The area of quadrilateral combined equation of whose sides are
`(x^(2) - y^(2)) (x^(2) - y^(2) - 8x + 16) = 0` is

Options

A. 0
B. `gt` 5
C. `2 sqrt(2)`
D. 9

Answer

To find the area of the quadrilateral defined by the equation \((x^2 - y^2)(x^2 - y^2 - 8x + 16) = 0\), we will follow these steps:

### Step 1: Factor the Equation
The given equation can be factored into two parts:
1. \(x^2 - y^2 = 0\)
2. \(x^2 - y^2 - 8x + 16 = 0\)

### Step 2: Solve the First Equation
From the first equation \(x^2 - y^2 = 0\), we can rewrite it as:
\[ y^2 = x^2 \]
This gives us two lines:
\[ y = x \quad \text{and} \quad y = -x \]

### Step 3: Solve the Second Equation
Now, let's simplify the second equation:
\[ x^2 - y^2 - 8x + 16 = 0 \]
Rearranging gives:
\[ x^2 - 8x + 16 = y^2 \]
Factoring the left side:
\[ (x - 4)^2 = y^2 \]
This gives us two more equations:
\[ y = x - 4 \quad \text{and} \quad y = -(x - 4) \quad \text{or} \quad y = -x + 4 \]

### Step 4: Find Intersection Points
Now we need to find the intersection points of these lines.

1. **Intersection of \(y = x\) and \(y = x - 4\)**:
\[ x = x - 4 \implies \text{No solution} \]

2. **Intersection of \(y = x\) and \(y = -x + 4\)**:
\[ x = -x + 4 \implies 2x = 4 \implies x = 2 \implies y = 2 \]
So, point B is \((2, 2)\).

3. **Intersection of \(y = -x\) and \(y = x - 4\)**:
\[ -x = x - 4 \implies -2x = -4 \implies x = 2 \implies y = -2 \]
So, point D is \((2, -2)\).

4. **Intersection of \(y = -x\) and \(y = -x + 4\)**:
\[ -x = -x + 4 \implies \text{No solution} \]

5. **Intersection of \(y = x - 4\) and \(y = -x + 4\)**:
\[ x - 4 = -x + 4 \implies 2x = 8 \implies x = 4 \implies y = 0 \]
So, point C is \((4, 0)\).

### Step 5: Identify the Vertices of the Quadrilateral
Now we have the vertices of the quadrilateral:
- Point A: \((0, 0)\) (origin)
- Point B: \((2, 2)\)
- Point C: \((4, 0)\)
- Point D: \((2, -2)\)

### Step 6: Calculate the Area of the Quadrilateral
The quadrilateral can be divided into two triangles: Triangle ABC and Triangle ACD.

1. **Area of Triangle ABC**:
Using the formula for the area of a triangle given vertices \((x_1, y_1)\), \((x_2, y_2)\), \((x_3, y_3)\):
\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]
Substituting the points:
\[ \text{Area}_{ABC} = \frac{1}{2} \left| 0(2 - 0) + 2(0 - 0) + 4(0 - 2) \right| = \frac{1}{2} \left| 0 + 0 - 8 \right| = \frac{1}{2} \times 8 = 4 \]

2. **Area of Triangle ACD**:
Using the same formula:
\[ \text{Area}_{ACD} = \frac{1}{2} \left| 0(2 - (-2)) + 4(-2 - 0) + 2(0 - 2) \right| = \frac{1}{2} \left| 0 - 8 - 4 \right| = \frac{1}{2} \times 12 = 6 \]

### Step 7: Total Area of Quadrilateral ABCD
\[ \text{Area}_{ABCD} = \text{Area}_{ABC} + \text{Area}_{ACD} = 4 + 6 = 10 \]

### Final Answer
The area of the quadrilateral is \(10\) square units.

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