XAT 2015 QA Question 10 · Geometry - Triangles

​​​​​​​

m∠ECD = m∠BCF = 60°
Also, m∠AFB = 60°, m∠BFC = 30°
∴ m∠AFC = 90°
In a 30° - 60° - 90° triangle, sides are in the ratio $1:\sqrt{3}:2.$

So, in ∆EDC, ED = 10√3 units

Also, in ∆FBC, BF = 10 units

$\Rightarrow \mathrm{FC}=\frac{20}{\sqrt{\mathrm{3 }}}$​​​​​​​ units and BC =  ​​​​​​​$\frac{10}{\sqrt{3}}$

In ∆AFC,

$\mathrm{FC}=\frac{20}{\sqrt{3}} \mathrm{units}\Rightarrow \mathrm{AC}=\frac{40}{\sqrt{3}} \mathrm{units}$

$\therefore \mathrm{AD}=\left(10+\frac{40}{\sqrt{3}}\right) \mathrm{units}$

$\mathrm{A}(∆\mathrm{ADE})=\frac{1}{2}\times 10\sqrt{3}\times \left(10+\frac{40}{\sqrt{3}}\right)$

= 50(√3 + 4) sq.units

Hence, option (d).