In a triangle ABC, medians AD and BE are drawn.

Question

In a triangle $ABC$, medians $AD$ and $BE$ are drawn. If $AD=4$,

$$\angle DAB = {\pi  \over 6}$$ and $$\angle ABE = {\pi  \over 3}$$, then the area of the $$\angle \Delta ABC$$ is :

Accepted Answer

$$AP = {2 \over 3}AD = {8 \over 3};\,\,PD = {4 \over 3};\,\,$$

Let $PB=x$

$$\tan {60^ \circ } = {{8/3} \over x}$$

or $$x = {8 \over {3\sqrt 3 }}$$

Area of $$\Delta ABD$$

$$ = {1 \over 2} \times 4 \times {8 \over {3\sqrt 3 }} = {{16} \over {3\sqrt 3 }}$$

$$\therefore$$ Area of $$\Delta ABC$$

$$ = 2 \times {{16} \over {3\sqrt 3 }} = {{32} \over {3\sqrt 3 }}$$

$$\left[ \, \right.$$ As median of a $$\Delta $$ divides it into two $$\Delta 's$$ of equal area. $$\left. \, \right]$$

Solutions · 1