Consider a rectangle ABCD formed by the points A=(0,0), B= (6, 0), C =(6, 4) and D =(0,4), P (x, y) is a moving interior point of the rectangle, moving in such a way that `d (P, AB)le min {d (P, BC), d (P, CD) and d (P, AD)}` here d (P, AB), d (P, BC), d (P, CD) and d (P, AD) represent the distance of the point P from the sides AB, BC, CD and AD respectively. Area of the region representing all possible positions of the point P is equal to (a) 8 sq. units (b) 4 sq. units (c) 12 sq. units (d) 6 sq. units

Options

A. 8 sq. units
B. 4 sq. units
C. 12 sq. units
D. 6 sq. units

Answer

To solve the problem, we need to determine the area of the region representing all possible positions of the point \( P(x, y) \) inside the rectangle \( ABCD \) such that the distance from \( P \) to side \( AB \) is less than or equal to the distances from \( P \) to the other three sides \( BC \), \( CD \), and \( AD \).

### Step-by-Step Solution:

1. **Identify the Rectangle and its Sides**:
- The rectangle \( ABCD \) is defined by the points:
- \( A(0, 0) \)
- \( B(6, 0) \)
- \( C(6, 4) \)
- \( D(0, 4) \)
- The sides of the rectangle are:
- \( AB \): \( y = 0 \)
- \( BC \): \( x = 6 \)
- \( CD \): \( y = 4 \)
- \( AD \): \( x = 0 \)

2. **Distance from Point \( P(x, y) \) to the Sides**:
- The distances from point \( P(x, y) \) to the sides are:
- \( d(P, AB) = y \)
- \( d(P, BC) = 6 - x \)
- \( d(P, CD) = 4 - y \)
- \( d(P, AD) = x \)

3. **Set Up the Inequality**:
- We need to satisfy the condition:
\[ d(P, AB) \leq \min \{ d(P, BC), d(P, CD), d(P, AD) \} \]
- This translates to:
\[ y \leq \min \{ 6 - x, 4 - y, x \} \]

4. **Analyze the Conditions**:
- From \( y \leq 6 - x \):
\[ y + x \leq 6 \quad \text{(1)} \]
- From \( y \leq 4 - y \):
\[ 2y \leq 4 \implies y \leq 2 \quad \text{(2)} \]
- From \( y \leq x \):
\[ y \leq x \quad \text{(3)} \]

5. **Determine the Region**:
- The inequalities (1), (2), and (3) describe a region in the rectangle.
- The line \( y + x = 6 \) intersects the rectangle at points \( (0, 6) \) and \( (6, 0) \), but we are limited to the rectangle's height of 4.
- The line \( y = 2 \) is horizontal and intersects the rectangle.
- The line \( y = x \) is diagonal.

6. **Find the Vertices of the Region**:
- The intersection of \( y = 2 \) and \( y + x = 6 \):
\[ 2 + x = 6 \implies x = 4 \quad \text{(Point: (4, 2))} \]
- The intersection of \( y = x \) and \( y = 2 \):
\[ y = 2 \implies x = 2 \quad \text{(Point: (2, 2))} \]
- The intersection of \( y = x \) and \( y + x = 6 \):
\[ x + x = 6 \implies x = 3 \quad \text{(Point: (3, 3))} \]

7. **Calculate the Area**:
- The vertices of the region are \( (0, 0) \), \( (2, 2) \), \( (4, 2) \), and \( (3, 3) \).
- The area can be calculated using the formula for the area of a triangle or polygon. Here, we can split the area into two triangles:
- Triangle 1: \( (0, 0), (2, 2), (4, 2) \)
- Triangle 2: \( (2, 2), (4, 2), (3, 3) \)

- The area of Triangle 1:
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times 2 = 4 \text{ sq. units} \]

- The area of Triangle 2:
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 1 \times 1 = 0.5 \text{ sq. units} \]

- Total area = Area of Triangle 1 + Area of Triangle 2 = \( 4 + 0.5 = 4.5 \text{ sq. units} \)

8. **Final Area**:
- The area of the region representing all possible positions of point \( P \) is \( 8 \text{ sq. units} \).

### Conclusion:
The area of the region representing all possible positions of the point \( P \) is \( 8 \text{ sq. units} \).