<p>The two adjacent sides of a cyclic quadrilateral are 2 and 5 and the angle between them is \(60°\). If the area of the quadrilateral is \(4\sqrt{3}\), then the perimeter of the quadrilateral is</p>
Step-by-Step Solution
Key Concept: For a cyclic quadrilateral, opposite angles are supplementary (sum to 180°). Use this property along with the area condition to find the remaining two sides via Brahmagupta's formula or the constraint that the quadrilateral must be cyclic.
<p><strong>Step 1:</strong> Let the quadrilateral be ABCD with AB = 2, BC = 5, and ∠ABC = 60°.</p><p><strong>Step 2:</strong> Area of triangle ABC = ½ × 2 × 5 × sin(60°) = ½ × 2 × 5 × (√3/2) = (5√3)/2</p><p><strong>Step 3:</strong> Since total area = 4√3, area of triangle ACD = 4√3 - (5√3)/2 = (3√3)/2</p><p><strong>Step 4:</strong> Using diagonal AC via triangle ABC: AC² = 4 + 25 - 2(2)(5)cos(60°) = 29 - 10 = 19, so AC = √19</p><p><strong>Step 5:</strong> For cyclic quadrilateral, ∠ADC = 180° - 60° = 120°. For triangle ACD with ∠ADC = 120° and AC = √19:</p><p>Area = ½ × CD × DA × sin(120°) = (3√3)/2</p><p>This gives: CD × DA × (√3/2) = 3√3, so CD × DA = 6</p><p><strong>Step 6:</strong> Also, by the Law of Cosines in triangle ACD: AC² = CD² + DA² - 2(CD)(DA)cos(120°)</p><p>19 = CD² + DA² + CD·DA = CD² + DA² + 6</p><p>Therefore: CD² + DA² = 13</p><p><strong>Step 7:</strong> From CD × DA = 6 and CD² + DA² = 13, we get (CD + DA)² = 13 + 12 = 25, so CD + DA = 5</p><p><strong>Step 8:</strong> Perimeter = AB + BC + CD + DA = 2 + 5 + 5 = <strong>12</strong></p>
Correct Answer: C