ABCD is a rectangle. There are two points P & Q on side AB and AD such that area of triangles PAQ, CDQ & PBC are equal. If the length of BP is 2 cm, find the length of AP.

Answer

To solve the problem, we will follow these steps:

### Step 1: Define the Variables
Let:
- \( AP = x \) (the length we need to find)
- \( BP = 2 \) cm (given)
- \( AQ = y \) (length of segment from A to Q)
- \( DQ = z \) (length of segment from D to Q)

Since \( AB \) is a rectangle, we know that:
- \( AB = AP + BP = x + 2 \)
- \( CD = AB = x + 2 \) (opposite sides of a rectangle are equal)
- \( BC = AQ + DQ = y + z \)

### Step 2: Area of the Triangles
The areas of the triangles are given to be equal:
1. Area of triangle \( PAQ \):
\[ \text{Area}_{PAQ} = \frac{1}{2} \times AP \times AQ = \frac{1}{2} \times x \times y \]
2. Area of triangle \( CDQ \):
\[ \text{Area}_{CDQ} = \frac{1}{2} \times DQ \times CD = \frac{1}{2} \times z \times (x + 2) \]
3. Area of triangle \( PBC \):
\[ \text{Area}_{PBC} = \frac{1}{2} \times BC \times PB = \frac{1}{2} \times (y + z) \times 2 = (y + z) \]

### Step 3: Set Up the Equations
Since the areas are equal, we can set up the following equations:
1. From \( \text{Area}_{PAQ} = \text{Area}_{CDQ} \):
\[ \frac{1}{2}xy = \frac{1}{2}z(x + 2) \implies xy = z(x + 2) \tag{1} \]
2. From \( \text{Area}_{CDQ} = \text{Area}_{PBC} \):
\[ \frac{1}{2}z(x + 2) = (y + z) \tag{2} \]

### Step 4: Solve the Equations
From equation (1):
\[ z = \frac{xy}{x + 2} \]

Substituting \( z \) into equation (2):
\[ \frac{xy}{x + 2}(x + 2) = y + \frac{xy}{x + 2} \]
This simplifies to:
\[ xy = y + \frac{xy}{x + 2} \]
Multiplying through by \( (x + 2) \):
\[ xy(x + 2) = y(x + 2) + xy \]
This simplifies to:
\[ xyx + 2xy = yx + 2y + xy \]
Rearranging gives:
\[ xyx + 2xy - yx - xy - 2y = 0 \]
This simplifies to:
\[ xyx + xy - yx - 2y = 0 \]
Factoring out \( y \):
\[ y(x - 2) + xy = 0 \]
Thus, we can express \( z \) in terms of \( y \):
\[ z = \frac{2y}{x} \]

### Step 5: Substitute Back
Substituting \( z \) back into equation (1):
\[ xy = \frac{2y}{x}(x + 2) \]
This simplifies to:
\[ x^2y = 2y + 4y \implies x^2y = 6y \]
Assuming \( y \neq 0 \):
\[ x^2 = 6 \implies x = \sqrt{6} \]

### Conclusion
Thus, the length of \( AP \) is \( \sqrt{6} \) cm.