Question

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

Answer 1

`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}}`