Question

In the diagram below, $CD = BF = 10$ units and $\angle CED = \angle BAF = 30^\circ$. What would be the area of $\triangle AED$?

• A. $100\;(\sqrt{2}+3)$
• B. $100\;(\sqrt{3}+4)$
• C. $50\;(\sqrt{2}+4)$
• D. $50\;(\sqrt{3}+4)$
• E. None of the above

Hint:

When a fold/height makes $30^\circ\!-\!60^\circ\!-\!90^\circ$ triangles, pin down the two legs quickly from the short leg $k$: the long leg is $k\sqrt{3}$ and the hypotenuse is $2k$. Use perpendiculars as altitudes to split the big area cleanly.

Answer 1

The Correct Option is D

Step 1: Work with $\triangle CED$.
At $D$ there is a right angle and $\angle CED=30^\circ$. Hence $\,\triangle CED\,$ is a $30^\circ\!-\!60^\circ\!-\!90^\circ$ triangle with the short leg $CD=10$.
Therefore \[ ED=10\sqrt{3},\qquad CE=2\cdot 10=20. \] Consequently \[ [\triangle CED]=\tfrac12\cdot CD\cdot ED=\tfrac12\cdot 10\cdot 10\sqrt{3}=50\sqrt{3}. \] Step 2: Use $\angle BAF=30^\circ$ in $\triangle AEF$.
The segment $BF$ is perpendicular to $AE$ (shown in the figure), so $BF$ is the altitude from $F$ to the side $AE$ of $\triangle AEF$. With $\angle BAF=30^\circ$ and $BF=10$ we have \[ \tan 30^\circ=\frac{BF}{AB}=\frac{10}{AB}\ \Rightarrow\ AB=\frac{10}{\tan 30^\circ}=10\sqrt{3}. \] Step 3: Area of $\triangle AEC$.
Since $CE$ is perpendicular to $AE$ (from Step 1), $CE$ is the altitude to the base $AE$ in $\triangle AEC$. Thus \[ [\triangle AEC]=\tfrac12\cdot AE\cdot CE=\tfrac12\cdot AE\cdot 20=10\,AE. \] Also $AE=AB+BE$, and the vertical drop from $B$ to $E$ equals $CE$ (same perpendicular), hence $BE=CE=20$. So \[ AE=AB+BE=10\sqrt{3}+20 \quad\Longrightarrow\quad [\triangle AEC]=10(10\sqrt{3}+20)=100\sqrt{3}+200. \] Step 4: Area of $\triangle AED$.
The big triangle splits into two right-triangle parts along $CE$: \[ [\triangle AED]=[\triangle AEC]+[\triangle CED] =(100\sqrt{3}+200)+(50\sqrt{3}) =50(\sqrt{3}+4). \] \[ \boxed{[\triangle AED]=50\bigl(\sqrt{3}+4\bigr)} \]