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

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