A triangle has sides with lengths of 3, 7, and 6. What is the radius of the triangles inscribed circle?

Jan 19, 2016

$r = \frac{\sqrt{5}}{2}$

Explanation:

To find the radius, we use the fact that the triangle contains three internal triangles whose height is the radius $r$. The sum of the areas of these triangles equals the area of the triangle ABC, which is calculated using Heron's formula
$A = \sqrt{p \left(p - a\right) \left(p - b\right) \left(p - c\right)}$ where $p = \frac{a + b + c}{2}$

$p = \frac{3 + 7 + 6}{2} = 8$
$\therefore A = \sqrt{8 \cdot 5 \cdot 1 \cdot 2} = \sqrt{80} = \sqrt{16 \cdot 5} = 4 \sqrt{5}$
The three internal triangles have areas
${a}_{1} = \frac{1}{2} \cdot 3 \cdot r = \frac{3 r}{2}$
${a}_{2} = \frac{1}{2} \cdot 7 \cdot r = \frac{7 r}{2}$
${a}_{3} = \frac{1}{2} \cdot 6 \cdot r = 3 r$

${a}_{1} + {a}_{2} + {a}_{3} = A$
$r \left(\frac{3}{2} + \frac{7}{2} + 3\right) = 4 \sqrt{5}$
$\therefore r = 4 \frac{\sqrt{5}}{8} = \frac{\sqrt{5}}{2}$