Circles
Inscribed and circumscribed circles
Grade 11
Question:
<p>The sum of the radii of inscribed and circumscribed circles for an \(n\) sided regular polygon of side \(a\), is</p>
<p>\(a \cot\left(\dfrac{\pi}{n}\right)\)</p>
<p>\(\dfrac{a}{2} \cot\left(\dfrac{\pi}{2n}\right)\)</p>
<p>\(a \cot\left(\dfrac{\pi}{2n}\right)\)</p>
<p>\(\dfrac{a}{4} \cot\left(\dfrac{\pi}{2n}\right)\)</p>
Step-by-Step Solution
Key Concept: For a regular n-sided polygon with side a, the inradius r = a/(2tan(π/n)) and circumradius R = a/(2sin(π/n)). Their sum uses the identity involving complementary trigonometric angles.
<p><strong>Step 1:</strong> For a regular n-sided polygon with side length a, identify the circumradius R and inradius r.</p><p><strong>Step 2:</strong> The circumradius R is the distance from center to vertex. Using the formula: R = a/(2sin(π/n))</p><p><strong>Step 3:</strong> The inradius r is the distance from center to the midpoint of a side (apothem). Using the formula: r = a/(2tan(π/n))</p><p><strong>Step 4:</strong> Sum the radii:</p><p>r + R = a/(2tan(π/n)) + a/(2sin(π/n))</p><p>= (a/2)[1/tan(π/n) + 1/sin(π/n)]</p><p>= (a/2)[cos(π/n)/sin(π/n) + 1/sin(π/n)]</p><p>= (a/2)[(cos(π/n) + 1)/sin(π/n)]</p><p><strong>Step 5:</strong> Using the identity cos(π/n) + 1 = 2cos²(π/(2n)) and sin(π/n) = 2sin(π/(2n))cos(π/(2n)):</p><p>r + R = (a/2) · [2cos²(π/(2n))]/[2sin(π/(2n))cos(π/(2n))]</p><p>= (a/2) · cos(π/(2n))/sin(π/(2n))</p><p>= <strong>a·cot(π/(2n))/2</strong> or equivalently <strong>a/(2tan(π/(2n)))</strong></p><p>∴ Answer: B</p>
Correct Answer: B