In a triangle ABC , the median AD (with D on BC ) and the angle...

Let $A D$ and $B E$ meet at $F$ Now $\angle \mathrm{ABF}=\angle \mathrm{FBD}=\frac{\mathrm{B}}{2}$ and $\angle \mathrm{AFB}=\angle \mathrm{BFD}=30^{\circ}$ and BF is common to triangles ABF and BFD hence the two triangles are Cengruent so $\mathrm{AF}=\mathrm{FD}=\frac{7}{2}$ Also, $\mathrm{AB}=\mathrm{BD}, \Rightarrow \mathrm{AB}: \mathrm{BC}=1: 2$ $\Rightarrow \mathrm{AE}: \mathrm{EC}=1: 2$ So area of triangle $A B C=3$ (Area of $\triangle A B E$ ) $$\begin{aligned} & =3\left(\frac{1}{2} \times \mathrm{AF} \times \mathrm{BE}\right) \\ & =\frac{3}{2} \times \frac{7}{2} \times 9=\frac{189}{4}=47.25 \end{aligned}$$ Hence nearest integer = 47