Area Under the Curve
Area of triangle inscribed in circle
Grade 12
Question:
<p>If <em>ABC</em> is an isosceles triangle inscribed in a circle of radius <em>r</em>. If <em>AB</em> = <em>AC</em> and <em>h</em> is altitude from <em>A</em> to <em>BC</em> then the triangle <em>ABC</em> has perimeter \(P = 2(\sqrt{2hr - h^2} + \sqrt{2hr})\), calculate area <em>A</em> and \(\lim_{h \to 0} \dfrac{A}{P^3}\).</p>
<p>\(1/64r\)</p>
<p>\(1/128r\)</p>
<p>\(1/32r\)</p>
<p>None of these</p>
Step-by-Step Solution
Key Concept: Express the triangle's dimensions using the altitude h and circle radius r through the constraint that the triangle is inscribed in a circle of radius r. The circumradius relation R = abc/(4K) combined with isosceles geometry allows expressing area A in terms of h, then evaluate the limit using L'Hôpital's rule or series expansion.
<p><strong>Step 1: Express BC and AB in terms of h and r</strong></p><p>For isosceles triangle ABC inscribed in circle of radius r with altitude h from A to BC:</p><p>Distance from center O to BC: d = r − h</p><p>Half of BC: BM = √(r² − d²) = √(r² − (r−h)²) = √(2hr − h²)</p><p>So BC = 2√(2hr − h²)</p><p><strong>Step 2: Find AB = AC</strong></p><p>AB² = h² + (√(2hr − h²))² = h² + 2hr − h² = 2hr</p><p>AB = √(2hr)</p><p><strong>Step 3: Calculate Area A</strong></p><p>A = ½ × BC × h = ½ × 2√(2hr − h²) × h</p><p><strong>A = h√(2hr − h²)</strong></p><p><strong>Step 4: Verify Perimeter expression</strong></p><p>P = 2√(2hr) + 2√(2hr − h²) = 2(√(2hr − h²) + √(2hr)) ✓</p><p><strong>Step 5: Evaluate the limit</strong></p><p>As h → 0: A ≈ h√(2hr) = h·√(2r)·√h = √(2r)·h^(3/2)</p><p>As h → 0: P ≈ 2√(2hr) = 2√(2r)·√h</p><p>Therefore: P³ ≈ 8(2r)^(3/2)·h^(3/2)</p><p>∴ lim(h→0) A/P³ = (√(2r)·h^(3/2))/(8(2r)^(3/2)·h^(3/2)) = 1/(8·2√(2r)) = <strong>1/(16√(2r))</strong></p><p>Or simplified: <strong>√(2r)/64r</strong></p>
Correct Answer: B