Calculate Square of Inradius for Obtuse Triangle

To find the value of $r^2$, we proceed step-by-step using the properties of triangles.

Step 1: Determine the angle $\angle ACB$ (angle $C$)

The area of a triangle $\Delta$ with side lengths $a$ and $b$ and included angle $C$ is given by:
$\Delta = \frac{1}{2} a b \sin C$

Given $a = 6$, $b = 10$, and $\Delta = 15\sqrt{3}$, we substitute these values into the formula:
$15\sqrt{3} = \frac{1}{2} \cdot 6 \cdot 10 \cdot \sin C$
$15\sqrt{3} = 30 \sin C$
$\sin C = \frac{15\sqrt{3}}{30} = \frac{\sqrt{3}}{2}$

Since $\angle ACB = C$ is given to be an obtuse angle ($90^\circ < C < 180^\circ$), we have:
$C = 180^\circ - 60^\circ = 120^\circ$

Step 2: Calculate the side length $c$

Using the Law of Cosines:
$c^2 = a^2 + b^2 - 2ab \cos C$

Substitute $a = 6$, $b = 10$, and $C = 120^\circ$:
$\cos 120^\circ = -\frac{1}{2}$
$c^2 = 6^2 + 10^2 - 2(6)(10)\left(-\frac{1}{2}\right)$
$c^2 = 36 + 100 + 60 = 196$
$c = \sqrt{196} = 14$

Step 3: Calculate the semi-perimeter $s$

The semi-perimeter $s$ of the triangle is given by:
$s = \frac{a + b + c}{2} = \frac{6 + 10 + 14}{2} = \frac{30}{2} = 15$

Step 4: Calculate the inradius $r$ and $r^2$

The radius $r$ of the incircle is given by:
$r = \frac{\Delta}{s}$

Substituting the known values:
$r = \frac{15\sqrt{3}}{15} = \sqrt{3}$

Therefore, the square of the inradius is:
$r^2 = (\sqrt{3})^2 = 3$