Consider the traingle having vertices `O(0,0),A(2,0)`, and `B(1,sqrt3)`. Also `b le"min" {a_1,a_2,a_3....a_n}` means ` b le a_1` when `a_1` is least, `b le a_2` when `a_2` is least, and so on. Form this, we can say `b le a_1,b le a_2,.....b le a_n`.
Let R be the region consisting of all those points P inside `DeltaOAB` which satisfy `d(P,OA)le "min"[d(P,OB),d(P,AB)]`, where d denotes the distance from the point to the corresponding line. then the area of the region R is

Options

A. `sqrt(3)`sq,units
B. `(2+sqrt3)` sq.units
C. `sqrt(3)//2` sq.units
D. `1//sqrt(3)` sq.units

Explanation

To find the area of the region \( R \) inside triangle \( OAB \) that satisfies the given distance conditions, we will follow these steps:

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

### Step 2: Determine the equations of the lines
We need to find the equations of the lines \( OA \), \( OB \), and \( AB \).

1. **Line OA** (from \( O \) to \( A \)):
- The slope \( m_{OA} = \frac{0 - 0}{2 - 0} = 0 \)
- The equation is \( y = 0 \).

2. **Line OB** (from \( O \) to \( B \)):
- The slope \( m_{OB} = \frac{\sqrt{3} - 0}{1 - 0} = \sqrt{3} \)
- The equation is \( y = \sqrt{3}x \).

3. **Line AB** (from \( A \) to \( B \)):
- The slope \( m_{AB} = \frac{\sqrt{3} - 0}{1 - 2} = -\sqrt{3} \)
- The equation is \( y - 0 = -\sqrt{3}(x - 2) \) or \( y = -\sqrt{3}x + 2\sqrt{3} \).

### Step 3: Determine the distances from a point \( P(x, y) \) to the lines
The distances from a point \( P(x, y) \) to the lines \( OA \), \( OB \), and \( AB \) are given by:

1. **Distance to line OA**:
\[ d(P, OA) = |y| \]

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

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

### Step 4: Set up the inequalities
We need to satisfy the condition:
\[ d(P, OA) \leq \min(d(P, OB), d(P, AB)) \]
This implies:
1. \( |y| \leq \frac{|y - \sqrt{3}x|}{2} \)
2. \( |y| \leq \frac{|y + \sqrt{3}x - 2\sqrt{3}|}{2} \)

### Step 5: Solve the inequalities
1. From \( |y| \leq \frac{|y - \sqrt{3}x|}{2} \):
- This leads to two cases: \( y \leq \frac{y - \sqrt{3}x}{2} \) and \( -y \leq \frac{y - \sqrt{3}x}{2} \).

2. From \( |y| \leq \frac{|y + \sqrt{3}x - 2\sqrt{3}|}{2} \):
- This leads to two cases: \( y \leq \frac{y + \sqrt{3}x - 2\sqrt{3}}{2} \) and \( -y \leq \frac{y + \sqrt{3}x - 2\sqrt{3}}{2} \).

### Step 6: Analyze the region defined by the inequalities
The region defined by these inequalities will be bounded by the lines and will form a smaller triangle within triangle \( OAB \).

### Step 7: Calculate the area of the region \( R \)
The area of triangle \( OAB \) can be calculated using the formula:
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \]
Here, the base \( OA = 2 \) and the height from \( B \) to line \( OA \) is \( \sqrt{3} \):
\[ \text{Area}_{OAB} = \frac{1}{2} \times 2 \times \sqrt{3} = \sqrt{3} \]

The region \( R \) is a triangle with vertices at the centroid of triangle \( OAB \) and the midpoints of sides \( OA \) and \( OB \). The area of \( R \) is thus:
\[ \text{Area}_{R} = \frac{1}{3} \text{Area}_{OAB} = \frac{1}{3} \sqrt{3} \]

### Final Answer
The area of the region \( R \) is:
\[ \frac{1}{3} \sqrt{3} \]