# A triangle has sides with lengths of 6, 4, and 3. What is the radius of the triangles inscribed circle?

Jun 7, 2016

$r = \sqrt{\frac{\left(p - a\right) \left(p - b\right) \left(p - c\right)}{p}}$ with $p = \frac{a + b + c}{2}$
$r = 0.820413$

#### Explanation:

From Heron's formula, the area of a triangle giving their sides $a , b , c$ is

$A = \sqrt{p \left(p - a\right) \left(p - b\right) \left(p - c\right)}$ with $p = \frac{a + b + c}{2}$

Let $o$ be the triangle orthocenter, then the distance between each side and $o$ is $r$ which is the inscribed circle radius. So

$A = \frac{a \times r}{2} + \frac{b \times r}{2} + \frac{c \times r}{2}$ then
$r \left(\frac{a + b + c}{2}\right) = A = r \times p$

Finally

$r = \frac{\sqrt{p \left(p - a\right) \left(p - b\right) \left(p - c\right)}}{p} = \sqrt{\frac{\left(p - a\right) \left(p - b\right) \left(p - c\right)}{p}} = 0.820413$

Jun 13, 2016

$\cong 0.820$

#### Explanation:

Suppose
$A B = 6$
$B C = 4$
$C A = 2$

$m + l = 6$ [1]
$l + n = 4$ [2]
$m + n = 3$ [3]

[3]-[2]
$m - l = - 1$ [4]

[4]+[1]
$2 m = 5$ => $m = \frac{5}{2}$
$\to l = \frac{7}{2}$
$\to n = \frac{1}{2}$

Applying law of cosines:
$A {B}^{2} = B {C}^{2} + C {A}^{2} - 2 \cdot B C \cdot C A \cdot \cos \left(A \hat{C} B\right)$
$36 = 16 + 9 - 2 \cdot 4 \cdot 3 \cos \left(A \hat{C} B\right)$
$24 \cos \left(A \hat{C} B\right) = - 11$ => $A \hat{C} B \cong {117.28}^{\circ}$

$\tan \left(\frac{A \hat{C} B}{2}\right) = \frac{r}{n}$
$r = \frac{1}{2} \cdot \tan \left({117.28}^{\circ} / 2\right) = 0.820$