# A triangle has corners at (2 , 1 ), (3 ,3 ), and (1 ,2 ). What is the radius of the triangle's inscribed circle?

Oct 2, 2016

Inradius$= 0.5097$

#### Explanation: Inradius formula : $r = 2 \frac{A}{p}$, where $A$ = Area of the triangle and $p$=perimeter of the triangle.

1) calculate d the distance between the corners of the triangles, use the distance formula :
d=sqrt((x2−x1)^2+(y2−y1)^2

Given $A \left(3 , 3\right) , B \left(1 , 2\right) , C \left(2 , 1\right)$
$\implies A B = \sqrt{{\left(1 - 3\right)}^{2} + {\left(2 - 3\right)}^{2}} = \sqrt{{2}^{2} + {1}^{2}} = \sqrt{5}$
$\implies A C = \sqrt{{\left(1 - 3\right)}^{2} + {\left(2 - 3\right)}^{2}} = \sqrt{{2}^{2} + {1}^{2}} = \sqrt{5}$
$\implies B C = \sqrt{{\left(2 - 1\right)}^{2} + {\left(1 - 2\right)}^{2}} = \sqrt{{1}^{2} + {1}^{2}} = \sqrt{2}$

perimeter $p = A B + A C + B C = \sqrt{5} + \sqrt{5} + \sqrt{2} = 2 \sqrt{5} + \sqrt{2}$

Since $A B = A C$, the triangle is isosceles.

Since triangle ABD is right-angled,
=> AD=sqrt((AB^2-(BD)^2); $\left(B D = \left(\frac{1}{2}\right) B C\right)$
$\implies A D = \sqrt{{\left(\sqrt{5}\right)}^{2} - {\left(\frac{\sqrt{2}}{2}\right)}^{2}} = \sqrt{5 - \frac{2}{4}} = \sqrt{\frac{9}{2}} = \frac{3}{\sqrt{2}}$

Area of triangle $A = \left(\frac{1}{2}\right) \cdot A D \cdot B C = \left(\frac{1}{2}\right) \cdot \left(\frac{3}{\sqrt{2}}\right) \cdot \sqrt{2} = \frac{3}{2}$

Inradius $r = \frac{2 A}{p} = \frac{2 \cdot \left(\frac{3}{2}\right)}{2 \sqrt{5} + \sqrt{2}} = 0.5097$