Question

A triangle has corners at `(2 , 6 )`, `(3 ,9 )`, and `(4 ,8 )`. What is the radius of the triangle's inscribed circle?

Answer 1

`r=(sqrt8xxsqrt2)/(sqrt8+sqrt2+sqrt10)~~0.54`

Explanation:

Name the points `A(2,6), B(3,9) and C(4,8)`.

By the Distance Formula,

`AB^2=(3-2)^2+(9-6)^2=10, => AB=sqrt10`
`AC^2=(4-2)^2+(8-6)^2=8, => AC=sqrt8`
`BC^2=(4-3)^2+(8-9)^2=2, => BC=sqrt2`
`=> AB^2=AC^2+BC^2`
This means that `DeltaABC` is a right triangle, having hypotenuse `AB`.

The in-radius `r` of the inscribed circle of a triangle is : `r=(2A)/p`,
where `A and p` are the area and perimeter of the triangle.

For a right triangle, the in-radius `r` takes the simple form :
`r=(a*b)/(a+b+c)`
where `a, b` are sides and `c` is the hypotenuse.

Therefore, the in-radius `r` of `DeltaABC` is :
`r= (AC*BC)/(AC+BC+AB)`

`=(sqrt8*sqrt2)/(sqrt8+sqrt2+sqrt10)~~0.54`