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

Question

1207 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-06-15T16:24:47

#r= 0.716#

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#