A triangle has sides with lengths of 6, 4, and 3. What is the radius of the triangles inscribed circle?

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Answer 1 · 2016-06-07T23:55:44

#r=sqrt(((p-a)(p-b)(p-c))/p) # with #p = (a+b+c)/2#
#r = 0.820413#

From Heron's formula, the area of a triangle giving their sides #a,b,c# is

#A=sqrt(p(p-a)(p-b)(p-c))# with #p = (a+b+c)/2#

Let #o# be the triangle orthocenter, then the distance between each side and #o# is #r# which is the inscribed circle radius. So

#A = (a xx r)/2+(b xx r)/2 + (c xx r)/2# then
#r((a+b+c)/2)=A = r xx p#

Finally

#r = (sqrt(p(p-a)(p-b)(p-c)))/p=sqrt(((p-a)(p-b)(p-c))/p) = 0.820413#

Answer 2 · 2016-06-13T21:39:14

#~=0.820#

I created this figure using MS Excel

Suppose
#AB=6#
#BC=4#
#CA=2#

#m+l=6# [1]
#l+n=4# [2]
#m+n=3# [3]

[3]-[2]
#m-l=-1# [4]

[4]+[1]
#2m=5# => #m=5/2#
#-> l=7/2#
#-> n=1/2#

Applying law of cosines:
#AB^2=BC^2+CA^2-2*BC*CA*cos(A hat C B)#
#36=16+9-2*4*3cos(AhatCB)#
#24cos(AhatCB)=-11# => #AhatCB~=117.28^@#

#tan((AhatCB)/2)=r/n#
#r=1/2*tan(117.28^@/2)=0.820#