In a triangle ABC with altitude AD, `angleBAC=45^(@), DB=3 and CD=2`. The area of the triangle ABC is :

Options

A. 6
B. 15
C. `15//4`
D. 12

Answer

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

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Angle \( \angle BAC = 45^\circ \)
- \( DB = 3 \)
- \( CD = 2 \)

2. **Calculate the Length of Base BC:**
- Since \( DB + CD = BC \), we have:
\[ BC = DB + CD = 3 + 2 = 5 \]

3. **Set Up the Triangle:**
- Let \( AD \) be the altitude from point \( A \) to line \( BC \).
- Since \( \angle BAC = 45^\circ \), we can denote \( \angle BAD = \alpha \) and \( \angle CAD = \beta \) such that \( \alpha + \beta = 45^\circ \).

4. **Use the Tangent Function:**
- From the triangle, we know:
\[ \tan \alpha = \frac{AD}{DB} = \frac{AD}{3} \]
\[ \tan \beta = \frac{AD}{CD} = \frac{AD}{2} \]

5. **Apply the Tangent Addition Formula:**
- Using the tangent addition formula:
\[ \tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} \]
- Since \( \tan(45^\circ) = 1 \):
\[ 1 = \frac{\frac{AD}{3} + \frac{AD}{2}}{1 - \left(\frac{AD}{3}\right)\left(\frac{AD}{2}\right)} \]

6. **Multiply Through by the Denominator:**
- This gives:
\[ 1 - \frac{AD^2}{6} = \frac{AD}{3} + \frac{AD}{2} \]
- Finding a common denominator for the right side:
\[ 1 - \frac{AD^2}{6} = \frac{2AD + 3AD}{6} = \frac{5AD}{6} \]

7. **Rearranging the Equation:**
- Rearranging gives:
\[ 6 - AD^2 = 5AD \]
- Rearranging further leads to:
\[ AD^2 + 5AD - 6 = 0 \]

8. **Solve the Quadratic Equation:**
- Using the quadratic formula \( AD = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ AD = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 1 \cdot (-6)}}{2 \cdot 1} \]
\[ AD = \frac{-5 \pm \sqrt{25 + 24}}{2} \]
\[ AD = \frac{-5 \pm \sqrt{49}}{2} \]
\[ AD = \frac{-5 \pm 7}{2} \]
- This gives us two potential solutions:
\[ AD = 1 \quad \text{(not valid since height cannot be negative)} \]
\[ AD = 6 \quad \text{(valid)} \]

9. **Calculate the Area of Triangle ABC:**
- The area \( A \) of triangle ABC is given by:
\[ A = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times BC \times AD \]
- Substituting the values:
\[ A = \frac{1}{2} \times 5 \times 6 = \frac{30}{2} = 15 \]

### Final Answer:
The area of triangle ABC is \( 15 \) square units.

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