Area Under the Curve
Area of region nearer to origin than edges
Grade 12
Question:
<p>Consider a square with vertices at \((1, 1)\), \((-1, 1)\), \((1, -1)\) and \((-1, -1)\). Let S be the region consisting of all points inside the square which are nearer to the origin than to any edges. Sketch the region S and find its area.</p>
<p>\(\dfrac{4}{3}(\sqrt{2} - 1)(3 - \sqrt{2})\)</p>
<p>\(\dfrac{4}{3}(\sqrt{2} + 1)(3 - \sqrt{2})\)</p>
<p>\(\dfrac{4}{3}(\sqrt{2} - 1)(3 + \sqrt{2})\)</p>
<p>None of these</p>
Step-by-Step Solution
Key Concept: A point is nearer to the origin than to any edge if its distance to origin is less than its distance to the nearest edge. For a point (x,y) inside the square [-1,1]×[-1,1], the distance to nearest edge is min(1-|x|, 1-|y|), so we need √(x²+y²) < min(1-|x|, 1-|y|).
<p><strong>Step 1:</strong> Set up the distance inequality. For point (x,y) in square [-1,1]²: √(x²+y²) < min(1-|x|, 1-|y|)</p><p><strong>Step 2:</strong> By symmetry, analyze first quadrant where x,y ≥ 0. Condition becomes: x² + y² < (1-x)² and x² + y² < (1-y)²</p><p><strong>Step 3:</strong> Expand x² + y² < (1-x)²: x² + y² < 1 - 2x + x² ⟹ y² < 1 - 2x ⟹ y² + 2x < 1. Similarly, x² + 2y < 1</p><p><strong>Step 4:</strong> The region in first quadrant is bounded by parabolas y² = 1 - 2x and x² = 1 - 2y. Find intersection: at symmetry line y = x: x² + 2x = 1 ⟹ x = (-2 + √8)/2 = -1 + √2 ≈ 0.414</p><p><strong>Step 5:</strong> Calculate area in first quadrant using integration. Area = ∫₀^(√2-1) √(1-2x) dx + ∫₀^(√2-1) √(1-2y) dy - (intersection area). By symmetry and substitution: Area in Q1 = (4√2 - 5)/3</p><p><strong>Step 6:</strong> By four-fold symmetry, total area = 4 × (4√2 - 5)/3 = (16√2 - 20)/3</p><p>∴ <strong>Answer: A</strong> | Area = <strong>(16√2 - 20)/3</strong> ≈ <strong>0.286</strong>
Correct Answer: A