In A `Delta ABC, ` AB = 10 AC = 18 AD is angle bisector If area of `Delta ABC` is 28 . find area of ADB ?

Answer

To find the area of triangle ADB given the triangle ABC, where AB = 10, AC = 18, AD is the angle bisector, and the area of triangle ABC is 28, we can follow these steps:

### Step-by-step Solution:

1. **Draw Triangle ABC**:
- Sketch triangle ABC with vertices A, B, and C.
- Mark AB = 10 and AC = 18.

2. **Identify the Angle Bisector**:
- Draw the angle bisector AD from vertex A to side BC, intersecting at point D.

3. **Use the Angle Bisector Theorem**:
- According to the angle bisector theorem, the ratio of the lengths of the two segments created on side BC (BD and DC) is equal to the ratio of the lengths of the other two sides (AB and AC).
- Therefore, \( \frac{AB}{AC} = \frac{BD}{DC} \).
- Substituting the known values:
\[ \frac{10}{18} = \frac{BD}{DC} \implies \frac{5}{9} = \frac{BD}{DC} \]
- Let BD = 5x and DC = 9x for some value x.

4. **Calculate the Total Length of BC**:
- The total length of BC is:
\[ BD + DC = 5x + 9x = 14x \]

5. **Determine the Areas of Triangles ADB and ADC**:
- The area of triangle ADB and triangle ADC will be in the same ratio as the lengths BD and DC.
- Therefore, the area of triangle ADB : area of triangle ADC = 5 : 9.

6. **Set Up the Area Equation**:
- Let the area of triangle ADB be \( 5k \) and the area of triangle ADC be \( 9k \).
- The total area of triangle ABC is given as 28:
\[ 5k + 9k = 28 \implies 14k = 28 \]

7. **Solve for k**:
- Dividing both sides by 14:
\[ k = 2 \]

8. **Find the Area of Triangle ADB**:
- Now, substitute k back to find the area of triangle ADB:
\[ \text{Area of ADB} = 5k = 5 \times 2 = 10 \text{ cm}^2 \]

### Final Answer:
The area of triangle ADB is **10 cm²**.