quantity, in particular your one, for this special triangle—usually a simple task.