# A triangle has sides with lengths of 5, 8, and 3. What is the radius of the triangles inscribed circle?

Jan 22, 2016

There exists no triangle with the given measurements because the given measurements form a straight line.

If you still want to calculate the are then I have explained below.

If $a , b \mathmr{and} c$ are the three sides of a triangle then the radius of its inscribed circle is given by

$R = \frac{\Delta}{s}$

Where $R$ is the radius $\Delta$ is the are of the triangle and $s$ is the semi perimeter of the triangle.

The area $\Delta$ of a triangle is given by

Delta=sqrt(s(s-a)(s-b)(s-c)

And the semi perimeter $s$ of a triangle is given by
$s = \frac{a + b + c}{2}$

Here let $a = 5 , b = 8 \mathmr{and} c = 3$

$\implies s = \frac{5 + 8 + 3}{2} = \frac{16}{2} = 8$

$\implies s = 8$

$\implies s - a = 8 - 5 = 3 , s - b = 8 - 8 = 0 \mathmr{and} s - c = 8 - 3 = 5$

$\implies s - a = 3 , s - b = 0 \mathmr{and} s - c = 5$

$\implies \Delta = \sqrt{8 \cdot 3 \cdot 0 \cdot 5} = \sqrt{0} = 0$

$\implies R = \frac{0}{8} = 0$

Mathematically, since the area comes out to be zero therefore it proves that it is a straight line it is not a triangle, and since there is no triangle so there is no in-circle and thus no in-radius.