In a triangle $ABC$, if $a:b:c = 4:5:6$, then $\frac{1}{4R}[r_1+r_2+r_3] =$

The perimeter of $\triangle ABC$ is $36 \text{ cm}$ and its inradius is $8 \text{ cm}$. Then,the area of the triangle is (in $\text{ cm}^2$)

In $\triangle ABC$,if $b=2, c=\sqrt{3}$,and $A=30^{\circ}$,then its inradius $r=$

In $\triangle ABC$,if $a=8, b=10, c=12$,then $\frac{r}{R}=$

In $\triangle ABC$,suppose the exradii opposite to angles $A, B$ and $C$ are denoted by $r_1, r_2$ and $r_3$ respectively. If $r_1=2, r_2=3, r_3=6$ and $R$ is the radius of the circumcircle,then the value of $r_1+r_2+r_3-r$ is:

If $a, b$ and $c$ are the sides of $\triangle ABC$ for which $r_1=8, r_2=12$ and $r_3=24$,then the ordered triad $(a, b, c)=$