<p>The region represented by |x − y| ≤ 2 and |x + y| ≤ 2 is bounded by a</p>
<p>square of side length \(2\sqrt{2}\) units.</p>
<p>rhombus of side length 2 units.</p>
<p>square of area 16 sq. units.</p>
<p>rhombus of area \(8\sqrt{2}\) sq. units.</p>
Step-by-Step Solution
Key Concept: Transform the absolute value inequalities |x − y| ≤ 2 and |x + y| ≤ 2 into a system of linear inequalities, then identify the vertices of the resulting polygon.
<p><strong>Step 1:</strong> Expand |x − y| ≤ 2 into −2 ≤ x − y ≤ 2, giving: x − y ≤ 2 and y − x ≤ 2</p><p><strong>Step 2:</strong> Expand |x + y| ≤ 2 into −2 ≤ x + y ≤ 2, giving: x + y ≤ 2 and x + y ≥ −2</p><p><strong>Step 3:</strong> The four bounding lines are: x − y = 2, x − y = −2, x + y = 2, x + y = −2</p><p><strong>Step 4:</strong> Find intersection vertices:</p><ul><li>x − y = 2 and x + y = 2 → (2, 0)</li><li>x − y = −2 and x + y = 2 → (0, 2)</li><li>x − y = −2 and x + y = −2 → (−2, 0)</li><li>x − y = 2 and x + y = −2 → (0, −2)</li></ul><p><strong>Step 5:</strong> These four vertices form a square with diagonals along the coordinate axes, each of length 4. The region is bounded by a <strong>square</strong>.</p><p>∴ Answer: A (Square)</p>
Correct Answer: A