Trigonometry & Inverse Trigonometry
Properties of Triangles
Grade 11
Question:
<p>Let \(S_1\) and \(S_2\) be the areas of inscribed and circumscribed polygons of 10 sides respectively and \(S_3\) is the area of regular polygon of 20 sides inscribed in a circle, then</p>
<p>\(2S_3 = S_1 + S_2\)</p>
<p>\(S_3^2 = S_1 S_2\)</p>
<p>\(\dfrac{1}{S_3} = \dfrac{1}{S_1} + \dfrac{1}{S_2}\)</p>
<p>\(\dfrac{2}{S_3} = \dfrac{1}{S_1} + \dfrac{1}{S_2}\)</p>
Step-by-Step Solution
Key Concept: For a circle of radius r, the inscribed n-gon has area (nr²/2)sin(2π/n), the circumscribed n-gon has area nr²tan(π/n), and increasing sides approximates the circle's area πr². Compare S₁ (10-sided inscribed), S₂ (10-sided circumscribed), and S₃ (20-sided inscribed).
<p><strong>Step 1:</strong> For a circle of radius r, the area of a regular n-sided inscribed polygon is:</p><p>S₁ = (10r²/2)sin(2π/10) = 5r²sin(36°) ≈ 2.939r²</p><p><strong>Step 2:</strong> Area of 10-sided circumscribed polygon:</p><p>S₂ = 10r²tan(π/10) = 10r²tan(18°) ≈ 3.249r²</p><p><strong>Step 3:</strong> Area of 20-sided inscribed polygon:</p><p>S₃ = (20r²/2)sin(2π/20) = 10r²sin(18°) ≈ 3.090r²</p><p><strong>Step 4:</strong> Comparing values: S₁ ≈ 2.939r² < S₃ ≈ 3.090r² < S₂ ≈ 3.249r²</p><p>Note: As n increases, inscribed polygons approach πr² ≈ 3.142r² from below, while circumscribed approach from above. With 20 sides inscribed, we get closer to π than 10-sided inscribed, but circumscribed still exceeds both.</p><p>∴ Answer: S₁ < S₃ < S₂ (Option B)</p>
Correct Answer: B