Xavier Aptitude Test 2015 Paper

(d)
In right $\mathrm{\Delta }\mathit{C}\mathit{D}\mathit{E},$
$\tan {30}^{°}=\mathrm{}\frac{\mathrm{}\mathit{C}\mathit{D}}{\mathrm{}\mathit{E}\mathit{D}}$
$\Rightarrow \frac{1}{\mathrm{}\sqrt{3}}=\frac{10}{\mathit{E}\mathit{D}}\Rightarrow \mathit{E}\mathit{D}=10\sqrt{3}$
In right $\mathrm{\Delta }\mathit{A}\mathit{B}\mathit{F}$
$\tan {30}^{°}=\frac{\mathit{B}\mathit{F}}{\mathrm{}\mathit{A}\mathit{B}}\Rightarrow \mathrm{}\frac{1}{\sqrt{3}}=\frac{10}{\mathit{A}\mathit{B}}$
$\Rightarrow \mathit{A}\mathit{B}=10\mathrm{}\sqrt{3}$
Since both BF and ED are perpendicular on AD,
therefore BF||ED and hence ∠BFC = ∠CED = 30°
[Alternate interior angles]
Now in right ΔCBF,
$\tan {30}^{°}=\mathrm{}\frac{\mathrm{}\mathit{B}\mathit{C}}{\mathrm{}\mathit{B}\mathit{F}}\Rightarrow \frac{1}{\sqrt{3}}=\frac{\mathrm{}\mathit{B}\mathit{C}}{10}$
$\Rightarrow \mathit{B}\mathit{C}=\frac{10}{\sqrt{3}}$
Now, Area AED $=$ $\frac{1}{2}\times \mathit{D}\mathit{E}\times \mathit{A}\mathit{D}$
$=\frac{1}{2}\times 10\sqrt{3}\left(10\sqrt{3}+\mathrm{}\frac{10}{\sqrt{3}}+10\mathrm{}\right)$
$=50\left(\sqrt{3}+4\right)$