A triangle has corners at #(3 , 5 )#, #(6 ,1 )#, and #(2 ,4 )#. What is the radius of the triangle's inscribed circle?

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Answer 1 · 2018-07-30T18:30:49

The radius of triangle's inscribed circle is:

#r=Delta/s=(3.5/(5+sqrt2/2))~~0.6133" units"#

Let , #triangleABC " # be the triangle with corners at

#A(3,5) , B(6,1) and C(2,4)#

#a=BC=sqrt((6-2)^2+(1-4)^2)=sqrt(16+9)=5#

#b=CA=sqrt((3-2)^2+(5-4)^2)=sqrt(1+1)=sqrt2#

#c=AB=sqrt((3-6)^2+(5-1)^2)=sqrt(9+16)=5#

So, the semiperimeter of triangle is :

#s=(a+b+c)/2=(5+sqrt2+5)/2=5+sqrt2/2~~5.7071#

#:. "The area of "triangleABC:#

#Delta=sqrt(s(s-a)(s-b)(s-c))#

#:.Delta=sqrt((5+sqrt2/2)(sqrt2/2)(5-sqrt2/2)(sqrt2/2))#

#Delta=sqrt([5^2-(sqrt2/2)^2][(sqrt2/2)^2]#

#=sqrt([25-1/2][1/2]#

#=sqrt(49/4)#

#:.Delta=7/2=3.5#

So , the radius of triangle's inscribed circle is:

#r=Delta/s=(3.5/(5+sqrt2/2))~~0.6133" units"#