In a `Delta ABC, AB=5 cm, AC=7 cm, BC=6cm`, If `AD _|_ BC` BC. Find the length of BD.

Answer

To find the length of \( BD \) in triangle \( ABC \) where \( AB = 5 \, \text{cm} \), \( AC = 7 \, \text{cm} \), and \( BC = 6 \, \text{cm} \) with \( AD \perp BC \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the triangle and the sides**:
- We have triangle \( ABC \) with sides \( AB = 5 \, \text{cm} \), \( AC = 7 \, \text{cm} \), and \( BC = 6 \, \text{cm} \).

2. **Calculate the semi-perimeter \( s \)**:
\[ s = \frac{AB + AC + BC}{2} = \frac{5 + 7 + 6}{2} = \frac{18}{2} = 9 \, \text{cm} \]

3. **Use Heron's formula to find the area of triangle \( ABC \)**:
\[ \text{Area} = \sqrt{s(s - AB)(s - AC)(s - BC)} \]
\[ = \sqrt{9(9 - 5)(9 - 7)(9 - 6)} = \sqrt{9 \times 4 \times 2 \times 3} \]
\[ = \sqrt{216} = 6\sqrt{6} \, \text{cm}^2 \]

4. **Set up the area using the base and height**:
- Here, \( BC \) is the base and \( AD \) is the height.
\[ \text{Area} = \frac{1}{2} \times BC \times AD \]
\[ 6\sqrt{6} = \frac{1}{2} \times 6 \times AD \]
\[ 6\sqrt{6} = 3 \times AD \]
\[ AD = 2\sqrt{6} \, \text{cm} \]

5. **Use the Pythagorean theorem in triangle \( ABD \)**:
- Since \( AD \perp BC \), we can apply the Pythagorean theorem.
\[ AB^2 = AD^2 + BD^2 \]
\[ 5^2 = (2\sqrt{6})^2 + BD^2 \]
\[ 25 = 24 + BD^2 \]
\[ BD^2 = 25 - 24 = 1 \]
\[ BD = \sqrt{1} = 1 \, \text{cm} \]

### Final Answer:
The length of \( BD \) is \( 1 \, \text{cm} \).