Question

In the diagram below, $CD = BF = 10	ext{ units}$ and $\angle CED = \angle BAF = 30^\circ$. What would be the area of triangle AED? (Note: Diagram below may not be proportional to scale.)

Accepted Answer

D. From right triangle $ABF$ with $\angle BAF = 30^\circ$ and $BF = 10$: $AB = 10 \sqrt{3}$. Since $ED = AB = 10 \sqrt{3}$. From right triangle $BCF$ with $\angle BCF = 60^\circ$: $BC = \frac{10}{\sqrt{3}}$. Total height $AD = AB + BC + CD = 10\sqrt{3} + \frac{10}{\sqrt{3}} + 10 = \frac{40 + 10\sqrt{3}}{\sqrt{3}}$. Area of $	riangle AED = \frac{1}{2} 	imes AD 	imes ED = \frac{1}{2} 	imes \frac{40 + 10\sqrt{3}}{\sqrt{3}} 	imes 10\sqrt{3} = 5(40 + 10\sqrt{3}) = 50(\sqrt{3} + 4)$.

Area of Triangle with Trigonometric Relations

Explanation

Practise this under exam conditions