Question
In a triangle $$ABC,$$ 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
A.
$$\frac{{64}}{3}$$
B.
$$\frac{{8}}{3}$$
C.
$$\frac{{16}}{3}$$
D.
$$\frac{{32}}{{3\sqrt 3 }}$$
Answer :
$$\frac{{32}}{{3\sqrt 3 }}$$
Solution :
$$\eqalign{ & AP = \frac{2}{3}AD = \frac{8}{3};\,\,PD = \frac{4}{3};\,\,{\text{Let }}PB = x \cr & \tan {60^ \circ } = \frac{{\frac{8}{3}}}{x}\,\,\,{\text{or }}x = \frac{8}{{3\sqrt 3 }} \cr} $$
Area of $$\Delta ABD = \frac{1}{2} \times 4 \times \frac{8}{{3\sqrt 3 }} = \frac{{16}}{{3\sqrt 3 }}$$
∴ Area of $$\Delta ABC = 2 \times \frac{{16}}{{3\sqrt 3 }} = \frac{{32}}{{3\sqrt 3 }}$$
[ $$\because $$ Median of a $$\Delta $$ divides it into two $$\Delta '{\text{s}}$$ of equal area. ]
$$\eqalign{ & AP = \frac{2}{3}AD = \frac{8}{3};\,\,PD = \frac{4}{3};\,\,{\text{Let }}PB = x \cr & \tan {60^ \circ } = \frac{{\frac{8}{3}}}{x}\,\,\,{\text{or }}x = \frac{8}{{3\sqrt 3 }} \cr} $$
Area of $$\Delta ABD = \frac{1}{2} \times 4 \times \frac{8}{{3\sqrt 3 }} = \frac{{16}}{{3\sqrt 3 }}$$
∴ Area of $$\Delta ABC = 2 \times \frac{{16}}{{3\sqrt 3 }} = \frac{{32}}{{3\sqrt 3 }}$$
[ $$\because $$ Median of a $$\Delta $$ divides it into two $$\Delta '{\text{s}}$$ of equal area. ]