Mathematics - Properties of Triangles Question with Solution | TestHub

Let $AB=AC$ and $\angle A=120^{\circ }$ $\therefore$ Area of $\Delta =\frac{1}{2}{a}^2\sin 120^{\circ }$ where, $a=AD+BD$ $=\sqrt{3}\tan 30^{\circ }+\sqrt{3}\cot 15^{\circ }$ $=1+\sqrt{3}\left(\frac{1+\tan 45^{\circ }\tan 30^{\circ }}{\tan 45^{\circ }-\tan 30^{\circ }}\right)$ $=1+\sqrt{3}\left(\frac{\sqrt{3}+1}{\sqrt{3}-1}\right)$ $\therefore a=4+2\sqrt{3}$ $\therefore$ Area of $\Delta =\frac{1}{2}(4+2\sqrt{3}{)}^2\left(\frac{\sqrt{3}}{2}\right)=12+7\sqrt{3}$