SAT Mathematics Practice Test 2

The regular hexagon can be divided into 6 equilateral triangles of side length a by drawing the six segments from the center of the regular hexagon to each of its 6 vertices. Since the area of the hexagon is $384\sqrt{3}$ square inches, the area of each equilateral triangle will be $\frac{384\sqrt{3}}{6}=64\sqrt{3}$ square inches.
Drawing any altitude of an equilateral triangle divides it into two $30°-60°-90°$ triangles. If the side length of the equilateral triangle is a, then the hypotenuse of each $30°-60°-90°$ triangle is a, and the altitude of the equilateral triangle will be the side opposite the 60° angle in each of the $30°-60°-90°$ triangles. Thus, the altitude of the equilateral triangle is $\frac{\sqrt{3}}{2}\mathit{a}$ and the area of the equilateral triangle is $\left(\frac{1}{2}\right)\left(\mathit{a}\right)\left(\frac{\sqrt{3}}{2}\mathit{a}\right)=\frac{\sqrt{3}}{4}{\mathit{a}}^2$. Since the area of each equilateral triangle is $64\sqrt{3}$ square inches, it follows that ${\mathit{a}}^2=\frac{4}{\sqrt{3}}\left(64\sqrt{3}\right)=256$ square inches. And since the area of the square with side length $\mathit{a}$ is ${\mathit{a}}^2$ , it follows that the square has area 256 square inches.
Choices B, C, and D are incorrect and may result from calculation or conceptual errors.