Question

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

• A. $\frac{25}{3}$
• B. $\frac{1}{3}$
• C. $\frac{11}{3}$
• D. $\frac{19}{11}$

Answer 1

The correct option is A $\frac{25}{3}$


Let medians $AD$ and $CE$ meet at centroid $G$
Given: $AD=5$
$\Rightarrow AG=\frac{10}{3},GD=\frac{5}{3}$
Also, in $\Delta AGC,\angle AGC=\frac{5\pi }{8}$
Now, using sine law in $\Delta AGC$, we have
$\frac{CG}{\sin \left(\frac{\pi }{8}\right)}=\frac{\frac{10}{3}}{\sin \left(\frac{\pi }{4}\right)}$
$\Rightarrow CG=\frac{10\sqrt{2}\sin \left(\frac{\pi }{8}\right)}{3}\cdots \left(i\right)$
Now, $\Delta \left(AGC\right)=\frac{1}{2}\left(AG\right)\left(CG\right)\sin \left(\angle AGC\right)$
$=\frac{1}{2}\times \frac{10}{3}\times \frac{10\sqrt{2}}{3}\sin \left(\frac{\pi }{8}\right)\sin \left(\frac{5\pi }{8}\right)$
$=\frac{100\times \sqrt{2}}{2\times 9}\times \frac{1}{2}\times 2\sin \left(\frac{\pi }{8}\right)\sin (\frac{\pi }{2}+\frac{\pi }{8})$
$=\frac{100\times \sqrt{2}}{2\times 9}\times \frac{1}{2}\times 2\sin \left(\frac{\pi }{8}\right)\cos \left(\frac{\pi }{8}\right)$
$=\frac{100\times \sqrt{2}}{2\times 9}\times \frac{1}{2}\times \sin \left(\frac{\pi }{4}\right)$
$=\frac{25\sqrt{2}}{9}\times \frac{1}{\sqrt{2}}=\frac{25}{9}$
$\therefore \Delta ABC=3\Delta AGC=\frac{25}{3}$