In `Delta` ABC, AB = 6 cms, BC = 10 cms, AC = 8cm and `AD_I_BC`. Find the value of the ratio of `BD: DC`.

Options

A. `3:4`
B. `9:16 `
C. `4:5 `
D. `16:25 `

Answer

To solve the problem, we need to find the ratio of \( BD : DC \) in triangle \( ABC \) where \( AD \) is perpendicular to \( BC \).

### Step-by-Step Solution:

1. **Identify the Triangle and Given Values:**
- We have triangle \( ABC \) with sides:
- \( AB = 6 \) cm
- \( AC = 8 \) cm
- \( BC = 10 \) cm
- \( AD \) is perpendicular to \( BC \).

2. **Verify Right Triangle:**
- Check if triangle \( ABC \) is a right triangle using the Pythagorean theorem:
\[ AB^2 + AC^2 = 6^2 + 8^2 = 36 + 64 = 100 \]
\[ BC^2 = 10^2 = 100 \]
- Since \( AB^2 + AC^2 = BC^2 \), triangle \( ABC \) is a right triangle with \( \angle A = 90^\circ \).

3. **Use Cosine Rule to Find \( BD \):**
- In triangle \( ABD \), we can use the cosine of angle \( B \):
\[ \cos B = \frac{AB}{BC} = \frac{6}{10} = \frac{3}{5} \]
- In triangle \( ABD \), we can also express \( \cos B \) in terms of \( BD \):
\[ \cos B = \frac{BD}{AB} = \frac{BD}{6} \]
- Setting these equal gives:
\[ \frac{3}{5} = \frac{BD}{6} \]
- Cross-multiplying:
\[ 3 \cdot 6 = 5 \cdot BD \implies 18 = 5 \cdot BD \implies BD = \frac{18}{5} \text{ cm} \]

4. **Calculate \( DC \):**
- Since \( BC = 10 \) cm, we can find \( DC \):
\[ DC = BC - BD = 10 - \frac{18}{5} \]
- Convert \( 10 \) to a fraction:
\[ 10 = \frac{50}{5} \]
- Now subtract:
\[ DC = \frac{50}{5} - \frac{18}{5} = \frac{32}{5} \text{ cm} \]

5. **Find the Ratio \( BD : DC \):**
- Now we can find the ratio:
\[ \text{Ratio } BD : DC = \frac{BD}{DC} = \frac{\frac{18}{5}}{\frac{32}{5}} = \frac{18}{32} = \frac{9}{16} \]

### Final Answer:
The ratio \( BD : DC = 9 : 16 \).