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

2 Answers
Jun 7, 2016

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

Explanation:

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

Jun 13, 2016

~=0.820

Explanation:

I created this figure using MS ExcelI 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