Three circles of radii 4 cm, 6 cm and 8 cm touch other pairwise externally. The area of the triangle formed by the line segments joining the centres of the three circle is

Options

A. 144√13 sq. cm
B. 12√105 sq. cm
C. 6√6 sq. cm
D. 24√6 sq. cm

Answer

To find the area of the triangle formed by the centers of three circles with radii 4 cm, 6 cm, and 8 cm that touch each other externally, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the lengths of the sides of the triangle:**
- The distance between the centers of the first circle (radius 4 cm) and the second circle (radius 6 cm) is:
\[ 4 + 6 = 10 \text{ cm} \]
- The distance between the centers of the second circle (radius 6 cm) and the third circle (radius 8 cm) is:
\[ 6 + 8 = 14 \text{ cm} \]
- The distance between the centers of the first circle (radius 4 cm) and the third circle (radius 8 cm) is:
\[ 4 + 8 = 12 \text{ cm} \]

Thus, the sides of the triangle are:
- \( a = 10 \) cm
- \( b = 14 \) cm
- \( c = 12 \) cm

2. **Calculate the semi-perimeter (s) of the triangle:**
\[ s = \frac{a + b + c}{2} = \frac{10 + 14 + 12}{2} = \frac{36}{2} = 18 \text{ cm} \]

3. **Use Heron's formula to find the area (A) of the triangle:**
Heron's formula states:
\[ A = \sqrt{s(s-a)(s-b)(s-c)} \]
Substituting the values we have:
\[ A = \sqrt{18(18-10)(18-14)(18-12)} \]
\[ A = \sqrt{18 \times 8 \times 4 \times 6} \]

4. **Simplify the expression under the square root:**
\[ A = \sqrt{18 \times 8 \times 4 \times 6} = \sqrt{18 \times 192} \]
We can break this down:
\[ 18 = 9 \times 2 \quad \text{and} \quad 192 = 64 \times 3 = 8^2 \times 3 \]
So,
\[ A = \sqrt{(9 \times 2) \times (64 \times 3)} = \sqrt{576 \times 6} = \sqrt{3456} \]

5. **Calculate the square root:**
\[ A = \sqrt{576} \times \sqrt{6} = 24\sqrt{6} \]

### Final Answer:
The area of the triangle formed by the line segments joining the centers of the three circles is:
\[ \boxed{24\sqrt{6}} \text{ cm}^2 \]