The ends of the base of an isosceles triangle are `(2sqrt(2), 0)` and `(0, sqrt(2))`. One side is of length `2sqrt(2)`. If `Delta` be the area of triangle, then the value of `[Delta]` is (where [.] denotes the greatest integer function)

Answer

To solve the problem step by step, we need to find the area of the isosceles triangle with the given vertices and then apply the greatest integer function to that area.

### Step 1: Identify the vertices of the triangle
The ends of the base of the isosceles triangle are given as points \( A(2\sqrt{2}, 0) \) and \( B(0, \sqrt{2}) \). We need to find the third vertex \( C \) such that the triangle remains isosceles.

### Step 2: Determine the length of the base
The length of the base \( AB \) can be calculated using the distance formula:
\[ AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Substituting the coordinates of points \( A \) and \( B \):
\[ AB = \sqrt{(0 - 2\sqrt{2})^2 + (\sqrt{2} - 0)^2} = \sqrt{(2\sqrt{2})^2 + (\sqrt{2})^2} = \sqrt{8 + 2} = \sqrt{10} \]

### Step 3: Use the given side length
We know one side of the triangle is \( 2\sqrt{2} \). Since the triangle is isosceles, we can assume that the two equal sides are of length \( 2\sqrt{2} \).

### Step 4: Find the height of the triangle
To find the height, we can drop a perpendicular from point \( C \) to the base \( AB \). The height \( h \) can be calculated using the Pythagorean theorem in triangle \( AOC \) (where \( O \) is the foot of the perpendicular from \( C \) to \( AB \)):
\[ AC^2 = AO^2 + OC^2 \]
Here, \( AC = 2\sqrt{2} \) and \( AO = \frac{AB}{2} = \frac{\sqrt{10}}{2} \). Therefore:
\[ (2\sqrt{2})^2 = \left(\frac{\sqrt{10}}{2}\right)^2 + h^2 \]
\[ 8 = \frac{10}{4} + h^2 \]
\[ 8 = 2.5 + h^2 \]
\[ h^2 = 8 - 2.5 = 5.5 \]
\[ h = \sqrt{5.5} = \sqrt{\frac{11}{2}} = \frac{\sqrt{22}}{2} \]

### Step 5: Calculate the area of the triangle
The area \( \Delta \) of the triangle can be calculated using the formula:
\[ \Delta = \frac{1}{2} \times \text{base} \times \text{height} \]
Substituting the values of base \( AB = \sqrt{10} \) and height \( h = \frac{\sqrt{22}}{2} \):
\[ \Delta = \frac{1}{2} \times \sqrt{10} \times \frac{\sqrt{22}}{2} = \frac{\sqrt{10} \cdot \sqrt{22}}{4} = \frac{\sqrt{220}}{4} = \frac{\sqrt{4 \cdot 55}}{4} = \frac{2\sqrt{55}}{4} = \frac{\sqrt{55}}{2} \]

### Step 6: Apply the greatest integer function
Now we need to find the value of \( [\Delta] \):
\[ [\Delta] = \left[\frac{\sqrt{55}}{2}\right] \]
Calculating \( \sqrt{55} \) gives approximately \( 7.416 \), so:
\[ \frac{\sqrt{55}}{2} \approx \frac{7.416}{2} \approx 3.708 \]
Thus, the greatest integer less than or equal to \( 3.708 \) is \( 3 \).

### Final Answer
The value of \( [\Delta] \) is \( 3 \).