Question

In a triangle $ABC$, median $AD$ and $CE$ are drawn. If $AD=5,\angle DAC=\frac{\pi }{8}$ and $\angle ACE=\frac{\pi }{4},$ and the area of the triangle $ABC$ is equal to $\Delta$, then $3\Delta$ is equal to:

• A. $25$
• B. $23$
• C. $40$
• D. $17$

Answer 1

The correct option is A $25$


Let $O$ be the point of intersection of the medians of triangle $ABC$. Then the area of $\Delta ABC$ is three times that of $\Delta AOC$
Now, in $\Delta AOC,AO=\frac{2}{3}\times AD=\frac{2}{3}\times 5=\frac{10}{3}$
Therefore applying the sine rule to $\Delta AOC$, we get:
$\frac{OC}{\sin \frac{\pi }{8}}=\frac{AO}{\sin \frac{\pi }{4}}$
$\Rightarrow OC=\frac{10}{3}\times \frac{\sin \frac{\pi }{8}}{\sin \frac{\pi }{4}}$
Now, area of $\Delta AOC=\frac{1}{2}AO.OC.\sin \angle AOC$
$=\frac{1}{2}\times \frac{10}{3}\times \frac{10}{3}\cdot \frac{\sin \frac{\pi }{8}}{\sin \frac{\pi }{4}}\times \sin (\frac{\pi }{4}+\frac{\pi }{8})$
$=\frac{50}{9}\times \frac{\sin \frac{\pi }{8}\cos \frac{\pi }{8}}{\sin \frac{2\pi }{8}}=\frac{25}{9}$
$\therefore$Area of $\Delta ABC=3\times \frac{25}{9}=\frac{25}{3}$
Hence $3\Delta =25$