Let $S$ be the set of all points $(x, y)$ in the x-y plane such that $|x|+|y| \le 2$ and $|x| \ge 1$. Then, the area, in square units, of the region represented by $S$ equals:

We need to find the area of region S defined by two conditions:

$|x| + |y| \leq 2$

$|x| \geq 1$

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The condition $|x| + |y| \leq 2$ represents all points inside and on a diamond shape (or square rotated 45°) centered at the origin.

When we consider all four cases:

If $x \geq 0, y \geq 0$: $x + y \leq 2$

If $x \geq 0, y \leq 0$: $x - y \leq 2$

If $x \leq 0, y \geq 0$: $-x + y \leq 2$

If $x \leq 0, y \leq 0$: $-x - y \leq 2$

The vertices of this diamond are at $(2,0), (0,2), (-2,0), (0,-2)$.

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The condition $|x| \geq 1$ means $x \geq 1$ or $x \leq -1$.

This excludes the vertical strip where $-1 < x < 1$.

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The region S is where both conditions are satisfied simultaneously.

Starting with our diamond $|x| + |y| \leq 2$, we remove the middle strip where $-1 < x < 1$.

This leaves us with four triangular regions:

Right Side ($x \geq 1$):

Upper right triangle: vertices $(1,1), (2,0), (1,0)$

Lower right triangle: vertices $(1,0), (2,0), (1,-1)$

Left Side ($x \leq -1$):

Upper left triangle: vertices $(-1,1), (-2,0), (-1,0)$

Lower left triangle: vertices $(-1,0), (-2,0), (-1,-1)$

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Each triangle has:

Base = 1 (horizontal distance from $x = \pm 1$ to $x = \pm 2$)

Height = 1 (vertical distance from $y = 0$ to $y = \pm 1$)

Area of each triangle = $\tfrac{1}{2} \times \text{base} \times \text{height} = \tfrac{1}{2} \times 1 \times 1 = \tfrac{1}{2}$

Total area = $4 \times \tfrac{1}{2} = 2$ square units

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This approach is efficient because we visualized the geometric shapes instead of complex algebraic manipulation, identified the symmetry where all four triangles are congruent, and used the basic triangle area formula rather than integration.

Answer: 2