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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.

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3 Answers 3

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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.

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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.

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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.

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