Consider a triangle ABC and let a, b and c denote the lengths of the sides opposite to vertices A, B and C, respectivelu. Suppose `a = 6, b = 10` and the triangle is `15 sqrt3`. If `angle ACB` is obtus and if r denotes than radius of the incircle of the triangle, then the value of `r^(2)` is _____

Explanation

To solve the problem, we need to find the value of \( r^2 \) where \( r \) is the radius of the incircle of triangle \( ABC \) with given side lengths \( a = 6 \), \( b = 10 \), and area \( \Delta = 15\sqrt{3} \).

### Step-by-Step Solution:

1. **Identify the given values**:
- \( a = 6 \)
- \( b = 10 \)
- Area \( \Delta = 15\sqrt{3} \)

2. **Use the formula for the area of a triangle**:
The area \( \Delta \) can also be expressed in terms of two sides and the sine of the included angle:
\[ \Delta = \frac{1}{2}ab \sin C \]
Plugging in the known values:
\[ 15\sqrt{3} = \frac{1}{2} \cdot 6 \cdot 10 \cdot \sin C \]

3. **Solve for \( \sin C \)**:
\[ 15\sqrt{3} = 30 \sin C \]
\[ \sin C = \frac{15\sqrt{3}}{30} = \frac{\sqrt{3}}{2} \]
This implies that \( C = 120^\circ \) since angle \( C \) is obtuse.

4. **Find the length of side \( c \)** using the cosine rule:
\[ c^2 = a^2 + b^2 - 2ab \cos C \]
Here, \( \cos 120^\circ = -\frac{1}{2} \):
\[ c^2 = 6^2 + 10^2 - 2 \cdot 6 \cdot 10 \cdot \left(-\frac{1}{2}\right) \]
\[ c^2 = 36 + 100 + 60 = 196 \]
\[ c = \sqrt{196} = 14 \]

5. **Calculate the semi-perimeter \( s \)**:
\[ s = \frac{a + b + c}{2} = \frac{6 + 10 + 14}{2} = \frac{30}{2} = 15 \]

6. **Use the formula for the radius of the incircle**:
The radius \( r \) of the incircle is given by:
\[ r = \frac{\Delta}{s} \]
Substituting the known values:
\[ r = \frac{15\sqrt{3}}{15} = \sqrt{3} \]

7. **Calculate \( r^2 \)**:
\[ r^2 = (\sqrt{3})^2 = 3 \]

### Final Answer:
The value of \( r^2 \) is \( \boxed{3} \).