The region represented by `|x-y| le 2` and `|x+y| le 2` is bounded by a

Options

A. rhombus of side length 2 units
B. square of side length `2sqrt(2)` units.
C. square of area 16 sq. units
D. rhombus of area `8sqrt(2)` aq. Units.

Explanation

To solve the problem represented by the inequalities \( |x-y| \leq 2 \) and \( |x+y| \leq 2 \), we will follow these steps:

### Step 1: Rewrite the inequalities
The absolute value inequalities can be rewritten as:
1. \( -2 \leq x - y \leq 2 \)
2. \( -2 \leq x + y \leq 2 \)

### Step 2: Break down the inequalities
From the first inequality \( -2 \leq x - y \leq 2 \), we can break it down into two separate inequalities:
- \( x - y \leq 2 \) (or \( x \leq y + 2 \))
- \( x - y \geq -2 \) (or \( x \geq y - 2 \))

From the second inequality \( -2 \leq x + y \leq 2 \), we can also break it down:
- \( x + y \leq 2 \) (or \( x \leq 2 - y \))
- \( x + y \geq -2 \) (or \( x \geq -2 - y \))

### Step 3: Identify the lines
Now we have four lines:
1. \( x = y + 2 \)
2. \( x = y - 2 \)
3. \( x = 2 - y \)
4. \( x = -2 - y \)

### Step 4: Find the intersection points
Next, we will find the intersection points of these lines to determine the vertices of the bounded region.

1. **Intersection of \( x = y + 2 \) and \( x = 2 - y \)**:
\[ y + 2 = 2 - y \implies 2y = 0 \implies y = 0 \implies x = 2 \]
Intersection point: \( (2, 0) \)

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

3. **Intersection of \( x = y - 2 \) and \( x = 2 - y \)**:
\[ y - 2 = 2 - y \implies 2y = 4 \implies y = 2 \implies x = 0 \]
Intersection point: \( (0, 2) \)

4. **Intersection of \( x = y - 2 \) and \( x = -2 - y \)**:
\[ y - 2 = -2 - y \implies 2y = 0 \implies y = 0 \implies x = -2 \]
Intersection point: \( (-2, 0) \)

### Step 5: Plot the points and identify the region
The points we found are:
- \( (2, 0) \)
- \( (0, 2) \)
- \( (-2, 0) \)
- \( (0, -2) \)

Plotting these points on a Cartesian plane, we can see that they form a square (or diamond shape) centered at the origin.

### Step 6: Calculate the area of the bounded region
The distance between the points along the axes gives us the side length of the square:
- The distance from \( (2, 0) \) to \( (0, 2) \) is \( 2\sqrt{2} \) (the diagonal of the square).

The area \( A \) of the square is given by:
\[ A = \text{side}^2 = (2\sqrt{2})^2 = 8 \]

### Conclusion
Thus, the region represented by the inequalities \( |x-y| \leq 2 \) and \( |x+y| \leq 2 \) is bounded by a square with an area of \( 8 \).

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