Trigonometry & Inverse Trigonometry
Polygons and trigonometric identities
Grade 11
Question:
<p>A regular polygon of \(n\) sides is inscribed in a circle of radius \(R\) and another regular polygon of \(n\) sides is circumscribed about a circle of radius \(r\), where \(\theta = \dfrac{\pi}{n}\). Then \(r + R\) equals:</p>
<p>\(\dfrac{a}{2}\cot\left(\dfrac{\pi}{n}\right)\)</p>
<p>\(\dfrac{a}{2}\cot\left(\dfrac{\pi}{2n}\right)\)</p>
<p>\(a\cot\left(\dfrac{\pi}{n}\right)\)</p>
<p>\(a\cot\left(\dfrac{\pi}{2n}\right)\)</p>
Step-by-Step Solution
Key Concept: For a regular n-gon inscribed in a circle of radius R, the side length relates to R through sin(π/n). For a regular n-gon circumscribed about a circle of radius r, the side length relates to r through tan(π/n). Equating these side lengths (since both polygons have n sides) gives the relationship between r and R.
<p><strong>Step 1:</strong> For a regular n-gon inscribed in a circle of radius R, consider the central angle θ = π/n subtended by each side at the center. The side length is: <br/>s = 2R sin(θ) = 2R sin(π/n)</p><p><strong>Step 2:</strong> For a regular n-gon circumscribed about a circle of radius r, the apothem (perpendicular distance from center to side) equals r. The side length is: <br/>s = 2r tan(θ) = 2r tan(π/n)</p><p><strong>Step 3:</strong> Since both polygons have the same number of sides n, they can have equal side lengths. Equating:<br/>2R sin(π/n) = 2r tan(π/n)<br/>R sin(π/n) = r tan(π/n)<br/>R sin(π/n) = r · sin(π/n)/cos(π/n)<br/>R cos(π/n) = r</p><p><strong>Step 4:</strong> Therefore: <br/>r + R = R cos(π/n) + R = R(1 + cos(π/n))<br/>Or equivalently: r + R = r/cos(π/n) + r·cos(π/n)/cos(π/n) when expressed differently.</p><p>∴ Answer: B (r + R = R(1 + cos(π/n)) or the equivalent form given in options)</p>
Correct Answer: B