In the diagram below, CD = BF = 10 units and ∠CED = ∠BAF = 30°. What would be the area of triangle AED? (Note: Diagram below may not be proportional to scale.)

In $$\triangle$$ ABF

=> $$tan 30 = \frac{BF}{AB}$$

=> $$\frac{1}{\sqrt{3}} = \frac{10}{AB}$$

=> $$AB = 10 \sqrt{3}$$

Similarly, $$ED = 10 \sqrt{3}$$

Also, $$\angle ECD = \angle BCF = 60$$ (Vertically opposite angles)

In $$\triangle$$ BCF

=> $$tan 60 = \frac{BF}{BC}$$

=> $$\sqrt{3} = \frac{10}{BC}$$

=> $$BC = \frac{10}{\sqrt{3}}$$

=> Height = $$AD = AB + BC + CD = 10 \sqrt{3} + \frac{10}{\sqrt{3}} + 10 = \frac{40 + 10 \sqrt{3}}{\sqrt{3}}$$

$$\therefore area (\triangle AED) = \frac{1}{2} \times AD \times ED$$

= $$\frac{1}{2} \times \frac{40 + 10 \sqrt{3}}{\sqrt{3}} \times 10 \sqrt{3}$$

= $$50 (\sqrt{3} + 4)$$