In a `DeltaABC`, `a=6, b=10,` area of `Delta=15sqrt3` sq units, if `/_ACB` is obtuse and `r` denotes the radius of inscribed circle then value of `r^2=`

Options

A. 4
B. 5
C. 6
D. 3

Answer

To solve the problem step by step, we will start with the given information and apply the appropriate formulas.

### Step 1: Given Information
We are given:
- \( a = 6 \)
- \( b = 10 \)
- Area of triangle \( \Delta = 15\sqrt{3} \) square units
- Angle \( \angle ACB \) is obtuse.

### Step 2: Use the Area Formula
The area of a triangle can be expressed using the formula:
\[ \Delta = \frac{1}{2} \times AC \times BC \times \sin(\angle ACB) \]
Here, we can denote:
- \( AC = b = 10 \)
- \( BC = a = 6 \)

Substituting the values into the area formula:
\[ 15\sqrt{3} = \frac{1}{2} \times 10 \times 6 \times \sin(\angle ACB) \]
This simplifies to:
\[ 15\sqrt{3} = 30 \sin(\angle ACB) \]

### Step 3: Solve for \( \sin(\angle ACB) \)
Dividing both sides by 30:
\[ \sin(\angle ACB) = \frac{15\sqrt{3}}{30} = \frac{\sqrt{3}}{2} \]

### Step 4: Determine Angle \( C \)
Since \( \angle ACB \) is obtuse, we find:
\[ \angle ACB = 120^\circ \]

### Step 5: Use the Cosine Rule
Using the cosine rule to find side \( c \):
\[ c^2 = a^2 + b^2 - 2ab \cos(C) \]
Substituting the known values:
\[ c^2 = 6^2 + 10^2 - 2 \times 6 \times 10 \times \cos(120^\circ) \]
Since \( \cos(120^\circ) = -\frac{1}{2} \):
\[ c^2 = 36 + 100 + 60 \]
\[ c^2 = 196 \]
Thus, \( c = \sqrt{196} = 14 \).

### Step 6: Calculate the Semi-Perimeter \( s \)
The semi-perimeter \( s \) is given by:
\[ s = \frac{a + b + c}{2} = \frac{6 + 10 + 14}{2} = \frac{30}{2} = 15 \]

### Step 7: Find the Radius of the Inscribed Circle \( r \)
The radius \( r \) of the inscribed circle is given by:
\[ r = \frac{\Delta}{s} \]
Substituting the values:
\[ r = \frac{15\sqrt{3}}{15} = \sqrt{3} \]

### Step 8: Calculate \( r^2 \)
Finally, squaring \( r \):
\[ r^2 = (\sqrt{3})^2 = 3 \]

### Final Answer
Thus, the value of \( r^2 \) is:
\[ \boxed{3} \]