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  4. In a triangle A B C, medians A D and B E are drawn. If A D=4, angle D A B=(π/6) and angle A B E=(π/3), then the area of the Δ A B C is :

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Q. In a triangle $A B C,$ medians $A D$ and $B E$ are drawn. If $A D=4, \angle D A B=\frac{\pi}{6}$ and $\angle A B E=\frac{\pi}{3}$, then the area of the $\Delta A B C$ is :

AIEEEAIEEE 2003

Solution:

Given $AD = 4$ and $BD = DC$
image
In $\triangle ABG, tan \frac{\pi}{3} = \frac{AG}{BG}$
$BG = AG \,cot\frac{\pi}{3}$
$B G= \frac{8}{3} \times \frac{1}{\sqrt{3}}=\frac{8}{3 \sqrt{3}} $
Area of $\triangle A D B =\frac{1}{2} \times B G \times A D $
$=\frac{1}{2} \times \frac{8}{3 \sqrt{3}} \times 4=\frac{16}{3 \sqrt{3}}$
Since, median divide a triangle into two triangles of equal area.
Therefore Area of $\triangle A B C=2 \times$ area of $\triangle A D B$
$=2 \times \frac{16}{3 \sqrt{3}}=\frac{32}{3 \sqrt{3}}$