Question

In the figure above, ABCD is a square and triangle $AED$ is isosceles. If $AE=4$, find the area of the square.

• A. $4\sqrt{3}$
• B. $16$
• C. $16\sqrt{3}$
• D. $32$
• E. $48$

Answer 1

The correct option is E $48$

First, note that $AE=4$ and $ED=4$. (Hint: fill in the diagram with information that you are given or can figure out.) When a ${120}^{\circ }$ angle is used in a triangle, it is often a clue that a $30-60-90$ triangle (shown on the first page of each SAT math section) is involved in the question. If you draw a line from E perpendicular to the bottom side of the square, you should see that two of these special triangles are formed, as shown in the figure above.
Using the $30-60-90$ triangle, the side opposite the ${30}^{\circ }$ angle is $2$, and the side opposite the ${60}^{\circ }$ angle is $2\sqrt{3}$. Then, the side of the square is $2\sqrt{3}+2\sqrt{3}=4\sqrt{3}$ so that the area of the square is $(4\sqrt{3}{)}^2=48$.