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 isosceles with `AC=BC and 5(AB)^(2)=2(AC)^(2)` then

Options

A. `ab=1/4`
B. `ab=1/8`
C. `ab=1/16`
D. `ab=1/2`

Answer

To solve the problem, we need to find the relationship between the variables \( a \) and \( b \) given the conditions of the triangle \( \Delta ABC \) with vertices \( A(a, 1) \), \( B(1, b) \), and \( C(0, 0) \). The triangle is isosceles with \( AC = BC \) and \( 5(AB)^2 = 2(AC)^2 \).

### Step 1: Calculate the distances \( AC \) and \( BC \)

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

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

### Step 2: Set up the equation for isosceles triangle condition \( AC = BC \)

Since \( AC = BC \):
\[ \sqrt{a^2 + 1} = \sqrt{1 + b^2} \]
Squaring both sides:
\[ a^2 + 1 = 1 + b^2 \]
This simplifies to:
\[ a^2 = b^2 \]

### Step 3: Solve for \( a \) and \( b \)

Taking the square root of both sides gives:
\[ a = b \quad \text{(since both are between 0 and 1)} \]

### Step 4: Calculate \( AB \) and \( AC \)

1. **Distance \( AB \)**:
\[ AB = \sqrt{(a - 1)^2 + (1 - b)^2} \]
Expanding this:
\[ AB^2 = (a - 1)^2 + (1 - b)^2 = (a^2 - 2a + 1) + (1 - 2b + b^2) = a^2 + b^2 - 2a - 2b + 2 \]

2. **Distance \( AC \)** (already calculated):
\[ AC^2 = a^2 + 1 \]

### Step 5: Use the second condition \( 5(AB)^2 = 2(AC)^2 \)

Substituting the values we have:
\[ 5(a^2 + b^2 - 2a - 2b + 2) = 2(a^2 + 1) \]
Since \( a = b \), we can replace \( b \) with \( a \):
\[ 5(a^2 + a^2 - 2a - 2a + 2) = 2(a^2 + 1) \]
This simplifies to:
\[ 5(2a^2 - 4a + 2) = 2(a^2 + 1) \]
Expanding both sides:
\[ 10a^2 - 20a + 10 = 2a^2 + 2 \]
Rearranging gives:
\[ 10a^2 - 2a^2 - 20a + 10 - 2 = 0 \]
This simplifies to:
\[ 8a^2 - 20a + 8 = 0 \]

### Step 6: Solve the quadratic equation

Dividing the entire equation by 4:
\[ 2a^2 - 5a + 2 = 0 \]
Using the quadratic formula \( a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ a = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 2 \cdot 2}}{2 \cdot 2} = \frac{5 \pm \sqrt{25 - 16}}{4} = \frac{5 \pm 3}{4} \]
This gives us:
\[ a = \frac{8}{4} = 2 \quad \text{(not valid since } a, b < 1\text{)} \]
\[ a = \frac{2}{4} = \frac{1}{2} \]

### Conclusion

Thus, we find that:
\[ a = b = \frac{1}{2} \]