Question

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

Answer 1

Inradius`=0.5097`

Explanation:

The radius of a triangle's inscribed circle is called inradius.

Inradius formula : `r=2A/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(3,3), B(1,2),C(2,1)`
`=> AB=sqrt((1-3)^2+(2-3)^2)=sqrt(2^2+1^2)=sqrt5`
`=> AC=sqrt((1-3)^2+(2-3)^2)=sqrt(2^2+1^2)=sqrt5`
`=> BC=sqrt((2-1)^2+(1-2)^2)=sqrt(1^2+1^2)=sqrt2`

perimeter `p=AB+AC+BC=sqrt5+sqrt5+sqrt2=2sqrt5+sqrt2`

Since `AB=AC`, the triangle is isosceles.

Since triangle ABD is right-angled,
`=> AD=sqrt((AB^2-(BD)^2)`; `(BD=(1/2)BC)`
`=> AD=sqrt((sqrt5)^2-(sqrt(2)/2)^2)=sqrt(5-2/4)=sqrt(9/2)=3/sqrt2`

Area of triangle `A= (1/2)*AD*BC=(1/2)*(3/sqrt2)*sqrt2=3/2`

Inradius `r=(2A)/p=(2*(3/2))/(2sqrt5+sqrt2)=0.5097`