P and Q are the points of intersection of the curves `y^(2)=4x` and `x^(2)+y^(2)=12`. If `Delta` represents the area of the triangle OPQ, O being the origin, then `Delta` is equal to `(sqrt(2)=1.41)`.

Answer

To find the area of triangle OPQ formed by the points of intersection of the curves \( y^2 = 4x \) and \( x^2 + y^2 = 12 \), we will follow these steps:

### Step 1: Find the points of intersection of the curves

We have two equations:
1. \( y^2 = 4x \)
2. \( x^2 + y^2 = 12 \)

Substituting \( y^2 = 4x \) into the second equation:

\[ x^2 + 4x = 12 \]

Rearranging gives:

\[ x^2 + 4x - 12 = 0 \]

### Step 2: Solve the quadratic equation

Using the quadratic formula \( x = \frac{-b \pm \sqrt{d}}{2a} \), where \( a = 1, b = 4, c = -12 \):

1. Calculate the discriminant \( d = b^2 - 4ac = 4^2 - 4 \cdot 1 \cdot (-12) = 16 + 48 = 64 \).
2. Now, substituting into the formula:

\[ x = \frac{-4 \pm \sqrt{64}}{2 \cdot 1} = \frac{-4 \pm 8}{2} \]

This gives us two solutions:

\[ x = \frac{4}{2} = 2 \quad \text{and} \quad x = \frac{-12}{2} = -6 \]

### Step 3: Find corresponding y-values

Now, we find the y-values using \( y^2 = 4x \):

1. For \( x = 2 \):
\[ y^2 = 4 \cdot 2 = 8 \implies y = \pm 2\sqrt{2} \]
Thus, we have the point \( P(2, 2\sqrt{2}) \) and \( Q(2, -2\sqrt{2}) \).

2. For \( x = -6 \):
\[ y^2 = 4 \cdot (-6) = -24 \quad \text{(not valid, as y would be complex)} \]

Thus, the valid points of intersection are \( P(2, 2\sqrt{2}) \) and \( Q(2, -2\sqrt{2}) \).

### Step 4: Calculate the area of triangle OPQ

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 \( O(0,0), P(2, 2\sqrt{2}), Q(2, -2\sqrt{2}) \):

\[ \text{Area} = \frac{1}{2} \left| 0(2\sqrt{2} - (-2\sqrt{2})) + 2(-2\sqrt{2} - 0) + 2(0 - 2\sqrt{2}) \right| \]

This simplifies to:

\[ = \frac{1}{2} \left| 0 + 2(-2\sqrt{2}) + 2(-2\sqrt{2}) \right| \]
\[ = \frac{1}{2} \left| -4\sqrt{2} - 4\sqrt{2} \right| = \frac{1}{2} \left| -8\sqrt{2} \right| = 4\sqrt{2} \]

### Step 5: Substitute the value of \( \sqrt{2} \)

Given \( \sqrt{2} \approx 1.41 \):

\[ \text{Area} \approx 4 \cdot 1.41 = 5.64 \]

Thus, the area \( \Delta \) of triangle OPQ is:

\[ \Delta = 5.64 \]