A triangle has corners at #(3 , 4 )#, #(6 ,3 )#, and #(5 ,8 )#. What is the radius of the triangle's inscribed circle?

1 Answer
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Feb 17, 2016

#R=sqrt{2* (sqrt5 - sqrt13 + 3) * (sqrt5 + sqrt13 - 3) * (-sqrt5 + sqrt13 + 3)}/{sqrt{sqrt5 + sqrt13 + 3}}#

Explanation:

First, we find the length of each side of the triangle using the Pythagorean Theorem.

#sqrt{(6 - 3)^2 + (3 - 4)^2} = sqrt10#

#sqrt{(6 - 5)^2 + (3 - 8)^2} = sqrt26#

#sqrt{(3 - 5)^2 + (4 - 8)^2} = 3sqrt2#

There is a short way to find the radius of the incircle. It is given by the following formula.

#"Radius of Incircle" = frac{2 * "Area of "Delta}{"Perimeter of "Delta}#

The perimeter is #sqrt10 + sqrt26 + 3sqrt2#.

The area is found quickly using Heron's Formula.

The semi-perimeter is #frac{sqrt5 + sqrt13 + 3}{sqrt2}#

The area is

#sqrt{(frac{sqrt5 + sqrt13 + 3}{sqrt2}) * (frac{sqrt5 - sqrt13 + 3}{sqrt2}) * (frac{sqrt5 + sqrt13 - 3}{sqrt2}) * (frac{-sqrt5 + sqrt13 + 3}{sqrt2})}#

#=sqrt{(sqrt5 + sqrt13 + 3) * (sqrt5 - sqrt13 + 3) * (sqrt5 + sqrt13 - 3) * (-sqrt5 + sqrt13 + 3)}/2#

The radius is

#=sqrt{2* (sqrt5 - sqrt13 + 3) * (sqrt5 + sqrt13 - 3) * (-sqrt5 + sqrt13 + 3)}/{sqrt{sqrt5 + sqrt13 + 3}}#

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