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

Question

3252 views around the world

You can reuse this answer
Creative Commons License

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#