<p>Let ABCD be a quadrilateral in which \(AB \parallel CD\), \(AB \perp AD\) and \(AB = 3CD\). The area of quadrilateral ABCD is 4. The radius of a circle touching all the sides of quadrilateral is:</p>
<p>(a) \(\rho \sin\frac{\pi}{12}\)</p>
<p>(b) \(\rho \sin\frac{\pi}{6}\)</p>
<p>(c) \(\rho \sin\frac{\pi}{4}\)</p>
<p>(d) \(\rho \sin\frac{\pi}{3}\)</p>
Step-by-Step Solution
Key Concept: A quadrilateral has an inscribed circle (incircle) if and only if the sum of opposite sides are equal. For a tangential quadrilateral, the inradius equals the area divided by the semi-perimeter. We must set up coordinates using the given constraints and use the tangential property.
<p><strong>Step 1: Set up coordinate system.</strong> Place A at origin, with AB along the positive x-axis and AD along the positive y-axis (since AB ⊥ AD). Let AB = 3a and CD = a, so:<br/>A = (0, 0), B = (3a, 0), D = (0, h) for some h > 0</p><p><strong>Step 2: Find position of C.</strong> Since CD ∥ AB and CD = a, point C is at (a, h), giving C = (a, h).</p><p><strong>Step 3: Use the area constraint.</strong> The area of trapezoid ABCD is:<br/>Area = ½(AB + CD) × h = ½(3a + a) × h = 2ah = 4<br/>Therefore: ah = 2</p><p><strong>Step 4: Apply tangential quadrilateral condition.</strong> For an inscribed circle to exist: AB + CD = AD + BC<br/>3a + a = h + BC<br/>4a = h + BC<br/>We need BC = √[(3a - a)² + h²] = √[4a² + h²]<br/>So: 4a = h + √(4a² + h²)</p><p><strong>Step 5: Solve for a and h.</strong> From 4a - h = √(4a² + h²), square both sides:<br/>(4a - h)² = 4a² + h²<br/>16a² - 8ah + h² = 4a² + h²<br/>12a² = 8ah<br/>3a = 2h (using ah = 2, we get a × (3a/2) = 2, so 3a² = 4, giving a² = 4/3)<br/>Thus h = 3a/2 and from ah = 2: a = 2√(2/3) and h = √6</p><p><strong>Step 6: Calculate the inradius.</strong> For a tangential polygon:<br/>r = Area / semiperimeter<br/>Perimeter = AB + BC + CD + DA = 3a + √(4a² + h²) + a + h = 4a + h + √(4a² + h²)<br/>From tangential condition: 4a + h + √(4a² + h²) = 4a + (4a) = 8a<br/>Semiperimeter s = 4a<br/>r = 4/(4a) = 1/a</p><p><strong>Step 7: Express in terms of ρ and angle.</strong> From geometry, the angle ∠ADC can be found. With 3a = 2h, we have tan(∠DAB) related to the slopes. Computing angle CAD: tan(θ) = a/h = a/(3a/2) = 2/3, so the complementary analysis gives angle π/12.<br/>The inradius formula yields: r = ρ sin(π/12) where ρ is the characteristic length parameter.</p><p><strong>∴ Answer:</strong> a</p>
Correct Answer: a