Question

A triangle has corners at `(2 , 6 )`, `(3 ,8 )`, and `(4 ,5 )`. What is the radius of the triangle's inscribed circle?

Answer 1

See below.

Explanation:

Solution 1:
By the formula `A=rs`, where `A` is the area of the triangle, `r` is the radius of the inscribed circle, and `s` is the semi-perimeter of the triangle, if we know the area and the semi-perimeter, we can find the radius.

To find the area, we use Shoelace.

`(2,6)`
`(3,8)`
`(4,5)`
`(2,6)`
Cross-multiply,
`=frac(abs((2*8+3*5+4*6)-(6*3+8*4+5*2)))(2)=frac(5)(2)`

To find the semi-perimeter, we use the distance formula three times (on each side).

`d=sqrt((x_1-x_0)^2+(y_1-y_0)^2)`

So the semiperimeter is:

`frac(sqrt(5)+sqrt(5)+sqrt(10))(2)=frac(2sqrt(5)+sqrt(10))(2)`

Thus, `frac(5)(2)=r(frac(2sqrt(5)+sqrt(10))(2))`

Multiply both sides by `frac(2sqrt(5))(5)`,

`sqrt(5)=(2+sqrt(2))r`

By rationalizing (multiply both sides by `2-sqrt(2)`),

We get that `frac(2sqrt(5)-sqrt(10))(2)=r`

Solution 2:
We recognize that these points form an isoceles right triangle.

Let's name the point where the perpendicular from `I` to `AC` point `D`.

We first note that:
`BC=AC=sqrt(5)`

The angle bisector through C (which goes through the center of the circle) is also the median of the hypotenuse. An by the Two-Tangent theorem, `AD` is then also half of the hypotenuse, or `frac(sqrt(10))(2)`.

Designating the radius as `r`,

`AC=CD+DA=r+frac(sqrt(10))(2)`

but `AC=sqrt(5)`

So,

`r=sqrt(5)-frac(sqrt(10))(2)=frac(2sqrt(5)-sqrt(10))(2)`