Question

A triangle has corners at `(5 , 5 )`, `(9 ,7 )`, and `(6 ,8 )`. What is the radius of the triangle's inscribed circle?

Answer 1

`r=10sqrt5(sqrt2-1)` unit

Explanation:

Let for `DeltaABC` the coordinates of vertices are

• `Ato(5,5)`

• `Bto(9,7)`

• `Cto(6,8)`

So sides are

`a=BC=sqrt((9-6)^2+(7-8)^2)=sqrt10`

`b=CA=sqrt((6-5)^2+(8-5)^2)=sqrt10`

`c=AB=sqrt((9-5)^2+(7-5)^2)=2sqrt5`

So it is obvious that

`a^2+b^2=c^2=20`

This means `DeltaABC` is a right isosceles triangle ,where `angleACB=90^@`.

So its area `Delta=1/2*a*b=50` squnit

If radius of in center of `DeltaABC` be `r` then

`1/2a*r+1/2b*r+1/2c*r=Delta`

`=>r=(2*Delta)/(a+b+c)`

`=(2*50)/(sqrt10+sqrt10+sqrt20)`

`=(2*50)/(2sqrt10+2sqrt5)`

`=(2*50)/(2sqrt5(sqrt2+1))`

`=(10sqrt5)/(sqrt2+1)`

`=10sqrt5(sqrt2-1)` unit