Find the area of the region formed by the points satisfying
`|x| + |y| + |x+y| le 2.`

Explanation

To find the area of the region formed by the points satisfying the inequality \( |x| + |y| + |x+y| \leq 2 \), we can break it down into cases based on the signs of \( x \) and \( y \).

### Step 1: Analyze the inequality
The expression \( |x| + |y| + |x+y| \) can be analyzed by considering the different cases for \( x \) and \( y \).

### Step 2: Case 1: \( x \geq 0 \) and \( y \geq 0 \)
In this case, \( |x| = x \), \( |y| = y \), and \( |x+y| = x+y \).
Thus, the inequality becomes:
\[ x + y + (x + y) \leq 2 \implies 2x + 2y \leq 2 \implies x + y \leq 1 \]

### Step 3: Case 2: \( x \geq 0 \) and \( y < 0 \)
Here, \( |x| = x \), \( |y| = -y \), and \( |x+y| = x+y \) (since \( x+y \) will be non-negative).
The inequality becomes:
\[ x - y + (x + y) \leq 2 \implies 2x \leq 2 \implies x \leq 1 \]
Also, since \( y < 0 \), we have \( y \geq -1 \) from the previous case.

### Step 4: Case 3: \( x < 0 \) and \( y \geq 0 \)
In this case, \( |x| = -x \), \( |y| = y \), and \( |x+y| = x+y \) (since \( x+y \) will be non-negative).
The inequality becomes:
\[ -x + y + (x + y) \leq 2 \implies 2y \leq 2 \implies y \leq 1 \]
Also, since \( x < 0 \), we have \( x \geq -1 \) from the previous case.

### Step 5: Case 4: \( x < 0 \) and \( y < 0 \)
Here, \( |x| = -x \), \( |y| = -y \), and \( |x+y| = -(x+y) \) (since \( x+y \) will be negative).
The inequality becomes:
\[ -x - y - (x + y) \leq 2 \implies -2x - 2y \leq 2 \implies x + y \geq -1 \]

### Step 6: Plotting the inequalities
Now we have the following inequalities to plot:
1. \( x + y \leq 1 \) (1st quadrant)
2. \( x \leq 1 \) and \( y \geq -1 \) (2nd quadrant)
3. \( y \leq 1 \) and \( x \geq -1 \) (3rd quadrant)
4. \( x + y \geq -1 \) (4th quadrant)

### Step 7: Finding the vertices of the region
The vertices of the region formed by these inequalities can be found at the intersections of the lines:
- \( (1, 0) \)
- \( (0, 1) \)
- \( (-1, 0) \)
- \( (0, -1) \)

### Step 8: Area of the region
The region formed is a square with vertices at \( (1, 0) \), \( (0, 1) \), \( (-1, 0) \), and \( (0, -1) \). The side length of the square is \( \sqrt{(1 - (-1))^2 + (0 - 0)^2} = 2 \).

The area \( A \) of the square is given by:
\[ A = \text{side}^2 = 2^2 = 4 \]

### Conclusion
Thus, the area of the region formed by the points satisfying \( |x| + |y| + |x+y| \leq 2 \) is \( 4 \) square units.