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.
[Math] How to find the inradius of a triangle with given side lengths
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Related Solutions
Let's denote the given side and the equivalent height by $a$ and $h$ respectively, then the area of the triangle $A$ is
- $A=\cfrac 12 a\cdot h=\cfrac 12 b\cdot c \cdot \sin Â$
Using law of cosine
- $a^2=b^2+c^2-2b\cdot c\cdot \cos Â$
You already have the sum of the two sides
- $b+c=k$
These three equations should help in finding $b$, $c$ and $Â$.
$$b\cdot c = \cfrac {2A} {\sin Â}\\(b+c)^2 = b^2 +c^2 +2bc = k^2 \implies b^2+c^2=k^2-2bc\\ a^2=k^2-2bc(1+\cosÂ)=k^2- 4A \cfrac {1+\cos Â} {\sin Â}=k^2- 4A \cdot\cot \cfrac Â2 \implies \tan \cfrac Â2=\cfrac {4A} {k^2-a^2}$$
We found $Â$!
The rest should be easy.
$$\begin{eqnarray*}r_1\left(p_{AFE}+p_{BDF}+p_{CED}\right)&=&2\left([AFE]+[BDF]+[CED]\right)\\&=&2\left([ABC]-[DEF]\right)\\&=&r\,p_{ABC}-r_{DEF}\,p_{DEF}\\&=&r_1\left(p_{ABC}+p_{DEF}\right)\end{eqnarray*}$$ leads to $$ r_{DEF}\, p_{DEF} = (r-r_1) p_{ABC} - r_1 p_{DEF}. $$ Now the interesting fact is that the perimeter of $DEF$ is fixed by $r,r_1,p_{ABC}$.
By considering the distances of $D,E,F$ from the tangency points marked with a red cross we have that $p_DEF$ equals the perimeter of $I_A I_B I_C$, with $I_A,I_B,I_C$ being the incenters of $AEF,BDF,CDE$. It follows that
$$ p_{DEF} = \frac{r-r_1}{r} p_{ABC} $$ and
$$ r_{DEF} = r- r_1.$$
Another interesting fact is that $D,E,F$ and $I_A,I_B,I_C$ belong to the same ellipse.
Best Answer
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.