Question

In triangle $ABC$,medians $AD$ and $CE$ are drawn. If $AD=5$, $\angle DAC=\pi /8$, and $\angle ACE=\pi /4$, then the area of the triangle $ABC$ is equal to

• A. $\frac{25}{9}$
• B. $\frac{25}{3}$
• C. $\frac{25}{18}$
• D. $\frac{10}{3}$

Answer 1

Answer: (B)

$\frac{25}{3}$

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, O being
the centroid of $\Delta$ABC, divides the median through B in the ratio
2 : 1 and the height $\Delta$AOC is one-third that of $\Delta$ABC.
Now in $\Delta$AOC, AO $=$ (2/3) AD $=$ 10/3. Therefore, applying
the sine rule to $\Delta$AOC, we get
$\frac{OC}{\sin \left(\pi /8\right)}=\frac{AO}{\sin \left(\pi /4\right)}\Rightarrow OC=\frac{10}{3}\frac{\sin \left(\pi /8\right)}{\sin \left(\pi /4\right)}$
area of $\Delta AOC=\frac{1}{2}.AO.OC.\sin \angle AOC$
$=\frac{1}{2}.\frac{10}{3}.\frac{10}{3}.\frac{\sin \left(\pi /8\right)}{\sin \left(\pi /4\right)}.\sin (\frac{\pi }{2}+\frac{\pi }{8})$
$=\frac{50}{9}.\frac{\sin \left(\pi /8\right)\cos \left(\pi /8\right)}{\sin \left(\pi /4\right)}=\frac{50}{18}=\frac{25}{9}$
$\Rightarrow$ area of $\Delta ABC=3.\frac{25}{9}=\frac{25}{3}$