A triangle has corners at #(5 , 5 )#, #(9 ,7 )#, and #(6 ,8 )#. What is the radius of the triangle's inscribed circle?

Question

1360 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-02T11:29:12

#r=10sqrt5(sqrt2-1)# unit

Let for #DeltaABC# the coordinates of vertices are

So sides are

#a=BC=sqrt((9-6)^2+(7-8)^2)=sqrt10#

#b=CA=sqrt((6-5)^2+(8-5)^2)=sqrt10#

#c=AB=sqrt((9-5)^2+(7-5)^2)=2sqrt5#

So it is obvious that

#a^2+b^2=c^2=20#

This means #DeltaABC# is a right isosceles triangle ,where #angleACB=90^@#.

So its area #Delta=1/2*a*b=50# squnit

If radius of in center of #DeltaABC# be #r# then

#1/2a*r+1/2b*r+1/2c*r=Delta#

#=>r=(2*Delta)/(a+b+c)#

#=(2*50)/(sqrt10+sqrt10+sqrt20)#

#=(2*50)/(2sqrt10+2sqrt5)#

#=(2*50)/(2sqrt5(sqrt2+1))#

#=(10sqrt5)/(sqrt2+1)#

#=10sqrt5(sqrt2-1)# unit