Consider a `DeltaOAB` formed by the point `O(0,0),A(2,0),B(1,sqrt(3)).P(x,y)` is an arbitrary interior point of triangle moving in such a way that `d(P,OA)+d(P,AB)+d(P,OB)=sqrt(3),` where `d(P,OA),d(P,AB),d(P,OB)` represent the distance of P from the sides OA,AB and OB respectively
Area of region reperesenting all possible position of point P is equal to

Options

A. `2sqrt(3)`
B. `sqrt(6)`
C. `sqrt(3)`
D. None of these

Answer

To solve the problem, we need to find the area of the region representing all possible positions of point \( P(x,y) \) inside triangle \( OAB \) where the distances from \( P \) to the sides \( OA \), \( AB \), and \( OB \) sum up to \( \sqrt{3} \).

### Step 1: Identify the vertices of triangle \( OAB \)
The vertices of triangle \( OAB \) are:
- \( O(0,0) \)
- \( A(2,0) \)
- \( B(1,\sqrt{3}) \)

### Step 2: Find the equations of the sides of the triangle
1. **Line \( OA \)**: This is the line along the x-axis, so its equation is:
\[ y = 0 \]

2. **Line \( OB \)**: The slope \( m \) of line \( OB \) from \( O(0,0) \) to \( B(1,\sqrt{3}) \) is:
\[ m = \frac{\sqrt{3} - 0}{1 - 0} = \sqrt{3} \]
Therefore, the equation of line \( OB \) is:
\[ y = \sqrt{3}x \]

3. **Line \( AB \)**: The slope \( m \) of line \( AB \) from \( A(2,0) \) to \( B(1,\sqrt{3}) \) is:
\[ m = \frac{\sqrt{3} - 0}{1 - 2} = -\sqrt{3} \]
Using point-slope form from point \( A(2,0) \):
\[ y - 0 = -\sqrt{3}(x - 2) \implies y = -\sqrt{3}x + 2\sqrt{3} \]

### Step 3: Calculate the distances from point \( P(x,y) \) to the sides
1. **Distance from \( P \) to line \( OA \)**:
\[ d(P, OA) = y \]

2. **Distance from \( P \) to line \( OB \)** (using the formula for distance from a point to a line):
\[ d(P, OB) = \frac{|-\sqrt{3}x + y|}{\sqrt{(\sqrt{3})^2 + (-1)^2}} = \frac{|-\sqrt{3}x + y|}{2} \]

3. **Distance from \( P \) to line \( AB \)**:
\[ d(P, AB) = \frac{|-\sqrt{3}x + y - 2\sqrt{3}|}{\sqrt{(-\sqrt{3})^2 + 1^2}} = \frac{|- \sqrt{3}x + y + 2\sqrt{3}|}{2} \]

### Step 4: Set up the equation based on the given condition
According to the problem:
\[ d(P, OA) + d(P, OB) + d(P, AB) = \sqrt{3} \]
Substituting the distances:
\[ y + \frac{|- \sqrt{3}x + y|}{2} + \frac{|- \sqrt{3}x + y + 2\sqrt{3}|}{2} = \sqrt{3} \]

### Step 5: Simplify the equation
Multiply through by 2 to eliminate the fractions:
\[ 2y + |-\sqrt{3}x + y| + |-\sqrt{3}x + y + 2\sqrt{3}| = 2\sqrt{3} \]

### Step 6: Analyze the absolute values
To analyze the absolute values, we consider different cases based on the signs of the expressions inside the absolute values. This will yield different linear inequalities that describe the region where \( P \) can lie.

### Step 7: Find the area of the region
The area can be found by integrating the upper boundary of the region defined by the inequalities derived from the absolute values. The limits of integration will be determined by the intersection points of the lines.

### Step 8: Calculate the area using integration
The area \( A \) can be calculated as:
\[ A = \int_{0}^{2} (-\sqrt{3}x + 2\sqrt{3}) \, dx \]
Evaluating this integral:
\[ A = \left[-\frac{\sqrt{3}}{2}x^2 + 2\sqrt{3}x\right]_{0}^{2} = \left[-\frac{\sqrt{3}}{2}(4) + 4\sqrt{3}\right] = -2\sqrt{3} + 4\sqrt{3} = 2\sqrt{3} \]

### Final Answer
The area of the region representing all possible positions of point \( P \) is:
\[ \boxed{2\sqrt{3}} \text{ square units} \]