The co-ordinates of the vertices of a `DeltaABC` are `A(1, 0) , B(3, 6) and C(3, 2)`. Find the length of its medians.

Answer

To find the lengths of the medians of triangle ABC with vertices A(1, 0), B(3, 6), and C(3, 2), we will follow these steps:

### Step 1: Determine the Midpoints of the Sides
The medians of a triangle connect a vertex to the midpoint of the opposite side. We need to find the midpoints of sides BC, AC, and AB.

1. **Midpoint D of BC**:
\[ D = \left( \frac{x_B + x_C}{2}, \frac{y_B + y_C}{2} \right) = \left( \frac{3 + 3}{2}, \frac{6 + 2}{2} \right) = \left( \frac{6}{2}, \frac{8}{2} \right) = (3, 4) \]

2. **Midpoint E of AC**:
\[ E = \left( \frac{x_A + x_C}{2}, \frac{y_A + y_C}{2} \right) = \left( \frac{1 + 3}{2}, \frac{0 + 2}{2} \right) = \left( \frac{4}{2}, \frac{2}{2} \right) = (2, 1) \]

3. **Midpoint F of AB**:
\[ F = \left( \frac{x_A + x_B}{2}, \frac{y_A + y_B}{2} \right) = \left( \frac{1 + 3}{2}, \frac{0 + 6}{2} \right) = \left( \frac{4}{2}, \frac{6}{2} \right) = (2, 3) \]

### Step 2: Calculate the Lengths of the Medians
Now we will calculate the lengths of the medians AD, BE, and CF using the distance formula:
\[ \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

1. **Length of Median AD**:
\[ AD = \sqrt{(3 - 1)^2 + (4 - 0)^2} = \sqrt{2^2 + 4^2} = \sqrt{4 + 16} = \sqrt{20} \]

2. **Length of Median BE**:
\[ BE = \sqrt{(2 - 3)^2 + (1 - 6)^2} = \sqrt{(-1)^2 + (-5)^2} = \sqrt{1 + 25} = \sqrt{26} \]

3. **Length of Median CF**:
\[ CF = \sqrt{(2 - 3)^2 + (3 - 2)^2} = \sqrt{(-1)^2 + (1)^2} = \sqrt{1 + 1} = \sqrt{2} \]

### Final Answer
The lengths of the medians of triangle ABC are:
- Median AD: \(\sqrt{20}\)
- Median BE: \(\sqrt{26}\)
- Median CF: \(\sqrt{2}\)