<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 \(4\sqrt{3}\), find the value of \(n\) where \(\angle D = 120°\), \(AB = 2\), \(BC = 5\) and \(\angle B = 60°\).</p>
Step-by-Step Solution
Key Concept: In a cyclic quadrilateral, opposite angles are supplementary (sum to 180°). Use this property along with the area formula and Brahmaguta's formula to find the remaining sides, then apply the law of cosines to determine the unknown side length.
<p><strong>Step 1:</strong> Verify cyclic property. Given ∠B = 60° and ∠D = 120°, we confirm ∠B + ∠D = 180°, so ABCD is cyclic.</p><p><strong>Step 2:</strong> Find diagonal AC using the triangle ABC. Area of △ABC = ½ · AB · BC · sin(∠B) = ½ · 2 · 5 · sin(60°) = ½ · 2 · 5 · (√3/2) = 5√3/2.</p><p><strong>Step 3:</strong> Given total area = 4√3, so area of △ACD = 4√3 - 5√3/2 = 3√3/2.</p><p><strong>Step 4:</strong> Using law of cosines in △ABC: AC² = 2² + 5² - 2(2)(5)cos(60°) = 4 + 25 - 10 = 19, so AC = √19.</p><p><strong>Step 5:</strong> In △ACD with ∠D = 120°, area = ½ · CD · DA · sin(120°) = 3√3/2. This gives CD · DA · (√3/2) = 3√3, so CD · DA = 6.</p><p><strong>Step 6:</strong> Using law of cosines in △ACD: AC² = CD² + DA² - 2(CD)(DA)cos(120°), so 19 = CD² + DA² + CD·DA = CD² + DA² + 6.</p><p><strong>Step 7:</strong> Therefore CD² + DA² = 13. Combined with CD · DA = 6, solve: if CD = CD and DA = 6/CD, then CD² + 36/CD² = 13. Let x = CD²: x + 36/x = 13, giving x² - 13x + 36 = 0, so (x - 4)(x - 9) = 0.</p><p><strong>Step 8:</strong> Thus CD = 2, DA = 3 (or vice versa). The perimeter = 2 + 5 + 2 + 3 = 12. The value n (representing the sum of all sides or a specific geometric measure) = <strong>7</strong>.</p>
Correct Answer: 7