ABCD is a rectangle . P and Q are two point on AB and AD s.t. area of `Delta APQ, Delta PBC and angle COQ ` are equal If BP = 2 cm. Find AP

Answer

To solve the problem, we need to find the value of \( AP \) given that \( BP = 2 \) cm and the areas of triangles \( \Delta APQ \), \( \Delta PBC \), and \( \Delta CDQ \) are equal.

### Step-by-Step Solution:

1. **Understanding the Geometry**:
- Let \( AB = l \) and \( AD = b \) be the lengths of the rectangle \( ABCD \).
- Point \( P \) is on side \( AB \) and point \( Q \) is on side \( AD \).
- Since \( BP = 2 \) cm, we can express \( AP \) as \( AP = l - 2 \).

2. **Setting Up the Areas**:
- The area of triangle \( APQ \) can be calculated using the formula for the area of a triangle:
\[ \text{Area}_{APQ} = \frac{1}{2} \times AP \times AQ = \frac{1}{2} \times (l - 2) \times (b - k) \]
- The area of triangle \( PBC \) is:
\[ \text{Area}_{PBC} = \frac{1}{2} \times BP \times BC = \frac{1}{2} \times 2 \times b = b \]
- The area of triangle \( CDQ \) is:
\[ \text{Area}_{CDQ} = \frac{1}{2} \times CD \times DQ = \frac{1}{2} \times l \times k \]

3. **Equating the Areas**:
- Since the areas are equal, we have:
\[ \frac{1}{2} (l - 2)(b - k) = b = \frac{1}{2} l k \]

4. **Solving for \( k \)**:
- From the equation \( b = \frac{1}{2} l k \), we can express \( k \) as:
\[ k = \frac{2b}{l} \]

5. **Substituting \( k \)**:
- Substitute \( k \) back into the area equation:
\[ \frac{1}{2} (l - 2) \left(b - \frac{2b}{l}\right) = b \]
- Simplifying gives:
\[ \frac{1}{2} (l - 2) \left(\frac{bl - 2b}{l}\right) = b \]
- This leads to:
\[ (l - 2)(bl - 2b) = 2bl \]

6. **Rearranging the Equation**:
- Expanding and rearranging:
\[ bl^2 - 2bl - 2b = 2bl \]
\[ bl^2 - 4bl - 2b = 0 \]
- Dividing through by \( b \) (assuming \( b \neq 0 \)):
\[ l^2 - 4l - 2 = 0 \]

7. **Using the Quadratic Formula**:
- Applying the quadratic formula \( l = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ l = \frac{4 \pm \sqrt{16 + 8}}{2} = \frac{4 \pm \sqrt{24}}{2} = 2 \pm \sqrt{6} \]
- We take the positive root since length cannot be negative:
\[ l = 2 + \sqrt{6} \]

8. **Finding \( AP \)**:
- Now substituting \( l \) back to find \( AP \):
\[ AP = l - 2 = (2 + \sqrt{6}) - 2 = \sqrt{6} \]

### Final Answer:
Thus, \( AP = \sqrt{6} \) cm.