In a `Delta ABC, AB=10 cm, AC=15, AD ` is angle bisector . If the area of `Delta ABD=18cm^(2)`. Find area of `Delta ABC`

Answer

To find the area of triangle ABC given the area of triangle ABD, we can follow these steps:

### Step 1: Understand the Given Information
We have triangle ABC with:
- AB = 10 cm
- AC = 15 cm
- Area of triangle ABD = 18 cm²
- AD is the angle bisector of angle A.

### Step 2: Use the Angle Bisector Theorem
According to the angle bisector theorem, the angle bisector divides the opposite side (BC) into segments that are proportional to the other two sides. Thus, we have:
\[ \frac{BD}{CD} = \frac{AB}{AC} = \frac{10}{15} = \frac{2}{3} \]

### Step 3: Assign Variables to Segments
Let:
- BD = 2x
- CD = 3x

Then, the total length of BC is:
\[ BC = BD + CD = 2x + 3x = 5x \]

### Step 4: Relate the Areas of Triangles ABD and ABC
The area of a triangle can be expressed as:
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \]
Since both triangles ABD and ABC share the same height (let's call it H), we can express their areas as:
- Area of triangle ABD = \(\frac{1}{2} \times BD \times H = \frac{1}{2} \times 2x \times H = xH\)
- Area of triangle ABC = \(\frac{1}{2} \times BC \times H = \frac{1}{2} \times 5x \times H = \frac{5}{2} xH\)

### Step 5: Set Up the Ratio of Areas
The ratio of the areas of triangles ABD and ABC can be expressed as:
\[ \frac{\text{Area of } \triangle ABD}{\text{Area of } \triangle ABC} = \frac{xH}{\frac{5}{2} xH} = \frac{2}{5} \]

### Step 6: Use the Given Area of Triangle ABD
We know that:
\[ \frac{18}{\text{Area of } \triangle ABC} = \frac{2}{5} \]

### Step 7: Cross Multiply to Solve for Area of Triangle ABC
Cross multiplying gives us:
\[ 2 \times \text{Area of } \triangle ABC = 18 \times 5 \]
\[ 2 \times \text{Area of } \triangle ABC = 90 \]
\[ \text{Area of } \triangle ABC = \frac{90}{2} = 45 \text{ cm}^2 \]

### Final Answer
The area of triangle ABC is **45 cm²**.

---