a and b are real numbers between 0 and 1 `A(a,1),B(1,b)and C(0,0)` are the vertices of a triangle.
If `DeltaABC` is equilateral its area is

Options

A. `(7sqrt3+12)/(4)`
B. `(8sqrt3-12)/(4)`
C. `(sqrt3)/(4)`
D. `(sqrt3)/(2)`

Answer

To find the area of the equilateral triangle formed by the points A(a, 1), B(1, b), and C(0, 0) where a and b are real numbers between 0 and 1, we can follow these steps:

### Step 1: Calculate the lengths of the sides of the triangle
1. **Length of AB**:
\[ AB = \sqrt{(a - 1)^2 + (1 - b)^2} \]

2. **Length of BC**:
\[ BC = \sqrt{(1 - 0)^2 + (b - 0)^2} = \sqrt{1 + b^2} \]

3. **Length of AC**:
\[ AC = \sqrt{(a - 0)^2 + (1 - 0)^2} = \sqrt{a^2 + 1} \]

### Step 2: Set the lengths equal for an equilateral triangle
Since triangle ABC is equilateral, we have:
\[ AB = BC = AC \]

### Step 3: Equate AB and BC
\[ \sqrt{(a - 1)^2 + (1 - b)^2} = \sqrt{1 + b^2} \]
Squaring both sides:
\[ (a - 1)^2 + (1 - b)^2 = 1 + b^2 \]
Expanding:
\[ (a^2 - 2a + 1) + (1 - 2b + b^2) = 1 + b^2 \]
This simplifies to:
\[ a^2 - 2a + 2 - 2b = 1 \]
Thus:
\[ a^2 - 2a - 2b + 1 = 0 \quad \text{(1)} \]

### Step 4: Equate BC and AC
\[ \sqrt{1 + b^2} = \sqrt{a^2 + 1} \]
Squaring both sides:
\[ 1 + b^2 = a^2 + 1 \]
This simplifies to:
\[ b^2 = a^2 \quad \Rightarrow \quad b = a \quad \text{(since } a, b \text{ are positive)} \]

### Step 5: Substitute b = a into equation (1)
Substituting \( b = a \) into equation (1):
\[ a^2 - 2a - 2a + 1 = 0 \]
This simplifies to:
\[ a^2 - 4a + 1 = 0 \]

### Step 6: Solve the quadratic equation
Using the quadratic formula:
\[ a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} \]
\[ = \frac{4 \pm \sqrt{16 - 4}}{2} = \frac{4 \pm \sqrt{12}}{2} = \frac{4 \pm 2\sqrt{3}}{2} = 2 \pm \sqrt{3} \]

### Step 7: Determine valid values for a
Since \( a \) must be between 0 and 1:
- \( 2 + \sqrt{3} \) is not valid (greater than 1).
- \( 2 - \sqrt{3} \) is valid (approximately 0.268).

### Step 8: Calculate the area of the triangle
Using the formula for the area of an equilateral triangle:
\[ \text{Area} = \frac{\sqrt{3}}{4} \times \text{side}^2 \]
Where the side length can be calculated as:
\[ AC = \sqrt{(2 - \sqrt{3})^2 + 1} = \sqrt{(2 - \sqrt{3})^2 + 1} = \sqrt{(4 - 4\sqrt{3} + 3 + 1)} = \sqrt{8 - 4\sqrt{3}} \]
Thus, the area becomes:
\[ \text{Area} = \frac{\sqrt{3}}{4} \times (8 - 4\sqrt{3}) \]

### Final Area Calculation
After simplifying:
\[ \text{Area} = \frac{\sqrt{3}}{4} \times (8 - 4\sqrt{3}) = \frac{8\sqrt{3}}{4} - \frac{12}{4} = 2\sqrt{3} - 3 \]

### Conclusion
The area of triangle ABC is:
\[ \text{Area} = 2\sqrt{3} - 3 \]