AIEEE 2003 (Offline)Question 1
In a triangle $ABC$, medians $AD$ and $BE$ are drawn. If $AD=4$, $\angle DAB=\frac{\pi}{6}$ and $\angle ABE=\frac{\pi}{3}$, then the area of the $\triangle ABC$ is :
Trigonometric Ratio & Identities
Solution
Concept. In triangle $ABG$ ($G$ the centroid) the angles at $A$ and $B$ are $30^\circ$ and $60^\circ$, so $\angle AGB=90^\circ$. The three triangles $ABG$, $BCG$, $CAG$ have equal areas. Formulas used. (1) $AG=\frac23AD$(2) Area $\triangle ABC=3\cdot$ area $\triangle ABG$Step 1 - The triangle $ABG$.Step 2 - Its legs. By (1), $AG=\frac23\cdot4=\frac83$. In the right triangle, $BG=AG\tan30^\circ=\frac{8}{3\sqrt3}$. Step 3 - Area of $ABG$. $\frac12\cdot\frac83\cdot\frac{8}{3\sqrt3}=\frac{32}{9\sqrt3}$. Step 4 - Area of $ABC$. By (2), $3\cdot\frac{32}{9\sqrt3}=\frac{32}{3\sqrt3}$. Answer: (D)