A triangle has corners at (5 , 8 )(5,8), (2 ,3 )(2,3), and (3 ,1 )(3,1). What is the radius of the triangle's inscribed circle?

1 Answer
Jun 15, 2016

r= 0.716r=0.716

Explanation:

Given the coordinates of three vertices of DeltaABC are

"Corodinate of A " x_A=5,y_A=8

"Corodinate of B " x_B=2,y_B=8

"Corodinate of C " x_C=3,y_C=1

Lengths of three sides

AB=sqrt((x_A-x_B)^2+(y_A-y_B)^2)

=sqrt((5-2)^2+(8-3)^2)=sqrt34

BC=sqrt((x_B-x_C)^2+(y_B-y_C)^2)

=sqrt((2-3)^2+(3-1)^2)=sqrt5

CA=sqrt((x_C-x_A)^2+(y_C-y_A)^2)

=sqrt((3-5)^2+(1-8)^2)=sqrt53

Now area of DeltaABC

=|1/2(y_A(x_B-x_C)+y_B(x_C-x_A)+y_C(x_A-x_B))|

=|1/2(8(2-3)+3(3-5)+1(5-2))|

=|1/2(-8-6+3)|=5.5

If the radius of the incircle of DeltaABC " "be" "r then

1/2(AB+BC+CA)xxr="area "DeltaABC

=>1/2(sqrt34+sqrt5+sqrt53)*r=5.5

=>r=(2xx5.5)/(sqrt34+sqrt5+sqrt53)

:.r= 0.716