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

In a $\triangle ABC$,which of the following formulae are correct?
$I. r = 4R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$
$II. r_1 = (s-a) \tan \frac{A}{2}$
$III. r_3 = \frac{\Delta}{s-c}$

If the vertices of a triangle are $(2, -2)$,$(8, -2)$,and $(8, 6)$,what is the excenter opposite to the vertex $(2, -2)$?

In a $\triangle ABC$,$a: b: c = 4: 5: 6$. The ratio of the radius of the circumcircle to that of the incircle is

Corresponding to a triangle $ABC$,match the items given in List-$I$ with the items given in List-$II$.

List-$I$	List-$II$
$(A)$ $rr_2 = r_1r_3$	$(I)$ $\angle A = 90^{\circ}$
$(B)$ $r_1 + r_2 = r_3 - r$	$(II)$ $b^2 = c^2 + a^2$
$(C)$ $r_1 = r + 2R$	$(III)$ $\angle C = 90^{\circ}$
	$(IV)$ $\angle B = 120^{\circ}$

The correct match is:

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: