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:

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

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