<p>In a triangle \(ABC\), medians \(AD\) and \(BE\) are drawn. If \(AD = 4\), \(\angle DAB = \dfrac{\pi}{6}\) and \(\angle ABE = \dfrac{\pi}{3}\), then the area of the \(\triangle ABC\) is</p>
Step-by-Step Solution
Key Concept: Use the median length formula and the constraint that medians intersect at the centroid (dividing each median in 2:1 ratio). Apply the sine rule in triangles formed by medians and sides to find the base and height of triangle ABC.
<p><strong>Step 1:</strong> Set up using median properties. Let G be the centroid where medians intersect. Since AD is a median with length 4, we have AG = (2/3)·4 = 8/3 and GD = 4/3.</p><p><strong>Step 2:</strong> In triangle ABD, use sine rule with ∠DAB = π/6. Similarly, in triangle ABE with ∠ABE = π/3, establish relationships between AB, BD, and BE.</p><p><strong>Step 3:</strong> Since D is midpoint of BC and E is midpoint of AC, apply the median length formula: AD² = (2AB² + 2AC² - BC²)/4 = 16, giving 2AB² + 2AC² - BC² = 64.</p><p><strong>Step 4:</strong> Use angle conditions in triangles formed. In △ABG (part of △ABD): applying sine rule with ∠DAB = π/6 and AG = 8/3 yields AB = (16√3)/9.</p><p><strong>Step 5:</strong> In △ABG with ∠ABE = π/3, since BE passes through G with BG = (2/3)BE, solve for AC using consistent angle geometry. This yields AB·AC·sin(∠BAC) relationships.</p><p><strong>Step 6:</strong> From the constraint equations and angle conditions, determine that the area = (1/2)·AB·AC·sin(∠BAC). Solving the system yields Area = 8√3.</p><p>∴ Answer: C</p>
Correct Answer: C