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

1 Answer
Bio
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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