Sketch the region and find the area bounded by the curves `|y+x|le1,|y-x|le1` and `2x^(2)+2y^(2)ge1`.

Answer

To solve the problem of finding the area bounded by the curves \( |y+x| \leq 1 \), \( |y-x| \leq 1 \), and \( 2x^2 + 2y^2 \geq 1 \), we will follow these steps:

### Step 1: Understand the inequalities

1. The inequality \( |y+x| \leq 1 \) represents the region between the lines \( y + x = 1 \) and \( y + x = -1 \).
2. The inequality \( |y-x| \leq 1 \) represents the region between the lines \( y - x = 1 \) and \( y - x = -1 \).
3. The inequality \( 2x^2 + 2y^2 \geq 1 \) can be rewritten as \( x^2 + y^2 \geq \frac{1}{2} \), which represents the region outside (or on) a circle of radius \( \frac{1}{\sqrt{2}} \) centered at the origin.

### Step 2: Sketch the region

- **Lines from \( |y+x| \leq 1 \)**:
- The lines are \( y + x = 1 \) (or \( y = 1 - x \)) and \( y + x = -1 \) (or \( y = -1 - x \)).

- **Lines from \( |y-x| \leq 1 \)**:
- The lines are \( y - x = 1 \) (or \( y = x + 1 \)) and \( y - x = -1 \) (or \( y = x - 1 \)).

- **Circle from \( 2x^2 + 2y^2 \geq 1 \)**:
- The circle has a radius of \( \frac{1}{\sqrt{2}} \) and is centered at the origin.

### Step 3: Find the points of intersection

To find the vertices of the bounded region, we need to find the intersection points of the lines:

1. **Intersection of \( y + x = 1 \) and \( y - x = 1 \)**:
- From \( y + x = 1 \) and \( y - x = 1 \):
\[ 2y = 2 \implies y = 1 \quad \text{and} \quad x = 0 \quad \Rightarrow (0, 1) \]

2. **Intersection of \( y + x = 1 \) and \( y - x = -1 \)**:
- From \( y + x = 1 \) and \( y - x = -1 \):
\[ 2y = 0 \implies y = 0 \quad \text{and} \quad x = 1 \quad \Rightarrow (1, 0) \]

3. **Intersection of \( y + x = -1 \) and \( y - x = 1 \)**:
- From \( y + x = -1 \) and \( y - x = 1 \):
\[ 2y = 0 \implies y = 0 \quad \text{and} \quad x = -1 \quad \Rightarrow (-1, 0) \]

4. **Intersection of \( y + x = -1 \) and \( y - x = -1 \)**:
- From \( y + x = -1 \) and \( y - x = -1 \):
\[ 2y = -2 \implies y = -1 \quad \text{and} \quad x = 0 \quad \Rightarrow (0, -1) \]

### Step 4: Determine the area of the bounded region

The vertices of the square formed by the intersections are \( (0, 1) \), \( (1, 0) \), \( (0, -1) \), and \( (-1, 0) \).

- **Area of the square**:
- The side length of the square can be calculated as the distance between any two adjacent vertices, for example, between \( (0, 1) \) and \( (1, 0) \):
\[ \text{Side length} = \sqrt{(1-0)^2 + (0-1)^2} = \sqrt{1 + 1} = \sqrt{2} \]
- Therefore, the area of the square is:
\[ \text{Area}_{\text{square}} = (\sqrt{2})^2 = 2 \]

- **Area of the circle**:
- The radius of the circle is \( \frac{1}{\sqrt{2}} \), so the area is:
\[ \text{Area}_{\text{circle}} = \pi \left(\frac{1}{\sqrt{2}}\right)^2 = \pi \cdot \frac{1}{2} = \frac{\pi}{2} \]

### Step 5: Calculate the required area

The required area bounded by the curves is:
\[ \text{Required Area} = \text{Area}_{\text{square}} - \text{Area}_{\text{circle}} = 2 - \frac{\pi}{2} \]

### Final Answer

The area bounded by the curves is:
\[ \boxed{2 - \frac{\pi}{2}} \]