question_answer

In \[\Delta ABC\], 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 \[\Delta ABC\] is     AIEEE  Solved  Paper-2003

A)                         8/3                        

B)       16/3                      

C)       32/3                      

D)       64/3

Correct Answer: D

Solution :

None of the given option is correct. Median divides a triangle into two parts of equal area. So, we have to find area of one triangle. Given, AD = 4 and SD = DC We know that the centroid G divides the line AD in the ratio 2 : 1 .             \[AG=\frac{8}{3}\] and \[DG=\frac{4}{3}\] In \[\Delta ABG\],                     \[\tan \frac{\pi }{3}=\frac{AG}{BG}\] \[\Rightarrow \]   \[BG=AG\cot \frac{\pi }{3}\]                     \[BG=\frac{8}{3}\times \frac{1}{\sqrt{3}}=\frac{8}{3\sqrt{3}}\] Area of MDS \[\Delta ADB=\frac{1}{2}\times BG\times AD\]                                     \[=\frac{1}{2}\times \frac{8}{3\sqrt{3}}\times 4=\frac{16}{3\sqrt{3}}\] Since, median divide a triangle into two triangles of equal area. Therefore, Area of \[\Delta ABC=2\times \] Area of \[\Delta ADB\]                                     \[=2\times \frac{16}{3\sqrt{3}}=\frac{32}{3\sqrt{3}}\] None of the given option is correct.