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

Assuming a triangle has sides #a#, #b# and #c#, its area can be expressed in two ways.

1. Using Heron's formula for an area of a triangle:
   let #p=(a+b+c)/2#, then
   #S = sqrt(p(p-a)(p-b)(p-c))#

2. Area of a triangle equals to half of a product of its perimeter by a radius of an inscribed circle:
   let #p=(a+b+c)/2# and #r# is a radius of an inscribed circle, then
   #S=p*r#

The derivation of both formulas for area of a triangle can be found in the course of advanced mathematics at Unizor

These two expressions for an area must be equal, therefore
#sqrt(p(p-a)(p-b)(p-c)) = p*r#

From this we derive #r#:
#r=sqrt(((p-a)(p-b)(p-c))/p)#

For this particular problem we can get the lengths of all sides using the coordinates of the vertices:

if vertices are #A(2,9)#, #B(3,9)# and #C(4,8)#

#a=BC=sqrt((4-3)^2+(8-9)^2)=sqrt(2)#
#b=AC=sqrt((4-2)^2+(8-9)^2)=sqrt(5)#
#c=AB=sqrt((3-2)^2+(9-9)^2)=1#
#p=(a+b+c)/2=(sqrt(2)+sqrt(5)+1)/2#
#p-a=(-a+b+c)/2=(-sqrt(2)+sqrt(5)+1)/2#
#p-b=(a-b+c)/2=(sqrt(2)-sqrt(5)+1)/2#
#p-c=(a+b-c)/2=(sqrt(2)+sqrt(5)-1)/2#

Putting all these values into a formula above for a radius of an inscribed circle, we calculate the radius.
These calculations we leave to a student who submitted a problem.