A triangle has corners at #(1 , 3 )#, #(4 ,8 )#, and #(9 ,7 )#. What is the radius of the triangle's inscribed circle?

1 Answer
Sep 21, 2017

Radius of incircle is #r=(2A)/(a+b+c)#
#A=# area of Triangle
#a,b,c# are side length of Triangle

Points of Traingle
#A(1,3), B(4,8), C(9,7)#
Side length #AB=c=sqrt((4-1)^2+(8-3)^2)=sqrt(9+25)=sqrt34#
Side length #BC=a=sqrt((9-4)^2+(7-8)^2)=sqrt(25+1)=sqrt26#
Side length #CA=b=sqrt((9-1)^2+(7-3)^2)=sqrt(64+16)=sqrt80#

Semi-perimeter of triangle #s=(a+b+c)/2=(sqrt34+sqrt26+sqrt80)/2#
Area of Triangle #A=sqrt(s(s-a)(s-b)(s-c))#
#A=sqrt(((sqrt34+sqrt26+sqrt80)/2)((sqrt34+sqrt26+sqrt80)/2-sqrt26)((sqrt34+sqrt26+sqrt80)/2-sqrt80)((sqrt34+sqrt26+sqrt80)/2-sqrt34)#

#A=sqrt(((sqrt34+sqrt26+sqrt80)/2)((sqrt34-sqrt26+sqrt80)/2)((sqrt34+sqrt26-sqrt80)/2)((-sqrt34+sqrt26+sqrt80)/2)#

#A=sqrt((((sqrt34+sqrt80)^2-(sqrt26)^2)/4)((-sqrt34^2-sqrt34sqrt26+sqrt80sqrt34+sqrt26sqrt34+sqrt26^2-sqrt26sqrt80+sqrt80sqrt34+sqrt80sqrt26-sqrt80^2)/2))#

#A=sqrt((((sqrt34+sqrt80)^2-(sqrt26)^2)/4)((-88+2sqrt80sqrt34)/2))#
#A=sqrt(((88+2sqrt34sqrt80)/4)((-88+2sqrt80sqrt34)/2))#
#A=sqrt(((-88^2+4(34)(80))/8)#
#A=sqrt(((-7744+10880)/8)#
#A=sqrt(((3136)/8))=sqrt(392)=14sqrt2#

Radius of in circle is #r=(2A)/(a+b+c)=(28sqrt2)/(sqrt34+sqrt26+sqrt80)#
#=(28)/(sqrt17+sqrt13+sqrt40)approx1.98#