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

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Answer 1 · 2018-03-29T13:35:55

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

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,

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

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