The area off `triangleABC=12sqrt(3)`, side `a=6,` and side `b=9` side c=

Options

A. `2sqrt(37)`
B. `2sqrt(13)`
C. `2sqrt(37)` or `2sqrt(13)`
D. 10

Explanation

To find the side \( c \) of triangle \( ABC \) given the area \( 12\sqrt{3} \), side \( a = 6 \), and side \( b = 9 \), we can use Heron's formula.

### Step-by-Step Solution:

1. **Calculate the semi-perimeter \( s \)**:
\[ s = \frac{a + b + c}{2} = \frac{6 + 9 + c}{2} = \frac{15 + c}{2} \]

2. **Apply Heron's formula**:
The area \( A \) of the triangle can be expressed using Heron's formula:
\[ A = \sqrt{s(s-a)(s-b)(s-c)} \]
Given that \( A = 12\sqrt{3} \), we can substitute the values:
\[ 12\sqrt{3} = \sqrt{s(s-a)(s-b)(s-c)} \]

3. **Substituting the values of \( s-a \), \( s-b \), and \( s-c \)**:
\[ s - a = \frac{15 + c}{2} - 6 = \frac{3 + c}{2} \]
\[ s - b = \frac{15 + c}{2} - 9 = \frac{-3 + c}{2} \]
\[ s - c = \frac{15 + c}{2} - c = \frac{15 - c}{2} \]

4. **Substituting into Heron's formula**:
\[ 12\sqrt{3} = \sqrt{\left(\frac{15 + c}{2}\right) \left(\frac{3 + c}{2}\right) \left(\frac{-3 + c}{2}\right) \left(\frac{15 - c}{2}\right)} \]

5. **Squaring both sides**:
\[ (12\sqrt{3})^2 = \left(\frac{15 + c}{2}\right) \left(\frac{3 + c}{2}\right) \left(\frac{-3 + c}{2}\right) \left(\frac{15 - c}{2}\right) \]
\[ 432 = \frac{(15+c)(3+c)(-3+c)(15-c)}{16} \]

6. **Multiply both sides by 16**:
\[ 6912 = (15+c)(3+c)(-3+c)(15-c) \]

7. **Expanding the right-hand side**:
\[ (15+c)(15-c) = 225 - c^2 \]
\[ (3+c)(-3+c) = c^2 - 9 \]
Thus,
\[ 6912 = (225 - c^2)(c^2 - 9) \]

8. **Expanding the equation**:
\[ 6912 = 225c^2 - 2025 - c^4 + 9c^2 \]
\[ 0 = c^4 - 234c^2 + 8837 \]

9. **Letting \( y = c^2 \)**:
\[ y^2 - 234y + 8837 = 0 \]

10. **Using the quadratic formula**:
\[ y = \frac{234 \pm \sqrt{234^2 - 4 \cdot 1 \cdot 8837}}{2 \cdot 1} \]
\[ y = \frac{234 \pm \sqrt{54756 - 35348}}{2} \]
\[ y = \frac{234 \pm \sqrt{19408}}{2} \]

11. **Calculating the square root**:
\[ y = \frac{234 \pm 139.43}{2} \]
Thus,
\[ y_1 = \frac{373.43}{2} \approx 186.715 \quad \text{and} \quad y_2 = \frac{94.57}{2} \approx 47.285 \]

12. **Finding \( c \)**:
\[ c = \sqrt{y_1} \quad \text{or} \quad c = \sqrt{y_2} \]
We take the positive values:
\[ c \approx \sqrt{186.715} \quad \text{or} \quad c \approx \sqrt{47.285} \]

### Final Values:
- \( c \approx 13.65 \) or \( c \approx 6.87 \)