A triangle has corners at #(2 , 1 )#, #(1 ,3 )#, and #(3 ,4 )#. What is the radius of the triangle's inscribed circle?

1 Answer
Mar 29, 2018

Answer:

# (sqrt5(2-sqrt2))/2#.

Explanation:

Let the vertices of #DeltaABC# be #A(2,1), B(1,3) and C(3,4)#, and

let #r# be the inradius of #DeltaABC#.

Then, #a^2=BC^2=(1-3)^2+(3-4)^2=4+1=5#,

#b^2=CA^2=(3-2)^2+(4-1)^2=1+9=10#, and,

#c^2=AB^2=(2-1)^2+(1-3)^2=1+4=5#.

#:. c^2+a^2=b^2 rArr /_B=90^@#.

#:. DeltaABC# is a right-triangle.

We know from Geometry, then,

#r=(c+a-b)/2=(sqrt5+sqrt5-sqrt10)/2=(sqrt5(2-sqrt2))/2#.